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September 10

Answer to a Problem

Ok, this is a question about a specific aspect of an extra credit thing that my math teacher gave us. We were given the following problem:

"Find an easy and useful way to determine the sum of all of the number between 1 and 500. Then, using this, determine the sum."

Now, I used the summation symbol to accomplish this as such:

$\sum _{i=1}^{n}i$

where n=500. and then using the sum sequence feture of my graphing calculator I calculated the sum because I didn't feel like typing out all of the numbers between 1 and 500 and adding them together. Turns out, she didn't like this particular solution and asked me "well how does the summation symbol work." I was taught (by a different though very qualified teacher) that the summation symbol is just a short hand way of writing out 1+2+3 and so on. Any help would be very much appreciated and if I'm not specific enough, let me know.

Deltacom1515 01:54, 10 September 2006 (UTC)

I think your problem is that the teacher is trying to get you to use your mathematical skills, not your skills at taking advantage of the computational abilities of your graphing calculator. What you've done makes it easier for you to compute that sum. As you've noted, your calculator is doing, internally, exactly what you would have done if you'd actually typed in "1+2+3+....", which is just as much work.

What your teacher almost certainly wanted you to do is describe can be rewritten as a closed form; this is very useful for two reasons. Firstly, the closed form can be calculated much more easily (an addition, a multiplication, and a division, no matter how big "i" is). The second is that rewriting in this way allows you to do all sorts of things you can't do with a summation, which come in very useful in more advanced maths.

If you're really clever you might consider using proof by induction to show that the closed form you work out is correct :)--Robert Merkel 02:24, 10 September 2006 (UTC)

Wow, thanks for your help. I appreciate that. One thing though, which link do I click on at the colsed form page? Deltacom1515 02:34, 10 September 2006 (UTC)

Well, let's put it this way - you're not using any calculus, are you? Try Solution in closed form. Confusing Manifestation 02:53, 10 September 2006 (UTC)

Nope, no calcus is necessary. Deltacom1515 03:12, 10 September 2006 (UTC) Ok, I got it on the page [[1]] I didn't even know such a thing existed. You guys were a BIG help. Deltacom1515 03:30, 10 September 2006 (UTC)

IIRC, Gauss first noticed the technique you are trying to find, when he was a child. Dysprosia 05:12, 10 September 2006 (UTC)

Is that the story about the teacher giving the class busywork? --jpgordon^∇∆∇∆ 05:21, 10 September 2006 (UTC)

Yes. Dysprosia 06:43, 10 September 2006 (UTC)

Are you saying that Gauss was the first to discover it (which I strongly doubt), or simply that he came up with it when he was a child (which is indeed well known)? -- Meni Rosenfeld (talk) 06:52, 10 September 2006 (UTC)

The latter. Dysprosia 07:33, 10 September 2006 (UTC)

Happy Numbers

About Happy Numbers, do they serve any practicable function or are they merely another way for mathematicians to humor themselves. (this really is a serious question, I myself came across this article and was laughing all the way through it) Autopilots 08:48, 10 September 2006 (UTC)

I have no prior knowledge of happy numbers, but this quote from the Wikipedia article suggests the latter: "The study of happy numbers is an example of recreational mathematics in that it can involve extensive mathematical knowledge, but the topic is not a central part of serious research." Note that base 10 (or any other base, other than 2) is completely arbitrary, so definitions that are based on manipulating the decimal digits of a number are very unlikely to have serious uses. -- Meni Rosenfeld (talk) 10:51, 10 September 2006 (UTC)

I have met this concept some years ago in the context of teaching UK school students about mathematical investigations. It was chosen as something which would be unlikely (at that time) to have ben encountered previously by students, so giving no advantage to students who might have read more widely than others. Apart from that pedagogical use, I think they only have recreational interest to mathematicians. Madmath789 11:03, 10 September 2006 (UTC)

Number of permutations with at least one 2-cycle

I was asked this question: "There are 70 people at a party and they put their names into a basket. What is the probability that two people get each-other's, if each person then picks one name from the basket." I thought this should be a high-school math puzzle, but I can't seem to find any solution to it. Mathematically, I'm asking for the number of permutations of length 70 with at least one cycle of length 2. I've looked at Stirling numbers and searched the encyclopaedia of integer sequences, but I can't find anything about this ... Any ideas? — Preceding unsigned comment added by 88.196.100.193 (talk • contribs)

Do you mean the probability that at least one pair get each other's name? Exactly one pair get each other's name? One pair in particular, and only they, get each other's name? One pair in particular, regardless of other possible matches, get each other's name? There is something of a difference.—86.132.167.165 13:00, 10 September 2006 (UTC)

It seems like the asker (a reminder: it's recommended to sign your posts by typing --~~~~) is referring to the first variant (the number of permutations ... with at least one cycle of length 2). That's an interesting question! But let's try to build a recursion. Let there be $n$ people; let the asked number of permutations be $f_{n}$. If we number the people $1,\ldots ,n$, then for any such permutation there is a unique person who is part of one such cycle (a "pair") and whose number ($a$) is the lowest. Let's call him/her the "lowest paired person" and the pair he/she is part of the "lowest pair". For every combination of $n$ and $a$ there are $(n-a)$ different ways of choosing the other person to form the lowest pair together with $a$. The $(a-1)$ people with lower numbers than $a$ may be permuted, provided they don't form any pairs with anyone (by assumption). Thus, if we denote the number of permutations with the lowest paired person $a$ and total length $n$ by $g_{n,\,a}$, there is a relation:

$g_{n,\,a}=\left[(n-2)!-\sum _{i=1}^{a-1}g_{n-2,\,i}\right](n-a)$,

because removing the lowest pair from a permutation with the lowest person $a$ (and lowering the people's indices as necessary to fill in the two gaps), we get a new permutation either without any pairs or with a lowest paired person $\geq a$; there are $(n-a)$ possible places for the person $a$ is paired with. Now we need some initial conditions. Clearly, $\forall n\ g_{n,\,n}=0$. Also, for any $n\geq 2$, $g_{n,\,1}=(n-2)!(n-1)=(n-1)!$, because if we say that the first person is paired with anyone, then there are $(n-1)$ people who can be his/her "companions" and the other people may permute freely. To sum up:

$g_{n,\,a}={\begin{cases}0,&{\mbox{if }}n=a\\(n-1)!,&{\mbox{if }}a=1\land n>1\\\left[(n-2)!-\sum _{i=1}^{a-1}g_{n-2,\,i}\right](n-a)&{\mbox{otherwise}}\end{cases}}.$

To get the asked total number, we have to sum over all the possible $a$-s:

$f_{n}=\sum _{a=1}^{n-1}g_{n,\,a}.$

All that can be done in a table:

n	a = 1	2	3	4	5	6	$f_{n}$
1	0						0
2	1	0					1
3	2	1	0				3
4	6	2	1	0			9
5	24	12	6	3	0		45
6	120	72	48	30	15	0	285

So, the sequence $0,1,3,9,45,285,\ldots$ answers your question. But is there a way to calculate it more easily (such tables are quite error-prone on paper and need $O(n^{2})$ of computer memory)? Let's see... (sequence A027616 in the OEIS)... we have the answer! The formula is:

$f_{n}=n!\left[1-\sum _{k=0}^{\lfloor n/2\rfloor }{\frac {(-1)^{k}}{2^{k}k!}}\right].$

Don't ask me to prove that right now. ${\ddot {\smile }}$ Anyhow, combinatorics is fun! — Pt _(T) 19:24, 10 September 2006 (UTC)

Since the question was about probabilities, don't we actually want ${\frac {f_{n}}{n!}}$? -GTBacchus^(talk) 19:33, 10 September 2006 (UTC)

That's true. So, the probability is:

$p_{n}=1-\sum _{k=0}^{\lfloor n/2\rfloor }{\frac {(-1)^{k}}{2^{k}k!}},$

a sum which Mathematica (but not yet I) can handle:

$p_{n}=1-{\frac {\Gamma (\lfloor n/2\rfloor +1;-1/2)}{{\sqrt {e}}\lfloor n/2\rfloor !}}$

Here in the numerator, we're dealing with the incomplete gamma function. Putting in $n=70$, we get:

$p_{70}={\frac {10746079308304741391988333080709297478383616359407}{27311096972527630634113530877844185625395200000000}}\approx 0.3934693403.$

By the way, there exists a limit: $\lim _{n\to \infty }p_{n}=1-{\frac {1}{\sqrt {e}}}\approx 0.3934693403$. $p_{70}$ differs from it by only $9.81\cdot 10^{-51}\%$! — Pt _(T) 19:53, 10 September 2006 (UTC)

(somewhat prolonged edit conflict, rendering my addition rather moot) He means, the probability that at least one pair of people got each other's names. I think I've figured out a way to solve it. It's like Pascal's triangle, but more complicated. Take the number of people, n. The nth level of the triangle (more like a pyramid) is an (n+1)xfloor(n/2+1) matrix. Each element (r,c) is the number of ways r of those people could have gotten their own name while c pairs of people switched names. The pyramid is recursive, then, with each level determined from the information in the level above. Imagine lining a group of five people up and numbering them 1-5. Mix up their names. Now add a sixth person to the line. Every possible permutation of the 6-group is identical to a permutation of the 5-group or a permutation of the 5-group plus a switch between the sixth person and one of the others. Take a particular permutation of the 5-group, with a particular (r,c) category. Adding a sixth person and allowing them to keep their name adds 1 to the (r+1,c) bin of the 6-level for each such permutation. Adding a sixth person and switching their name with someone who had kept their own name adds 1 to the (r-1,c+1) bin of the 6-level, for a total of r per permutation. Adding a sixth person and switching their name with someone who had already traded names adds 1 to the (r,c-1) bin of the 6-level, for a total of 2c per permutation. Adding a sixth person and switching their name with someone who hadn't traded names but didn't have their own name adds 1 to the (r,c) bin of the 6-level, for a total of n-r-2c per permutation. And that produces every possible 6-group permutation. So, each element (r,c)_n+1 = A(r-1,c)_n + B(r+1,c-1)_n + C(r,c+1)_n + D(r,c)_n where A=1, B=r+1, C=2c+2, and D=n-r-2c. Anything that involves coordinates outside the matrix (coordinates less than zero, in this case), can be ignored or set to zero, like in Pascal's regular triangle. The level 1 matrix would be {{0},{1}}, making level 2 {{0,1},{0,0},{1,0}}, level 3{{2,0},{0,3},{0,0},{1,0}}, etc. You can get the number of permutations involving no switches by adding up column zero (0+1, 0+0+1, 2+0+0+1, etc). Subtracting that from n! and dividing by n! gives the probability that there will be at least one switch. It might be possible to get this in closed form, but I don't see how. Black Carrot 19:59, 10 September 2006 (UTC)

Searching the first few terms in the OEIS yields A027616. – b_jonas 22:10, 10 September 2006 (UTC)

Oh, sorry, I just see Pt has already found that. For the record, I calculated the terms using brute force with the following J snippet:

      (+/@:(2&e.@:(#@>)@:C."1)@:(i.@:!A.i.))"0>:i.9

which gave this answer:

      0 1 3 9 45 285 1995 15855 142695

– b_jonas 22:13, 10 September 2006 (UTC)

One more question on the question: If a person draws their own name, do they return their name and draw again ? StuRat 22:38, 10 September 2006 (UTC)

On the subject of drawing one's own name, the probability of at least one person doing this tends to 1 - 1/e or 0.63212... as n tends to infinity, a result possibly better known from the "letter in envelope" equivalent problem. The similarity in result, with the square root missing, is noteworthy.—86.132.234.126 18:19, 11 September 2006 (UTC)

Thanks to a friend of mine I now know a way to prove and even generalize this property. But firstly let's prove the formula for pairs:

$f_{n}=n!\left[1-\sum _{k=0}^{\lfloor n/2\rfloor }{\frac {(-1)^{k}}{2^{k}k!}}\right]$.

$f_{n}$ gives us the number of the permutations of $n$ people, where there is at least one 2-cycle. Numbering the people as I earlier did, let $A_{ij}$ be the set of all the permutations of $n$ people, where people with numbers $i$ and $j$ are paired:

$A_{ij}:=\{\varphi \in P(n)\,|\,\varphi (i)=j\land \varphi (j)=i\}$.

Here $P(n)$ denotes the set of all the permutations of the first $n$ natural numbers. $f_{n}$ is the number of different elements in the set of all possible $A_{ij}$-s. Without loss of generality we may assume that $i<j$, because that way we already go through all the possible pairs. Let $|\cdot |$ denote the cardinality of a set (that equals the number of elements in it, since we'll only be dealing with finite sets); let $\mathbb {N} _{n}$ be the set of all the natural numbers $1,\ldots ,n$. Then:

$f_{n}=\left|\bigcup _{i<j\in \mathbb {N} _{n}}A_{ij}\right|$.

Clearly, there can be at most $\left\lfloor {\frac {n}{2}}\right\rfloor$ pairs among $n$ people. Let's make (for every $k\in \mathbb {N} _{\lfloor n/2\rfloor }$) the set of all the unique collections of exactly $k$ pairs of people (within a collection every person can be in at most one pair) and denote it by $S_{k}$:

$S_{k}:=\{\{(i_{\alpha },j_{\alpha })\in \mathbb {N} _{n}^{2}\,|\,\alpha \in \mathbb {N} _{k}\}\,|\,\forall l<m,p\neq q\in \mathbb {N} _{k}:i_{l}<i_{m}\land i_{p}<j_{p}\land i_{p}\neq j_{q}\land j_{l}\neq j_{m}\}$.

Now, by the inclusion-exclusion principle, we can write $f_{n}$ as:

$f_{n}=\left|\bigcup _{i<j\in \mathbb {N} _{n}}A_{ij}\right|=\sum _{k=1}^{\lfloor n/2\rfloor }\left[(-1)^{k+1}\sum _{\{(i_{1},j_{1}),\ldots ,(i_{k},j_{k})\}\in S_{k}}\left|\bigcap _{l=1}^{k}A_{i_{l}j_{l}}\right|\right]$.

$\bigcap _{l=1}^{k}A_{i_{l}j_{l}}$ is simply the set of all possible permutations of the people, where the people $(i_{1},j_{1})$, $(i_{2},j_{2})$, …, $(i_{k},j_{k})$ are paired. Others may or may not be paired. Thus, the cardinality of this intersection of sets is equal to the number of possible permutations of "the others", the people not fixed with the combination of $i$-s and $j$-s. (As we fixed earlier that $\forall p:i_{p}<j_{p}$, there is no way we could permute the paired people anymore.) There are, in total, $k$ $i$-s and $k$ $j$-s. So we are left with:

$\left|\bigcap _{l=1}^{k}A_{i_{l}j_{l}}\right|=(n-2k)!$,

not depending on the particular choice of $i$-s and $j$-s! So far we have:

$f_{n}=\sum _{k=1}^{\lfloor n/2\rfloor }\left[(-1)^{k+1}|S_{k}|(n-2k)!\right]$.

$|S_{k}|$ is the number of ways to choose $k$ non-overlapping pairs from among our $n$ people. Let's calculate it! Firstly, for the first people of the pairs (the $i$-s) there are ${n \choose k}$ different possibilities. If we, for one moment, omit the requirement that the pairs must be ordered for counting, we count each pair twice, choosing the $j$-s completely freely from among the people not referred to by any $i$. As long as we take care of it a moment later, we may do so. Thus, we can choose the $k$ $j$-s from $(n-k)$ people, giving us ${\frac {(n-k)!}{(n-2k)!}}$ possibilities (since order matters here, we must use the formula for permutations, not combinations). This must be now divided by the total number of different orderings of the $k$ pairs, $2^{k}$, giving us:

$|S_{k}|={n \choose k}{\frac {(n-k)!}{(n-2k)!}}{\frac {1}{2^{k}}}={\frac {n!}{k!(n-k)!}}{\frac {(n-k)!}{(n-2k)!}}{\frac {1}{2^{k}}}={\frac {n!}{k!(n-2k)!2^{k}}}$.

Therefore:

$|S_{k}|(n-2k)!={\frac {n!}{k!2^{k}}}$

and:

$f_{n}=\sum _{k=1}^{\lfloor n/2\rfloor }\left[(-1)^{k+1}|S_{k}|(n-2k)!\right]=\sum _{k=1}^{\lfloor n/2\rfloor }\left[(-1)^{k+1}{\frac {n!}{k!2^{k}}}\right]=n!\sum _{k=1}^{\lfloor n/2\rfloor }\left[{\frac {(-1)^{k+1}}{k!2^{k}}}\right]$.

For $k=0$ the summand is $(-1)$, so after taking a $(-1)$ out of the sum, we get what we wanted to:

$f_{n}=n!\left[1-\sum _{k=0}^{\lfloor n/2\rfloor }{\frac {(-1)^{k}}{2^{k}k!}}\right]$,

Q.E.D. This derivation can be quite readily be generalized from 2-cycles to $N$-cycles, where $N$ is an abitrary natural number. Then we simply observe the patitions into $N$-tuples etc., the same principles apply. Thus, the number of such pemutations of $n$ people that contain at least one $N$-cycle is:

${}_{N}f_{n}=n!\left[1-\sum _{k=0}^{\lfloor n/N\rfloor }{\frac {(-1)^{k}}{N^{k}k!}}\right]$.

The probability for $n$ people to contain an $N$-cycle is:

${}_{N}p_{n}={\frac {{}_{N}f_{n}}{n!}}=1-\sum _{k=0}^{\lfloor n/N\rfloor }{\frac {(-1)^{k}}{N^{k}k!}}$.

I'm sure this is a known result, because, for example, the formula for $N=3$ lies neatly in the OEIS (sequence A027617 in the OEIS). Now what about the limit $\lim _{n\to \infty }{}_{N}p_{n}$ of this probability? Well, writing the limit of the sum as $\sum _{k=0}^{\infty }{\frac {(-{\frac {1}{N}})^{k}}{k!}}$, it's evidently just the Taylor series for $e^{x}$ taken at $x=-{\frac {1}{N}}$, that is, $e^{-1/N}={\frac {1}{\sqrt[{N}]{e}}}$. Thus:

$\lim _{n\to \infty }{}_{N}p_{n}=1-{\frac {1}{\sqrt[{N}]{e}}}.$

Really, it's fun! — Pt _(T) 16:15, 12 September 2006 (UTC)

Hello, I'm the one who originally posted the question, and I'm just so impressed with the quality of the response here that I have to say a big THANK YOU!!!! I was quite surprise that the integer sequence entry have almost the same keywords as I have in the title of this post, yet I was not able to find it! Anyway, thanks again for the explanations, and sorry for not signing, I will create an account :) --193.40.37.156 10:52, 15 September 2006 (UTC)

SPPS graph

Hi: Is it possible to do a histogram or bar chart with SPSS for a grouped frequency distribution table? Thanks much.-- Hersheysextra 20:22, 10 September 2006 (UTC)

Area Problem. Shapes etc.

Ok well I encountered this problem (I added the y to simplify some calculations a bit, but otherwise its exactly the same information i got). I tried to solve but I wasnt sure wether I had done it correctly. So I thought I'd post here for someone to maybe review it and find the faults or confirm wether its correct. So heres what I went about doing. I numbered the steps for reference in your replies.

1. Well firstly isnce the triangles in the corners have x on both sides, we can assume they are isoceles right angled triangles, therefore the angles are 90, 45, and 45, thus meaning that all the angles in any of the shapes in this case are either 45 or 90.
2. $y={\sqrt {x^{2}+x^{2}}}$ (pythagorus)
3. The white triangle in the centre of the right hand edge is (assuming the two sides identical in length are variable z)
   1. $2006-2x=z^{2}+z^{2}$ (pythagorus)

comment - this line should be $(2006-2x)^{2}=z^{2}+z^{2}$ Richard B 22:10, 10 September 2006 (UTC)

1. 1. $2006-2x=2z^{2}$
   2. ${\frac {2006-2x}{2}}=z^{2}$
   3. ${\sqrt {\frac {2006-2x}{2}}}=z$
2. the L shape can now be divided into 3 segments, one square in the centre which is $y^{2}$, and 2 identical rectangles which are y by z. And as such the grey area consists of
   • $2(y({\sqrt {\frac {2006-2x}{2}}}))+y^{2}$
3. As such the entire grey area works out to equal this when the left part of the shape is included
   • $2({\sqrt {x^{2}+x^{2}}})({\sqrt {\frac {2006-2x}{2}}})+({\sqrt {x^{2}+x^{2}}})^{2}+2006x$
4. Therefore, the entire size of the shape should equal
   1. $8({\frac {2({\sqrt {2x^{2}}})({\sqrt {\frac {2006-2x}{2}}})+2x^{2}+2006x}{5}})$
   2. $8({\frac {2(x{\sqrt {2}})({\sqrt {\frac {2006-2x}{2}}})+2x^{2}+2006x}{5}})$
   3. $8({\frac {2x({\sqrt {2006-2x}})+2x^{2}+2006x}{5}})$
5. The two triangles with their sides against the x by 2006 shaded areas two identiacal sides can be calculated (assuming the two sides identical in length are variable a)
   1. $(z+y)^{2}=a^{2}+a^{2}$
   2. $(z+y)^{2}=2a^{2}$
   3. ${\frac {(z+y)^{2}}{2}}=a^{2}$
   4. ${\sqrt {\frac {(z+y)^{2}}{2}}}=a$
6. Substituting in the other formulae, to give in terms of x leaves
   • ${\sqrt {\frac {({\sqrt {\frac {2006-2x}{2}}}+2(x{\sqrt {2}}))^{2}}{2}}}=a$
7. So the formula for the entire square equals
   • $2006({\sqrt {\frac {({\sqrt {\frac {2006-2x}{2}}}+2(x{\sqrt {2}}))^{2}}{2}}})+2x)$
8. So now we have 2 formula for the entire shape, so we can substitute them together
   • $2006({\sqrt {\frac {({\sqrt {\frac {2006-2x}{2}}}+2(x{\sqrt {2}}))^{2}}{2}}})+2x)=8({\frac {2x({\sqrt {2006-2x}})+2x^{2}+2006x}{5}})$

Unfortunately this is where I got stuck. So if anyone knows where to go from here... The help would be appreciated. Philc _T^E_C^I 21:15, 10 September 2006 (UTC)

You could work out the total unshaded area. It's just the sum of 5 triangles. The leftmost two triangles have a total area of 1003² - they're isosceles - and they meet at exactly half way down the shape (i.e. 2006/2). The 2 small triangles at top and bottom right corners have a total area of x². The larger triangle on the right side has an area of (1003-x)². So the sum of all unshaded area = x² + 1003² + (1003-x)².

Total area must be 2006(1003 + 2x) - so

• ${\frac {x^{2}+1003^{2}+(1003-x)^{2}}{2006(1003+2x)}}={\frac {3}{8}}$
• $8x^{2}-14042x+5030045=0$

Now it's just a quadratic in x - and straightforward to solve - one of the roots of the equation will lead to a negative length - so choose the other one. Richard B 21:55, 10 September 2006 (UTC)

You can simplify the arithmetic by introducing y = x/1003, which gives

$8y^{2}-14y+5=0$

Both roots are positive, but you can discard the root that is greater than 1, as this would make x greater than 1003. Gandalf61 10:45, 11 September 2006 (UTC)

I don't get the same number for the 5th white triangle as Richard B. Rather than an area of (1003-x), I have ${\frac {1}{2}}*({\frac {2006-2*x}{\sqrt {2}}})^{2}$ (based on the 1-1-root2 ratio for 45-degree right triangles). Plugging this into the resulting quadratic, I get imaginary roots, as my c term is roughly double the size of Richard's. I also attempted finding the area of the shaded stuff and had imaginary roots there, too. Am I missing something re: that (1003-x) area triangle? — Lomn | Talk 15:35, 11 September 2006 (UTC)

But you do get the same as me for the 5th white triangle. My area is (1003-x)² - yours is;

${\frac {1}{2}}*({\frac {2006-2*x}{\sqrt {2}}})^{2}$

$={\frac {1}{2}}*{\frac {(2006-2*x)^{2}}{2}}$

$={\frac {(2006-2*x)^{2}}{4}}$

$={\frac {(2*(1003-x))^{2}}{4}}$

$={\frac {4*(1003-x)^{2}}{4}}$

$=(1003-x)^{2}$

So we actually do agree on the area of the 5th triangle Richard B 16:42, 11 September 2006 (UTC)

I seem to have skimmed right over the "squared" term in your original math. That clears it up. — Lomn | Talk 16:57, 11 September 2006 (UTC)

Take the lower half of your diagram. By symmetry, it's 5/8ths shaded as well. Label the unshaded (middle) segment on the lower edge u. Labe the unshaded upper segment on the right edge v. Then we know the following things:

1. u = 1003. It's the height of the half we kept.
2. x+v = 1003. Same reason.
3. (2x + u)(x+v) = the area. This is (2x + 1003)(1003).
4. 1/2 (u^2 + v^2 + x^2) is the unshaded area. This is 1/2 (1003^2 + (x - 1003)^2 + x^2).
5. 1 - 5/8 = 3/8 = unshaded fraction = 1/2 (1003^2 + (x - 1003)^2 + x^2) / ((2x + 1003)(1003))

Solving the resulting quadratic ( 8x^2 - 14042x + 5030045 = 0) gives x = 1003/2 and x = 5014/4. -- Fuzzyeric 23:04, 11 September 2006 (UTC)

How can you prove the top half and bottom half are symmetrical, or that they're even halves for that matter?

Earwig apartment complex

This is a bit mathsy, I guess, so it goes here. If there were some tiny people whose eyes were at the level of an earwig's, and we built them a building that was the same height as the average human male (average for us, not for the earwig-people :P), how many stories would it have? Vitriol 21:17, 10 September 2006 (UTC)

As many as you give it. You didnt define a storey height. Philc _T^E_C^I 21:32, 10 September 2006 (UTC)

I thought there was a standard size or something. More fool me, I guess. Vitriol 22:05, 10 September 2006 (UTC)

Well, we can approximate something, I suppose. Let's say an earwig's eyes are at 1cm, and that an average human's eyes are at 150 cm. So we're at 1/150 scale. So, if we were earwig people, your question could be rephrased as: how many storeys is a building that's the height of a giant person 150 times normal size. If normal is around 5'6" (switching from metric to imperial), then 150 x normal would be 825 feet, which is like an 80 storey building or so, roughly? -GTBacchus^(talk) 22:12, 10 September 2006 (UTC)

That sounds good, except for the average story being 5'6". I'd say the interior distance from floor to ceiling is typically more like 7 feet, with an additional foot for the thickness of the floor, for a total story height of 8 feet. This is for residential structures, industrial structures tend to have stories more like 10 feet high. StuRat 22:23, 10 September 2006 (UTC)

I didn't take the average storey to be 5'6", that was the height of an average human. I took the average story to be about 10 feet: hence 825' ~ 80 storeys. If it's more like 8 feet per storey, that'd be a 100 storey apartment building. -GTBacchus^(talk) 00:54, 11 September 2006 (UTC)

calculate

there is $100,000 invested in a c.d. at 5.1 interest for 5 months....how is the money earned calculated....thanks

More info is needed, is that 5.1% interest annually, compounded monthly ? StuRat 22:27, 10 September 2006 (UTC)

Sounds very much like a homework question. We do not solve such questions for people (they would learn nothing that way), but we are prepared to give hints if people tell us how far they have got, and where they hit a snag ... Madmath789 22:25, 10 September 2006 (UTC)

See compound interest and interest Richard B 22:32, 10 September 2006 (UTC)

September 11

No questions today

Aw shucks! hydnjo talk 00:05, 12 September 2006 (UTC)

That's the second day in the last two months!! — [Mac Davis] (talk) (Desk|Help me improve)

Must be the no homework policy. Tends to drive away the lazy bastards. 202.168.50.40 00:29, 12 September 2006 (UTC)

Experience shows that the "no homework policy" does not tend to drive away the lazy bastards. -- Meni Rosenfeld (talk) 17:13, 12 September 2006 (UTC)

September 12

Fermat's Factoring Method

Suppose that n is odd composite. Then, Fermat assures us that it ay be written $m^{2}-d^{2}=n$ for some m and d integers. Suppose $n\cong 1\ ({\mbox{mod 4}})$. Then we may show that $m^{2}\cong 1\ ({\mbox{mod 4}})\land d^{2}\cong 0\ ({\mbox{mod 4}})$. Equivalently, $m\cong 1\ ({\mbox{mod 2}})\land d\cong 0\ ({\mbox{mod 2}})$. For what moduli does a theorem of this form hold, and how do we lift from a statement for a small modulus to a statement about a larger modulus without a quadratic increase in the number of cases to be retained? (I.e. if we lift to the modulus 16, then depending on the residue of n we find that one of m and d is constrained to one value (mod 4) and the other is constrained to two values (mod 8) (that are not congruent (mod 4). If we then lift (mod 3) then we get two or four cases (depending on whether either of m or d can be congruent to zero (mod 3)), and using the Chinese remainder theorem to glue these cases to the cases derived from n (mod 16), we end up with four or eight cases -- some from residue classes (mod 12) and some from residue classes (mod 24).)

So, how do we encode the retained cases without creating an exponential explosion in the encoding of the retained cases and retaining the ability to perform additinoal lifting and additional applications of the CRT. -- Fuzzyeric 03:33, 12 September 2006 (UTC)

STATISTICS

Require a method to calculate confidence intervals for weighted sums. The sum is of the form SUM wi xi and SUM wi = 1.

Thank You

Gert Engelbrecht Pretoria South Africa

This is not possible without further information about the distributions of the random variables x_i.  --Lambiam^Talk 15:50, 12 September 2006 (UTC)

square centimeters

need to know how many square centimeters are in a piece of tisse that measures 4cm X16cm --need the answer - not the formula. I am not a student needing assistance with homework ---

• 4 x 16. --jpgordon^∇∆∇∆ 15:38, 12 September 2006 (UTC)

Why do they teach pupils to do multiplication before they teach them to read the large notice at the top of the page saying 'NO HOMEWORK'? They also seem to have omitted the lesson in which pupils are taught to thank people who answer their questions. —Daniel (‽) 20:03, 12 September 2006 (UTC)

The answer is 4 * 16 = 2101₃ square centimeters. There I have just given you the exact answer. 202.168.50.40 23:06, 12 September 2006 (UTC)

octahedral rotational symmetry

Can you draw a graph to show octahedral rotational symmetry

1. rotation about an axis from the center of a face to the center of the opposite face by an angle of 90°: 3 axes, 2 per axis, together 6
2. ditto by an angle of 180°: 3 axes, 1 per axis, together 3
3. rotation about a body diagonal by an angle of 120°: 4 axes, 2 per axis, together 8

Many thanks!--82.28.195.12 20:17, 12 September 2006 (UTC)Jason

No. What variables would be plotted on the graph?

There is some discussion of symmetries in octahedron. ColinFine 23:21, 12 September 2006 (UTC)

Better still, try octahedral symmetry. It has many figures. Consider whether you wish to deliberately exclude any reflection symmetry; most simple examples naturally include it. --KSmrq^T 23:28, 12 September 2006 (UTC)

Calculus, Limits and First Principles.

Hello. This is one of those problems which hits you hard when realise you don't know how to do it.

-So- much mathematics is based on the result that d/dx(e^x) is e^x, or written in a different way.. the integral of 1/x is ln(x). The question is, how do we prove this?

We can go back to first principles easily enough, and say that the derivative of a^x, as h tends to zero, is:

(a^(x+h) - a^x)/h

Factorise out a^x, and get:

a^x(a^h - 1)/h

Now, we know from basic calculus that differentiating this should go ln(a).a^x, so we're looking to show that the below limit is true:

(a^h - 1) / h = ln(a), as h tends to zero.

This doesn't look tricky, does it? But remember that we're trying to prove a result fundamental to calculus, so what we can use is limited (no pun), we can't use l'hopital's rule (which would give the right answer), as it relies on differentiating an exponential - our thing to be proven.

So, basically, I would really, really appreciate it if somebody could attempt to prove my limit is true, or comment that they could not (so I know the ability levels it's going to take).

Thank you, and remember: No circular reasoning! No proving that e^x differentiates to itself by assuming it in the first place!

Michael.blackburn 20:56, 12 September 2006 (UTC)

The first question, I suppose, is the definition of e. If you use the Taylor series for e^x, as the definition of e, you can use term by term differentiation of a polynomial:

$e^{x}=\sum _{i=0}^{\infty }{\frac {x^{i}}{i!}}$

Assuming (without proof here, although it can be proven) that

${\frac {d}{dx}}\sum _{i=0}^{\infty }{\frac {x^{i}}{i!}}=\sum _{i=0}^{\infty }{\frac {d}{dx}}\left({\frac {x^{i}}{i!}}\right)$

you can differentiate and find the result. --TeaDrinker 21:01, 12 September 2006 (UTC)

I found an easy solution:

${\frac {d}{dx}}e^{x}=\lim _{h\to 0}e^{x}\left({\frac {e^{h}}{h}}-{\frac {1}{h}}\right)$
Using definition $e=\lim _{h\to 0}(1+h)^{\frac {1}{h}}$
$=\lim _{h\to 0}(e^{x}){\frac {\left((1+h)^{\frac {1}{h}}\right)^{h}}{h}}-{\frac {1}{h}}$
$=\lim _{h\to 0}(e^{x}){\frac {1+h}{h}}-{\frac {1}{h}}=\lim _{h\to 0}(e^{x}){\frac {h}{h}}=e^{x}?$
so ${\frac {d}{dx}}e^{x}=e^{x}$ iff $\lim _{h\to 0}{\frac {h}{h}}=1$
This would seem to require L'Hôpital's rule, but what do you all think? M.manary 21:11, 12 September 2006 (UTC)

For reference: The proof of the fact that $\lim _{h\to 0}{\frac {h}{h}}=1$ does not require l'Hôpital. There is a rather elementary proof, directly based on the definition of the limit of an expression. Hint: Can you simplify ${\frac {h}{h}}$? JoergenB 18:02, 18 September 2006 (UTC)

I also believe that TeaDrinker's solution will rely on a formula already using d/dx e^x, as Taylor series can ONLY be derived from that notion (try it yourself and see), so that solution is no-go. M.manary 21:15, 12 September 2006 (UTC)

Formally, I have actually used the Taylor series as the definition of e^x, so no derivation of the Taylor series is needed. The proof from first principles does depend on your definition of e. --TeaDrinker 21:25, 12 September 2006 (UTC)

M.manary, I think using L'Hôpital's rule is okay here, as long as we don't use it with exponentials. It can be proven from fairly basic principles. Michael.blackburn 21:17, 12 September 2006 (UTC)

I just found a proof that you can read at: http://www.ltcconline.net/greenl/courses/106/ApproxOther/lhop.htm so: ${\frac {d}{dx}}e^{x}=e^{x}$ Q.E.D. M.manary 21:19, 12 September 2006 (UTC)

Nice. The way we learned it in Calculus, and the article agrees, the natural logarithm is defined as the area under the graph of 1/x from 1 to b. And the natural exponent is defined as the inverse of the natural logarithm. So,

• f(x) = ln(x)
• f'(x) = 1/x
• g(x) = e^x
• f(g(x)) = x
• f'(g(x))g'(x) = 1
• g'(x) = 1/f'(g(x))
• g'(x) = 1/(1/(e^x))
• g'(x) = e^x

The fifth step uses the chain rule, which makes no assumptions about the functions themselves, other than that they can be differentiated in the first place. Black Carrot 01:15, 13 September 2006 (UTC)

This is fine if you replace "e^x" by "exp(x)", where exp is defined to be the inverse function of ln. Next, it has to be shown that there exists a constant e such that exp(x) = e^x.  --Lambiam^Talk 03:41, 13 September 2006 (UTC)

I was under the impression such a proof existed. And that it in no way compromised this one. Do you know it? Black Carrot 06:14, 13 September 2006 (UTC)

After defining "exp", you define power as a^b = exp (b lna ). You need to show that this agrees with the definition of powers with rational exponents, which isn't hard. It is also easy to show that ln is bijective, so you define e as the preimage of 1. Then you're pretty much good to go. -- Meni Rosenfeld (talk) 08:08, 13 September 2006 (UTC)

(Another) on Calculus, with Radians.

Okay, thank you (Very) much for the awnsers to the previous question. This one is probably a lot more obvious, yet through searching Wikipedia and the internet I've still failed to find a solution. My mathematics teachers were also uncertain.

Basically, the question is "Why radians?"

From thinking about it, it's obvious that the Calculus can only work with a single angle measure, and obviously it's the radian. But the best explaination I've recieved is that solving Taylor series of Trig. functions works in radians only... but those Taylor series surely required Radians in the first place.

Can -anybody- offer a sensible 'proof' or reasoning that you have to use 2pi 'units' per circle for the Calculus to work, without starting in radians in the first place?

Thank you again!

It's not a proof at all, but consider a small angle segment of a circle. The arclength = rθ - with theta in radians. But this is approximately equal to r * sin (θ). So sin(θ) ~ θ, if theta is in radians (for small angles). You can show this by simple geometry Richard B 21:36, 12 September 2006 (UTC)

Oh, so.. if you used degrees for example, and arclength is.. kθ, so sin(θ) ~ kθ in degrees (where k isn't one), and so the results go icky. Is that it? Michael.blackburn 21:44, 12 September 2006 (UTC)

So the definition of a radian is that in a circle of radius r, a central angle of one radian will define an arclength of the circle that is exactly r long. If the radius of a circle is r, then there are (by circumference) 2π radii in the arclength of the circle. SO when we say an angle = 2 radians, it inscribes an arclength of 2r out of its circle.

The numbers actually don't work out that well, as you may notice, because we always have to say an angle is π/3 or 2π/7 radians, which really aren't that great numbers... M.manary 23:12, 12 September 2006 (UTC)

2π/7 radians, eh? How'd you write that in degrees, 360/7°, or 51.43° or 51°25' or (51+3/7)°? If you write that, can you tell faster how many of those you need for a full cycle? Personally, I think even π/5 is easier to understand than 36 degrees. – b_jonas 11:23, 13 September 2006 (UTC)

If you look at the Taylor series' defining sin and cos, you'll notice that there aren't any unnecessary coefficients. If you wanted to express the same thing in degrees, you would need to put π/180 in front of every x term (and square it for x^2, etc...). This will give you a complicated sequence of coefficients to keep track of. The radian is the simplest unit to use in this kind of calculation. There are also interesting geometric properties, and probably other reasons as well, but the main reason (to me) is that it avoids keeping track of unnecessary coefficients in calculation, or when using calculus. - Rainwarrior 23:27, 12 September 2006 (UTC)

Exactly. It's the same as why we have to use e as the base of exponentials. If you want to solve the linear differential equation $a_{n}{\frac {d^{n}y}{dx^{n}}}+\dots +a_{1}{\frac {dy}{dx}}+a_{0}y=0$, you find the roots of the polynomial $a_{n}z^{n}+\dots +a_{1}z+a_{0}$ in the form of $z=p+qi$ and then the base solutions are $y=e^{px}\sin(qx)$ (not counting the cases of roots with multiplicity). Here, you have to use $e$ as the base of the exponent and the sine function using radians. – b_jonas 11:15, 13 September 2006 (UTC)

Inverse

We are learning determinants and inverses in matrices in math. My teacher said that if you have a determinant of "0' then there is no inverse. I went on to say that in some remote field of mathematics, there is probably a way to fond the inverse of matrix with d=0. She said, "Maybe, but I doubt it. Why don't you look that up and tell us tomorrow." I took look at abstract algebra, and it was kind of confusing. I did some relevant Google searches, but to no avail. My question is: am I right? Is there, in some field of mathematics, a way to find the inverse of a matrix that has a determinant of 0? schyler 23:37, 12 September 2006 (UTC)

I don't think so. If a matrix A is invertible, then AA^-1=I where I is the identity matrix. You can check by simuntaneous equations that you get two contradictions:

${\begin{pmatrix}1&2\\2&4\end{pmatrix}}{\begin{pmatrix}a&b\\c&d\end{pmatrix}}={\begin{pmatrix}1&0\\0&1\end{pmatrix}}$

If you go on to more difficult mathematics, you might understand a bit more about matrices when you learn systems of linear equations and related topics. x42bn6 Talk 02:12, 13 September 2006 (UTC)

The articles on Determinant, Invertible matrix, Identity matrix, and Matrix multiplication look nice, and link to other stuff. I don't know much about matrices, but I think I can give some general advice. First, if they say it's true, there's probably a good reason, especially in something as exhaustively studied and widely used as matrices. Even if in some obscure branch of mathematics someone decided they could do it, or made up a system in which it worked, that doesn't change that if you wind up with {{1,1},{1,1}}A={{1,0},{0,1}}, and you need A, you're screwed. I wish you luck finding the exception, though. While I was in Junior High I decided it was ridiculous to say that division by 0 was impossible, as everyone kept repeating, so I spent a few years figuring out how it works. I was, as you might imagine, rather upset to find out that limit notation and functional analysis have existed for a few centuries and nobody told me. But still, that doesn't change that 1/0 is meaningless in arithmetic and algebra. Black Carrot 02:21, 13 September 2006 (UTC)

There is a thing called a Pseudoinverse, which is the closest thing to an inverse matrix which a given matrix (even a singular or non-square one) has. If the determinant of some matrix is not 0 but an infinitesimal, you will be able to talk of an inverse with infinite entries (of course, this does not work with real numbers). -- Meni Rosenfeld (talk) 05:46, 13 September 2006 (UTC)

A nice way to see why d = 0 is a problem is to look at the following. Simplifying this somewhat, if A is a matrix and d its determinant, its inverse can be given by 1/d X, where X is some other matrix (see Application section on Adjugate matrix). Clearly if d = 0, you will have problems -- division by zero. Dysprosia 05:53, 13 September 2006 (UTC)

Another nice way to understand why an inverse matrix does not exist if det(A)=0 is to think of an nxn matrix A as representing a linear transformation of n dimensional space. So a 2x2 matrix is a linear transformation of 2-d space (i.e. the plane). The inverse matrix (if it exists) then represents the inverse of this transformation. But the inverse transformation is only defined if the original transformation is 1-1. Some linear transformations are not 1-1 because they map n dimensional space onto a linear sub-space of itself in a many-1 way - for example, the transformation represented by the matrix

${\begin{pmatrix}1&2\\2&4\end{pmatrix}}$

maps the plane onto the line y=2x, because

${\begin{pmatrix}1&2\\2&4\end{pmatrix}}{\begin{pmatrix}x\\y\end{pmatrix}}={\begin{pmatrix}x+2y\\2x+4y\end{pmatrix}}={\begin{pmatrix}t\\2t\end{pmatrix}}$

and so all the points along each of the parallel lines x+2y=t are mapped to the single point (t,2t). A many-1 linear transformation (for which an inverse transformation cannot, by definition, exist) is always represented by a matrix with determinant 0, which in turn does not have a matrix inverse. Gandalf61 10:50, 13 September 2006 (UTC)

The question is provocative, and fruitful. Multiplication is defined for two compatible matrices, where compatibility means that the number of columns of the left matrix equals the number of rows of the right matrix. The usual algebraic definition of inverse depends on a definition of identity. If a matrix is not square, we might have two different identities (left and right), suggesting the possibility of a left or right inverse. For example, the 2×3 matrix

$A={\begin{bmatrix}1&0&0\\0&0.6&0.8\end{bmatrix}}$

can be said to have a right inverse 3×2 matrix

$B={\begin{bmatrix}1&0\\0&0.6\\0&0.8\end{bmatrix}}$

because the product AB is the 2×2 identity matrix. This also shows that A is a left inverse for B.

Determinants, however, are only defined for square matrices. It is impossible for a square matrix to have a left inverse but not a right inverse (or vice versa), because the row rank and column rank are always equal. The determinant of a square matrix is nonzero precisely when the matrix has full rank, which means that if we look at the image space of n-vectors under the action of the n×n matrix, it also has dimension n.

Put more geometrically, a singular matrix collapses one or more dimensions, smashing them flat. The determinant measures the ratio of output volume to input volume, so a zero determinant tells us that such a collapse has occurred. And because the flattening has thrown away information in at least one dimension, we can never construct an inverse to recover that information. The rank of a matrix is simply the number of dimensions of the image space, while the nullity is the number of dimensions that are flattened. (Thus we have the rank-nullity theorem, which says that the sum of the two is the size n.)

Thus your teacher's skepticism is justified. But in some practical situations we will be satisfied with less than a full inversion. If we can only recover the dimensions that are not flattened, that's OK. The singular value decomposition of a matrix (of any shape) reveals both the rank and the nullspace of a matrix beautifully.

$A=U\Sigma V^{T},\,\!$

where A is m×n, U is an m×m orthogonal matrix, V is an n×n orthogonal matrix, and Σ is an m×n diagonal matrix with non-negative entries. (The diagonal entries of Σ are called the singular values of A.) Let Σ⁺ be Σ with each nonzero entry replaced by its reciprocal. Then we may define the Moore–Penrose pseudoinverse of A to be

$A^{+}=V\Sigma ^{+}U^{T}.\,\!$

This accomplishes what we hoped, and perhaps a bit more. For rectangular matrices whose columns (or rows) are linearly independent, we get the unique left (or right) inverse discussed earlier. For invertible square matrices, we get the ordinary inverse. And for singular matrices, we indeed get a matrix that inverts as much as we can. (This discussion also extends to matrices with entries that are complex rather than real.)

In applied mathematics, use of the pseudoinverse is both valuable and common. We need not wander into some remote and esoteric realm of pure mathematics to find it. So congratulations on your instincts; you may have a promising career ahead of you. (And, of course, congratulations to your teacher, whose instincts were also correct.) --KSmrq^T 10:36, 13 September 2006 (UTC)

Estimating a contour map from known spot heights.

Please tell me if I've got this right, or is there a better formula I could use?

I have ten spot heights scattered irregularly over a retangular map. I intend to estimate the height of every point on this map so that I can create a contour map.

I am going to estimate the height of each point by the weighted average of all the spot heights. The weight I am going to use is the inverse of the distance squared.

In fact I am going to use a further refinement - instead of just using the square, I am going to estimate the exact power to use by disregarding each of the ten spot heights in turn, and finding what the power is that best predicts the disregarded spot height from the nine known spot heights, and then calculate the arithmetic average of these ten powers.

So my weights will be 1/d^n where d is distance and n is the power. My questions are-

a) is there any better (ie more accurate estimator) formula to use for the weights than 1/d^n?

b) should I use an arithmetic average of the ten powers, or some other kind of average?

c) is there any better approach I could use, even though the scheme described above is easy to program? Thanks 62.253.52.8 23:55, 12 September 2006 (UTC)

I find the above idea interesting, but here's a totally different one. Get the delaunay triangulation of the ten points, then for every other point find the triangle it's in and just find the height from the plane the triangle defines. Your contour map would be all straight lines. A problem is that anything outside the convex hull of your spots is undefined, using the plane of the closest triangle should work. (You might be able to fix that and add curvature to the triangles using the adajacent triangles and spline interpolation, maybe.)

Are the points with known heights just random samples or were they chosen because they are relative maxima/minima ? Your method would work well in the latter case. However, the inability of your method to give elevations above the highest sample point or below the lowest sample point would be a problem if you are just using random points, resulting in a flatter geography than is really the case. StuRat 05:25, 13 September 2006 (UTC)

The spot heights are not max/min, so they must be the other choice. Actually they are not actually heights: I am interested in creating a contour map of house prices. The spot heights represent the prices in various towns. The house prices in the surrounding country are sometimes more, sometimes less. Another possibility would be to also weight the 'spot heights' by the population of each town, so I suppose I would get something like $w=p/d^{n}$ where p is population.

I see three problems here:

• There is no reason to think that house prices are a continuous function. In fact, houses prices frequently vary dramatically on the other side of some barrier, such as a river, highway, railroad tracks, or city/school district boundary. So, you are using a method that depends on having a continuous function when you don't actually have one, which will lead to poor results.

I am working with a large area, so the fine detail is unimportant. If I was working with just a town or city, then such step changes are the very things I would like my map to make clear.

• You can only compare prices on comparable houses. For example, 20 year old, 2000 square foot, 3 bedroom, 2 car garage houses. Otherwise, you are comparing "apples and oranges", so this doesn't tell you much about the premium/penalty of placing a home in that location. To distinguish between this "location premium" and the size and quality of houses built in an area, you might want to compare vacant lot prices. This should only give you the "location premium".

The statistical series I am working with are for a very large number of sales, not individual houses. The sub-series are for different types of house. I agree that prices may vary both by the quality of the house, and the favourableness of the location. This is less of a problem when comparing year on year price changes.

• If you only look at houses offered for sale, there might be a built in bias there, in that people will want to move more when something is wrong (basement floods continuously ?). Therefore, houses offered for sale may not be typical of the true value of houses in the area. StuRat 02:07, 16 September 2006 (UTC)

All the statistical series are for sold houses. I am in the UK: here the statistical series are I'm sure different from what you have in what I assume is the US.

Another suggestion (though I have no idea if it is a good one) is to match your data points to a polynomial of the form

$a+bx+cy+dx^{2}+exy+fy^{2}+gx^{3}+hx^{2}y+ixy^{2}+jy^{3}$

which, conveniently, has 10 coefficients. -- Meni Rosenfeld (talk) 11:04, 13 September 2006 (UTC)

Sorry, I don't quite understand this. Would a.....j be the spot heights, and x, y the position on the plane, or what please?

Thanks. Although Delaunay triangulation is attractive, I think it would be far too difficult to program - it would require several days I expect.

I did wonder if there would be any advantage in using weights based on formulas such as:

$w=1/(x+d^{n})$ where x is a constant. I think the magnitude of x would determine to what extent the estimated local height was based on the average of all heights, rather than those of the nearer spot heights. Are there any better formulas I could use?

I don't actually know how I would estimate x and n by regression or otherwise - could any help me please? Thanks.

There's a fundamental problem here. To pick a method of fitting to data, some additional requirements have to be applied. In sciences where there is a theory, the theory predicts a form for the data. Just add data, fit the form, and then you know the values of the (typically unspecified) constants. However, in the absence of a model (i.e. a form with some parameters identified as free for fitting), onehas a vastly wider range of equally bad solutions. Examples:

1. Construct the Voronoi diagram of the locations of the data points, then for each sample, raise it's height to the sampled altitude. This is a solution that from (at least) one point of view assumes the least: no intermediate altitudes are inferred and each point is the height indicated by the nearest sample.
2. Fit the bivariate polynomial suggested by Meni, above. This has the advantage that the result is smooth. It has the disadvantage that it can "blow up" going rapidly to "unreasonable" values outside the region where the data is provided.
3. Take the logs of your input heights, fit the bivariate polynomial to the logs, then take $\exp({\mbox{fit }})$.

This much freedom should indicate that there are a lot of ways to do this, and which one is right depends on what a priori information you have (or even suspect) about the result. This is similar to trying to capture imprecise or subjective priors in Bayesian analysis.

The method you describe of omitting some data to evaluate a proposed model is a good one (q.v. Cross-validation), but... Ten data points is small to begin with, so one should expect large random effects caused by small sample sizes using this method. It may turn out that your data is "very nice" meaning that the method (surprisingly) works well.

Given that, the method for gluing the (slightly) mismatched exponents is also a problem in model choice. Perhaps a better method would be to use Bayesian inference on your 10 subsets to estimate the maximum likelihood choice for the exponent. -- Fuzzyeric 01:58, 15 September 2006 (UTC)

I have been wondering how I should best average ten different formulas of the form $1/(x+d^{n})$. I'm not sure if taking the arithmetic average of x and perhaps the geometric average of n would necessarily be the best thing to do.

To clarify my polynomial, if you choose to give it a try - x and y are indeed the coordinates on the map, but a...j are coefficients you need to find. By substituting in the polynomial the x, y coordinates of a known point, and equating it to its known height, you get an equation in a...j. By doing it for all 10 points, you'll get ten equations, which you can solve. -- Meni Rosenfeld (talk) 04:40, 15 September 2006 (UTC)

September 13

ANSWERED -> Friday the 13th

Here is a question. What proportion of the 13th of a month is a Friday? Is it 1/7 as most people would expect? My gut feelings tells me that I need to find a formulae that turns (YYYY,MM,DD) into the day of the week. Is there such a formulae? 202.168.50.40 04:50, 13 September 2006 (UTC)

There's calculating the day of the week. Frencheigh 05:11, 13 September 2006 (UTC)

In the long run, of all the 13th days of the month, 1/7th will be Fridays, yes. However, when looking at shorter time periods, like a year, the ratio may be quite a bit different. StuRat 05:18, 13 September 2006 (UTC)

That's not true. I've made the calculation, and the proportion is actually 43 / 300. -- Meni Rosenfeld (talk) 05:32, 13 September 2006 (UTC)

OK, that makes it 43/300 instead of 43/301 (one seventh). A small, but significant, difference. I stand corrected. StuRat 15:09, 13 September 2006 (UTC)

That's because not only the structure of the years repeat every 400 years, but, as it it happens (I've checked), the days of the week also repeat. So you don't get a true random distribution. For reference, the number of times sunday through saturday are the 13th of the month in a cycle of 400 years is {687, 685, 685, 687, 684, 688, 684}. -- Meni Rosenfeld (talk) 05:38, 13 September 2006 (UTC)

And, as you'll notice, Friday actually has a tiny advantage over the other days ... spooky, no? Confusing Manifestation 10:30, 13 September 2006 (UTC)

I've done the calculation once (with a computer of course), and indeed, as User:Meni Rosenfeld says, I've got that Friday was the most frequent day for 13th. – b_jonas 10:51, 13 September 2006 (UTC)

Perhaps it's also worth noting that the cause of all this mess is the fact that the number of days in the 400-year cycle, 146,097, is divisble by 7. Otherwise we would get a perfect 1/7 for questions like this. -- Meni Rosenfeld (talk) 19:49, 13 September 2006 (UTC)

proability

I need some help with a problem that has been bugging me for a while: Suppose the odds of an event happening are 1/n. If I repeat the event n times, what are the odds that the given event will happen at least once? For example, if I flip a coin twice, the odds are 3/4 that at least once I get heads. If I roll six dice, what are the odds of at least one landing on 1? If I pull one card at random from 52 different decks, what are the odds that at least one will be the ace of spades? I know the answer involves calculating the odds that it WON'T happen every trial, but that's as far as I got. Thanks! Duomillia 15:32, 13 September 2006 (UTC)

You're on the right track. The odds of "getting something at least once" are the same as one minus the odds of not getting it at all, and that's an easier thing to compute. For the coins, the odds of not getting heads are 1/2 per flip, so the odds of that happening twice are (1/2)*(1/2) = 1/4, so the odds of getting heads at least once is 1 - 1/4 = 3/4 (as you said). For the dice, the odds of not getting a one are 5/6, so the odds of that happening six times are (5/6)^6 ~ 0.335, so the odds of getting at least 1 one is 1 - (5/6)^6 ~ 0.665. You should be able to see the pattern from here. -- SCZenz 15:40, 13 September 2006 (UTC)

(Edit conflicted.) Yes, you're exactly right. This is classic introductory probability theory. If the odds of something happening are p the odds of it not happening are q = 1 - p. The odds of it not happening twice are q². The odds of it not happening n times are qⁿ. So to get the odds of it happening at least once (but possibly more) it is 1 - qⁿ = 1 - (1 - p)ⁿ. The reason it is easier to calculate with the odds of it not happening is that if you calculate the odds of something happening in multiple independent trials, you have to account for when it happens (did it happen the first time or the fourteenth?) and how many times it happened. But to calculate the odds of it not happening, you just calculate the odds of the same thing (not) happening at each trial. See binomial distribution for (much) more. –Joke 15:41, 13 September 2006 (UTC)

To put it all together, the general answer is $1-\left(1-{\frac {1}{n}}\right)^{n}$. For large n, this is roughly equal to $1-1/e\approx 0.63212\dots$. -- Meni Rosenfeld (talk) 15:43, 13 September 2006 (UTC)

I think you mean 1-1/e, which would be about 0.63212. Black Carrot 19:36, 13 September 2006 (UTC)

(getting rid of evidence) Yeah, that's what I said, 1 - 1/e. (evil grin) :-) -- Meni Rosenfeld (talk) 19:44, 13 September 2006 (UTC)

So, as n approaches infinity, the odds of n in n trials approaches 63% and a little? Duomillia 21:31, 13 September 2006 (UTC)

Yeah. That's cool—and see how close it is already for n=6... -- SCZenz 22:16, 13 September 2006 (UTC)

Well, 6 is a big number in the "1, 2, 3, lots, many" system ;-) -- AJR | Talk 23:18, 13 September 2006 (UTC)

In the Discworld troll counting system, they go "1, 2, 3, many, many-one, many-two, ..., many-many-many-three, lots". Although the article does point out the alternative of "one, two, many, lots" and not bothering about the rest of the numbers. Confusing Manifestation 03:45, 14 September 2006 (UTC)

September 14

Unexpected symbol behvior w/ LaTex

Would someone please take a look at the following page:

http://en.wikipedia.org/wiki/Beer-Lambert_law

I created a new section entitled "Derivation" with several equations. Some look fine, but there are two types of problems with others ...

1. The equation font size differs depending on whether there is a fraction in the equation or not. When there is a fraction, the equation looks fine (e.g. 'Absorbance' or 'Transmittance"). In equations without a fraction, however, the symbols are almost too small to read. How can I make all of the math fonts the same?

2. Several of the equations have a small hyphen "-" at the end, and I can't figure out how to get rid of them.

I haven't used LaTex before, so that may be the problem, except that I don't see the problem occurring in the first section, 'Equations' that was written by someone else.

Thanks for any help,

Axewiki 00:25, 14 September 2006 (UTC)

I've fixed the first problem - the thing is, the software Wikipedia runs on has a default setting that if it can turn an expression in <math> tags to regular text and still look ok, it will, otherwise it will turn it into a PNG image. However, the definition of "looking ok" doesn't always work, especially if you're trying to make a bunch of equations look the same, so sometimes you have to trick it into making the equation into a PNG by adding a small space in the form of \, to the end of the equation. I got that trick from m:Help:Formula#Forced PNG rendering. As to the second, I seem to remember seeing that kind of thing happen elsewhere, but I can't remember the fix. Confusing Manifestation 01:32, 14 September 2006 (UTC)

Thanks ConMan! Parts of it are much better ...

Axewiki 02:16, 14 September 2006 (UTC)

Cylindrical sections

I am looking for information on cylindrical sections, i.e., ellipses; specifically, I want to determine what kind of a curve the ellipse will map to if the cylinder is unrolled after a cut in its long dimension (it appears to be a sine or cosine, based on a pencil rubbing of a wood model I have). I have searched the web and Wikipedia, and have not found much on this subject.

If the angle of the cut is 45 degrees (the case I am interested in), then the long-axis, or height, dimension (assuming the cylinder is standing on its end) is a simple function of the ellipse's x coordinate, when plotted in the ellipse's own plane, but the horizontal dimension on the unrolled cylinder is much more difficult, and I have not been able to visualize it well enough to calculate the formula for the transformation. It equals the length of the circular arc subtended at that height, but that depends on the angle, which is difficult for me to see. (I hope that makes sense.) Thanks in advance for any help you can give.

-- Ed

On the cylinder, we can pick our coordinates so the z-axis is coincident with the axis of the cylinder and the $r-\theta \$ plane is at the bottom face of the cylinder. Suppose the radius of the cylinder is R, and the height of the cut is $a+b\sin(\theta )\$. Converting to Cartesian coordinates with the same z-axis and (positive) x-axis pointing along the ray $\theta =0\$, we find that the point $\{R,\theta ,z\}\$ is mapped to the point $\{x,y,z\}=\{R\cos(\theta ),R\sin(\theta ),z\}\$. So the cut is mapped as $\{R,\theta ,a+b\cos(\theta )\}\rightarrow \{R\cos(\theta ),R\sin(\theta ),a+b\cos(\theta )\}\$. But that's just $\{x,y,a+{b \over R}x\}\$ (where we've used a notational dodge to roll it for us). This tells us that the cut is straight (not wavy) (equivalently planar) and sinusoidal when unrolled. -- Fuzzyeric 04:58, 14 September 2006 (UTC)

Fun Game

Someone in my math club proposed a variation on Nim: instead of removing n objects from a heap, you remove n objects from each of m heaps, where m and n are at least 1. Anyone have suggestions on solving it? Black Carrot 04:49, 14 September 2006 (UTC)

I'm assuming you mean "in a given step, from each remaining heap remove n objects", "there are initially m heaps", "the m heaps may have different starting sizes". Use Bouton's method, described in the article you link, and find a solution driving t to zero in each heap simultaneously. In general, this will require taking large counts to make most of the heaps vanish so that the remaining heaps can be simultaneously optimized.

If you mean one takes n objects, distributed as one likes from m heaps, then the distinction into m heaps is superfluous since there's effectively only one heap.

If you mean one must take >1 object from each of m heaps and the first player to deplete any heap is the loser, then use Bouton's method on each heap independently.

If n is upper bounded, one may not be able to reach a t=0 state in one move. For instance if the bound is <m and you must draw an object from every heap to reach a winning intermediate state. Then, similar to some more complicated games, the first moves are sort of random and the loser is the first player to make a mistake noticed by his opponent. In fact the early play is probably to mini-max the ability of the other player to drive the game out of the non-winning equilibrium. -- Fuzzyeric 05:10, 14 September 2006 (UTC)

No, I meant "in a given step", from each of "as many heaps as you feel like (m)" remove n objects. The starting condition could be whatever the players agree upon. So, for instance, from (3,2,1) I would take one from each (2,1), they might take the entire first heap (1), then I would take the remaining one and win. Or from (7,3,3,1,1) I would take three from each of the first three heaps (4,1,1), they might take one from the first and last (3,1), I would take one from the first (2,1), and it would end the same way as before. Black Carrot 15:32, 15 September 2006 (UTC)

Then you mean the variation described above as "If you mean one takes n objects, distributed as one likes from m heaps, then the distinction into m heaps is superfluous since there's effectively only one heap." This is equivalent to one heap as there is no consequence to heap boundaries. Use Bouton's method on the total number of objects, regardless of heap membership. -- Fuzzyeric 04:02, 16 September 2006 (UTC)

If that was what I meant, that's what I would have said, and I wouldn't have had any trouble solving it myself. Where did I say it could be distributed as you like? The same amount (n) must be removed from each of as many stacks as you like (m). To put it visually, it's like yanking out a rectangle of arbitrary dimensions, and it's been devilishly difficult to solve. If you had stacks (7,3,3,1,1) as above, for instance, under your interpretation, I could just remove all of them and have done with it. Under the one I've been trying to explain, you could remove 7, 6, 5, or 4 from the first one, or 3 or 2 each from any of the first three of them, or 1 each from any of them. Reduced to two stacks, it simplifies to Wythoff's Game. Black Carrot 22:02, 18 September 2006 (UTC)

Recurrence Relations

The recurrence relation article doesn't talk about the cases of nonlinear recurrence relations. Could anyone indicate me what would be standard methods for solving relations such as $u_{n}=f(u_{n-1})$ where $f(x)={\frac {1}{1-x}}$ (just an example, I don't really want to know what that particular solution is). What could the function f(x) be that would leave the recurrence relation relatively easy to solve (no Z / Laplace transforms or others...). I'm also interested in the inhomogeneous cases and thoses where the degree is greater, but I'll start with this... --Xedi 21:01, 14 September 2006 (UTC)

There should be an article on the "cycle structure" of solutions of $u_{n}=f(u_{n-1})$ ... perhaps somewhere near Chaos theory? In any case, for Möbius transformations, such as the one you mention, the solution is fairly simple, and contained within that article. As $u_{n}=f^{(n)}(u_{0})$, there are "obvious" properties; e.g., if f is injective, so is f⁽ⁿ⁾, and there are various techniques that work in the vicinity of a fixed point of f. Other than that, solutions usually depend on a change of coordinates; finding a function g such that, e.g. f(g(x)) = g(x + 1) (at least locally), leading to f⁽ⁿ⁾(x) = g(g^-1(x) + n).

Thanks for the swift reply, I did actually notice the periodicity of the recurrence relation I gave where $f(x)={\frac {1}{1-x}}$. I'll try to understand clearly everything and then come back. Thanks again ! --Xedi 21:42, 14 September 2006 (UTC)

September 15

Quadratic forms

What is the typical way of working with quadratic forms with added linear terms? Diagonalization of the associated matrix, and then folding in the linear terms through translation?

I have not spent too much time thinking about this yet, but I wonder whether it would be more elegant to consider x^T = (1 x y), and defining a 3 by 3 matrix M so that x^T M x = 0 is what we want to work with. Is this what is typically done for such things?--HappyCamper 00:33, 15 September 2006 (UTC)

Depends what you're doing. Examples:

• projective (homogeneous coordinates): ${\begin{pmatrix}x&y&1\end{pmatrix}}\ {\begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix}}\ {\begin{pmatrix}x\\y\\1\end{pmatrix}}$ , where the matrix is usually restricted to be symmetric or triangular
• polynomial: ${\begin{pmatrix}x&y\end{pmatrix}}\ {\begin{pmatrix}a&b\\d&e\end{pmatrix}}\ {\begin{pmatrix}x\\y\end{pmatrix}}+{\begin{pmatrix}j\\k\end{pmatrix}}\ (x\ \ y)+m$ , where the matrix is almost always symmetric.
• eliminate the linear terms -- the method is usually taught in analytic geometry as the method of moving conic sections into standard form -- i.e. rotate to eliminate cross terms, complete squares, and represent the result as, for example, ${\frac {(x-h)^{2}}{a^{2}}}+{\frac {(x-k)^{2}}{b^{2}}}\ =\ 0$ .

From the point of view of the theory of quadratic forms inclusion of linear or constant terms is an affine transformation, suggesting that the first or third is more natural when extending from that theory. However, it is equivalent to the polynomial form and sometimes the polynomial form is more convenient for expressing certain proofs. Despite this, both of the displayed forms above are inconvenient for cubes and higher, so the Einstein notation is commonly used. This notation is fundamentally equivalent to the polymer form above, but is both more compact and easier to work with (with some practice, because initially you don't realize just how many relations you've written down with one equation). -- Fuzzyeric 02:28, 15 September 2006 (UTC)

By definition, a quadratic form has no linear terms. But suppose we have a quadratic polynomial in two variables, such as one used for the implicit equation of a rotated and displaced ellipse,

$0=481x^{2}-216xy+544y^{2}-1920x-1440y-6400.\,\!$

Using homogeneous coordinates we can rewrite this so all terms have total degree 2.

$0=481x^{2}-216xy+544y^{2}-1920xw-1440yw-6400w^{2}.\,\!$

Et voila! We have a polynomial of degree 2 which is homogeneous in the sense that all terms have the same degree, and thus a valid candidate for a quadratic form in three variables. As with any quadratic form, we can convert this to a bilinear form, and hence to a symmetric matrix, your proposed M.

$0\,\!$	${}={\mathbf {x}}^{T}M{\mathbf {x}}\,\!$
	${}={\begin{bmatrix}w&x&y\end{bmatrix}}{\begin{bmatrix}6400&-960&-720\\-960&481&-108\\-720&-108&544\end{bmatrix}}{\begin{bmatrix}w\\x\\y\end{bmatrix}}$

Indeed, this is a convenient and popular technique in applications like computer graphics. The matrix has uses that are not immediately obvious. For example, suppose p is a point outside the ellipse; then

$0={\mathbf {p}}^{T}M{\mathbf {x}}\,\!$

is the equation of a line intersecting the ellipse in the two points where a ray from p is tangent. (This line is called the "polar" of "pole" p.) If p is on the ellipse, then this line goes through p and is tangent to the ellipse there. Or consider the last column of M; it gives the homogeneous coordinates for the center of the ellipse.

So congratulations, your instincts in this are quite fruitful. --KSmrq^T 05:29, 15 September 2006 (UTC)

Thanks for the elaborate responses - this is quite helpful. I think I am most interested in the homogeneous coordinate representation. What I am unsure of, is the interpretation of this result...

From above, we have ${\mathbf {x}}^{T}M{\mathbf {x}}=0\,\!$. Now, since M is symmetric, there exists an orthogonal matrix O such that ${\mathbf {x}}^{T}{\mathbf {O}}^{T}{\mathbf {D}}{\mathbf {O}}{\mathbf {x}}=0\,\!$. D is a diagonal matrix which will tell us everything about the conical intersection centered at the origin. Now, what does the product ${\mathbf {O}}{\mathbf {x}}$ mean? This seems to be an affine transformation on the original coordinate system. However, is it not true that we are transforming the identity element as well when this is done? This is what seems odd. --HappyCamper 16:05, 15 September 2006 (UTC)

From Morse theory every quadratic equation is equivilent to Ax²+By² when translation and rotation are removed. The equation will generally have one critical point, ${\mathbf {p}}$ say. What happens with the transformation is that ${\mathbf {O}}{\mathbf {p}}=0$, ${\mathbf {D}}{\mathbf {O}}{\mathbf {p}}=0$ and ${\mathbf {O} }^{T}{\mathbf {D} }{\mathbf {O} }{\mathbf {p} }={\mathbf {p} }$, that is it leaves the critical point stationary.--Salix alba (talk) 19:10, 15 September 2006 (UTC)

Equivalently, the matrix O rotates the coordinate unit vectors to be parallel to the eigenvectors of the matrix. Equivalently, rotates the matrix to be diagonal (each ellipsoid axis is parallel to a coordinate axis). -- Fuzzyeric 04:05, 16 September 2006 (UTC)

The term "identity element" is misplaced. If (w:x:y) are the homogeneous coordinates of a point, each point in the plane is defined by an infinite number of triples, all non-zero scalar multiples, (σw:σx:σy), σ ≠ 0. That's the reason we often use colons (":") rather than commas (","); only the ratios matter. We are definitely not obliged to have w = 1. In fact, we have a whole line of points with w = 0, sometimes called "points at infinity".

Although we can diagonalize M with an orthogonal matrix Q, we can also use any invertible matrix A. For example, let

$A=RTS={\begin{bmatrix}4/5&-3/5&0\\3/5&4/5&0\\0&0&1\end{bmatrix}}{\begin{bmatrix}1&0&3\\0&1&0\\0&0&1\end{bmatrix}}{\begin{bmatrix}5&0&0\\0&4&0\\0&0&1\end{bmatrix}}.$

This will diagonalize σM, where σ = 10⁻⁴ and M is the matrix that I used as my example, producing a circle x²+y²−w². (Note that the scalar σ has no effect in homogeneous coordinates.) We see that A is merely an affine change of coordinates, combining rotation, translation, and non-uniform scaling. The matrix has been reduced to a diagonal form involving only +1, −1, and 0, revealing the signature of the quadratic. Acting on points p, the effect of A is to transform them into a coordinate system where points on the ellipse become points on an origin-centered unit circle. --KSmrq^T 05:37, 17 September 2006 (UTC)

This description is quite useful. The note about Morse theory is quite intriguing too. So it seems, a better way to think of this, is that in homogeneous coordinates, the introduction of w induces a convenient parameterization of the problem. I must read up on this more. This is far, far, far too interesting to be missing out on. --HappyCamper 03:05, 18 September 2006 (UTC)

ANSWERED -> The red car or the blue car?

The terrorists has stolen a nuclear device and escaped with the help of a truck. Chasing after them in his Lotus is James and his two gadgeteers Q and R.

Catching up to the truck in a back alley way, James watched helplessly as the terrorists abandoned the truck for two getaway cars. The red car sped away to the east while the blue car sped away to the west.

Luckily Q and R aim their prototype radiation detector (disguised as CANON and NIKON digital SLR cameras respectively) at the two separate cars as they disappeared in the distance. Q aimed at the red car while R aimed at the blue car.

Since there is only one nuclear device, it's not certain which car contains the nuke.

Q said "My detector has detected the nuke in the red car. If the red car do indeed have the nuke, my detector will say so in 85% of the time!"

R injected "But if the red car do not have the nuke, Q's detector will give a false positive 70% of the time!"

R said "My superior detector did not detected any nuke in the blue car. If the blue car does not have the nuke, my detector will give a negative (or true reading) 98% of the time. On the otherhand if the blue car do have the nuke, my detector will also give a false negative 60% of the time."

Question: Which car should James go after?

210.49.155.134 10:04, 15 September 2006 (UTC)

Here's the chances each result would be found under each scenario:

Scenario 1     Q results   R results
==========     ==========  ==========
Red has nuke   85% chance
Blue doesn't               98% chance

Scenario 2 
========== 
Blue has nuke              60% chance
Red doesn't    70% chance

I would multiply the results, to get a (0.85)(0.98) or 83.3% chance we would get both the Q and R results if Scenario 1 is correct, and a (0.60)(0.70) or 42% chance if Scenario 2 is correct. Let's normalize those results to get 83.3/(83.3+42) or around a 66.5% chance the red car has the nuke and 42/(83.3+42) or around a 33.5% chance the blue car has the nuke. StuRat 10:25, 15 September 2006 (UTC)

Recommend using Bayesian analysis to estimate the posterior probability that the bomb is in each car given the data. This explains the method used by StuRat. -- Fuzzyeric 04:12, 16 September 2006 (UTC)

Since the problem involves the event 'positive Q' and 'negative R readings', we need to know whether Q and R detectors are independent in order to solve the problem. (Igny 16:13, 17 September 2006 (UTC))

I things it is implied that the detectors are independent. M.manary 17:20, 17 September 2006 (UTC)

September 16

Exponential functions

Take the simple equation of an exponential function.

y = a^x

why can't "a" equal 1 or be a negative real number?

a can be 1 or negative. 1^x is always 1. (-3)^x = (-1)^x3^x, for example. If you're working off the definition of the Exponential function article, we just call exponential functions with base a those with positive base a; it doesn't mean it can't happen. See also on that page "On the complex plane", which goes into more technical reasons for this. Dysprosia 05:35, 16 September 2006 (UTC)

I thought when a = 1 (the equation now being y = 1^x), if this was now drawn onto a graph there would only be a straight line no matter what x equals. Also if the equation becomes y = (-a) ^x, what happens then? Please answer this on a maths B level.(medium level)i don't want to know the technical side.

What's wrong with a straight line? It's a perfectly legitimate function. However, if you have the function (-a)^x = (-1)^xa^x (a positive of course), you will find that it will only be a real number if x is an integer, but that doesn't mean that there's anything really wrong, because we can still find values for the function (that's the advanced bit). Dysprosia 06:25, 16 September 2006 (UTC)

In more detail, consider two specific examples of a.

• If a = 1, then y = a^x becomes y = 1^x, which is the constant function y = 1. This is a function whose graph is a horizontal line, which is both reasonable and useful.
• If a = −1, then for x = 1⁄2 we have y = √(−1). As you may be aware, and can easily verify, no real number squares to −1, because negative times negative yields positive, as does positive times positive. Although we can switch to more sophisticated tools, especially complex numbers, the conclusion is that negative values of a break the definition for most values of x.

This latter issue comes up in the context of one generalization of the Pythagorean distance function, (x²+y²)^1/2. Everywhere a "2" appears, replace it with a real number, p, with the stipulation that p is at least 1. This almost works, but breaks for the reason we just saw. Instead we use

$(|x|^{p}+|y|^{p})^{1/p}.\,\!$

By ensuring non-negativity, we get a well-defined family of interesting distance functions, including Pythagorean distance as the special case p = 2. --KSmrq^T 04:48, 17 September 2006 (UTC)

Actually, if I'm interpreting the question right, if you have a = 1, it's just a straight line and doesn't really exhibit any of the properties of an exponential function. Like others said, if you have negative a, you get weird things happening in-between the "nice numbers". Even if you just look at the integers (using (-2)^x as an example), you get something like "1, (-2), 4, (-8), 16..." which also isn't really so "exponential". —AySz88\^-^ 05:22, 17 September 2006 (UTC)

The only thing that fails for 1^x is the derivative function; everything else holds trivially. You could interpret (-2)^x at the integers to be "exponential-like" in that the terms oscillate exponentially, but that's not really something formally used. Dysprosia 08:36, 17 September 2006 (UTC)

... also the flatness of the graph when a=1 means that the inverse function $x=log_{a}(y)$ is not well-defined for a=1 i.e. you cannot take logs to base 1. Gandalf61 09:55, 17 September 2006 (UTC)

Well, don't forget we're dealing with "medium level" math which is probably pre-calc, and their description of "exponential function" is probably more based on the really representative case. But yeah, it's kinda misleading to completely exclude someting like 1^x from the set of "exponential functions", instead of saying something like "1^x is just a really boring exponential function"... —AySz88\^-^ 19:23, 17 September 2006 (UTC)

a⁰ is equal to one, right? --AstoVidatu 19:52, 17 September 2006 (UTC)

$\lim _{x\to \infty }1^{x}$ and $\lim _{x\to \infty }x^{0}$ are both indeterminate forms and needs to be evaluated. M.manary 21:12, 17 September 2006 (UTC)

Rephrasing the problem to coincide with the domain of definition provided in Exponential function, i.e. that $\!\,a^{x}=e^{x\ln a}$, for a > 0. The requirement that a>0 is equivalent to the requirement that natural logarithm in the definition yields a Real number. This comes to two cases:

Case a = 0... Consider 0^x. As x -> 0, this has the value 0. Contrariwise, a^0 -> 1 as a -> 0. Thus, the function a^x is discontinuous at 0^0 and so the value at 0^0 is a matter of choice. We may associate this behaviour with the essential singularity in the complex logarithm at zero.

Case a < 0... The complex logarithm has a branch cut along the negative real axis. A consequence is that the value of the logarithm may be continuously continued (analytic continuation) from its values along the positive real axis to values along the negative real axis. But, you get different answers if you go clockwise or counterclockwise. So there's not one answer on the negative real axis. Again, there's a choice. Typically, the "resolution" is to switch from the Argand plane (complex plane) to the Riemann surface for the logarithm, which stacks an infinite number of values over each point in the complex plane. This method extends logarithm to a multi-valued function. Then, which value you take for the value of the logarithm on the negative real axis depends on which branch of the Riemann surface you're on -- i.e. whether you continue in the clockwise or counterclockwise direction and how many times you go around zero. (This method of going from one copy of the complex plane to another induces a group structure on the Riemann surface modulo the complex plane. For the logarithm function, this group is infinite and isomorphic to addition on the integers.)

To sum up... We require a > 0 so that we can have the result be single-valued (under the conventional orientation of log's branch point and branch cut). We may extend to all complex a (except zero), but then logarithm generates an infinite number of values per input and the result of the exponentiation is multiple-valued. (... unless we take another step back in generality and don't project the Riemann surface onto the Argand plane.) -- Fuzzyeric 21:20, 17 September 2006 (UTC)

M.manary : No. Both $\lim _{x\to \infty }1^{x}$ and $\lim _{x\to \infty }x^{0}$ are equal to 1. What you meant to say is that $\lim _{x\to a}f(x)^{g(x)}$, where $\lim _{x\to a}f(x)=0$ and $\lim _{x\to a}g(x)=0$ (or $\lim _{x\to a}f(x)=1$ and $\lim _{x\to a}g(x)=\infty$, or $\lim _{x\to a}f(x)=\infty$ and $\lim _{x\to a}g(x)=0$) is indeterminate. Regardless, it is very common to define 0⁰ = 1, so a⁰ = 1 for every a. About a^x for negative a, a natural definition can be given if we restrict ourselves to real x:

$a^{x}=(-a)^{x}(\cos {\pi x}+i\sin {\pi x})\,\!$

-- Meni Rosenfeld (talk) 05:02, 18 September 2006 (UTC)

Nope. I had exactly the limits I wanted. Of course, neither of them were limits at infinity... -- Fuzzyeric 14:41, 18 September 2006 (UTC)

Is there any clearer way to specify that I am addressing User:M.manary than writing "M.manary : " at the beginning? -- Meni Rosenfeld (talk) 15:16, 18 September 2006 (UTC)

<sigh> Sorry. Maybe if it had been linked? -- Fuzzyeric 15:28, 18 September 2006 (UTC)

Fixed :-) -- Meni Rosenfeld (talk) 15:34, 18 September 2006 (UTC)

September 17

Multidimensional stochastic processes

I am studying stochastic processes in applications to biology. I need to find textbook/publications/other sources regarding multidimensional random walks and multidimensional diffusion. When I say multidimensional I mean that it is really multi: dimension is in thousands. The most popular textbooks, like Cramer-Leadbetter, Feller, Gardiner, Van Kampen, etc. usually after pronouncing word "multidimensional" immediately reduce the problem to dimesion 2, sometimes 3, cite the Polya result regardting recurrence and they are done. So far I failed to find a good treatment of even such simple problem as computation of diffusion coefficient in Fokker Plank equation in N dimensions, Can anybody help? Thank you

How about searching for things like "multidimensional random walk protein"? Does this help? Fokker-Planck equation might have something, but I suspect this is not what you need. --HappyCamper 02:52, 18 September 2006 (UTC)

(Edit conflict) Are you asking about a multidimensional generalization of $u_{t}=u_{xx}+u+x\,u_{x}$? This is called the Fokker-Planck equation in Bluman and Kumei and bears an obvious relation to the diffusion equation. This doesn't answer your question, but have you seen this book?:

• Berg, Howard C. (1983). Random Walks in Biology. Princeton University Press. ISBN 0-691- 00064-6. (1993 reprint)

---CH 03:00, 18 September 2006 (UTC)

(I added a header to the question so that it shows up in the contents. Feel free to improve the title if you can. – b_jonas 18:36, 18 September 2006 (UTC))

Currency matters

Hello sir/mam

what is yen? and how it is converted in rupees ? can u please explain me.

Yours faithfully,

ASHOK.

For yen, see our article Yen. To convert yen to rupees, go to a large bank with a bag of yen and have them convert to rupees. Or if you just want to know the equivalent value, use any of several currency converters on the Internet. --Lambiam^Talk 19:21, 17 September 2006 (UTC)

The easiest one is simply by using google. Type in "100 Japanese yen in Indian rupees" and you'll get a good result. See [2]. Oskar 01:14, 18 September 2006 (UTC)

Google can do that now? Neat. --HappyCamper 02:19, 18 September 2006 (UTC)

That's amazing. I wish I had known about that last week, when I was in Canada, and I could have easily done things like this. Chuck 16:15, 21 September 2006 (UTC)

Sinc Filter?

Could someone please explain to me the concepts behind a Sinc Filter especally how they relate to Anti-Aliasing and the function sin(x)/(x). Any input is appreciated, thanks. HP 50g 23:56, 17 September 2006 (UTC)

The sinc function in the time domain is a windowing function in the frequency domain. Browse around Fourier transform for some goodies - Continuous Fourier transform might be of use too. To prevent aliasing, you want to fit the frequency content of your signal of interest inside the entire window. --HappyCamper 02:19, 18 September 2006 (UTC)

A sinc filter is a lowpass filter (with a really, really sharp cutoff). If you are resampling something below a certain frequency (i.e. scaling an image down), applying a sinc convolution filter beforehand to the original would prevent aliasing of frequencies greater than the new nyquist frequency. - Rainwarrior 07:11, 18 September 2006 (UTC)

It's hard to answer briefly, since this topic is part of a much broader discussion. In digital signal processing, a core theorem is that the frequency response of a (time-invariant, linear) filter is the Fourier transform of the filter impulse response. That is, if we feed the filter a Dirac delta function, essentially an instantaneous unit spike, and record the output for all time, then the Fourier transform of that output describes the effect of the function across frequencies. (To get a sense impression of an impulse response, stand in a large empty room with eyes closed, sharply and loudly clap your hands once, and listen.) Another core fact is that such a filter can only scale or phase shift a sinusoidal input, but not change its frequency.

When a continuous signal is digitally sampled at periodic intervals, the frequency spectrum of the input is replicated periodically as well, with period inversely proportional to the sampling period. This can produce the phenomenon called "aliasing", where high frequencies masquerade as low ones. When this occurs, we can no longer reliably reproduce the continuous signal from the sampled one. To avoid this, we need to select a single period in the frequency domain, to filter the input before we sample. The ideal filter has a simple frequency response, one that looks like a square box.

So now the question is, what filter has a frequency response that looks like a box? Yep, you guessed it. And what is its impulse response? Right again. --KSmrq^T 11:17, 18 September 2006 (UTC)

Thanks for the responses. I be sure to read about the topics you told me about. Thanks. HP 50g 14:52, 18 September 2006 (UTC)

I award KSmrq with this mathematical equation for his comprehensive contributions to the Mathematics Reference Desk. :-) --HappyCamper 15:41, 18 September 2006 (UTC)

$f\left(t\right)={\frac {1}{\sqrt {2\pi }}}\int \limits _{-\infty }^{\infty }F\left(\omega \right)e^{i\omega t}\,d\omega .$

September 18

Arithmetic pre-Fibonacci

If Fibonacci introduced the Hindu-Arabic numeral system to Europe in the 13th century, what was used before? I presume that it was Roman numerals and Roman arithmetic. Was roman arithmetic actually done as listed in our article, or is that a "modern way that would work"? Are things like the Domesday book recorded in Roman numerals? What did bookkeeping look like in those days? -- SGBailey 09:36, 18 September 2006 (UTC)

I believe the Domesday Book used words for numbers ("one hundred and forty") instead of either Arabic or Roman numerals. StuRat 09:43, 18 September 2006 (UTC)

ANSWERED -> math questoin

if anything divided by 0 is infinity how come 0 divided by 0 is 1?

0/0 is undefined, as we have three different rules providing a paradox:

• Anything divided by zero is either positive or negative infinity.

• Anything divided by itself is one.

• Zero divided by anything is zero.

So, since the answer can't be simultaneously equal to all of those values, the answer is undefined. StuRat 12:20, 18 September 2006 (UTC)

[Edit conflict] More generally, if x is any number, then for all a other than 0, ${\frac {ax}{a}}=x$. If we want it to hold for a = 0 as well, we will get ${\frac {0}{0}}={\frac {0x}{0}}=x$. Hence you will sometimes see 0/0 mentioned as being equal to anything. This is usually little more than a memory trick - as a stand-alone expression, as StuRat explained, it is almost never defined. More details can be found in the article division by zero. -- Meni Rosenfeld (talk) 12:33, 18 September 2006 (UTC)

However, when the top and bottom of a fraction both approach zero, there may very well be an answer:

• X/X, as X approaches zero from either side, is equal to 1.

• X/2X, as X approaches zero from either side, is equal to 1/2.

• X/X^2, as X appoaches zero from the positive side, equals positive infinity.

• X/X^2, as X appoaches zero from the negative side, equals negative infinity.

• X^2/X, as X appoaches zero from either side, equals zero.

See L'Hopital's Rule for more details.

StuRat 12:27, 18 September 2006 (UTC)

Klein bottle is its own 2-fold covering?

There is such information on the main page of Mathematics portal ("Did You know..."), but there is no further information in the article "Klein bottle". Anybody has any further information? At first glance it seems impossible for a space with nontrivial fundamental group to cover itself...

Double cover indication

I've sketched up an easy construction of a double cover and it should be visible on the right. Note that the square on the right is one patch, just drawn twice. The square on the left can be put in bijection with the square on the right by translation. This situation is simple enough to demonstrate by diagram. Note that this can be continued to any n-covering by stretching even further (or by replacing sub-diagrams on the right by copies of the diagram on the right). This is equivalent to the n-covering of the torus by the torus. Unlike the torus, this construction only demonstrates odd coverings in the other direction (vertically in the drawing) but not even. -- Fuzzyeric 15:51, 18 September 2006 (UTC)

Abstract -> Practical

I guess this could go either here or at the science desk, but I imagine you guys will be of more help. Basically I'd like to know some more abstract mathematical concepts that have been discovered and understood mathematically before we knew there was a practical application. Sort of like non-euclidean geometry was worked out before we discovered that there actually existed the configuration in the universe. Thanks--152.23.204.76 15:00, 18 September 2006 (UTC)

How about things from quantum mechanics? Much of the theory can be quite abstract, but still useful. Stuff from operator theory say. Group theory is also quite useful in chemistry - you can classify the types of spectra molecules will have with group theory.

On the other hand, sometimes things go the other way around. Fourier series is one example. We have a whole field called Harmonic analysis which deals with understanding this marvelous thing. --HappyCamper 15:34, 18 September 2006 (UTC)

Yes, Fourier series seem like a perfect example (even though the order seems a bit off). I'd like specific examples from Quantum mechanics. I understand there is some complex math involved, but I'd like to read some more about examples of this. And in reply to Maelin, yep primes are also a good example along with anything involved with cryptography.--152.23.204.76 17:32, 18 September 2006 (UTC)

Prime numbers were studied extensively ever since the days of Pythagoras, but they didn't really have any practical applications until the invention of computers and public-key cryptography. Maelin 15:46, 18 September 2006 (UTC)

• Boolean logic didn't have much practical application until computers were invented. StuRat 18:26, 18 September 2006 (UTC)

• Wave packet theory, describing the location of electrons, is a good example. Mathematically, it says the position of electrons is indeterminate. Figuring out how this can be physically the case is a bit more challenging. StuRat 18:17, 18 September 2006 (UTC)

• More recently, string theory basically came from the math, not from observations. StuRat 18:17, 18 September 2006 (UTC)

For me, in addition to group theory, one of the classic examples is good old Hilbert space. Quantum mechanics is basically a theory about linear operators in Hilbert space, and many fields of applied mathematics (e.g. control, optimization and filtering problems in time series) are now formulated using Hilbert space techniques. –Joke 20:15, 18 September 2006 (UTC)

x+2^x=37

How do you do problems like these without using the guess and check method?

$x+2^{x}=37$

I know the answer. I just want to know if there is a better way to get is other than substituting random numbers for x and checking it. If, in other problems, x was irrational, guess and check would be useless in getting an excact answer. --Yanwen 20:54, 18 September 2006 (UTC)

This kind of problems cannot be solved with elementary functions. You'll have to use functions like Lambert W function (substitute $t=(37-x)\ln 2\,\!$). However, you can use Newton's method to quickly find numerical solutions. Also, if the numerical solution turns out to appear to be a nice number, you can check that this number actually is a solution. -- Meni Rosenfeld (talk) 21:24, 18 September 2006 (UTC)

We can solve equations algebraically by applying the inverse of the operation in the equation. If the x is being added onto, we add the opposite number to both sides of the equation to solve for x. If the x is being multiplied, we multiply both sides of the equation by the reciprocal number. If x is an exponent in an exponential equation, we take the logarithm of the appropriate base of both sides of the equation. If x is the base of a power equation, we take the appropriate root of both sides of the equation. If x is the angle in a trigonometric equation, we take the inverse trig function of both sides of the equation.

However if x occurs in more than one type of function in an equation, most of the time we have no way to solve it algebraically. However, sometimes there are methods to change a multifunction equation into a single-function equation. For example, in an equation with two different trig functions, we may be able to use a trig identity to rewrite it with only one trig function. In solving polynomial equations, we may be able to factor the polynomial in to linear factors and solve each one. The theory of polynomial equations has developed formulas and methods for solving any equation of degree 2, 3 or 4; and has shown that there are no algebraic methods to solve all polynomial equations of higher degree. MathMan64 23:27, 18 September 2006 (UTC)

About those high degree polynomial equations, what is really meant by "algebraic methods"? Or, from the other perspective, if it's impossible to solve it with "algebraic methods and taking roots" (as I believe it is, right?), what methods are left? —Bromskloss 11:22, 19 September 2006 (UTC)

"Algebraic methods" usually mean addition, subtraction, multiplication, division, powers and roots. The "other methods" which are applicable to solving algebraic equations are, for example, Jacobi's elliptic functions. -- Meni Rosenfeld (talk) 11:30, 19 September 2006 (UTC)

For integer solutions, Category:modular arithmetic can (sometimes) work. In the given instance, the equation has to be tru to any modulus, so we might be able to lift a solution all the way to the integers...

First, if x<0 and an integer, then the left-hand side of the equation is negative, so x>=0. Also, x=0 isn't a solution, so x is positive. Both terms on the left are >37 is x>37, so 0<x<=37.

$x+2^{x}\equiv 37{\pmod {2}}\$ implies

$x+0\equiv 1{\pmod {2}}\$, or "x is odd". Replace x with 2xx+1.

xor $x=0\land 0+1\equiv 1{\pmod {2}}\$ (which isn't a solution)

$(2xx+1)+2^{2xx+1}=37\$ is equivalent to

$xx+2^{2xx}=18\$, taking mods again...

$xx+2^{2xx}\equiv 18{\pmod {2}}\$ implies

$xx+0\equiv 0{\pmod {2}}\$, or xx is even. Replace xx with 2xxx.

$xxx+2^{4xxx-1}\equiv 9{\pmod {2}}\$ implies xxx is odd.

$xxxx+2^{8xxxx-1}\equiv 4{\pmod {4}}\$ implies xxxx is a multiple of 4.

Now x = 2xx+1, xx = 2xxx, xxx = 2xxxx+1, and xxxx=4k for some k. So, xxx = 8k+1, xx = 16k+2, and x = 32k+5. The only values of k compatible with 0<x<=37 are k=0 or k=1. Equivalently, x=5, or x=37. Five works. Thirty-seven doesn't.

Therefore, x = 5 is the solution in integers. I also know that there are an infinite number of solutions in the complex plane, but this method won't find them without considerable extension. -- Fuzzyeric 16:17, 19 September 2006 (UTC)

September 19

Outlier test for a inverse squared distribution

I'm analyzing the number of inbound links to news articles. The number of links to articles follows an inverse power law; the number with zero is impracticably large to count, the vast majority with any inbound links have only one link, far fewer have two, a rare handful have, say, 10 inbound links. The distribution changes from day to day, on a slow news day there will be few or no outstandingly popular articles, on a fast news day there may be many.

What'd I need is a statistical test for outliers on a given day. The distribution of quantities of inbound links is not normally distributed, and the depth of my statistical knowledge ends with normal distributions. I can't just pick an arbitrary threshold value because of the fast news day effect or other background effects like the internet growing larger, etc.

One crude idea I had was to copy the distribution for a given time period and reflect it over the y-axis, simulating the shape of a normal distribution, then calculate the standard deviation and use a standard test based on that. I can hear the mathematicians cringing.

Can you give me a better way?

--Phig newton 01:41, 19 September 2006 (UTC)

Your distribution is discrete. If you believe that your distribution satisfies a power law, then you want the geometric distribution. Find the mean number of links, q. Set p=1/q. The CDF of this distribution is 1-(1-p)^n. Setting this equal to x, your confidence level (typically x = 0.95, 0.99, or similar number near 1) and solving for n gives ln(1-x)/ln(1-p). This is the number of links required to exceed x "percent" of the population of pages for that set of data. -- Fuzzyeric 18:26, 19 September 2006 (UTC)

I don't see how you go from a power law probability distribution (as observed in scale-free networks) to a geometric distribution. Power law distributions are heavy-tailed, quite unlike geometric distributions. --Lambiam^Talk 12:08, 20 September 2006 (UTC)

Lambiam, you're right. I plead brain damage. The discrete power law distributions are the zeta distribution and its finitary counterpart the Zipf distribution. These are much more irritating to work with.

The best way to estimate the exponent, s, is to log-log plot your data, and find the slope of the line of best fit through your data. I.e., if your data is {(a,b),(c,d), ... }, find the slope of the line of best fit through {(log(a), log(b)), (log(c), log(d)), ...}, skipping instances where either variable is zero, since you can't take its log. The use of the Riemann zeta function and (generalized) harmonic numbers makes this distribution awkward to work with. If you can estimate your s and your confidence level, x, I can expand the CDF and solve for n to use as your outlier criterion. I recommend doing this at my talk page. -- Fuzzyeric 05:45, 23 September 2006 (UTC)

You might want to have a look at this book: Barnett, V. (1984). Outliers in Statistical Data. New York: John Wiley & Sons. {{cite book}}: Unknown parameter |coauthors= ignored (|author= suggested) (help) Vectro 18:34, 19 September 2006 (UTC)

Basic Statistics, Lottery Odds

My generally math-capable brain takes a vacation when it comes to statistics. I can handle the first question but the second one is giving me pause. I could probably figure it by poking at the web but I'm much more interested in understanding why and how than what the actual answer is. Could anyone please help?

1. Say there is a lottery in which we have fifty balls, numbered 1 through 50. Five of the balls are selected, at random. What is the probability that an arbitrary set of numbers matches the randomly chosen numbers?

2. Say we have have a similar lottery, except that the arbitrarily chosen numbers much match the randomly chosen numbers in sequence, that is, the player must choose which number is selected first, which is selected second, etc. What are the odds then?

Thanks, VermillionBird 04:06, 19 September 2006 (UTC)

Given a set of 50 balls, choosing 5 of them, there are 50!/(5!(50-5)!) combinations and 50!/(50-5)! permutations, respectively. Given a specific set of 5 balls, there is 1 combination and there are 5! possible permutations. To figure the odds, divide the number of ways it can match the desired results by the total number of possibilities. Black Carrot 05:18, 19 September 2006 (UTC)

Permutations! That's the word I forgot. Thanks. VermillionBird 22:34, 19 September 2006 (UTC)

Please refer to Permutations and combinations

1. basic combination ${50 \choose 5}$

2. basic permutation ${\frac {50!}{45!}}$

210.49.155.134 11:06, 19 September 2006 (UTC)

As to your question' why': There is exactly 1 chance in 50 that the first ball matches. If it doesn't, we failed; but if it did match, there is exactly 1 chance in 49 that the next ball matches; etcetera.JoergenB 11:31, 19 September 2006 (UTC)

Probability and statistics is largely about learning to count. The usual path to learning is to start with small, discrete examples where we can count explicitly; then move to larger discrete examples where we apply formulae we developed on the small examples; then tackle continuous examples.

An important general rule is how to combine probabilities. Let's try your second example, scaled down. Assume we have 10 balls labeled with the numbers 1 through 10, and that we will draw 3 balls in sequence without replacement. That is, after we draw the first ball we do not replace it in the pool of 10, so it cannot be drawn a second time.

How many different 3-ball sequences can we draw? We have 10 choices for the first ball, so there are exactly 10 "sequences" of 1 ball. Now there are 9 balls left in the pool, any one of which may be the second ball in a 2-ball sequence. Finally, we have 8 balls left to choose from for the third ball in a 3-ball sequence.

To sum up, we observe that for each different first ball chosen, we have a set of 2-ball sequences. Furthermore, for each different second ball we have a set of 8 final choices. Therefore the tally is 10×9×8 distinct sequences of 3 balls chosen from 10.

Explicitly, let's try choosing 2 balls from 3. The possible sequences are

1 2

1 3

2 1

2 3

3 1

3 2

Our reasoning says we have 3 choices for the first ball, and 2 for the second ball, giving 3×2 possibilities, which agrees with the 6 sequences listed.

So the key idea here is the multiplication of possibilities. This occurs so often we have a "descending product" notation, the factorial function n! = n×(n−1)×(n−2)×⋯×2×1. To chop this after the first k numbers we divide by (n−k)!. Thus we have the general result that in choosing sequences of k balls from n distinct balls without replacement, the total number of distinct sequences (or "permutations") is

$P(n,k)=n\times (n-1)\times \cdots \times (n-k+1)=n!/(n-k)!\,\!$

For the 3-of-10 example, we have P(10,3) = 720 distinct sequences. Your second example essentially asks the question, "What are the chances we will pick the right one at random?". The answer for 3-of-10 is clearly 1 out of 720, or approximately 0.00139, so you'd better be very lucky!

Your first example actually requires a more sophisticated computation. In the tiny case 2-of-3, the equivalent question is, "How many ways can we eliminate 1 ball?", which is, of course, 3. But that won't work for 3-of-10, or for your example. (In the 3-of-10 case the answer is that we can draw 120 different collections, ignoring order.) Anyway, you say you can handle this (though I wonder), so I'll stop here. --KSmrq^T 12:54, 19 September 2006 (UTC)

No, you're quite right. I knew 2 should be less likely than 1 but I wasn't sure how. I thought I was underestimating the unlikelyhood of 2 when I was actually overestimating the unlikelyhood of 1. Thanks for your explanation. VermillionBird 22:34, 19 September 2006 (UTC)

Can the sine function be represented as an exponential function?

That is, $\sin x=N^{x}\,$

Solve for $N$.

--Ķĩřβȳ♥ŤįɱéØ 08:15, 19 September 2006 (UTC)

No, but you do have

$\sin {x}={\frac {e^{ix}-e^{-ix}}{2i}}$.

-- Meni Rosenfeld (talk) 08:26, 19 September 2006 (UTC)

And from this the "unhelpful" answer

${\frac {1}{2i}}\det({\begin{pmatrix}e^{i}&0\\0&1+e^{-2i}\end{pmatrix}}^{x})$

can be constructed. -- Fuzzyeric 18:37, 19 September 2006 (UTC)

The "unhelpful" answer evaluates to $(e^{i}+e^{-i})^{x}/2i=sin(1)^{x}$, which is the wrong thing. Here is a "better" unhelpful answer:

${\frac {1}{2i}}\operatorname {tr} \left[{\begin{pmatrix}e^{i}&0\\0&e^{-i}\end{pmatrix}}^{x}\right]$

–Joke 20:10, 19 September 2006 (UTC)

Nope, this evaluates to $(e^{ix}+e^{-ix})/2i=\cos {x}/i$. It's going to be difficult to get the minus sign required for sin. -- Meni Rosenfeld (talk) 20:24, 19 September 2006 (UTC)

I can't believe I wrote that exponentiation distributes over addition. <hangs head in shame>. Teach me to make last-minute "simplifications" before hittin the Save button. However, if it somehow did, this would work...

${\frac {1}{2i}}\det({\begin{pmatrix}e^{i}&0\\0&1+e^{-2i}\end{pmatrix}}^{x})$

$={\frac {1}{2i}}\det({\begin{pmatrix}e^{ix}&0\\0&1+e^{-2ix}\end{pmatrix}})$ (powers of a diagonal matrix)

$={\frac {1}{2i}}(e^{ix})(1+e^{-2ix})-(0)(0)$

$={\frac {1}{2i}}e^{ix}+e^{-ix}$

but it doesn't, so... (scribbles offline, and will be back in a bit) -- Fuzzyeric 23:37, 19 September 2006 (UTC)

Oh, that was dumb. At least I got $-i\cos(x)$. Not that this exercise has much point. –Joke 20:29, 19 September 2006 (UTC)

Finally remembered... Wrong trick. Should have used $\sin x={\mbox{Im}}(e^{ix})\approx {\mbox{Im}}(\ (0.540+0.841i)^{x}\ )$ -- Fuzzyeric 02:24, 20 September 2006 (UTC)

Ummm... anyway... there should be an infinite number of positive solutions for N, I think. Any N for which N^x slopes down, since N^0 starts at 1, and sin(x) continutally returns to 1, there's going to be a solution for x as long as it moves downward. And... x will be negative if N > 1. So, how about N >= 0 for a solution? As for x, there's going to be tons of solutions for it, but eventually they will all approach positions where sin(x)=0. - Rainwarrior 04:51, 20 September 2006 (UTC)

Or are you actually hoping to create a sine with an exponential function? At best it would be a terrible approximation. With several terms having coefficients it could be improved... but, ah, I suppose even a broken clock is correct twice a day, similar with this one. - Rainwarrior 05:00, 20 September 2006 (UTC)

Finding coefficients of a series

Suppose I have a function which can be presented in the form:

$f(x)=\sum _{k=0}^{\infty }a_{k}x^{-k}$

The coefficients a_k are unknown and need to be found. I can (using a different method) evaluate the function numerically for any integer x, but this is quite expensive computationally, especially if x is large. I also know that the a_k are rational numbers with reasonable denominators, so if I find that a coefficient is numerically close to a rational number, I can assume it is equal to it. Any ideas about an efficient method to find the coefficients? Thanks. -- Meni Rosenfeld (talk) 13:08, 19 September 2006 (UTC)

You can take

$g(x)=f(1/x)=\sum _{k=0}^{\infty }a_{k}x^{k}$,

and basically you need to find the coefficients of the Taylor series with the restriction of not evaluating function at small x. Maybe numerical differentiation will be helpful. Conscious 14:26, 19 September 2006 (UTC)

The formula you give is actually a Laurent series, which is similar to a Taylor series, but allows for negative powers as well. I have no idea if it makes any sense in your situation, but if you could extent your function to a holomorphic one, defined on the complex plane (or a subset thereof), perhaps you can use the standard procedure for finding the coefficients of Laurent series. —Bromskloss 15:14, 19 September 2006 (UTC)

Thank you for your response. Unfortunately, in practice this method seems to reduce to picking values x₀, ..., x_n, and solving the equations (for every 0 ≤ i ≤ n):

$\sum _{k=0}^{n}a_{k}x_{i}^{-k}=f(x_{i})$

and the numerical differentiation article does little to suggest which choice for the x_i will be optimal. Also, it seems to me that there should be a better method. Any additional ideas will be appreciated. -- Meni Rosenfeld (talk) 15:02, 19 September 2006 (UTC)

A regular grid is usually used for differentiation, so this would suggest $x_{i}={\frac {N}{i+1}}$ (N should be divisible by 1..n+1 if you can calculate f for integer arguments only). I'm not sure if it's optimal and if the method is efficient (though you get n coefficients from evaluating f in n points). As Fuzzyeric points out below, you can just use Lagrange polynomial (or any other form of interpolation polynomial) instead of solving the system. Conscious 15:45, 19 September 2006 (UTC)

Interpolation methods require that the function be smooth enough. If you know or assume this is the case, then...

If you can do some additional things to f(), then maybe:

• If you can (repeatedly) integrate and compute the residue of f() at zero, then you can use what amounts to the Taylor series expansion for this Laurent series. (Evaluate at zero, get a₀. Integrate. Evaluate at zero, multiply by a constant, get a₁. Iterate.)
• If you can analytically take the inverse Fourier transform of f() on a circle on the complex plane near enough to zero that no poles are contained in the circle, then you can do the above more or less all at once by application of Cauchy's integral formula.
• If you know that f() has only a semi-infinite or finite domain of definition, then fitting by polynomials, orthogonal on the domain of definition, is a good idea.

If you are unable to enable to augment the implementation of f(), then...

• You can replace x -> 1/x and use your polynomial interpolation method of choice. E.g. Lagrange polynomial.
• You can numerically sample f() on a circle around zero and, as above, apply Cauchy's theorem.

You may be able to bound the growth of the a_k if you find that f() doesn't converge inside a circle centered at the origin. (Equivalent to f(1/x) converges inside such a circle.)

You may be able to constrain the denominators of the a_k if you find that f() is simple p-adicly for some p. -- Fuzzyeric 15:16, 19 September 2006 (UTC)

Thanks. Regrettably, most of these suggestions won't work for me, as my ability to manipulate f is very limited - I can only evaluate it numerically for positive integers. Perhaps I was too optimistic about how efficient this process can be. I guess the most useful one will be the Lagrange polynomial, which can streamline the solution of my equations above. Any suggestions on how to pick the interpolation points optimally? -- Meni Rosenfeld (talk) 15:42, 19 September 2006 (UTC)

It sounds like you're limited to evaluating f(n). So to convert the Laurent series to a Taylor series, you'll be working with g(1/n) = f(n). You can only evaluate g() at a sequence of points between 0 and 1 (and accumulating at 0). If you actually had freedom to choose points, I'd recommend the Chebyshev roots, but you don't have this freedom. Since you don't really get to move your points around, I have two recommendations:

Rigorous: Polynomial interpolation refers to the Lagrange form of the interpolation polynomial (which I learned as a ratio of determinants). This is a direct method for acquiring the coefficients.

Less than rigorous: I'd recommend selecting evaluation points to be linearly independent in their prime power vectors. (By the Fundamental Theorem of Arithmetic, there's a bijection between integers and the sequence of exponents of the primes in their prime decomposition. For instance 15 = 2^0 3^1 5^1 <-> {0,1,1, 0,0,0,...}). Make sure these vectors are linearly independent. A way to do this is to select integers that are powers of an irrational. Historically, powers of the golden mean are popular. (Probably due to convenient convergence properties for subdividing root-finding intervals.) -- Fuzzyeric 16:36, 19 September 2006 (UTC)

Okay, I guess this is enough information to get me going. Thank you all for your help. -- Meni Rosenfeld (talk) 17:37, 19 September 2006 (UTC)

Just to chime in...I wonder if you could make use of the z-transform of the function somehow? --HappyCamper 03:05, 20 September 2006 (UTC)

This won't be practical for me, no. -- Meni Rosenfeld (talk) 05:33, 20 September 2006 (UTC)

TI 83 Substitute

I am in need of a website that has functions similar to a TI 83 Calculator...or better, a TI 83 Emulator-type site? Pretty much, I will not have my TI 83 until it gets shipped out here, and I need a better calculator than the damn windows one. Thanks ChowderInopa 22:22, 19 September 2006 (UTC)

Not sure about TI-83 functionality, but G-calc is a nice graphing calculator: [3]. StuRat 22:54, 19 September 2006 (UTC)

There is a flash applet you can download of the TI-83 or 82 out there somewhere. — [Mac Davis](talk) (SUPERDESK|Help me improve)00:08, 20 September 2006 (UTC)

There's an emulator called VTI you can download here. You will need a ROM image for a TI83+/84+, but you can download them. I think they're of dubious legality unless you own one, so I won't link to it. 206.124.138.153 00:19, 20 September 2006 (UTC)

Statistics Q

I have the following table:

                       Unemployed (Y=0)  Employed (Y=1)   Tot
Non Graduate (x = 0)      0.045            0.709          0.754
Graduate (x = 1)          0.005            0.241          0.246
Tot                       0.050            0.950          1

And I am asked for the E(Y), which is the expected value of Y. Is it merely 0.95?

Yes, for the total population. In general, for a discrete random variable X, the expected value E(X) = Σ (v × Pr(X=v)), summing over all possible values v for X. So for a r.v. Y that has a range {0,1}, we find an expected value E(Y) = (0 × Pr(Y=0)) + (1 × Pr(Y=1)) = Pr(Y=1).  --Lambiam^Talk 11:19, 20 September 2006 (UTC)

Could use some tips for modelling a hot air balloon

I'm trying to make a simple hot air balloon simulator of sorts (Actually, a Morrowind mod). I'd like to make it reasonably realistic. So, I need to find how hot the inside of the balloon is as a function of how long it has been being heated by a flame. I've already found an equation online for calculating the bouyant force, but that requires the temperature. If any physicsy types could help out I'd really appreciate it; even a rough approximation would be great. BungaDunga 00:31, 20 September 2006 (UTC)

Well, as a rough model I suppose you could work with the following ideas: depending on how strong you've got the fire going, say heat is going into the balloon at a constant rate $k_{1}$. Heat is also going out of the balloon at a rate proportional to the temperature difference between the air in the balloon and the surrounding air (Newton's law of cooling), and the temperature is roughly a linear function of the temperature, so essentially we have ${\frac {dT}{dt}}=k_{1}+k_{2}(T-T_{surr})$. You may have to vary $T_{surr}$ depending on the height, though. Confusing Manifestation 00:58, 20 September 2006 (UTC)

Here k₂ would be negative. --Lambiam^Talk 04:22, 20 September 2006 (UTC)

Suppose you meant 'and the temperature is roughly a linear function of the heat', ConMan. --CiaPan 05:45, 20 September 2006 (UTC)

Is that function the derivative of f(T) or something similar? I've yet to learn calculus in school, so I don't have much knowledge of it (though I shall next year). I'm not sure how I would apply that equation. Something like this, maybe?

temp = 5

k1 = 5

k2 = .1

tSurr = 3

xV = 0

onEnterFrame = function(){

temp += k1-k2*( temp - tSurr )

trace(temp)

}

September 20

Words for the terms in an implication/entailment

What are the words for the terms in an entailment? For example, if $A\models B$, is A the "implicand" or the "implicant" or the "antecedent" of this entailment? Similarly, what's the word for B? I'm looking for something similar to addend. Having this word on hand would really help me think about the mathematical logic work I'm doing. I checked the entailment article, and asked on the talk page there, but I haven't gotten any responses. Thanks! -- Creidieki 12:43, 20 September 2006 (UTC)

I'd call them antecedent (for A) and consequent (for B). You could also use premise and conclusion. "Implicant" means literally: "that which implies" and might be used for A, while "implicand" means literally: "what is to be implied" and might be used to refer to B, but not to A. The similarity of these two words in English is confusing, and as, furthermore, entailment is not the same as implication, it is better to avoid them for use in relation to entailment. --Lambiam^Talk 15:43, 20 September 2006 (UTC)

Thank you! On a related note, are "implicant" and "implicand" the best terms to use when talking about an implication (rather than an entailment)? -- Creidieki 16:44, 20 September 2006 (UTC)

I would still use "antecedent" and "consequent", if only because of the confusing similarity between "implicant" and "implicand"; moreover, these terms are not widely used. Our Wikipedia article on material implication also uses "antecedent" and "consequent". --Lambiam^Talk 20:19, 20 September 2006 (UTC)

San Juan

The lowest temperature ever recorded in San Juan, Puerto Rico, is 60 degrees Fahrenheit. Write an inequality for T representing San Juan's recorded temperatures.

Do your own homework: if you need help with a specific part or concept of your homework, feel free to ask, but please do not post entire homework questions and expect us to give you the answers. Letting someone else do your homework makes you learn nothing in the process, nor does it allow us Wikipedians to fulfill our mission of ensuring that every person on Earth, such as you, has access to the total sum of human knowledge. -- Meni Rosenfeld (talk) 16:36, 20 September 2006 (UTC)

In the problem statement, it is the intention that T stands for any temperature (in degrees Fahrenheit) that may have been recorded in San Juan, Puerto Rico. While we do not know which temperatures have been recorded, we do know that 14 °F is not one of them. But, for all we know – if we only use what is given – it may have been 451 °F on a particularly hot day in 1898. The inequality involving T must therefore be such that if you replace T in it by 451, it becomes a true statement. That already rules out many things, such as T > 666 and T ≤ 2. A possible solution is T ≥ -459.67. But that solution does not take account of everything that is given in the problem statement. The information from the first sentence is not used at all. You want not just any inequality that is true for all recorded temperatures: you want it to be as strong as possible. In particular, you want the inequality to be such that it rules out all temperatures that cannot have been recorded, namely those that are lower than 60 °F. So if you substitute any lower value, for example 59, for T, you want the inequality to become a false statement. I hope this helps. --Lambiam^Talk 20:40, 20 September 2006 (UTC)

This is a joke right? Or do you really have problems understanding the concept of lowest and the concept of greater than? 202.168.50.40 00:23, 21 September 2006 (UTC)

The first two responses are appropriate; this third one is absolutely not. We can reasonably expect someone to read and abide by the guidelines at the top of this page, and chide them when they do not. We can offer guidance in a subject area without providing homework answers.

We want people to ask questions here, and to feel comfortable revealing their areas of ignorance in order to learn. Mocking and insulting responses will not achieve these goals. Please do not do this again. --KSmrq^T 13:16, 21 September 2006 (UTC)

Wythoff's Game

Does anyone know where I can find the solution to Wythoff's Game, with a proof? There are a few sites scattered around that claim staunchly that it's all about the Golden Mean and the Fibonacci numbers, and as far as I can tell they're right, but nobody bothers to really explain it. Black Carrot 17:13, 20 September 2006 (UTC)

Whoa! I just read the page. Rounding down to integers?! That sounds like things Ramanujan came up with. Such occult phenomena! Surprising number sequence, in any case. Do you have more link tips? —Bromskloss 20:00, 20 September 2006 (UTC)

I am sure I read a pretty full analysis of the game in John Conway's book "On Numbers and Games" - but that was nearly 30 years ago, so I might have mis-remembered ... but I still have the book somewhere, so if you are really desperate, I will try and find it :-) (unless you have access to a copy) Madmath789 20:16, 20 September 2006 (UTC)

Sure. Iirc, a complete proof is given in this book:

Csákány Béla, Diszkrét matematikai játékok. Polygon, Szeged, 2nd ed, 2005.

This is a nice easily understandable book that requires little prior knowledge. I recommend it heartily. – b_jonas 11:31, 21 September 2006 (UTC)

Searching for variations on -wythoff nim- gives quite a bit. If it helps, I can easily prove the claim at [4], which gives a vague link to Fibonacci-eque numbers. And the whole rounding-down thing comes pretty naturally (I thought of it too, just didn't have a number to plug in) from writing the numbers out in order. They aren't going by intervals of 1, or 2, or 1.5, or 1.6, but seem to be jittering around in that area. Yeah, I'd appreciate it if you could find that book. Black Carrot 13:34, 21 September 2006 (UTC)

Liquid Mixture Puzzle

A while ago I found this puzzle that I cannot solve. It states that:

• There is a tank of water, with a capacity of 100 litres.
• This tank contains 100 litres of LiquidA
• This tank has 2 pipes connected to it, one at the top and one at the bottom
• From the top pipe, LiquidB flows into the tank at 1 litre per minute
• From the bottom pipe, the contents of the tank (both LiquidA and LiquidB - assume they mix perfectly throughout the tank) both flow out at 1 litre per minute

What is the ratio of LiquidA to LiquidB in the tank after one hour?

I do not seem to be able to solve this problem using algebra and after some searching, it seems like I need to know some calculus in order to solve it (which I don't). Can someone tell me the answer please, and how to work it out? If the answer requires knowledge of calculus, can someone please point me to a good reference that is easy to understand in order to learn it (if I wait for school to teach me it, I will have to wait around 3 to 4 years - I am currently 14)? Thanks for any help you can provide.

P.S. Side question: Does anyone know why in England, you have to wait until Year 13 to do logarithms? I have had a look at them and they don't seem particularly difficult. Also, they seem quite useful, but if you not take Maths for A Level, you will not learn about. Strange... -80.229.152.246 21:06, 20 September 2006 (UTC)

The exact symbolic solution to the problem is a differential equation. You can get an approximate numerical solution using repeated algebraic computations: see numerical analysis. --Serie 21:41, 20 September 2006 (UTC)

Here is how you solve it.

(1) Change the unit of time to minute. Having a consistent unit of time helps a lot.

(2) Have two variables VolA and VolB which represent the volume of each type of liquid within the Tank

(3) Write out the equation which represents the volume of each type of liquid in the Tank.

VolA(t) = input_VolA(t) - output_VolA(t)

VolB(t) = input_VolB(t) - output_VolB(t)

 determine output_VolA(t) and output_VolB(t)

(4) input_VolA(t) = 0

(5) input_VolB(t) = int(t=T0,t=T,1,dt)

(6) output_VolA(t) = int(t=T0,t=T,ratio_A(t),dt)

(7) output_VolB(t) = int(t=T0,t=T,ratio_B(t),dt)

(8) ratio_A(t) = VolA(t)/( VolA(t) + VolB(t) )

(9) ratio_B(t) = VolB(t)/( VolA(t) + VolB(t) )

(10) Solve the differential equation.

202.168.50.40 00:30, 21 September 2006 (UTC)

Let's begin with some bounds and a crude guess. If liquid B only was removed, after 60 minutes (1 hour) the ratio would be 0:100. If liquid A only was removed, the ratio would be 60:40. That is, B must be somewhere between 0% and 60% of the total volume, and a crude guess might be 30% (a ratio of 30:70). Now let's try to be more precise.

We can easily state exactly how much of B has been added after t minutes, namely t litres. We also know that the amount of B present is this minus the amount removed. And the rate at which B is removed is the amount present divided by 100. Of course, at the beginning the amount removed, r, is zero. Thus, with a little rearranging, we wish to solve the ordinary differential equation

$r'(t)+{\begin{matrix}{\frac {1}{100}}\end{matrix}}r(t)={\begin{matrix}{\frac {1}{100}}\end{matrix}}t$

with initial condition

$r(0)=0.\,\!$

The final amount of B can be recovered as 60−r(60), the amount added minus the amount removed after 60 minutes.

Here we have a first order linear differential equation, which can be solved quite easily, as our article explains. It turns out that after 60 minutes we will have just over 45 litres of B, and a B-to-A ratio of approximately 23:28. --KSmrq^T 00:08, 22 September 2006 (UTC)

Thanks very much for the answer everyone. It looks like I'll have to wait until I do calculus (3 years and counting) before I completely understand exactly what you are saying, but I think I get the gist. Thanks. --80.229.152.246 21:21, 22 September 2006 (UTC)

Math problem

I know you guys don't do homework, but I've got this question and our teacher just told us to use trial and error or the Pythagorean triplets to solve it because he said that the algebraic solution is too difficult. But im just curious, how do you solve this algebraically?:

a) Express the volume of a cone to the surface area of the same cone (includes the bottom)

b) Using the answer from part (a), make the ratio of volume to surface area equal to 1. What are the values of the radius, height and slant height?

The answer is radius=6, height=8 and slant height =10. But can someone post the solutions to part b? Jamesino 22:47, 20 September 2006 (UTC)

This requires that we work with dimensionless units. Equating the formulas for volume and surface area for a cone with radius r and height h gives us:

$\pi r(r+{\sqrt {r^{2}+h^{2}}})=\pi r^{2}h/3.$

Divide both sides by πr/3:

$3(r+{\sqrt {r^{2}+h^{2}}})=rh.$

Subtract $3r$ from both sides, and square:

$9(r^{2}+h^{2})=r^{2}(h-3)^{2}.\,$

Multiply out and bring everything to one side:

$(r^{2}-9)h^{2}-6r^{2}h=0.\,$

Divide by h, and then solve for h:

$h={\frac {6r^{2}}{r^{2}-9}}$.

Any value of r > 3 will give you now a value for h such that together they are a solution. For example, r = 21 gives h = 49/8, which results in a slant height of 175/8 and a volume and surface area of 7203π/8. If you want both r and h to be whole numbers, note that h − 6 = 54 / (r² − 9) must also be an integer, so r² − 9 must be an integer divisor of 54, leaving for r² only the possibilities 10, 11, 12, 15, 18, 27, 36, and 63. The only square is 36, giving r = 6 and h = 8. --Lambiam^Talk 00:11, 21 September 2006 (UTC)

Niggling note: The step "Divide both sides by πr/3" should be accompanied by a test that $\pi r/3\neq 0$. This test would show that r=0 is a solution (but not an interesting one). -- Fuzzyeric 15:47, 21 September 2006 (UTC)

Good point. It would also be necessary, because of the fifth step, to test h=0, but that's not a solution. Black Carrot 16:56, 21 September 2006 (UTC)

There is no need to test for r ≠ 0. In fact, how could we "test" for that? – we are trying to find the value of r here. What I'm doing is deriving a sufficient condition for the equation to hold. By reducing something of the form r·X = 0 to X = 0, I discarded the solution r = 0. However, since a cone is constrained to have a positive radius, it is harmless to drop that case, which I did tacitly. Likewise, I did not explicitly mention my throwing overboard the algebraic solutions with negative h. --Lambiam^Talk 19:53, 21 September 2006 (UTC)

Because any step that reads "divide both sides by X" had better ensure that X is not zero, or the step is invalid. Division by zero is undefined. Also, since you did not check for this, you did not discover that r=0 is also a solution for all values of h. -- Fuzzyeric 04:06, 22 September 2006 (UTC)

You do not understand. Suppose you have to solve X³−X² = 0 for X. Clearly, that follows from X²−X = 0, which has solutions X = 0 and X = 1. So I have divided here by X, replacing the equation to be solved by a simpler equation all of whose solutions will be valid solutions to the original equation. That is the meaning of "sufficient condition". One of the solutions of X²−X = 0 is X = 0, so I have as it were divided by 0, but that does not invalidate the step. In a logic formula, XY = 0 ← Y = 0 (where "←" stands for "follows from"). That is also true if X happens to be equal to 0. Of course I saw that algebraically r = 0 was a solution for all h; I just assumed the side condition (constraint on the solutions) r > 0 and h > 0 to be so obvious that it could be dealt with implicitly. --Lambiam^Talk 15:14, 22 September 2006 (UTC)

Would it somehow be clearer to point out that not checking for division by zero is inherently wrong, that it is incorrect method, and that it will permit you to prove that 1 = 2. It's a common enough error that there's an entire talk page about it at Talk:Division_by_zero. The examples I'm quoting are present at Talk:Division_by_zero#Wow.2C_2.3D1.3F -- Fuzzyeric 04:15, 23 September 2006 (UTC)

Oh, and your example is an excellent one for demonstrating why it is wrong to not check. You justify the step $x^{3}-x^{2}=0\implies x^{2}-x=0$ by "dividing by x". (Repeating this step will eliminate the solution x=0 completely, which is incorrect.) However, the valid implication is $x^{3}-x^{2}=0\implies x^{2}-x=0\lor x=0$ "by the definition of integral domain". Integral domains are distinguished by the property that you can divide by sides of an equality by any nonzero element you want. R[x,y] is an integral domain under addition and multiplication and therefore in your demonstration, it is required to verify that no divisor can be zero. -- Fuzzyeric 04:30, 23 September 2006 (UTC)

I do not justify the step $x^{3}-x^{2}=0\implies x^{2}-x=0$ because I do not take that step. I do not derive a NECESSARY condition but a SUFFICIENT condition. Sorry for the shouting, but I said this twice already and you just seem to ignore it. --Lambiam^Talk 15:45, 23 September 2006 (UTC)

Just in case Lambiam's explanation is still unclear: The step he takes is $x^{3}-x^{2}=0\Longleftarrow x^{2}-x=0$, which is obviously legitimate (via multiplication, not division, by x, which is always valid). -- Meni Rosenfeld (talk) 15:57, 23 September 2006 (UTC)

So which part of Lambiam's "Divide both sides by πr/3: $3(r+{\sqrt {r^{2}+h^{2}}})=rh.$" is a multiplication? -- Fuzzyeric 22:34, 23 September 2006 (UTC)

Is this a homework question? What about $\pi r(r+{\sqrt {r^{2}+h^{2}}})=\pi r^{2}h/3\Longleftarrow 3(r+{\sqrt {r^{2}+h^{2}}})=rh$? --Lambiam^Talk 23:04, 23 September 2006 (UTC)

However, the demonstration you wrote was $\pi r(r+{\sqrt {r^{2}+h^{2}}})=\pi r^{2}h/3\Longrightarrow 3(r+{\sqrt {r^{2}+h^{2}}})=rh$! -- Fuzzyeric 04:43, 24 September 2006 (UTC)

This is turning into an Achilles–tortoise dialogue. I did not "write" that. I showed the operation by which I obtained one equation from the other. I could instead have just presented the solution, for which it is easy enough to verify it satisfies the original equation, but I also wanted to sketch the steps that brought me there. Maybe the presentation was confusing, but it is you who made the mental leap to an entailment. --Lambiam^Talk 07:59, 24 September 2006 (UTC)

Thanks alot =) Jamesino 21:04, 21 September 2006 (UTC)

September 21

Ok, so we have the trigonometric sine, and then the hyperbolic sine...

The trigonometric sine is related to the unit circle, while the hyperbolic sine is related to the unit hyperbola. So, can there be sines for other conic sections as well? The parabolic sine and elliptic sine, perhaps?

Thanks for satisfying my mathematical urges 0_0 --Ķĩřβȳ♥ŤįɱéØ 06:48, 21 September 2006 (UTC)

The unit circle is described by the parametric equation x = cost, y = sint. The unit hyperbola is described by x = cosht, y = sinht. More generally, the equation x = acost, y = bsint describes an ellipse (so there is no need for separate "elliptic" sine and cosine) and the equation x = acosht, y = bsinht describes a hyperbola. Since the equation x = at, y = bt² describes a parabola, the natural candidates for "parabloic" sine and cosine are cospt = t, sinpt = t², which aren't at all interesting. So no. -- Meni Rosenfeld (talk) 09:36, 21 September 2006 (UTC)

Not quite — there's nothing "natural" about choosing x = t, y = t²; x = t³, y = t⁶ would do equally well.

If you use the hyperbolic functions to parametrise the hyperbole, by setting, say, $\gamma (t)=(\sinh(t),\cosh(t))$, the normalisation property is that $|\gamma '(t)|=|\gamma (t)|$ (and the equivalent holds for cos and sin).

I'm going to leave solving this for the parabolic curve as an exercise for the reader, but would like to caution about the choice of "focus" in a parabola. Using the origin might not be the right thing.

RandomP 14:47, 21 September 2006 (UTC)

Not directly relevant to your original question, but there are some elliptic functions: sn, cn, and dn covered in the article Jacobi's elliptic functions that you might find interesting. Madmath789 10:04, 21 September 2006 (UTC)

There are also an infinite family of "sines" and "cosines" based on q-exponentials. --HappyCamper 15:05, 21 September 2006 (UTC)

Parabolic trigometry arises in the Kleinian geometry described in the book by I. M. Iaglom, A simple non-Euclidean geometry and its physical basis: an elementary account of Galilean geometry and the Galilean principle of relativity, which bears the same relationship to "Newtonian spacetimes" as Minkowski geometry bears to Lorentzian manifolds, for which see the book by Misner, Thorne, and Wheeler, Gravitation. More technically:

• E^1,1 is a plane geometry (defined by an indefinite but nondegenerate quadratic form) which can serve as the model for tangent spaces to a two dimensional Lorentzian manifold,
• E^1,0 is a plane geometry (defined by a degenerate quadratic form) which can serve as the model for tangent spaces to a two dimensional Newtonian spacetime,
• E² is a plane geometry (defined by a positive definite quadratic form) which can serve as the model for tangent spaces to a two dimensional Riemannian spacetime.

There are algebraic formulations in which these three geometries arise from three kinds of Cayley-Klein algebras (a generalization of the complex number field considered as a two-dimensional real algebra). I wrote a Wikipedia article about this stuff once but (sigh) some ignorant mathcrank munged it out of existence. Unfortunately the book by Iaglom (a translation from the Russian) is rather hard to find; the only remaining copy which I know of is in the collection of the research library of the Los Alamos National Laboratory. You might be able to get it on interlibrary loan, even through your public library system, if you happen to live in the Western U.S. It's a suprisingly elementary book (originally written I think for Russian high school students), and parabolic trigonometry is significantly easier than circular or hyperbolic trig. (If you've seen books by Jaglom or Yaglom, this is the same author; at different times his name has been transliterated in many ways, which leads to all kinds of confusion).---CH 21:44, 21 September 2006 (UTC)

There are two Yagloms: A. M. Yaglom and I. M. Yaglom. I don't know their relationship, but they have published together.[5][6]  --Lambiam^Talk 22:38, 22 September 2006 (UTC)

What about sine and cossine which when: x = sqcos t, y = sqsin t, they define a square? ☢ Ҡi∊ff⌇↯ 04:17, 22 September 2006 (UTC)

The rule of 17

What is the rule of 17? It appears to be used in design considerations for equal temperament. --152.62.109.163 10:29, 21 September 2006 (UTC)

When placing frets on an equal tempered guitar (e.g.) the distance to the next fret (a minor second interval) can be calculated by multiplying the octave length (in inches, e.g.) with 1 minus the 12th root of 2 (1 - 2^(1/12), or approx. 0.059463. This is the same as dividing by 16.817, which is sort of close to 17. Maybe that's it.---Sluzzelin 13:40, 21 September 2006 (UTC)

Close, but no cigar. And two explanations I found on the web state that 17.817 is the 1/12th root of 2, which is nonsense.

The true explanation depends on a little physics, a little perceptual psychology, a little mathematics, and artistic taste.

A vibrating guitar string is not nearly as stiff as a metal bar, nor even a piano string; so half a string produces a frequency twice as high as a full string. Perceptually, a double frequency sounds similar to the original, and very harmonious. In terms of a musical scale, the higher note is said to be an "octave" above the lower one, from the eight steps in a Western major scale. (The notes of a C-major scale are C, D, E, F, G, A, B, c.) The scales we use in Western music are originally based on simple frequency relationships; for example, a "fifth" (C to G) was a ratio of 3:2, a "fourth" (C to F) was a ratio of 4:3, and a major "third" (C to E) was a ratio of 5:4. However, around the time of Johann Sebastian Bach composers and performers on instruments like the harpsichord began to find this tuning decidedly inconvenient. The problem was that what sounded good for C-major sounded awful for G-major or F-major. Thus a transition was made to a chromatic scale where the ratio between any two adjacent notes was exactly the same, with twelve steps in an octave. If an "A" had a frequency of 440 Hz, then the ratio of "C" to "B", say, had to be the 12th root of 2, approximately 1.059463:1. The fifth became 1.498:1 instead of 1.5:1, the fourth became 1.335:1 instead of 1.333:1, and the major third became 1.2599:1 instead of 1.25:1. Every interval except the octave has been perturbed a little, but now the intervals work equally well in every key.

The guitar is a fretted string instrument, and (to a first approximation) the ratio of full string length to fretted string length gives the frequency ratio of shorter to longer. Thus a fret halfway along the string doubles the frequency, producing a pitch an octave above the unfretted pitch. Where do we place the other frets? We expect the fifth to have a fret at about 1/3, producing a ratio of 3:2, but we really need something systematic.

If the "E" string, say, has a length of L = 650 mm, what proportion p of its length should we remove (by fretting) to produce an "F" note, the next chromatic note above it? Well, if we remove L/p we are left with L−L/p, and the ratio of L to this should be the twelfth root of 2, 2^1/12. The answer is easily found, and happens to be

2^1/12⁄(2^1/12−1) ,

which is approximately 17.8172 — hence the "rule of 17". (Musicians are not necessarily the best mathematicians!) And because the ratios are all equal, the next fret should cut off the same proportion of the remaining length, and so on. --KSmrq^T 15:22, 21 September 2006 (UTC)

KSmrq's answer is, of course, nearly equivalent to Sluzzelin's: 16/17 is approximately 2^-1/12, and 18/17 is approximately 2^1/12 (though Sluzzelin was closer, I wouldn't bet on many musicians realising that).

RandomP 15:43, 21 September 2006 (UTC)

Hey! :-/ —Bromskloss 10:21, 22 September 2006 (UTC)

Vincenzo Galilei was quite fond of 18/17 (and so were luthiers of his time), but there are even better rational approximations... 53/50, 71/67, etc... (I put a list up at Talk:Semitone recently.) - Rainwarrior 19:35, 21 September 2006 (UTC)

Probability Q

I dont understand how the correlation coefficient relates to finding the mean and variance of a summation. For example, Men make 40,000 a year with SD of 12,000, women make 45,000 with SD of 18,000. Now assuming I am given an Rsquared (CC) of 0.7 between male and female earnings, how does this affect the mean and variance of total household income for dual earner families? I assume the mean would be 85,000, but then I am afraid of multiplying the variances and square rooting, because this doesnt involve the 0.7...help?

Edit: I attempted to use the equation: Var(X+Y) = VarX + VarY + 2 Covariance (XY) but to find the covariance using the Correlation Coefficient (CC) I need to use the eqn: Corr(XY) = Cov(XY)/((SDX*SDY)) but that gave me a covariance of 151.2 million (because it becomes: 0.7 = Cov/(12000*18000)

...Help?

What's wrong with a covariance of 151.2 million? -- Meni Rosenfeld (talk) 19:17, 21 September 2006 (UTC)

Well...that would mean that the new variance of the total household income, using Var(X+Y) = VarX + VarY + 2 Covariance (XY), would be IMMENSE...and that would also mean that the Variances of the individual X and Y would barely factor in, which seems counterintuitive. Is the 151.2 million an accurate intermediate step? ChowderInopa 22:02, 21 September 2006 (UTC)

Var(X)+Var(Y) = 2169 468 million, so how can you say that would "barely factor in"? I'm not sure what you mean by "Rsquared (CC)". Correlation coefficients can be negative, so what is being squared here? --Lambiam^Talk 00:42, 22 September 2006 (UTC)

ChowderInopa, you have forgotten that the variance is the square of the SD. Btw, Lambiam, I got VarX + VarY = 468 million. -- Meni Rosenfeld (talk) 07:03, 22 September 2006 (UTC)

You are correct; I copied the wrong number into my mental calculator. --Lambiam^Talk 11:17, 22 September 2006 (UTC)

You are correct, I was using the standard deviations, which was why I was confused as to the small contribution to the final answer. Oh, and for the R squared CC, i thought i remembered from a stats class that the R^2 and Correlation coefficient are the same? or is that just R and the CC? Finally, my confusion stems from the fact that I can not visualize how such a huge variance for the total family income makes sense... I have always wondered the practical point (aside from calculations) of the variance...SD i understand, as showing the centeredness of the data, but variance is such a huge, (in my mind, completely abstract) number, compared to the rest of the data. Seriously, what use is it to know that the variance of family income is hundreds of millions? ChowderInopa 18:05, 22 September 2006 (UTC)

Variance is the square of the standard deviation. This also means that the units in which it is measured are the square of the units of the SD (and the data). So if the SD is 10,000 USD, then the variance is 100 million "USD squared". You cannot compare the 100 million with the 10,000 because they are not measured with the same units (the same way you cannot compare a mass of 100 kilograms with a length of 2 kilometers). Variance, just like SD, measures the centeredness (or lack thereof) of the data, but using a completely different scale from the data itself. -- Meni Rosenfeld (talk) 18:23, 22 September 2006 (UTC)

Histogram

Does any one know why a histogram is named such?--Willworkforicecream 17:51, 21 September 2006 (UTC)

The basic meaning of Greek histos is: "something that has been made to stand upright", with specific meanings of "mast", "beam" and "loom". The -gram part means: "something that has been written". Together: "something written with beams". --Lambiam^Talk 19:17, 21 September 2006 (UTC)

Thanks.--Willworkforicecream 13:38, 22 September 2006 (UTC)

Displacement, Velocity, Acceleration, Jerk, ...

The first derivative (with respect to time) of displacement is velocity, the second derivative is acceleration, and the third derivative is jerk. Is there a word for the fourth derivative of displacement? —Mets501 (talk) 20:02, 21 September 2006 (UTC)

This is discussed in the Jerk article. Chuck 20:30, 21 September 2006 (UTC)

Wow I'm blind! I thought I read the whole article... Thanks anyway Chuck :-) —Mets501 (talk) 20:49, 21 September 2006 (UTC)

Ah! I didn't know there was a word for it, but I have many times thought about it when going by metro – realising that the engineer never took it into account, but should have. —Bromskloss 10:16, 22 September 2006 (UTC)

…and I love the "snap, pop, crackle"! :-) —Bromskloss 10:22, 22 September 2006 (UTC)

See also Third derivative of position in the Usenet Physics FAQ. – b_jonas 18:13, 23 September 2006 (UTC)

Bijection from Z^2 to Z

Sorry for the notation in the subsection header. I'm looking for a simple bijection $\phi :\mathbb {Z} \times \mathbb {Z} \leftrightarrow \mathbb {Z}$. By the fact these two sets have the same cardinality, we know some bijection must exist, but I'm wondering if there's a fairly simple function that exhibits this property.

-- Braveorca 23:55, 21 September 2006 (UTC)

How about a square spiral with two arms? Melchoir 00:05, 22 September 2006 (UTC)

Well, I thought of that one (the conventional "put them on a string like so and then straighten the string"), I was wondering if there was something like $\phi (x,y)=xy-x+y$ - a simple, easy-to-define function. - Braveorca 02:26, 22 September 2006 (UTC)

There will not be a continuous map, in the sense that the map is continuous on the underlying RxR, like the polynomial you mention. I recall that this follows from the different dimensionalities of the two sets (points in ZxZ have too many neighbors to map continuously down to the line). There are discontinuous mappings that are simple. E.g. $(\sum _{n=0}^{\infty }2^{n}x_{n},\sum _{n=0}^{\infty }2^{n}y_{n})\leftrightarrow \sum _{n=0}^{\infty }2^{2n}x_{n}+\sum _{n=0}^{\infty }2^{2n+1}y_{n}$ which just interleaves the bits of x and y in the forward direction and demultiplexes them in the reverse direction. -- Fuzzyeric 04:51, 22 September 2006 (UTC)

Fuzzyeric, your example does not work. The assignment you've described is not a bijection. The reverse (right-to-left) function is not one-to-one, so the forward (left-to-right) function is not a surjection. For example, 1/6 is a repeating fraction 0.001...₂ in binary. When demultiplexed, it gives a pair of (0.01...₂, 0) = (0.1₂, 0) = (1/2, 0). This in turn interleaved gives 1/2, which is obviously not the starting number, 1/6.

You need either some extra restrictions for this method to work, or just redefining it. The latter can be done in a simple way—just interleave strings of ones terminated by zero, rather than single digits. This protects against 'gluing' multiple ones into an infinite string, and as a result guarantees a one-to-one mapping in both directions.

Examples:

(1/3, 14/15) = (0.01...₂, 0.1110...₂) = (0.010...₂, 0.1110...₂) maps to 0.0111010...₂ = 0.011101...₂ = 29/63

(1/3, 1/3) = (0.01...₂, 0.01...₂) = (0.010...₂, 0.010...₂) maps to 0.0010...₂ = 0.001...₂ = 1/6

(1/3, 55/64) = (0.01...₂, 0.110111₂) = (0.010...₂, 0.11011100...₂) maps to 0.0110101110100...₂ = 1507/3584

CiaPan 16:14, 22 September 2006 (UTC)

Ooops, sorry, I confused integer numbers with reals. Of course you method works fine for integers, as their binary representation as always finite. CiaPan 16:35, 22 September 2006 (UTC)

Uh... the question before us concerns integers, right? Melchoir 16:28, 22 September 2006 (UTC)

Melchoir is right that this is a question about integers, not rationals. Although, CiaPan, is right for the wrong reason. The recipe I gave only works for half of ZxZ because it doesn't handle the signs well. A better method would poke the sign of x into the least significant bit ("-" → "1", "+/0" → "0"), then take the sign of y to be the sign of the result. And this should patch up the omission. -- Fuzzyeric 16:40, 22 September 2006 (UTC)

Map integer numbers onto naturals with $m:\mathbb {Z} \to \mathbb {N}$ given by

$m(z)={\begin{cases}2z&{\textrm {for}}\ z\geq 0\\-2z-1&{\textrm {for}}\ z<0\end{cases}}$

then use bijection $b:\mathbb {N} ^{2}\leftrightarrow \mathbb {N}$ defined as

$b(n,m)=\left((n+m)^{2}+n+3m\right)/2$

CiaPan 06:25, 22 September 2006 (UTC)

It's interesting to note that CiaPan's map runs sequentially through the diagonals with slope -1, in the order (0,0), (1,0), (0,1), (2,0), (1,1), (0,2), ... . The other step does fold, spindle, and mutilate Z, though... -- Fuzzyeric 01:35, 23 September 2006 (UTC)

Yes, that's exactly what I wanted when inventing this function (of course I'm sure there are and were hundreds people, who invented it before me!). And the m(z) mapping is essentialy the same which you described above: it shifts an absolute value by one bit and inserts the sign into the ones' position. --CiaPan 06:22, 27 September 2006 (UTC)

My favourite bijection from the square of a countable set to itself uses the set $X=\mathbb {Z} ^{+}=\{1,2,\ldots \}$. Then define a bijection $f:X\times X\to X$ by $f(m,n)=2^{m-1}(2n-1)$.

Doctorbozzball 11:15, 23 September 2006 (EDT)

September 22

rational number is group under addition

the set of rational numbers (gcd(a,b)=1) whose denominator are odd is group under addition?(include zero).

Yes (as long as negatives are also included, of course). It is left as an easy exercise for the reader to prove it. -- Meni Rosenfeld (talk) 07:31, 22 September 2006 (UTC)

Original poster, ignore this: we can also look at that as $\mathbb {Z}$ localized at the prime ideal (2), right? Tesseran 08:48, 22 September 2006 (UTC)

To be honest, this is the first time I've heard about localization of a ring. If I understand that article correctly, then no - rather, it would be Z localized at the set of odd integers. -- Meni Rosenfeld (talk) 20:22, 22 September 2006 (UTC)

You can localize in multiplicative set, for example odd integers (here You are right), but if one sais "to localize in prime ideal" he means to localize in a multiplicative set consisting of all elements not in a given prime ideal. So "to localize at the prime ideal (2)" means exactly "to localize using multiplicative set of odd numbers".

Oh, okay. -- Meni Rosenfeld (talk) 06:02, 23 September 2006 (UTC)

Furthermore, it's even a ring, isn't it? – b_jonas 18:01, 23 September 2006 (UTC)

ANSWERED -> "difference between average mode mean?"

In statistics, what is exactly the difference between mode, mean and average?"

Usually "mean" and "average" mean the same thing. See our articles on Mode, Mean, and Average. You may also be interested in Median. --Lambiam^Talk 15:20, 22 September 2006 (UTC)

"Arithmetic mean" is the same as "average", but "geometric mean" is something else. StuRat 15:23, 22 September 2006 (UTC)

A distinction should be drawn between the precise meaning of these terms and the popular meaning. In the latter case, "average" nearly always refers to the arithmetic mean, but more precisely it refers to a "typical value", or with more jargon, a "measure of central tendency". Thus any mean, or mode, or median, is an average. (Though the mode, in particular, can be anything but central.)

Again, "mean" without qualification usually refers to the AM, but there are any number of different means which can be calculated from a set of figures, the commonenest are maybe the geometric and harmonic varieties.

It's a great pity that a subject concerned with the careful evaluation of data is cursed with such sloppy usage.--86.132.238.249 18:01, 22 September 2006 (UTC)

chi-square in statistics

how do I calculate the chi-square in very clear and simple steps?Mariesaintmichel 15:12, 22 September 2006 (UTC)thank you Marie Saint Michel

What is the nature of your data? Is it a 2 by 2 contingency table? --Lambiam^Talk 15:22, 22 September 2006 (UTC)

If you can calculate the expected values, then Pearson's chi-square test tells you the calculation ($\chi ^{2}=\sum _{i=1}^{N}{\frac {({O_{i}-E_{i}})^{2}}{E_{i}}}$) to use.

For a congintency table, here's a hint:

${\mbox{expected frequency of cell}}={\frac {{\mbox{row total where cell is}}\times {\mbox{column total where cell is}}}{\mbox{sum of all cell values in contingency table}}}$

x42bn6 Talk 01:39, 23 September 2006 (UTC)

September 23

Identities

What are identities involving nested hyperbolic functions and inverse trigonometric functions called? --HappyCamper 04:43, 23 September 2006 (UTC)

Say, something of this sort: $\tanh ^{-1}(\sin 2x))\equiv 2\tanh ^{-1}(\tan x).$

This is also equal to ${\mbox{gd}}^{-1}(2x),$ which uses the inverse Gudermannian function. I don't think the species has been named; perhaps call them "inverse-Gudermannian identities"? --Lambiam^Talk 15:57, 23 September 2006 (UTC)

Ah, I did not spot that quite so quickly. Actually, this came from a foreign language textbook that I found, there were other interesting identities invovling inverse sines and hyperbolic cosines, which was what prompted this question. --HappyCamper 20:32, 23 September 2006 (UTC)

Questions about the shape of the earth and distances

I was thinking, what is the relationship to a torus and a sphere? That is, let's assume that we have a torus and a sphere with equal surface area and equal volume. So let's say that the earth was a torus, with two equators, and the north pole and south pole coalesce into each other, in the middle. And let's say that the continents and oceans of the earth were proportionally mapped onto the torus. So my question would be, let's say someone wanted to get from Alaska to South America. In a spherical earth, a person would have to fly a plane south from Alaska to reach to South America. However, in a toroidal earth, a person could theoretically travel north, through the pole, and arrive at South America, approaching it from the south. Let's ignore the affects of weather. So would it be a shorter distance? And is there a formula to decide whether or not a distance from random points A and B on a torus would be longer or shorter than the proportional points A' and B' on a sphere? Thanks. --Ķĩřβȳ♥ŤįɱéØ 06:05, 23 September 2006 (UTC)

Let's begin by saying that there is not torus with equal surface area and volume as a sphere. Even if we just stick to "same surface area" (and lesser volume), I do not understand what your transformation from the sphere to the torus is, but I am inclined to believe that distances will typically be longer. -- Meni Rosenfeld (talk) 06:13, 23 September 2006 (UTC)

Ok, they have the same surface area then. I mean it the earth to look like this: [7]. --Ķĩřβȳ♥ŤįɱéØ 06:54, 23 September 2006 (UTC)

Here are some thoughts:

• The image is misleading, because the "second equator", the line where the poles are stitched together, has a discontinuity, where in the image it looks continuous.
• Suppose the earth has a radius of R, and the torus has an inner radius of a and an outer radius of b. Then:
  • The distance around the equator is 2πR on earth, (b+a)π on the torus.

What was I thinking? It's 2bπ. -- Meni Rosenfeld (talk) 08:01, 25 September 2006 (UTC)

• • The distance from the south pole, through the equator, to the north pole is πR on earth, π(b-a) on the torus (and of course, just the distance from the equator to a pole is half of that).
  • The surface area of earth is 4πR², of the torus is (b²-a²)π.
  • The distance around latitude θ is 2πRcos(θ) on earth, ((b-a)cos(2θ)+a+b)π on the torus. (a nicer formula: 2π(asin²θ + bcos²θ)
• We would want the first three of these to be equal for earth and the torus, but they are easily seen to be incompatible. If the area is to be equal, then at least one of the first two distances must be greater on the torus.
• The fourth distance, for θ = 90°, is 0 for earth but (b-a) for the torus. Not nice. This also means that whatever our choice for a and b, there will be near the poles points which are very close on earth but distant on the torus (and the ratio can be as large as we please).
• All that aside, regarding your original question: Yes, since we can "magically" leap from the south pole to the north pole, there will necessary be points which on earth are distant (one near the north pole, one near the south pole) but will be close on the torus.
• About a general formula, it could be complicated, but I have given above examples of some specific cases. Of course, since they are all circles, the corresponding arcs will have proportional lengths.

I hope this answers your question. -- Meni Rosenfeld (talk) 08:24, 23 September 2006 (UTC)

Yes, it answers many of my questions, but I don't understand one part. How is it the second equator discontinuous? --Ķĩřβȳ♥ŤįɱéØ 08:38, 23 September 2006 (UTC)

It is not continuous in the same sense as a sphere because you're mapping a sphere (the map of Earth) on a torus. The north pole is connected to the south pole in the torus, so if you did morphed Earth as a torus, you'd have a discontinuity in the landmasses. Here's a visualization:

That horizontal middle line is how it would look like in the second equator, and that's what I think he meant with discontinuity. Hope that helps. ☢ Ҡi∊ff⌇↯ 09:33, 23 September 2006 (UTC)

Yes, suppose the north pole would be painted blue and the south pole painted red. Then in the stitch, you'll see a discontinuous jump from red to blue. -- Meni Rosenfeld (talk) 10:03, 23 September 2006 (UTC)

In the most general terms, it's not possible to go from a surface without a hole, to one with a hole, without tearing the surface. StuRat 10:53, 23 September 2006 (UTC)

Iterated sine function

Is there a name and a documented use for an iterated sine function? Example (my notation): $\sin _{5}x=\sin \sin \sin \sin \sin x$

I've been playing with it and it's been kinda interesting: when i tends to infinity, $\sin _{i}x\times \left({\frac {1}{\sin _{i}\pi /2}}\right)$ approaches a smooth square wave $f(x)=\operatorname {sgn} \sin x$ (using the sign function), which looks better than one made by additive synthesis of sine waves. I was also wondering if there would be a way to get the triangle and sawtooth waves with a similar method. $\tan \sin x$ is pretty close from a triangle wave, but I can't find a way to make it converge to a true triangle wave. $\sin \tan x$ resembles a sawtooth wave, but it gets chaotic near the jumps... Still, pretty interesting behaviors. ☢ Ҡi∊ff⌇↯ 08:30, 23 September 2006 (UTC)

Well, the infinite iteration could simply reprsented as $\sin x=x$

sawtooth

And that could be represented as an infinite series: $\sin x,\sin(\sin x),\sin(\sin(\sin x))),\sin(\sin(...))$

So the iterated sine would be the nth term of this series, where n is the number of iterations. I don't think this has a name though.

I'm not sure about your second question though. My guess would be a piecewise function involving absolute value. I hope that helps --Ķĩřβȳ♥ŤįɱéØ 09:03, 23 September 2006 (UTC)

I'm not sure if I get your point. The infinite iteration doesn't fall back to simply $\sin x=x$. Just plot sin(sin(sin(...x))) yourself and take a look. About the second question, I really don't want to involve piecewise functions here. ☢ Ҡi∊ff⌇↯ 09:26, 23 September 2006 (UTC)

The iterated sine function converges pointwise (i.e., for each fixed x) to zero. I would not like to express this exactly as the infinite iteration could simply reprsented as $\sin x=x$; but there is a connection with the fact that that equation has no other (real) solution than zero. That makes the idea of Kieff to multiply with a 'normalising factor' rather interesting. (Note, that there is nothing magical connected with $\pi /2$ in this context. If you pick any number c strictly between 0 and $\pi$, and use $(\sin _{i}c)^{-1}\,$ as normalising factor, you'll get the same limit function.) The 'normalisation' counteracts a very slow convergence close to the origin.

I've no idea whether or not this is done - this is not exactly my branch of mathematics - but if it isn't, I think it would be worth doing. JoergenB 15:43, 23 September 2006 (UTC)

Experimentally, the normalizing factor appears to be asymptotically equal to ${\sqrt {\frac {i}{3}}}$. --Lambiam^Talk 16:33, 23 September 2006 (UTC)

How come you say any number c strictly between 0 and π will get the same limit function? I used π/2 because it is a point when the function is at its highest value, so the normalization would make its max be exactly 1 at any iteration. Does that detail become completely irrelevant when dealing with infinite iterations? ☢ Ҡi∊ff⌇↯ 17:17, 23 September 2006 (UTC)

OK, I'll answer, but I fear I can't without getting slightly technical.

Starting with c strictly between 0 and $\pi /2$ yields a positive number not greater than 1 in the next step, and continues with a strictly decreasing sequence of positive numbers:

$\sin c>\sin _{2}c>\sin _{3}c>\sin _{4}c>\ldots >0$

(since, by an exercise in elementary calculus, indeed $0<\sin x<x\,$ for $0<x<\pi /2\,$). Thus, this sequence has an infimum $L\geq 0\,$. Now, if L were strictly greater than 0, then for some i we would have $\sin _{i}c<\arcsin L\,$, yielding the contradiction $\sin _{i+1}c<L\leq \sin _{i+1}c\,$. Thus, instead, $L=0$, for any start value c in the open interval $(0,\pi /2)\,$.

In particular, for any two given (fixed) c and d in that interval; there is a natural number j, such that $\sin _{1+j}c<\sin d\,$ and $\sin _{1+j}d<\sin c\,$. Moreover, the sine function is strictly increasing in the interval from 0 to 1 (check its derivative!), whence these properties carry over to all higher iterations as well:

$(1)\ \ \sin _{i+2j}c<\sin _{i+j}d<\sin _{i}c,\ i=1,2,3,4,\ldots$.

If you've followed the argument this far, then you're ready for the main point: Every time we do the iteration, we are replacing the value of some x with this x times a factor ${\frac {\sin x}{x}}$, and these factors converge to 1 as x approaches 0. In fact, in this interval, $x-{\frac {x^{3}}{6}}<\sin x<x$, whence the factor is a number between $1-{\frac {x^{2}}{6}}$ and 1. Thus, also the j'th power of this factor tends to one. Summing up, if we compare the quotients of the terms in formula (1) with $\sin _{i}c\,$, then we find that the first quotient converges to 1, while the third one indeed is (the constant) 1; whence the squeezed middle quotient also converges to 1.

I hope I glossed over an appropriate amount of details (giving you just enough to fill in yourself) :-) JoergenB 18:59, 23 September 2006 (UTC)

For a triangle wave, try $\arcsin ~\sin x$ or $|x-\lfloor x\rfloor -1/2|$. For sawtooth, try $\arctan ~\tan x$ or $x-\lfloor x\rfloor$. – b_jonas 17:46, 23 September 2006 (UTC)

Thanks for those but they're not what I'm looking for right here. I'm trying to find smooth functions that approximate these ideal waveforms as some value increases, like the iterated sine with the square wave. Using the floor function is kinda bad in this case too, since it's a bit of a "hack". ☢ Ҡi∊ff⌇↯ 06:13, 24 September 2006 (UTC)

Well, the triangle one might be possible by modifying the square wave one you have... if you are transforming a sin into a sqaure gradually via some process, it seems that the triangle should be available, since the square is, I believe, $\sum _{i=1}^{\infty }{\frac {\sin {((2i-1)x})}{(2i-1)}}$, and the triangle the milder $\sum _{i=1}^{\infty }{\frac {\sin {((2i-1)x})}{(2i-1)^{2}}}$, a less severe transformation of your initial sin might get you there (or something with an equivalent spectrum, maybe). Saw waves might be harder, since you will need to find some way to generate the even harmonics as well. Do you know about Frequency modulation synthesis? It is similar to what you are doing, though it is not usually taken to infinity like what you are doing here. - Rainwarrior 18:20, 24 September 2006 (UTC)

But by the way, if you are trying to find a smooth function that approximates one of these, why not just use the Fourier series (the sums I just mentioned above) of them and stop after a certain number of harmonics? - Rainwarrior 18:30, 24 September 2006 (UTC)

Oh, I don't have a direct use for these stuff, it's just something I've been toying with, really. But the problem with Fourier series is that they have artifacts. The iterated sine gives a much better result than a square wave Fourier series, and it has no artifacts at all, so I thought "hey, maybe there are similar functions for the other waves, and they could be pretty handy"... That's pretty much it.

I already tried getting some info from the Fourier series, btw, but that didn't really help. :( ☢ Ҡi∊ff⌇↯ 20:53, 24 September 2006 (UTC)

What do you mean "artifacts"? A summation of sine waves is continuous... But, if you're interested in finding other curves which approximate various waveforms, why not construct a polynomial curve with suitable guidelines between 0 and 2pi? (For starters, make it pass through (0,0) and (2pi,0), and make both tangents at those points equal as well, then keep going, add point or tangent constraints until you have the curve you like.) - Rainwarrior 03:57, 25 September 2006 (UTC)

The artifacts I'm talking about is the Gibbs phenomenon. And no, a polynomial would take away all the fun of it (too easy), plus it would not be cyclic. ☢ Ҡi∊ff⌇↯ 07:49, 25 September 2006 (UTC)

Sorry for my stupidity, but what does an X with the L and the mirrored L around it mean? ChowderInopa 00:09, 24 September 2006 (UTC)

It denotes application of the floor function to the argument. For example, ⌊3.14⌋ = 3, because 3 is the largest integer that does not exceed 3.14. If we had a skyscraper with a floors at the height of every integer, and real numbers had to live at their own heights, then 3.14 would be living on the 3rd floor. --Lambiam^Talk 00:32, 24 September 2006 (UTC)

September 24

Random Variable X and its Mean Cancelling

If I have a random variable X with a mean mu(X), and both are in the same equation, one being negative, can the two simply cancel out?

Basically, how do the two relate?

Like in X – mu(X)? No, they won't simply cancel out. This is a new random variable, say Z. Have you read the articles Random variables and Expected value? Take for example that X is: the result of throwing two dice and adding the number of eyes. The outcomes of X may range from 2 to 12, with an expected value (arithmetic population mean) of 7. Suppose you throw the dice 10 times and observe for X this sample: [3, 7, 10, 9, 9, 7, 2, 7, 2, 7]. (I actually threw two dice ten times here.) That means for Z = X − 7 this sample: [−4, 0, 3, 2, 2, 0, −5, 0, −5, 0]. Occasionally the outcome of Z is 0, but the random variable Z itself is clearly not the constant 0. However, mu(Z) = 0. To see this, we need three facts: mu(V) = E(V) for a random variable V – which is true by definition of mu(.), E(X − Y) = E(X) − E(Y), and E(C) = C for constant C. Then mu(Z) = E(Z) = E(X – mu(X)) = E(X) – E(mu(X)) = mu(X) – mu(X) = 0. Indeed, the sample mean for our sample of 10 outcomes for Z is −0.7 – not quite 0 due to the random fluctuations, but fairly close. --Lambiam^Talk 01:13, 24 September 2006 (UTC)

Okay, but in the long run, they will cancel? Like if I have a function Y equal to the equation you said, Y = X - mu(X), the E(Y) will be 0, simply due to the rule that X can be transformed into mu (X)?

Yes, E(Y)=0, but I'm not sure what you meant by "X can be transformed into mu (X)". -- Meni Rosenfeld (talk) 16:38, 24 September 2006 (UTC)

Well in order for E(Y)=0, X must be equal to mu (X) to cancel...how does this come about? How do you show that E(Y)=0?

That's what I showed above, except that I called it Z instead of Y. --Lambiam^Talk 16:59, 24 September 2006 (UTC)

Wow Sorry you are completely correct, that makes perfect sense, thank you. Just as a final clarification, the E(muX) is the part that pertains to E(C) = C for constant C?

Exactly. E(X) (aka μ(X)) is a constant, therefore E(E(X)) = E(X). -- Meni Rosenfeld (talk) 17:11, 24 September 2006 (UTC)

the other delta function?

OK, so I am trying to understand how the fourier transform of a pulse train in the time domain gives also a pulse train in the frequency domain. Actually i'm stuck on a little detail. I get that: $F[\delta (t)]=\int \delta (t)\exp ^{-jwt}dt={1}$, that $F^{-1}{[1]}=\int \exp ^{jwt}dw={\frac {1}{jt}}\exp ^{jwt}$, and that $F^{-1}F[\delta (t)]=\int (\int \delta (t)\exp ^{-jwt}dt)\exp ^{jwt}dw={\frac {1}{jt}}\exp ^{jwt}$. And that in the end, you make the jump and say that $\delta (t)={\frac {1}{jt}}\exp ^{jwt}$...

I don't see it. Did I miss something? Anybody has some intuition (hopefully on physical grounds) on how the expression ${\frac {1}{jt}}\exp ^{jwt}$ is equivalent to a delta function? please? --crj

What makes you think a pulse train in the time domain should give a pulse train in the frequency domain? On the contrary, it should be all over the place in the frequency domain (see "Localization property" in the article Continuous Fourier transform). This should be intuitively obvious: you can't assign any one frequency to a single spike. It also fits with your result: 1, which is hardly a spike. In the expression you give for $F^{-1}{[1]}$, there is an occurrence of the variable $w$ (the traditional notation is $\omega$). That can't be right; the result should only depend on $t$ . This is not an indefinite integral (antiderivative; primitive function) but a definite integral for $w$ from $-\infty$ to $+\infty$. --Lambiam^Talk 21:30, 24 September 2006 (UTC)

Ooops. I meant to type t instead of w after the integration. There are some notes on the fourier transform of pulse train in the article Dirac comb. Maybe I am doing the problem the wrong way but something smells fishy here... because what happens at t=0? oh dear. Thanks anyway. --crj 00:48, 25 September 2006 (UTC)

Whoa! Careful there. The transform of a single impulse is a constant; the transform of a periodic sequence of impulses is again a periodic sequence of impulses, whose period is inversely proportional to the original period. --KSmrq^T 22:34, 24 September 2006 (UTC)

So the Dirac delta function is a tiny bit confusing? Since it's not really a normal function at all, that's not surprising. One way we approach it is through its properties within an integral. That is,

$\int _{-\infty }^{+\infty }\delta (t-t_{0})f(t)dt\,\!$

selects the value f(t₀). We can be more formal by using a limiting process. For example, we know that a Gaussian bell curve,

${\frac {1}{\sigma {\sqrt {2\pi }}}}\exp(-{\frac {x^{2}}{2\sigma ^{2}}}),\,\!$

integrates to 1, and can be made as narrow as we like by letting σ approach zero. We also know (and can easily verify) that the Fourier transform of a Gaussian is again a Gaussian, but with width inversely proportional to σ. As we pinch the Gaussian to its delta function limit, its transform spreads and flattens to a constant.

A delta function cannot be written as you propose, as a simple exponential, which may account for your confusion.

If making a single impulse requires some "funny business", making an impulse train requires more. But perhaps we should stop here for now. --KSmrq^T 23:24, 24 September 2006 (UTC)

Let me rephrase the question:
According to the Dirac comb article the fourier transform of an impulse train is:
$\sum _{n=-\infty }^{\infty }\delta (t-nT)\quad \Longleftrightarrow \quad {1 \over T}\sum _{k=-\infty }^{\infty }\delta \left(f-{k \over T}\right)\quad =\sum _{n=-\infty }^{\infty }e^{-i2\pi fnT}$.
My question is, under what grounds are the last two terms (sum with delta function in the frequency domain, sum with complex exponential) equivalent? -- Crj 02:34, 25 September 2006 (UTC)

Let's look at the sum. $\sum _{n=-\infty }^{\infty }e^{-i2\pi fnT}={\frac {e^{-2\pi ifT(N+1)}-e^{2\pi ifTN}}{e^{-2\pi ifT}-1}}={\frac {\sin 2\pi fT(N+{1 \over 2})}{\sin 2\pi fT}}$. Now let N approach infinity. It's possible to check that this sum approaches infinity if f = k/T (with integer k), approaches zero otherwise, and that its integral over any interval containing exactly one point where f = k/T approaches 1. So it's a sum of delta functions. Conscious 11:53, 25 September 2006 (UTC)

The question is based on an elementary slip. If the sum 1+2+3 equals the sum 2+2+2, we have no grounds for assuming equality between respective terms. This is true for infinite sums as well. --KSmrq^T 14:41, 25 September 2006 (UTC)

By the way, you are wrong in saying (in your initial post) that $F^{-1}{[1]}={\frac {1}{it}}\exp ^{i\omega t}$. You have to use the definite integral, i.e. $F^{-1}{[1]}=\int _{-\infty }^{+\infty }e^{i\omega t}d\omega$, to obtain the result. Conscious 17:02, 25 September 2006 (UTC)

wow this turned out to be a funny business. my difficulty really was in recovering the delta function intact after I take the fourier transform of an impulse train: "delta functions in... delta functions out", or so I thought. Thanks for all the responses! -- Crj 14:17, 26 September 2006 (UTC)

September 25

Dice-sum probability formula

It's September 24th here on the East Coast, but anyway: suppose I roll n s-sided dice. What is the theoretical probability of getting each possible sum, irrespective of the individual die values that make up that sum? I know a little bit of combinatorics and probability, but not quite enough to figure this one out on my own. Google has failed to enlighten me, so I turn to your collective wisdom for help. —Saric (Talk) 01:20, 25 September 2006 (UTC)

I think there is no truly simple formula for that. For one die the graph of the probabilities as a function of the sum is a horizontal line: a segment from a polynomial of order 0. For two dice you get a roof shape: two segments of polynomials of order 1. I guess that continues, like N segments of polynomials of order N−1 glued together. It should be possible to find a general formula for the i-th polynomial. I haven't given the issue much thought, though. --Lambiam^Talk 02:32, 25 September 2006 (UTC)

Lambiam is right. Here are several additional suggestions:

• First, we will denote by f (n, s, k) the number of possibilities (from which the probability is easily calculated) of receiving a sum of k when throwing n s-sided dice.
• For sufficiently large n, You can approximate the function with the normal distribution as:

$f(n,s,k)\approx s^{n}\exp {\left({\frac {3(n(s+1)-2k)^{2}}{2n(1-s^{2})}}\right)}{\sqrt {\frac {6}{\pi n(s^{2}-1)}}}$

• You can define f recursively: f (1, s, k) is 1 if k is an integer between 1 and s, and 0 otherwise; And

$f(n+1,s,k)=\sum _{i=1}^{s}f(n,s,k-i)$

• If I'm not mistaken, the piecewise polynomials from which the function is made are stitched at points of the form $s(1+j)+n/2\,\!$, where 0 ≤ j ≤ n-2. You can use this and the above formula to find, for every n, the polynomials that make up the nth function (though you shouldn't hope for a closed form for the nth polynomials).

I hope this helps. -- Meni Rosenfeld (talk) 07:42, 25 September 2006 (UTC)

Ah, yes, thank you both. I implemented that function in Perl and it seems to work nicely:

sub funkshun
 {my ($n, $s, $k) = @_;
  my ($i, $sum);
  if ($n == 1)
     {if ($k >= 1 and $k <= $s)
         {return 1;}
      return 0;}
  foreach $i (1 .. $s)
     {$sum += funkshun(($n - 1), $s, ($k - $i));}
  return $sum;}

All you do is divide the result by s to the nth power, and there's your probability. There is, however, one problem: it's really slow. f(5, 20, 55), for example, takes a couple of seconds. Maybe it's just my 300-megahertz processor, but I think the recursion makes this a relatively inefficient algorithm.

Now, here's the interesting part: some searching brought me to this site. The instructions page attempts the explain the math, but the explanation eluded me, which is why I came here to the reference desk for an algorithm I could understand (and thus implement myself). However, rereading it afterwards, I realized that the words "(k = 0) to the next lowest integer of…", which had puzzled me before, were implying the use of summation, like you, Meni Rosenfield, had used in your formula. That was enough for me to figure out the rest and code it thus:

sub funkshun
 {my ($n, $s, $k) = @_;
  my ($i, $sum);
  foreach $i ( 0 .. int(($k-$n)/$s) )
     {$sum += ( ((-1)**$i) * comb($n, $i) *
                comb($k-1-($s*$i),$n-1) );}
  return $sum;}

("Comb" is the combination function; "**" is the Perl exponentiation operator.) Looks crazy, doesn't it? It does to me, at least— but it works, and a lot faster than the other way, too. Who woulda thunk? —Saric (Talk) 00:11, 26 September 2006 (UTC)

I see. Rewriting this in terms I can understand, it's:

$f(n,s,k)=\sum _{i=0}^{\left\lfloor {\frac {k-n}{s}}\right\rfloor }(-1)^{i}{n \choose i}{k-1-si \choose n-1}$

Which is simpler than I imagined. Nice! -- Meni Rosenfeld (talk) 09:21, 26 September 2006 (UTC)

By the way, the reason your first implementation was so slow is because each value of f is calculated many times, making the complexity exponential in n. If you are able to store calculated values of f in memory, and add some more fine-tuning, the complexity will only be quadratical (that is, much faster). -- Meni Rosenfeld (talk) 09:32, 26 September 2006 (UTC)

Density Curve (?)

In my basic Stat class (Algebra Based) the teacher was explaining density curves. She then gave us various examples, including one that had uniform distribution. She described the density curve as being the base times the height, where the density curve would follow the box-shaped outline exactly. However, I don't understand how that could be a "curve" because of the sharp points. The only way I could see a function like that defined is with a piece-wise equation. Help me - just try to keep it within Calc II understanding. ;) --AstoVidatu 02:48, 25 September 2006 (UTC)

If you define the function as having result 0 for arguments outside the range of the random variable, you're right, technically speaking the shape of the graph of that function is not a curve, since it's not continuous. It does not "climb up" the sides of the box, but it jumps from 0 to this constant positive value. If you pay close attention mathematicians say very sloppy things all the time, like they say "the function f(x)" when they mean "function f" and think that is fine, which is a bit amazing for a field that is supposed to be exact and precise. --Lambiam^Talk 03:41, 25 September 2006 (UTC)

If she described the graph of the density function as following the vertical edges of the box, she must have been confused; a piece-wise definition for the density function is indeed what you're looking at here.

Of course, that leaves the question of the points where the function jumps, and which function value to choose there; however, it turns out that it doesn't matter.

Using "curve" to describe this graph might indeed be a bit problematic, but the graph of the absolute value function - despite having a corner, is usually considered a curve.

RandomP 16:32, 25 September 2006 (UTC)

Parallelepipeds' volume

I know that to get the volume of a parallelepiped, when given three vectors, you use the formula ${\vec {a}}\cdot [{\vec {b}}\times {\vec {c}}]$, which gives you a 3x3 determinant. However, does it matter, when evaluating the determinant, the order in which I put the rows? My Engineering and my Calculus textbooks put the row outside the parenthesis (e.g. $a_{x}$, $a_{y}$, $a_{z}$) on the top row of the 3x3 matrix, but I don't know if this is a coincidence, a convention, or if there is some mathematical reason this happens. And the reason I'm asking is because I can't ever remember which row goes where, so it would be quite relaxing to know it doesn't matter... Titoxd^(?!?) 03:04, 25 September 2006 (UTC)

Changing the order can only cause a sign flip: see Triple product for the relevant identity. If you're looking for a positive quantity, you'll want to take the absolute value anyway, so short answer: no, it doesn't matter. Melchoir 03:26, 25 September 2006 (UTC)

If, for any reason (such as determining orientation), you wish to preserve the sign of ${\vec {a}}\cdot [{\vec {b}}\times {\vec {c}}]$, then the order does matter, but it's not a biggie - if I'm not mistaken, the order of the rows is exactly the order in which they appear in this expression. -- Meni Rosenfeld (talk) 06:30, 25 September 2006 (UTC)

Yes, it matters. Since you already know the volume is a determinant, you should also know a basic definition of a determinant:

1. The determinant of the identity matrix is 1.
2. The determinant is a linear function of each column.
3. The determinant is an alternating function of the columns.

   (Two equal columns gives a zero result, so swapping two columns negates the result.)

We can show that transposing a matrix does not change its determinant, so the above facts can use rows instead of columns. Either way, these three rules uniquely define the determinant of any n×n matrix.

The cross product is also an alternating function of its arguments, so the sign change is already visible when swapping b and c.

But why should we care about the sign, when the magnitude is the same either way? Because when we are calculating more complicated volumes with this as a portion of a larger sum, we must get the sign right. Consider a complicated polygon in 2D, but one that does not intersect itself. (Technically, it is called a "simple" polygon.) We can compute the area inside the polygon by summing triangles each of which consists of a polygon edge and the origin — but only if we get the signs right. In 3D, using faces to give tetrahedra (sliced parallelepipeds) works similarly. This convenient method can be viewed as a specialization of Stokes' theorem, which itself is just a fancy version of the fundamental theorem of calculus, telling us we can find the integral over the interior of a region from what happens on the boundary.

So resist the urge to forget the order. The investment in good habits today will pay off in easier computations tomorrow. --KSmrq^T 15:58, 25 September 2006 (UTC)

But of course, "the identity matrix" has no special parallelepiped associated to it: see right-hand rule for one convention on it, though. (To mathematicians, this rarely matters, because we're usually just as happy to see our geometry mirrored. Physicists have that whole "real world" thing to take into account, though.)

KSmrq is right, of course: it's essential to get into the habit of getting things right.

RandomP 16:23, 25 September 2006 (UTC)

ANSWERED -> Arc length

I need to find the length of a circular arc (shown). I know the length of the straight line between the two end nodes (BD), and I know the height of the arc (AC), but cannot think of any way to find the length of the curve. Is there a formula I can use in this case? smurrayinchester^{(User), (Talk)} 11:03, 25 September 2006 (UTC)

Let O be the centre of the arc. OA=OB=r is the radius of the circle. From the OBC triangle: OB²=BC²+OC², or r²=BC²+(r-AC)²; you can find r from this equation. After this, the angle BOC is equal to arcsin(BC/r), and the arc length is 2rarcsin(BC/r). Conscious 11:24, 25 September 2006 (UTC)

After simplifications, it's $l={\frac {a^{2}+4h^{2}}{4h}}\arcsin {\frac {4ah}{a^{2}+4h^{2}}}$ where a=BD and h=AC. Conscious 11:32, 25 September 2006 (UTC)

Another method: Find the center, and hence the radius, by constructing tangent lines at both ends and constructing perpendicular lines to those. The intersection of those two lines will be the center point. Then measure the angle swept by the arc. You can then apply the simple formula l = piR(theta/360), where theta = the angle swept in degrees. StuRat 11:35, 25 September 2006 (UTC)

Thanks to both of you; I've gone for Conscious's method as I don't have an accurate sketch of the arc available. The result is also close enough to my very simple approximation $l\approx 2{\sqrt {AC^{2}+CD^{2}}}$. smurrayinchester^{(User), (Talk)} 11:53, 25 September 2006 (UTC)

My preferred approximation is l ≈ BD :) Conscious 11:57, 25 September 2006 (UTC)

OK, maybe not that simple... smurrayinchester^{(User), (Talk)} 12:10, 25 September 2006 (UTC)

ANSWERED -> Factoring A Polynomial

Yes, yes, I realize this is a math HW prob. But we're kinda stuck. How does one go about factoring this polynomial: y^2-cy-dy+cd=0  ? I don't need answer, just advice on how to solve it. Thanks in advance. John 17:00, 25 September 2006 (UTC)

Just try to look what you can factor out from the first two terms, and from the two last ones. Conscious 17:04, 25 September 2006 (UTC)

Yes, I see that now. John 17:23, 25 September 2006 (UTC)

ANSWERED -> Using the angle (in radians) and length to get coordinates

Hi, if I have a line, starting at the origin (0,0), 10 units long, at a certain angle in radians, how can I work out what the coordinates of the end of the line is?

Any help would be much appreciated!

--Mary

[note: this isn't for homework: I'm writing a program where circles move around on a screen, and I want to include a line in the circle that shows a current direction.]

Take a look at our article on the trigonometric functions. Conscious 17:54, 25 September 2006 (UTC)

Better still, look at the circle article. --KSmrq^T 18:19, 25 September 2006 (UTC)

This is a conversion issue, from polar coords to 2D rectangular coords. StuRat 18:30, 25 September 2006 (UTC)

Thanks all! --Mary Bold text

Whoever adds all these "ANSWERED -> " stuff to the titles, I kindly ask that you don't do it in the future, as it will break #-links. Thanks. – b_jonas 16:15, 26 September 2006 (UTC)

Hmm, I didn't think about that. Are there many links to these questions ? StuRat 16:34, 26 September 2006 (UTC)

What's with the "m"? Are there any links to these questions? -- Meni Rosenfeld (talk) 16:56, 26 September 2006 (UTC)

Well, I occasionally link to some of them, especially when someone asks a suspicously similar question on RD again. Of course, I'm not an authority here so if the community decides for these title changes, you can do that, then I will resort to adding explicit Anchor tags instead of linking to lables. – b_jonas 18:10, 26 September 2006 (UTC)

How about if we create a subsection titled "ANSWERED" instead ? This will still show up in the TOC but won't keep your link from working. Sound good ? StuRat 22:10, 26 September 2006 (UTC)

ANSWERED -> Central Angle of a Tetrahedron

According to VSEPR theory, molecules such as methane are arranged in a tetrahedral form where the angle from one hydrogen to the carbon to another hydrogen is approximately 109 degrees. Is there any way to derive this angle mathematically? I have tried using the Pythagorean Thm in three dimensions, but it was a complete mess. Thanks, --JianLi 21:08, 25 September 2006 (UTC)

It's the top angle α of an equilateral triangle with base (2/3)×√6 and sides of length 1. Then the law of cosines gives you cos α = −1/3, as stated in the VSEPR article. No idea where the number −1/3 comes from. --Lambiam^Talk 22:30, 25 September 2006 (UTC)

Things become easier if you consider that a tetrahedron can be embedded in a cube, each face of the cube has one edge of the tetrahedron across it. If the cube has sides of length 2 and is centered at the origin then the verticies of the cube will be (+/-1,+/-1,+/-1). Four of these will be verticies of the tetrahedron, say A (1,1,1), B (1,-1,-1), C (-1,-1,1), D (-1,1,-1). Consider the triangle OAB. This has sides a=√3, b=√3, c=2√2. The law of cosines gives $c^{2}=a^{2}+b^{2}-2ab\cos(\gamma )$ so $8=3+3-2*3*\cos(\gamma )$ so $\cos(\gamma )=-2/6$ and $\gamma =109.471$ degrees. --Salix alba (talk) 23:07, 25 September 2006 (UTC)

Or you can just take the dot product of OA and OB. Melchoir 00:11, 26 September 2006 (UTC)

Wow, this problem seems so easy now. Thanks! --JianLi 01:56, 26 September 2006 (UTC)

Should some of this be mentioned in the text of VSEPR? Before cos^-1(-1/3) is mentioned there now, there is nothing constraining the shape of the tetrahedron. Is there an argument showing that the tetrahedron should be one sixth part of a cube? --Lambiam^Talk 15:34, 26 September 2006 (UTC)

September 26

uncountable set

x is uncountable set, y is subset of x, if y is countable, is that x\y uncountable.

Yes but you did not ask for proof. Twma 01:55, 26 September 2006 (UTC)

yes, the proof may be like this, assume x\y is countable, so N ~ x\y, and x\y is finite set, so every element in N, there is corresponding element in x\y, and y is subset of x, and y is countable, so x must be finite set, but this is a contradiction with x is uncountable set, so x\y is uncountable.

That doesn't make sense. An easy proof notes that x is the union of y and x\y, and the union of two countable sets is countable, so one of the two must be uncountable. You seem to have many questions-- perhaps a textbook like [8] would help? Melchoir 05:12, 26 September 2006 (UTC)

countable set

P is the set of polynomial function with rational coefficient, is P is countable set?

Yes but you did not ask for proof. Twma 01:56, 26 September 2006 (UTC)

The quadratic formula to solve x

For quadratic equation ax² + bx +c:

x = ( -b +- sqrt( b² - 4ac ) ) / 2a

I asked my math teacher the other day how this equation was formulated, and he told me that you must complete the square of ax² + bx + c ... he showed me, but I didn't quite understand. When I formed two squares, it was the sum of two squares to equal zero, which has no solution. Could somebody take me through it?

See quadratic equation#Derivation. Conscious 04:35, 26 September 2006 (UTC)

To have a quadratic equation, we must have an equality:

$0=ax^{2}+bx+c.\,\!$

Perhaps before studying the formal manipulations of "completing the square" we should look at simpler cases. We are looking for values of x that make the quadratic polynomial (the right-hand side of the equation) equal to zero. In more geometric terms, we wish to find where the parabola defined by

$y=ax^{2}+bx+c\,\!$

crosses the x-axis. This may happen zero, one, or two places. Suppose b is zero:

$ax^{2}+c=0.\,\!$

Then if a is not zero, we can divide both sides by a to give

$x^{2}+c/a=0.\,\!$

Now subtracting c/a from both sides, we see that the square of x must equal −c/a. So long as a and c have opposite signs, we can take a square root to get one answer and take its negation for another.

However, if b is non-zero the parabola is shifted left or right so that the zero-crossings do not occur symmetrically on each side of the y-axis. For example, suppose c is zero:

$ax^{2}+bx=0.\,\!$

Since both terms of the polynomial contain a factor of x, we can rewrite this as

$(ax+b)x=0.\,\!$

If x is zero, clearly the polynomial is zero. Otherwise we must solve

$ax+b=0.\,\!$

Divide by a and subtract to reveal the unique solution, −b/a. So in this case the parabola is centered halfway between 0 and −b/a, at −b/2a. If we shift the x values by this amount, the parabola and its zero-crossings will again be centered around zero.

Another way to describe "completing the square" is, "centering the parabola". Conveniently, we have just found the solution! For, if we add c back in to the polynomial, it bumps the parabola up or down, but does not change its center. That is, the parabola will always cross zero at −b/2a plus or minus a square root.

All that remains is to decide what should be inside the square root. We make the shifting substitution

$x=X-{\frac {b}{2a}},\,\!$

to produced the centered form

$aX^{2}-{\frac {b^{2}}{4a}}+c=0.\,\!$

This has no X term, so we can adapt the solution from above. Divide by a, subtract the constant, and simplify to produce:

$X=\pm {\frac {\sqrt {b^{2}-4ac}}{2a}}.\,\!$

Thus the solutions for x, rather than X, are

$x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}.\,\!$

To recap, the quadratic formula combines centering with square roots.

The question then naturally pops into our minds: Can we do something similar for cubic polynomials or higher? The answer is not at all obvious, took centuries to resolve, and produced some remarkable new mathematics. We can, with a little more difficulty, use shifting, square roots, and cube roots to solve cubics. With a great deal more difficulty we can similarly solve all quartics as well, though the method is so complicated it finds little use. But we can go no further; a polynomial of degree five or higher generally has no such closed-form solution.

So enjoy the quadratic formula. It's remarkably useful, and rather special. --KSmrq^T 16:51, 26 September 2006 (UTC)

equivalent class

if S is any infinite set, and x(- S, is that S is equivalent class to S\{x}.

I'm not quite sure what you're trying to say by "is that S is equivalent class to S\{x}", but if you're trying to ask whether S is in the same equivalence class (or, more succintly, "is equivalent to") S\{x}, you need to state what your equivalence relation is - for example, S and S\{x} are equivalent if the relation is "has the same cardinality" (since S is infinite), but they are not equivalent if the relation is "contains x". Confusing Manifestation 07:50, 26 September 2006 (UTC)

solve equations simultaniously

Solve equations 2x+3y=2 and 6x-y=-4 simultaneously.

What does this mean? How do I do this? I've probably been doing problems like this for about 5 years, but never got it well enough to remember anything. Thanks. 71.231.150.146 05:18, 26 September 2006 (UTC)

It means the value for X is the same in both equations and the value for Y is the same in both equations. There are many ways to solve them, the substitution method, addition method, graphing, using matrices, etc. StuRat 10:08, 26 September 2006 (UTC)

See Simultaneous equations. Conscious 05:38, 26 September 2006 (UTC)

In this case, I recommend you to read System of linear equations in the first place. JoergenB 10:00, 26 September 2006 (UTC)

Each separate equation defines a straight line in the plane; its solutions are precisely the points (x, y) lying on that line. The solution to both equations simultaneously is the coordinate where the lines intersect. Fredrik Johansson 11:26, 26 September 2006 (UTC)

It's hard to know how best to answer your question, since you do not make it clear what you already understand and what you want. The specific example you give is two equations in two variables, where each equation has at most one variable in a term. Is your question about this limited class of problems? Or are you also interested in more equations with more variables? Or perhaps more complicated equations such as quadratics? If you can "do them", what is it you'd like to remember? Are you saying you have to look up the solution technique each time, so would like help retaining the procedure? Help us help you. --KSmrq^T 17:17, 26 September 2006 (UTC)

Square Root?

What is the square root of France?

${\sqrt {(}}France)$

Thank you in advance for your assistance in solving this highly complex mathematical expression?

24.39.182.101 17:22, 26 September 2006 (UTC)

The usual definition of France is not mathematically rigorous, so we cannot use it to determine if √France is meaningful and its value. If seen as a variable which can take positive real values, then there is no way to simplify √France without knowing more about the value it represents. -- Meni Rosenfeld (talk) 17:34, 26 September 2006 (UTC)

France does not represent much of value. For most practical purposes, its square root may be considered zero. Fredrik Johansson 18:20, 26 September 2006 (UTC)

Think carefully before answering, as the verbal question would properly be notated

${\sqrt {\operatorname {France} }},\,\!$

which is not what we see. Perhaps what is meant is, "What country, if it applied itself twice over, could equal France?" But if so, it's a trick question, because France is, of course, incomparable. --KSmrq^T 19:07, 26 September 2006 (UTC)

Approximately $1.64(evil)^{\frac {1}{4}}$, of course. Factor out the root e, then apply the principle that money is the root of (all) evil. —AySz88\^-^ 22:24, 26 September 2006 (UTC)

I belive in this case, evil is the root of all money. Which, considering the widespread notion that money is the root of all evil, poses an interesting pair of questions: what is money? and what is evil? so that such an interesing pair of relations can occur.

There exist only complex numbers that, when squared, equal each other. So, economics and morality must both be complex. Black Carrot 06:44, 27 September 2006 (UTC)

Not exactly. The equations a²=b, b²=a have solutions in reals, (0,0) and (1,1). We can thus safely conclude that money = evil. -- Meni Rosenfeld (talk) 15:04, 27 September 2006 (UTC)

Pi

For some reason, I just can't comprehend how the Bailey-Borwein-Plouffe_formula calculates Pi. Does it really calculate individual digits? Could I easily calculate the, say, 84th digit? The article isn't too clear to me. Maybe it's not possible to explain in simple terms? --Russoc4 20:18, 26 September 2006 (UTC)

I don't really have time to answer this now, but if nobody gets back to you I'll try to explain this evening. But I have a question: is there a known, simple formula for calculating the decimal digits of π? It seems that to get an individual digit with these binary formulae you need to calculate all the preceding digits. –Joke 20:41, 26 September 2006 (UTC)

No, I know of no cheat for decimal digits. For hexadecimal digits, we can — as requested — calculate the 84th digit, say, without first accumulating all the preceeding hex digits. The reason this works is that the summation formula has two lucky properties: (1) it expands in powers of 1/16, and (2) it limits the contributions to any selected hex digit. Read carefully about the use of the modulo operator, which is key to the latter. --KSmrq^T 21:12, 26 September 2006 (UTC)

Why can't the hex values be converted to decimal? --Russoc4 23:24, 26 September 2006 (UTC)

They can, but to convert a hexadecimal expansion to decimal you need to have all its hex digits to find the last decimal digit. If I tell you the number is 3.???????????????????????DEADBABE, where each ? is a hex digit I chose not to reveal, there's nothing you can do with the deceased infant at the end. --Lambiam^Talk 23:45, 26 September 2006 (UTC)

At the mathworld article on BBP formulas, it says that it has been proven that the BBP method for PI cannot be adapted to any base that is non-binary (power of two)m, though there may still be a different method, as yet undiscovered, for finding decimal digits. - Rainwarrior 00:21, 27 September 2006 (UTC)

Shortest algorithmn for calculating infinite digits of pi (written in python programming language)

 #!/usr/bin/python
 from sys import stdout
 k, a, b, a1, b1 = 2, 4, 1, 12, 4
 while 1:
     p, q, k = k*k, 2*k+1, k+1
     a, b, a1, b1 = a1, b1, p*a+q*a1, p*b+q*b1
     d, d1 = a/b, a1/b1
     while d == d1:
         stdout.write('%d' % d)
         a, a1 = 10*(a%b), 10*(a1%b1)
         d, d1 = a/b, a1/b1

Sample run

 $ python short_pi.py 
 3141592653589793238462643383279502884197169399375105820974944592307816406286208998
 6280348253421170679821480865132823066470938446095505822317253594081284811174502841
 0270193852110555964462294895493038196442881097566593344612847564823378678316 and so on

202.168.50.40 00:05, 27 September 2006 (UTC)

It's easy to reduce this by one line. Here's a short astounding obfuscated C program by Dik T. Winter and Achim Flammenkamp:

 a[52514],b,c=52514,d,e,f=1e4,g,h;main(){for(;b=c-=14;h=printf("%04d",
 e+d/f))for(e=d%=f;g=--b*2;d/=g)d=d*b+f*(h?a[b]:f/5),a[b]=d%--g;}

However, it gives only 15000 decimals instead of an infinitude like in the Python program. --Lambiam^Talk 01:50, 27 September 2006 (UTC)

If hex digits of π are fine, I've written a Python program for calculating them with the BBP formula. The page explains how you rewrite the formula to calculate isolated digits. If a base 10 BBP-type formula existed for π, the same could be done for decimals, but as was pointed out above, there is no (arctangent-based) BBP formula for π in any non-binary base. By the way, you can calculate isolated decimal digits of ln(9/10). - Fredrik Johansson 04:41, 27 September 2006 (UTC)

That last note is interesting...can you elaborate? :-) --HappyCamper 17:09, 27 September 2006 (UTC)

Straight Angle's interior

My math teacher doesn't seem to know. Does a straight angle have an interior or exterior? Reywas92 21:41, 26 September 2006 (UTC)

Okay, a straight angle is 180 degrees, or π radians. A 120 degree angle has an interior angle of 120 and an exterior angle of 60. A 150 degree angle has an interior angle of 150 and an exterior angle of 30. Therefore, it would make sense for a 180 degree angle to have an interior angle of 180 degrees, and an exterior angle of 0 degrees. Makes sense to me at least. --AstoVidatu 22:42, 26 September 2006 (UTC)

Huh? Wouldn't a 150 degree angle have an exterior angle of 210 deg, not 30? Reywas92 00:20, 27 September 2006 (UTC)

Check out the Internal angle page. The interior/exterior angle diagram was the type of thing that I was thinking of. I'm pretty sure interior angles and exterior angles add up to 180 degrees. --AstoVidatu 00:51, 27 September 2006 (UTC)

September 27

Math Question

Jody paid $44 bill at the hardware store using a combination of $10, $5, and $1 bills. If she paid with 13 bills in all, how many of each bill did she use?

Let i, j and k stand for the number of $10, $5, and $1 bills, respectively. Then we have 10i + 5j + k = 44 and i + j + k = 13. Subtracting the two (corresponding to the substitution k := 13 − i − j) gives 9i + 4j = 31. Modulo 4 this is i ≡ 3, so substitute i := 4m + 3, giving 36m + 27 + 4j = 31, or 36m + 4j = 4, or 9m + j = 1. Since m and j can't be negative, this is only solved by (m, j) = (0, 1). Backsubstitution gives i = 4·0 + 3 = 3 and k = 13 − 3 − 1 = 9, so (i, j, k) = (3, 1, 9). --Lambiam^Talk 01:23, 27 September 2006 (UTC)

Yo dude, I know that Lambian basically spoonfed you the answer, but make sure you can do this. An A on the homework doesn't make up for an F on a quiz. Representin', --AstoVidatu 02:46, 27 September 2006 (UTC)

Note that there aren't very many ways to get $44 from those 3 denominations, so you could just try all the possibilities, too. Obviously, you can exclude those possibilities with 14 or more $1 bills:

10+10+10+10+ 1+ 1+ 1+ 1
10+10+10+ 5+ 5+ 1+ 1+ 1+ 1
10+10+ 5+ 5+ 5+ 5+ 1+ 1+ 1+ 1
10+ 5+ 5+ 5+ 5+ 5+ 5+ 1+ 1+ 1+ 1
 5+ 5+ 5+ 5+ 5+ 5+ 5+ 5+ 1+ 1+ 1+ 1

10+10+10+ 5+ 1+ 1+ 1+ 1+ 1+ 1+ 1+ 1+ 1
10+10+ 5+ 5+ 5+ 1+ 1+ 1+ 1+ 1+ 1+ 1+ 1+ 1
10+ 5+ 5+ 5+ 5+ 5+ 1+ 1+ 1+ 1+ 1+ 1+ 1+ 1+ 1
 5+ 5+ 5+ 5+ 5+ 5+ 5+ 1+ 1+ 1+ 1+ 1+ 1+ 1+ 1+ 1

StuRat 09:44, 27 September 2006 (UTC)

Use a reasoning like: you need 4 $1 bills to get from $40 to $44 (why?), so the problem reduces to getting $40 from 9 bills. You can have 0 or 5 $1 bills (why?), 0 is impossible (why?), so you are left with $35 and 4 $5/$10 bills. Now it should be obvious.--gwaihir 10:00, 27 September 2006 (UTC)

Plural of 'series'

On the urging of another Wikipedia member, after reversions of my ideas regarding this topic, I have decided to post here and ask your opinions.

In his Principia Mathematica, Isaac Newton, when discussion series, used the latin plural 'seriei' instead of our ambiguous english habit. While the habit of singular/plural duality is tolerated for words of english origin such as 'deer', I do not believe this ambiguity belongs in Wikipedia mathematics articles.

Quite simply, the singular/plural duality of 'series', as we somewhat haphazardly have defined it, is confusing. It would not be as confusing if the word did not end with an 's', however it does. Anyone reading these articles has to do a double take to check the verb when reading an article on a series. I have asked several collegues about this already, and they voiced support to me for a clearer, less ambiguous, more grammatically correct spelling.

I thus propose the changing of 'series' in the plural to 'seriei'. One series, several seriei; again, a distinction made by Isaac Newton and lacking the ofttimes confusing singular/plural duality we have assigned to 'series' at present. We use 'criterion/criteria', and 'nucleus/nuclei', and 'formula/formulae', and 'curriculum/curricula', many of which are not exactly common distinctions made by your average english speaker, yet which go a long way towards the goal of precision and accuracy that we strive for here on Wikipedia. I propose simply extending this customary observance to another latin borrow-word which has its own well-established plural. Dbsanfte 05:19, 27 September 2006 (UTC)

I think there must be some mistake here. In Latin, the plural of series is series, and we preserve this in English. Seriei is the genitive singular. Latin nouns ending in -es do not form their plurals in -ei. Maid Marion 07:46, 27 September 2006 (UTC)

Now I feel absolutely silly. You're correct, I misread my dictionary. Please disregard this. Dbsanfte 13:03, 27 September 2006 (UTC)

Consider it thoroughly ignored. Black Carrot 14:37, 27 September 2006 (UTC)

What is this question called?

I heard this question a long time ago, but I have forgotten what it is called. Suppose you have 2 types of stamps, one with a denomination of p dollars, and another of q dollars. What is the largest amount that cannot be made with a linear combination of them? The solution is (p-1)(q-1)-1 if I recall, but I'd like to find more background about this. Thoughts? --HappyCamper 17:12, 27 September 2006 (UTC)

Coin problem. Chuck 17:28, 27 September 2006 (UTC)

Largest Number Contest.

What is the largest number you can fit into the width of the line below. Make it a single string of numbers (no second line). Exponents allowed.:

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