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April 5

A^⊕n

What does A^⊕n denote? 76.14.188.230 (talk) 10:27, 5 April 2011 (UTC)[reply]

(just a guess) Maybe

A\oplus \dots \oplus A

, n times? Staecker (talk) 11:28, 5 April 2011 (UTC)[reply]

Yes, this is correct. Invrnc (talk) 11:45, 5 April 2011 (UTC)[reply]

Thank you! 76.14.188.230 (talk) 12:12, 5 April 2011 (UTC)[reply]

BibTex Style File

Resolved

Can anyone recommend a nice BibTex style file, please? I like to use numbers for references, e.g. [1], [2], [3], instead of [Smith99]. I saw a paper where the numbers were in bold: [1] instead of just [1]. I usually put book titles and journal names in italics and in speech marks. To recap, I would like:

Bolded, number citations.
Book or journal titles it italics with speech marks.

Also, is there a way to personalise them? I have found one or two that are okay, but they put the journal number as no. 1 instead of No. 1, and I don't like that. How hard is it to write one from scratch? Any suggestions boys and girls? Fly by Night on Tour (talk) 16:15, 5 April 2011 (UTC)[reply]

You will probably find it much easier to tweak existing .bst files than to start from scratch. How easy it is depends wholly on your experience with coding in general, and the idiosyncrasies of *TeX in particular. The file pnas.bst does almost what you want, changing to bold shouldn't be too hard. See file here: [1] SemanticMantis (talk) 17:09, 5 April 2011 (UTC)[reply]

You are right that the coding is quite hard. But I managed to find an automated program that makes one for you. I run TeXnicCentre. I opened CMD (which is a DOS window) and typed latex makebst. It asked me a big long list of multiple choice questions. At the end it made a .dbj file, that it called a "batch file". It gave me the option to run that, which I did, and I now have my own personalised .bst file. However, I made a mistake and told it to put the year after the author. I don't like that and want to change that to putting the date at the end. How would I tweak the .bst file? Fly by Night on Tour (talk) 17:37, 5 April 2011 (UTC)[reply]

Deep in the bst file you will find something like FUNCTION{book}. You will see a list of things inside the curly brackets. That is the order that they are shown. You can move the date to a lower position. Notice that there is a completely different function for articles, manuals, injournals, etc... You may want to change more than just one of them. Personally, I download the bst file for the publication that I am submitting my article to. For example, I'm writing one for IEEE right now, so I'm using the IEEEtran.bst file. -- kainaw ™ 18:45, 5 April 2011 (UTC)[reply]

Does anyone know how I might tweak my current .bst file to change the position of the dates? I've run the auto-compiler three times and it won't even create a .dbj output any more. It is obviously unhappy about some of my choices, but can't tell me because it's a DOS window. Fly by Night on Tour (talk) 18:23, 5 April 2011 (UTC)[reply]

Well-behaved functions with respect to measure

Consider functions $f:X\rightarrow Y$ on the sets $X$ with measure $\mu _{1}$ and $Y$ with measure $\mu _{2}$ . We may say that $f$ is well behaved wrt measure if there exists a $\lambda$ such that whenever $S\subset X$ is a measurable set, $\mu _{2}(f(S))=\lambda \mu _{1}(S)$ .

Is there a name for such functions? Some special cases are linear maps between $\mathbb {R} ^{n}$ and $\mathbb {R} ^{m}$ , as well as any function which maps into a set of measure 0. I'm sure there are much more exotic examples. --COVIZAPIBETEFOKY (talk) 16:47, 5 April 2011 (UTC)[reply]

Did you mean a less than or equal instead of equal? Not many linear maps even would satisfy that. Dmcq (talk) 17:13, 5 April 2011 (UTC)[reply]

No, I meant equal. I was under the impression that for a linear map f between

\mathbb {R} ^{n}

and

\mathbb {R} ^{n}

, ~~the absolute value of the determinant~~

\lambda =|\det(f)|

behaves in exactly this way, and the wikipedia article gives a generalization of that property. Have I missed something? --COVIZAPIBETEFOKY (talk) 19:04, 5 April 2011 (UTC)[reply]

Well, you seem to have started with measures of arbitrary (presumably arbitrary measurable) sets, and retrenched to measures of parallelepipeds. I am not aware of any way to take the determinant of an arbitrary measurable subset of R^n. --Trovatore (talk) 19:09, 5 April 2011 (UTC)[reply]

The determinant of

f

, as a matrix! And I explicitly required that

S

be measurable, and by virtue of requiring that

f(S)

have a measure, I implicitly required that to be measurable as well. --COVIZAPIBETEFOKY (talk) 19:14, 5 April 2011 (UTC)[reply]

Right, sorry, I got confused. I thought you meant the determinant itself was the function you were calling "well-behaved". --Trovatore (talk) 19:19, 5 April 2011 (UTC)[reply]

Oh, I see. I was trying to avoid using awkward language. I'll go back and write it out. --COVIZAPIBETEFOKY (talk) 19:24, 5 April 2011 (UTC)[reply]

Reading what has been written it seems that you have a map ƒ : X → Y, where both X and Y have volume forms. A volume form on a k-dimensional manifold is a non-degenerate, skew-symmetric, differential k-form. Assume that X^m is equipped with the volume form ω, while Yⁿ is equipped with the volume form τ. A necessary condition for the map ƒ to be "well behaved" is that the pull back ƒ_{_*}τ = ω; which implies that m = n. After that, it seems that there is a lot of freedom, and it would take some thought. It's a bit late for me. Let me know what you come up with. I'd be interested to hear (read). Fly by Night on Tour (talk) 23:56, 5 April 2011 (UTC)[reply]

Without bothering about manifolds volume is transformed using the Jacobian determinant. Otherwise Measure-preserving dynamical system might be of interest, the constant there is 1 but that probably doesn't matter. But basically as Fly by Night you just have to make sure the measure of the volume form acts like you want. Dmcq (talk) 10:07, 6 April 2011 (UTC)[reply]

It depends how you want to measure volume in the image. If the volume forms are compatible with respect to a flat, torsion-free Koszul connection then you can use the Jacobian. You really have triples ƒ : (X,D,ω) → (Y,∇,τ) where D and ∇ are connections and ω and τ are volume forms. If Dω = ∇τ = 0 and D and ∇ are both flat and torsion free you're okay. If not then you're in trouble. The pull-back condition takes care of that because it expresses the condition in terms of the volume forms themselves, which are then free to vary as they choose. Fly by Night on Tour (talk) 14:25, 6 April 2011 (UTC)[reply]

Modifying encrypted data without decrypting it

This is a CS question, but it seems more suited towards the math desk than the computing desk.

I remember reading a story a while back in which some new encryption algorithms were designed that allowed other people to perform operations on your data without being able to see what it actually is. So, for a trivial example, you might encrypt a large number, send it to someone else, have them multiply it by two using this type of algorithm, and send back the encrypted result for you to decrypt, without them ever finding out what the number is. Can anyone dredge up this article for me, or link to more information? Thanks. « Aaron Rotenberg « Talk « 19:16, 5 April 2011 (UTC)[reply]

See Homomorphic encryption, especially the Fully section and its references. Invrnc (talk) 19:27, 5 April 2011 (UTC)[reply]

Looks like what I'm looking for. Thanks. « Aaron Rotenberg « Talk « 19:46, 5 April 2011 (UTC)[reply]

For some situations there are simpler method like blind signatures. 75.57.242.120 (talk) 07:11, 6 April 2011 (UTC)[reply]

Resolved

Recursive formula

I wanted to solve the recursive equation $r_{i}=br_{i-1}-r_{i-2}$ , where b is some constant. I got that

r_{i}={\frac {1}{\sqrt {b^{2}-4}}}\left[{\left({\frac {b+{\sqrt {b^{2}-4}}}{2}}\right)}^{i}-{\left({\frac {b-{\sqrt {b^{2}-4}}}{2}}\right)}^{i}.\right]

This should simplify into a simple polynomial, but all attempts at doing that have failed. I tried representing the term inside the brackets by as a difference of powers, but I still wasn't able to get rid of the square roots. Is there a way to simplify this? 74.15.137.130 (talk) 22:37, 5 April 2011 (UTC)[reply]

What exactly do you mean by a "simple polynomial" in this case? Normally if you speak of a polynomial function of i, you would not have i in the exponent. By "simple", do you mean expressible in terms of integers, so that the radicals wouldn't need to be there? If so, I'm inclined to suspect that such an expression does not exist. Michael Hardy (talk) 23:17, 5 April 2011 (UTC)[reply]

Use the formula

(x+y)^{i}=\sum _{n=0}^{i}{\binom {i}{n}}x^{n}y^{i-n}

If you use this on both powers, lots of things will cancel out. Looie496 (talk) 23:43, 5 April 2011 (UTC)[reply]

Thanks. 74.15.137.130 (talk) 23:48, 5 April 2011 (UTC)[reply]

Oh. Yes, that's right. But the degree of the thing gets bigger as i increases. Not what I think of when I see the word "polynomial". I guess the poster didn't mean a polynomial in the variable i, but rather, for each value of i separately, a polynomial in b. Michael Hardy (talk) 23:59, 5 April 2011 (UTC)[reply]

Generally

r_{i}=A_{1}\lambda _{1}^{i}+A_{2}\lambda _{2}^{i}

where

\lambda _{1}=a+{\sqrt {a^{2}-1}}

and

\lambda _{2}=a-{\sqrt {a^{2}-1}}

and

a={\frac {b}{2}}

. When using rational initial values, like

r_{-1}=1

and

r_{-2}=0

, then the square roots cancel out and all the

r_{i}

turns out rational as they should. The special case

a=1

has the solution

r_{i}=A_{1}+A_{2}i

which is indeed a polynomial in

i

. Bo Jacoby (talk) 11:24, 6 April 2011 (UTC).[reply]

The nth term is a polynomial in b of degree n. These polynomials are related to the Fibonacci polynomials

F_{n}(x)

as follows:

r_{n}(b)={\frac {F_{n}(ib)}{i^{n}}}

(here, i is the square root of -1). Gandalf61 (talk) 11:00, 6 April 2011 (UTC)[reply]

Polyhedron whose vertices correspond to unordered pairs

Is there a convex polyhedron whose vertices correspond to the 15 unordered pairs of elements of a 6-element set { a, b, c, d, e, f}, with an edge between two pairs precisely if they are intersecting subsets of { a, b, c, d, e, f} (thus there would be an edge between ab and ac but not between ab and cf)?

(Each vertex would thus have degree 8, so there would be 8 × 15/2 = 60 edges. Euler's formula V − E + F = 2 would then imply

15 − 60 + F = 2,

so F = 47, i.e. there would be 47 faces.) Michael Hardy (talk) 23:55, 5 April 2011 (UTC)[reply]

The answer is no. The graph you describe is a convex polyhedron only if it is a planar graph. But it's fairly straightforward to construct a subgraph that is a subdivision of the complete bipartite graph K_3,3. I can get one from the subgraph spanned by the eight vertices {ab, ac, cd, bd, ad, be, ef, df}. Specifically, take the complete bipartite graph on the pair of vertex sets {ab,cd,ef} and {ac,bd,df}, and then insert the vertex ad into the (ab,df) edge, and the vertex be into the (bd,ef) edge. The resulting graph appears as a subgraph of the graph in question. Sławomir Biały (talk) 01:18, 6 April 2011 (UTC)[reply]

Very nice. Your answer leaves me wondering why I didn't think of that. I must be getting rusty in some things. Michael Hardy (talk) 01:56, 6 April 2011 (UTC)[reply]

Next question: Is there some visually nice way to display the graph I described? Michael Hardy (talk) 02:03, 6 April 2011 (UTC)[reply]

The graph you've described is the complement of the Kneser graph KG_6,2, so it's quite closely related to the Petersen graph, which is KG_5,2. I'm sure someone has made a nice drawing of KG_6,2. —Bkell (talk) 03:04, 6 April 2011 (UTC)[reply]

I see that the complement of KG_6,2 is called the (6,2)-Johnson graph; there's a bit about Johnson graphs in the Kneser graph article (in the #Related graphs section), but I didn't notice that before. So now you have a name for your graph, at least. Wolfram Alpha will draw a picture of it, but there's probably a more beautiful way to draw it. —Bkell (talk) 03:31, 6 April 2011 (UTC)[reply]

Interesting. Probably Wolfram Alpha's picture will be far better than what I could draw by hand. Thank you, Bkell and Sławomir Biały.

(BTW, Sławomir, am I right in suspecting that the first syllable of your first name is pronounced something like "suave"?) Michael Hardy (talk) 04:11, 6 April 2011 (UTC)[reply]

That's a bit strange, probably worth a look, Mathworld has Triangular Graph with

T 6

corresponding to this whereas we have triangular graph redirecting to planar graph as composed of triangles. Dmcq (talk) 08:39, 6 April 2011 (UTC)[reply]

So that picture on Mathworld is it! Thank you. Michael Hardy (talk) 12:50, 6 April 2011 (UTC)[reply]

Another representation is given in [2] where they remove the complete graphs of 5 points. Dmcq (talk) 13:38, 6 April 2011 (UTC)[reply]

I've used Sage_(mathematics_software) in the past to draw graphs. StatisticsMan (talk) 15:48, 6 April 2011 (UTC)[reply]

April 6

Laplace

Hi,

Given that partial dw/dt = laplacian(w) and w(t,x) = f(x) Does anyone have any suggestions as to how I might show that (partial d/dt)(volume integral of (grad w)²) is ≤ 0? Thanks! 131.111.222.12 (talk) 12:44, 6 April 2011 (UTC)[reply]

When you say "w(t,x) = f(x)", did you mean that that holds when t = 0? Michael Hardy (talk) 12:48, 6 April 2011 (UTC)[reply]

....and is f some particular function? Michael Hardy (talk) 12:48, 6 April 2011 (UTC)[reply]

Hi Michael, yes it holds at t=0 (in fact for all t) and no, f is not specified... —Preceding unsigned comment added by 131.111.222.12 (talk) 13:03, 6 April 2011 (UTC)[reply]

I think maybe you need to assume some decay conditions at infinity? I'd suggest first proving it for the total heat energy, the volume integral of

w^{2}

. This you can do by interchanging the order of differentiation and integration, applying the equation, and then integrating by parts. For the integral of the gradient squared, use the fact that the partial derivatives each satisfy the heat equation as well. Sławomir Biały (talk) 14:13, 6 April 2011 (UTC)[reply]

Thanks, Sławomir Biały, I shall try that —Preceding unsigned comment added by 131.111.222.12 (talk) 15:30, 6 April 2011 (UTC)[reply]

This doesn't make sense: if w(t,x) = f(x) for ALL values of t, not just for t = 0, then w(t,x) does not change as t changes. The would mean dw/dt = 0 for all values of t. Michael Hardy (talk) 15:20, 6 April 2011 (UTC)[reply]

Hi, Michael, you are very right. I have just realized that I failed to specify that w(t,x)=f(x) for all t is a boundary condition not generally true on the entire volume. —Preceding unsigned comment added by 131.111.222.12 (talk) 15:29, 6 April 2011 (UTC)131.111.222.12 (talk) 15:31, 6 April 2011 (UTC)[reply]

I'm guessing that by (grad w)² you mean the square of the norm of the gradient.

The inequality certainly looks plausible. Michael Hardy (talk) 19:30, 6 April 2011 (UTC)[reply]

Let's try something in the one-dimensional case: We're trying to show that

{\frac {\partial }{\partial t}}\int \left({\frac {\partial w}{\partial x}}\right)^{2}\,dx\leq 0.

So integrate by parts with respect to x and the integral becomes:

{\frac {\partial }{\partial t}}\int -{\frac {\partial ^{2}w}{\partial x^{2}}}\cdot w\,dx

(provided there is no contribution from the boundaries of the x-space). If sufficient continuity/differentiability/whatever conditions hold to justify pushing that first differential operator under the integral sign, then we get

\int -{\frac {\partial ^{2}w}{\partial x^{2}}}\cdot {\frac {\partial w}{\partial t}}\,dx.

Now apply the differential equation you started with to get

\int -{\frac {\partial ^{2}w}{\partial x^{2}}}\cdot {\frac {\partial ^{2}w}{\partial x^{2}}}\,dx.

That is clearly non-positive. Michael Hardy (talk) 19:39, 6 April 2011 (UTC)[reply]

Michael, thanks a lot! :) 131.111.222.12 (talk) 20:52, 6 April 2011 (UTC)—Preceding unsigned comment added by 131.111.222.12 (talk) 20:50, 6 April 2011 (UTC)[reply]

Continuity

If f(x) is continuous on [0,1] and f(0)=f(1) and I have to show that for any n there exists a point a(n) in [0, 1-(1/n)] s.t. f(a+(1/n))=f(a) I have defined a new function, say g(x)= f(a+(1/n))-f(a) and am thinking of using the IVT to prove that there exists a point where g(x)=0 but am not quite sure how. Thanks in advance for any help! —Preceding unsigned comment added by 131.111.222.12 (talk) 14:56, 6 April 2011 (UTC)[reply]

Extend f periodically and then see that the integral of g from 0 to 1 is 0. Apply the mean value theorem to conclude that g vanishes as some point. Sławomir Biały (talk) 15:24, 6 April 2011 (UTC)[reply]

Thanks, Sławomir Biały. Could you possibly explain what is meant by expand f periodically? Thanks again —Preceding unsigned comment added by 131.111.222.12 (talk) 19:12, 6 April 2011 (UTC)[reply]

Extend f to a function on the real line such that

f(x+1)=f(x)

for all x. Sławomir Biały (talk) 19:27, 6 April 2011 (UTC)[reply]

Thanks, Sławomir Biały. There was a typo in my question. The closed interval should have read [0,1-1/n] instead of [1,1-1/n]. I have changed it in the question. Sorry about that.

This looks like homework, but anyway, ... there was another typo, so I assume you meant g(x)= f(x+(1/n))-f(x). I'm fairly sure the theorem only works when n is an integer, so your proof has to reflect this somehow. I suggest dividing the interval into n segments, then looking at g(m/n) with n fixed and m varying over the integers 0 to n. Then you will see that g(x) must change sign for one of these, unless it is already zero somewhere, when the problem is solved anyway. I'll leave the rest to you, that is, if you should chance back to have another look. It's been emotional (talk) 18:37, 9 April 2011 (UTC)[reply]

Actually, it's true for any (possibly irrational) period 1/n, provided you take the periodic extension of f as I suggested. In the approach I suggested, one actually needs to use induction on n to prove it for non-periodic f. Then if the point a is in the interval [0,1−1/n], we're done. Otherwise, if it's in the interval (1−1/n,1], we apply the induction hypothesis to f restricted to the subinterval [a+1/n−1, a] to obtain the required point there.

The proof you suggest is actually similar in spirit, since it ultimately involves observing the telescoping sum

g(0)+g(1/n)+\cdots +g((n-1)/n)=0

, and therefore at least two values

g(m/n)

and

g(k/n)

must have opposite sign, and so the intermediate value theorem can be invoked. Sławomir Biały (talk) 19:09, 9 April 2011 (UTC)[reply]

(ec) Good point. As a concrete counterexample for non-integer n, consider

f(x)=\cos(5\pi x)+2x

, which certainly is continuous with f(0)=f(1)=1. However,

f(a+2/5)-f(a)=4/5\neq 0

for all

a

. –Henning Makholm (talk) 19:08, 9 April 2011 (UTC)[reply]

Yes, and this is not periodic of period 1, so there is no reason my proof should work with this function. Sławomir Biały (talk) 19:14, 9 April 2011 (UTC)[reply]

How about

f(x)=\cos(5\pi (x-\lfloor x\rfloor ))+2(x-\lfloor x\rfloor )

, then? Then g does have zeroes, but they are outside the allowed interval for a. –Henning Makholm (talk) 19:21, 9 April 2011 (UTC)[reply]

In honesty, I had ignored the restriction on a. It doesn't really make sense from the periodic perspective (it's just to ensure that

f(a+1/n)

is well-defined in the non-periodic case). So I don't really consider this to be part of the problem. Sławomir Biały (talk) 19:30, 9 April 2011 (UTC)[reply]

Thanks to both of you for the interest. Accepting the problem as stated, the counterexample for all non-integer n > 1 is to draw an imaginary line y = x, and mark on this line the points x = 0, 1/n, 2/n, ... m/n as long as m/n < 1, m an integer. Do the same going backwards from the point (1,0), ie. draw an imaginary line y = x-1, and mark the points x = 1, 1-1/n, 1-2/n, .... etc until you run out. Then just join the dots, and that's your function f. I'm fairly sure g is a constant, equal to 1. Please correct me if I'm wrong, and brownie points (a cookie perhaps?) for anyone who can actually give a precise definition of the function I described. It's been emotional (talk) 19:49, 9 April 2011 (UTC)[reply]

Cool. (No cookie for me. :-) Sławomir Biały (talk) 20:03, 9 April 2011 (UTC)[reply]

That's how I constructed my example, too. I smoothed the triangles into a cosine wave for notational convenience when writing it down. :-) The triangle wave article suggests some expressions, neither of them pretty. For a formal proof, all we really need is some nontrivial continuous function h(x) with period 1/n such that h(0)≠h(1). –Henning Makholm (talk) 20:03, 9 April 2011 (UTC)[reply]

April 7

Pairwise distinct

What are "pairwise distinct" columns in a matrix? Widener (talk) 02:24, 7 April 2011 (UTC)[reply]

"Pairwise distinct" means that no two are the same. —Bkell (talk) 04:10, 7 April 2011 (UTC)[reply]

Widener - in context, is a single column described as "pairwise distinct", or are two columns being compared and described as "pairwise distinct" from one another ? Gandalf61 (talk) 08:10, 7 April 2011 (UTC)[reply]

The elements of an indexed family are pairwise distinct if no pair of elements of the family consists of identical elements. If the cardinality of the family is 0 or 1, then there are no pairs, and thus no pair consists of identical elements, and so the elements of the family are indeed pairwise distinct. Bo Jacoby (talk) 15:45, 7 April 2011 (UTC).[reply]

Gsndalf61 - You're asked to construct a matrix that has four pairwise distinct columns (and which satisfies other conditions). I've figured out what it means now. Widener (talk) 21:23, 7 April 2011 (UTC)[reply]

Does anyone ever speak of a family of elements being distinct in a non-pairwise manner? I can see sense in speaking of a family of "pairwise disjoint" sets (as opposed to the weaker condition that the intersection of all the sets must be empty), and a family of "pairwise relatively prime" numbers (as opposed to the common gcd of all numbers being unity), or even a "pairwise linearly independent" set of vectors (but don't ask me what use this one would have). But does pairwiseness add any semantics to "distinct"? There seems to be only one halfway reasonable way to generalize ≠ from 2 to n operands. –Henning Makholm (talk) 16:24, 7 April 2011 (UTC)[reply]

Well, if you just say "distinct", it might be possible to read it as "not all columns are the same". Maybe not a very likely reading. --Trovatore (talk) 16:29, 7 April 2011 (UTC)[reply]

Good point. This does even feel structurally similar to my other examples. Thanks. –Henning Makholm (talk) 17:03, 7 April 2011 (UTC)[reply]

I don't think it's an unlikely reading at all - it's how I'd read it. Would you not describe (1,2,3) and (1,2,4) as being distinct from each other? --Tango (talk) 15:59, 9 April 2011 (UTC)[reply]

Would you describe (1,2,3), (1,2,4), and (1,2,3) as being distinct? They are certainly not all the same, so Trovatore's "unlikely" reading would call them district. On the other hand, they are clearly not pairwise distrinct, because (1,2,3) and (1,2,3) are the same. –Henning Makholm (talk) 19:14, 9 April 2011 (UTC)[reply]

Convergence to zero

Suppose $|\pi -x_{j+1}|=(\tan ^{2}c_{j})|\pi -x_{j}|$ for some $c_{j}$ between $x_{j}$ and $\pi$ . Show that $\lim _{j\to \infty }|\pi -x_{j}|=0$ if $x_{0}$ is sufficiently close to $\pi$ .
The best I can say is that $|\pi -x_{j+1}|<|\pi -x_{j}|$ for $x_{0}$ sufficiently close to $\pi$ (such that $\tan ^{2}c_{j}<1$ ) but that's not a proof that it converges to zero. Widener (talk) 09:44, 7 April 2011 (UTC)[reply]

Hint:

\tan \pi =0

and tan is continuous, so for x sufficiently close to π,

\tan ^{2}c<1/2

. -- Meni Rosenfeld (talk) 11:01, 7 April 2011 (UTC)[reply]

So that means

c_{j}

approaches π as j approaches infinity so

\lim _{j\to \infty }(\tan ^{2}c_{j})|\pi -x_{j}|=0

. — Preceding unsigned comment added by Widener (talk • contribs) 11:30, 7 April 2011 (UTC)[reply]

I don't understand this argument. That

c_{j}

approaches π follows from

x_{j}\to \pi

, not the other way around. -- Meni Rosenfeld (talk) 12:10, 7 April 2011 (UTC)[reply]

Well, thanks for your help Meni. I'll have to think about this some more. Widener (talk) 22:12, 7 April 2011 (UTC)[reply]

Graph Ramsey number

I am trying to establish the following bound given as an exercise in my book: $r(K_{m}+{\overline {K_{n}}},K_{p}+{\overline {K_{q}}})\leq {\tbinom {m+p-1}{m}}n+{\tbinom {m+p-1}{p}}q$ . Here r(G,H) denotes the Graph Ramsey number of G and H. Can anyone suggest a way to prove this? Thanks-Shahab (talk) 11:34, 7 April 2011 (UTC)[reply]

Shahab, I don't know if you're still interested in the answer to this question, but if you are, can you please clarify the notation? I am interpreting

r(K_{m}+{\overline {K_{n}}},K_{p}+{\overline {K_{q}}})

to mean the smallest integer k such that if the edges of

K_{k}

are colored red and blue, then there is either a red

K_{m}

and a vertex-disjoint red

{\overline {K_{n}}}

, or a blue

K_{p}

and a vertex-disjoint blue

{\overline {K_{q}}}

. Is that right? If so, what is meant by a red

{\overline {K_{n}}}

? I would interpret that to refer to simply any collection of n vertices, regardless of how the edges between them are colored. If that's the right interpretation, then it seems to me that you can say

r(K_{m}+{\overline {K_{n}}},K_{p}+{\overline {K_{q}}})\leq {\tbinom {m+p+2}{m-1}}+\max\{n,q\}

, since we know

r(K_{m},K_{p})\leq {\tbinom {m+p+2}{m-1}}

, and then you just add on a few more vertices that you don't really care about in order to take care of the

{\overline {K_{n}}}

or

{\overline {K_{q}}}

. On the other hand, if a red

{\overline {K_{n}}}

is supposed to mean a blue

K_{n}

, then I don't think

r(K_{m}+{\overline {K_{n}}},K_{p}+{\overline {K_{q}}})

is well defined, since a complete graph of any order with all edges colored red has neither (a red

K_{m}

and a blue

K_{n}

) nor (a blue

K_{p}

and a red

K_{q}

). Or perhaps there is a third possible meaning that I am overlooking. —Bkell (talk) 09:21, 10 April 2011 (UTC)[reply]

Ah, in fact, if a red

{\overline {K_{n}}}

just means any n vertices, then we can say

r(K_{m}+{\overline {K_{n}}},K_{p}+{\overline {K_{q}}})\leq \max\{{\tbinom {m+p+2}{m-1}},m+n,p+q\}

: If

{\tbinom {m+p+2}{m-1}}

is at least as large as

m+n

and

p+q

, then after we find a red

K_{m}

or a blue

K_{p}

in it there will be enough remaining vertices for the red

{\overline {K_{n}}}

or the blue

{\overline {K_{q}}}

, and otherwise we just need to add a few extra vertices to make up the difference. —Bkell (talk) 19:28, 10 April 2011 (UTC)[reply]

Curvature

On reading about the curvature of a complex function $f(z)$ I have hit a section that I don't understand and was hoping that someone here may be able to help me out. Essentially, the text I'm reading says that to evaluate the curvature of $f(z)$ , we write $f(z)$ in terms of the angle $\phi =arg(z)$ to give a representation $\mathbf {x} =\mathbf {x} (\phi )$ of the curve $f(C_{r})$ , where $C_{r}$ is the circle radius r, centred at the origin, in terms of $\phi$ :

x(\phi )=Re[f(z(\phi ))]=Re[f(\phi )]

y(\phi )=Im[f(z(\phi ))]=Im[f(\phi )]

where $\mathbf {x} (\phi )=(x(\phi ),y(\phi ))$ .

From this, I am meant to show that the curvature |k| is given by

|k|={\frac {|\mathbf {x'} \times \mathbf {x''} |}{{|\mathbf {x'} |}^{3}}}

where

\mathbf {x'} =(Re[{\frac {df(z(\phi ))}{d\phi }}],Im[{\frac {df(z(\phi ))}{d\phi }}])

\mathbf {x''} =(Re[{\frac {d^{2}f(z(\phi ))}{d{\phi }^{2}}}],Im[{\frac {d^{2}f(z(\phi ))}{d{\phi }^{2}}}])

The reason I am confused is because I am unsure as to how to evaluate the expression for mod k. Somehow I am taking the cross product of two vectors that I am explicitly told are two dimensional. What meaning does this have and how do I go about computing mod k? Thanks. asyndeton talk 14:20, 7 April 2011 (UTC)[reply]

They must mean (a,b)×(c,d) = ad-bc. I'm sure I have seen this two-dimensional usage somewhere, though our cross product article does not mention it. Intuitively one can just imagine that there were a third dimension to take an ordinary cross product in, and then note that when the factors are in the xy-plane, there is only one nontrivial component of the product. (More abstractly, we can identify the 3D cross product with the antisymmetric tensor product -- up to vector space isomorphism -- and generalize from there).

In the complex plane, this works out to

z\times w={\tfrac {1}{2i}}({\bar {z}}w-{\bar {w}}z)\in \mathbb {R}

, for whatever that is worth. –Henning Makholm (talk) 16:49, 7 April 2011 (UTC)[reply]

OK, that does seem to be the best, and probably only, way of interpreting it. Thanks for the help. Any suggestions on how I might go about showing the following?

|k|={\frac {|\mathbf {x'} \times \mathbf {x''} |}{{|\mathbf {x'} |}^{3}}}

I've tried plunging into the algebra and that hasn't helped. Thanks. asyndeton talk 21:49, 7 April 2011 (UTC)[reply]

A fairly extensive derivation is given here [3]. If you compare equation 13 to your equation, it should be fairly easy where to go from there.

I've seen that derivation before but the problem is that I'm unsure how to unify the fact that in my problem, x and y are parameterised by

\phi

, which is also the tangential angle, but in that derivation they are parameterised by t and the tangential angle is still

\phi

. Since they also compute the derivative

{\frac {d\phi }{dt}}

, this is certainly not some simple change of variable from t to

\phi

. How do I rectify the two? asyndeton talk 14:02, 8 April 2011 (UTC)[reply]

The curvature measures the angular velocity of the tangent vector. I think that's what dφ/dt means. You have a curve parametrised by t; which we can assume to be arc length. Now, take a fixed unit vector in the plane, say v. By definition the dot product of v with the unit tangent vector T is the cosine of the angle between v and T. That is, v⋅T(t) = cos(φ(t)). Differentiate this identity with respect to t to give

{\mathbf {v} }\cdot \left[\kappa (t){\mathbf {N} }(t)\right]=-{\dot {\varphi }}(t)\sin(\varphi (t))\,.

Since cos(φ+π/2) = −sin(φ) for all φ and N is obtained from T by a rotation through π/2 radians; it follows that

\kappa (t)={\dot {\varphi }}(t)\,,

which means that the curvature measures the angular velocity of T. (That was an answer to what I thought was the question. If not, then run it by me again and see if I get get it the second time.) Fly by Night on Tour (talk) 18:27, 8 April 2011 (UTC)[reply]

P.S. The most geometric characterisation of curvature is terms of the radius of curvature. Away from inflection points, there is a unique circle having at least three point contact with the curve at that point, the so-called osculating circle. (For more than three point contact, see vertex point.) The radius of this circle is the radius of curvature, and the reciprocal of the radius of curvature is the curvature. At inflections, a curve has at least three point contact with its tangent line. You can think of that as the osculating circle whose centre has "gone to infinity". That means the radius of curvature is infinite, and the curvature is zero. Fly by Night on Tour (talk) 18:40, 8 April 2011 (UTC)[reply]

Thank you for trying but I don't think that's answered my question. Your answer still uses the parameter t, with

\phi

as a function of t. My question is simply how to modify the derivation of the curvature given here [4] so that I have x

(\phi )

and

y(\phi )

, as my question demands in the above extract from it, instead of

x(t)

and

y(t)

, which mathworld uses. The reason I can't just do it it myself is because it is more subtle than a straight change of variable from to

\phi

due to the presence of terms such as

{\frac {d\phi }{dt}}

in the mathworld derivation, which I don't know how to amend. asyndeton talk 22:19, 8 April 2011 (UTC)[reply]

MathWorld says that φ is the tangent slope, so it's just the same as my φ. I'm not sure it's possible, or useful to do it the other way. For example, consider the curve y = sin(x). Here –1 ≤ φ ≤ 1, and there are infinitely many curve points for each choice of φ. In this case they have the same curvatures because of the symmetry, but in general they won't. You'll end up with a one-to-many function from "tangent slope" to curve curvature. In case of a complex curve, consider z(θ) = θ·e^i·h(θ). This will be a one-to-infinity function. It's a really interesting problem; don't be me wrong. I just don't think the answer is too simple. Fly by Night on Tour (talk) 23:07, 8 April 2011 (UTC)[reply]

your phi is not the same as the phi in the mathworld derivation, yours is merely an input parameter. Your phi corresponds to the mathworld t, and the mathworld phi corresponds to the radius of curvature. You'll notice that if you write f[φ]=x[φ]+iy[φ] where x and y are real functions with real arguments, and then change all of your φ to t, without loss of generality you can reproduce the mathworld derivation.

That would make sense but I don't see at what point, in my initial question, I begin writing a

\phi

that can be replaced by a t. It all seems to flow pretty soundly with, in my opinion at least, no obvious place where I can just start to write t. Could you identify for me the point at which I can start writing t instead of

\phi

and why not earlier? asyndeton talk 12:02, 9 April 2011 (UTC)[reply]

What's so difficult about replacing all of your phi's with t's? Variable names are interchangeable; conventions for what to use when are just conventions. –Henning Makholm (talk) 14:28, 9 April 2011 (UTC)[reply]

Alernatively, keep your phis as they are, and instead rename Mathworld's phi to theta and Mathworld's t to phi. –Henning Makholm (talk) 14:36, 9 April 2011 (UTC)[reply]

The problem is that it won't work. Both the derivation given on mathworld and the above extract from my problem discuss

\phi

as an angle but then my problem proceeds to use some

\phi

, which appears to me to be indistinguishable from the angle

\phi

, as a parameterisation of the curve whereas mathworld uses t. If I just replace all of my '

\phi

's with 't's, I will end up with exactly the same problem: mathworld is using two variables where I am only using one. asyndeton talk 14:41, 9 April 2011 (UTC)[reply]

Your phi has nothing to do with Mathworld's one. Your phi is a parameter, the exact same thing as mathworld's t. Mathworld's phi is the direction of the tangent to the curve that is traced out as a function of your phi. The only thing they have in common is that you happened to choose the same Greek letter for one thing as Mathworld happens to use for another thing. More precisely,

\phi _{\text{mathworld}}={\text{arg}}\left({\frac {d}{d\phi _{\text{asyndeton}}}}f(z(\phi _{\text{asyndeton}}))\right)

–Henning Makholm (talk) 14:51, 9 April 2011 (UTC)[reply]

OK, this is going to take a bit of getting my head around. Thank you for your help and patience and please excuse my ignorance. asyndeton talk 14:16, 10 April 2011 (UTC)[reply]

More on "triangular graphs"

Following up on one of my questions above: We considered the set of all 15 unordered pairs of members of the 6-member set {a, b, c, d, e, f}. Regard each such pair as a vertex of a graph, and put an edge between any two pairs that have a member in common. Thus, each vertex has eight neighbors. This gets us the graph labeled T₆ at this page. Or so that page seems to tell us. My question: Is there an efficient way to match the unordered pairs with the vertices in the picture on that page? Michael Hardy (talk) 19:24, 7 April 2011 (UTC)[reply]

I suppose yes. Here's an attempt. First, choose a triangle from the picture and call its vertices {a,b}, {a,c}, {b,c} (you can do this for symmetry). Secondly find all vertices that are common neighbours of {a,b} and {a,c} but are not equal to {b,c}, these must be {a,d}, {a,e}, … {a,k}, and you can assign the names for them in any way to choose for symmetry again. Thirdly take any vertex not named yet (so it can't have a as a member), and find the two of its neighbours that have a as a member: if these are eg. {a,x} and {a,y} then you must call that vertex {x,y}. Finally verify that you've got a graph isomorphism. – b_jonas 09:06, 8 April 2011 (UTC)[reply]

On second thought, this simple method won't work for the picture of T_7 on that page because you can't make out the edges. – b_jonas 09:08, 8 April 2011 (UTC)[reply]

I see. I guess I was being lazy this time. It's not clear that that gives a felicitous way of associating alphabetical order with the way the graph is laid out, nor even that such a thing is possible, but I can reshuffle the labels afterwards.

Thanks. Michael Hardy (talk) 12:48, 8 April 2011 (UTC)[reply]

"Straight" line on a curved surface

Hi, say I have an arbitrary (sufficiently well-behaved) surface in 3D space. I want draw a line on it such that, for every point P on the line, the projection of the line onto a plane tangent to the surface at P is "as straight as possible" -- perhaps in the sense that the curvature is zero, or second derivative is zero, or something like that. Is this the same thing as a geodesic? 86.181.204.196 (talk) 22:03, 7 April 2011 (UTC).[reply]

I found this question confusing until I realized that by "line" you didn't mean "straight line", but rather a curve.

Differential geometry is something I'm only marginally acquainted with, and your specifications (perhaps in the sense that the curvature is zero, or....) are not too precise, but it does sound as if you're talking about geodesics. But don't take my comment as the final word. Michael Hardy (talk) 03:12, 8 April 2011 (UTC)[reply]

Later I came across this from Mathworld: "The normal vector to any point of a geodesic arc lies along the normal to a surface at that point." I think this is describing the same "straight as possible" property as I have in mind?? 86.181.206.30 (talk) 11:19, 8 April 2011 (UTC)[reply]

Think of the problem the other way: if you project a straight line through the origin of the tangent plane at a point, you get a curve on the surface. Hence "as straight as possible" can only mean "is a straight line" (in the tangent plane). But these will not, in general, be geodesics. An ellipsoid is a counterexample. Sławomir Biały (talk) 14:28, 8 April 2011 (UTC)[reply]

That's if you're only looking at one P at a time. If instead we ask for "the strongest approximation to "straight" that one can reasonably require to hold at all P for a general embedded surface", then I think the only answer must be that the curve is a geodesic, satisfying the property that its projection onto the tangent plane at any point of the curve has curvature 0 at the tangent point. (But that's just geometric intuition speaking -- caveat emptor!) –Henning Makholm (talk) 16:56, 8 April 2011 (UTC)[reply]

Ah, I thought the question was about a particular point. If we impose that the curvature of the projection into each tangent plane is zero, then the curve is a geodesic since by introducing an orthogonal basis adapted to the tangent space, it's easy to see that the normal to the curve must be normal to the tangent plane (by Frenet-Serret). So that is the right condition. Sławomir Biały (talk) 17:05, 8 April 2011 (UTC)[reply]

I think your idea of "as straight as possible" relates to geodesic curvature. Geodesics are curves in a surface for which the geodesic curvature is zero at each point. (Such points are also known as geodesic inflections.) Fly by Night on Tour (talk) 18:59, 8 April 2011 (UTC)[reply]

I think the geodesic curvature at P is the curvature of the projection of the curve onto the tangent plane at P. Our article doesn't seem to mention this very intuitive characterization, though. Sławomir Biały (talk) 19:24, 8 April 2011 (UTC)[reply]

Yeah, you're right there. Maybe we should add something. In fact, I noticed that all of the elementary differential geometry of plane curves, space curves and surfaces is lacking a lot. I don't even think that plane curves have their own coverage. There defiantly isn't any geometrical motivation either. Shall we mention this at the mathematics project page? Fly by Night on Tour (talk) 22:41, 8 April 2011 (UTC)[reply]

April 8

Hysterisis

I am looking for simple analytic examples of;

$f(x,y)=0\$

where the time derivatives of the variables x' and y' are related in a hysterisis curve relationship. SpinningSpark 16:36, 8 April 2011 (UTC)[reply]

Okay, so what would you like to discuss? Fly by Night on Tour (talk) 23:16, 8 April 2011 (UTC)[reply]

I was hoping that some examples would be forthcoming. I am not good enough at mathematics to find them for myself (assuming that such functions even exist). SpinningSpark 01:06, 9 April 2011 (UTC)[reply]

I don't know why you would expect them exist, the whole point of hysteresis is that it requires a system with memory. Hence the name. —Preceding unsigned comment added by 92.20.218.201 (talk) 01:50, 9 April 2011 (UTC)[reply]

I don't follow why memory causes the function not to exist. Is there a proof? If so, that answers my question. SpinningSpark 02:06, 9 April 2011 (UTC)[reply]

Could you try reformulating your question? It's not clear what it is that you're asking for. Do you have a particular f in mind, or are you asking for an example of an f that fits some conditions? Which conditions? The hysteresis article does not seem to present any canonical "hysteresis equation" that you could be thinking of. –Henning Makholm (talk) 14:33, 9 April 2011 (UTC)[reply]

I need examples of f which force the time derivatives of x and y to be related in a hysterisis-like curve, an example of which is shown in the graphic. That is,

\scriptstyle {\dot {y}}

depends on the previous values of

\scriptstyle {\dot {x}}

. Curves crossing through the origin are acceptable, even desirable for this. SpinningSpark 19:57, 9 April 2011 (UTC)[reply]

But what is f, and what does it have to do with x and y? Apparently x and y are inputs to it, but what comes out of it, and what will you use that output for? What do you mean by f "forcing" x and y to do something? –Henning Makholm (talk) 21:34, 9 April 2011 (UTC)[reply]

@Henning. I don't understand why you don't understand. If

\scriptstyle f=x+y=0

for instance, that forces a definite relationship between

x\

and

y\

(a straight line function) which in turn determines a definite relationship between

{\dot {x}}

and

{\dot {y}}

. SpinningSpark 22:11, 9 April 2011 (UTC)[reply]

But how do you think a relation between x and y (which does not have any history information avaialble to it) can possibly force a hysteresis relation? Can you give an example of how a function of only x and y can do that? –Henning Makholm (talk) 14:46, 10 April 2011 (UTC)[reply]

Umm...no, I can't. That was my question to this board I think. SpinningSpark 20:49, 10 April 2011 (UTC)[reply]

You can get a Hysterisis effect from a Cusp catastrophe.--Salix (talk): 20:38, 9 April 2011 (UTC)[reply]

A good introduction to a mathematical formulation can be found at Systems with Hysteresis. The simplest example there is a "Nonideal relay" which has two output values 0 and 1, this has two critical points a and b with a<b. At point a a falling input will give a change from 1 to 0, at b a rising input will give a change from 0 to 1. An operator to describe this behaviour requires an internal state, $\nu _{0}$ set to the previous value of y. A simplified formulation for the operator can be written as

y={\begin{cases}0,&x\leq a\\\nu _{0},&a<x<b\\1&x\geq b\end{cases}}

More complicated forms of hysteresis build upon this example but the all require some internal state.--Salix (talk): 07:04, 11 April 2011 (UTC)[reply]

Prime number

I discover a prime number form.i live in a smoll village in India.I publis my new invention but i dnt know what to do?plz slove my probler — Preceding unsigned comment added by Prime007 (talk • contribs) 17:33, 8 April 2011 (UTC)[reply]

How do you know it is really new? Robinh (talk) 20:24, 8 April 2011 (UTC)[reply]

I know quite a bit about prime number forms. If you post it here then I may be able to say whether it is already known or seems mathematically interesting. Thousands of more or less interesting prime number forms have been listed and often named. Your user name made me think of James Bond primes (less interesting). PrimeHunter (talk) 20:59, 8 April 2011 (UTC)[reply]

It's a tricky one. If he does list it here then how does he know that some dirty, low-down, good-for-nothing pirate like me might not steal it? I wouldn't trust ~~him~~ me if I were me him. Fly by Night on Tour (talk) 23:19, 8 April 2011 (UTC)[reply]

I doubt Prime007 has something a peer-reviewed journal will publish, or something others will consider "stealing". When amateur mathematicians say they "discover" a prime form it usually just means they pick some subset of the primes and maybe invent a name for it, as shown with the "James Bond primes". They may then compute some examples and guess or try to prove how many there are. Selfpublishing on the Internet is the way to go for most people dabbling with primes. It might as well start here. It will give a public diff with date and time to show first publication. If you are worried about "stealing" then post a real name with the publication or make sure not to forget the password if you want to stake a claim to the prime form later. PrimeHunter (talk) 23:38, 8 April 2011 (UTC)[reply]

I know… I was just trying to be funny. Sorry :-( At least I know that I can safely divulge my comedic material without fear of plagiarism. Fly by Night on Tour (talk) 00:20, 9 April 2011 (UTC)[reply]

OK, but fear of having their work stolen is a frequently expressed concern for amateur mathematicians who seek advice on the Internet. It gives 10 points at [5]. PrimeHunter (talk) 02:07, 9 April 2011 (UTC)[reply]

I assume that this would be your first article. If you have a new idea then you might like to subject it to the peer review process. This process is very difficult, and work needs to meet the highest standards. Why not read our article on the Journal of number theory? It is a nice journal that focuses on the area that you seemed to be interested in. Please be aware that it is very difficult for a piece of work to be accepted by such a journal. Even professional mathematicians, with numerous publication, will have work reject over their career. Please make sure that any work that you submit is of a sufficiently high standard. Rejection can be very upsetting. Fly by Night on Tour (talk) 00:44, 9 April 2011 (UTC)[reply]

It would be better to find a professional number theorist at a nearby university and get them to review it (or post it here and we can review it). Submitting something to a journal without having had someone else take a look at it first is probably a waste of time. --Tango (talk) 15:53, 9 April 2011 (UTC)[reply]

If it is rejected from a peer reviewed journal, he can submit it here. Count Iblis (talk) 17:44, 9 April 2011 (UTC)[reply]

If it is rejected then send it to another journal. I had a paper rejected that was snapped up by another journal. The refereeing process is quite subjective. Especially if, like me, you're working in a cross-over field, say applying A to B. If the referee is only an A-theorist or a B-theorist then the chances are they'll miss 50% of the meaning. Of course, Tango does that the best solution in the meantime. Fly by Night on Tour (talk) 19:03, 9 April 2011 (UTC)[reply]

Okay, let's stop dancing around the issue: With the grasp of written English that the OP demonstrates in his question, there is no chance at all that he'll be able to write something by himself that any English-language journal will accept, ever. I don't know whether he's a child, illiterate, dyslexic, not a native speaker, just stupid, a troll, or any combination of the above, but whatever the cause, the result is what we can see. To suggest that he submit to a journal himself, is at best meaningless, at worst a callous prank.

Assuming that he really has discovered something new, the only way he will get it out is to team up with someone else to write it up. Wikipedia may not be a good place to recruit such a partner, because it's a written medium and he will at least have to explain his idea to the volunteer. He should find some way to discuss his idea with a collaborator either in his native language or orally (depending on what it is his underlying problem is). –Henning Makholm (talk) 21:48, 9 April 2011 (UTC)[reply]

There are still non-english journals out there. Taemyr (talk) 21:52, 9 April 2011 (UTC)[reply]

Sure, which is why I explicitly qualified with "English-language". –Henning Makholm (talk) 22:00, 9 April 2011 (UTC)[reply]

What's happened to you recently Henning? You used to be a superstar. Now you're suggesting that someone's mathematics might not be worth publication because they are "a child, illiterate, [or] dyslexic…". It's not compulsory for you to write something. In this case you should not have written anything. And not just that it was rude, but that it was plain silly. History has many brilliant youthful mathematician, blind mathematicians (and hence illiterate), and dyslexic mathematicians. As for your pathetic claim that my journal comments were a prank, well: I gave someone the information that they asked for, while at the same time telling them how damn hard it was going to be. A generally polite and accurate answer I would think. What would you rather me do, be an intellectual snob, tell him he's not worthy? Fly by Night on Tour (talk) 00:53, 10 April 2011 (UTC)[reply]

No, it is not compulsory for me to write something; we are all volunteers. I chose to call you out on your mocking of the OP. Sure, his question was not eloquent, but he does not deserve how you're treating him here. –Henning Makholm (talk) 01:20, 10 April 2011 (UTC)[reply]

I'm mocking the OP?! I'm the one assuming good faith and answering his question. You're the one saying he's a child, illiterate, or dyslexic. Hmmm… Fly by Night on Tour (talk) 13:25, 10 April 2011 (UTC)[reply]

Yes, you're mocking him. The only thing we know about him is that he is apparently unable to produce English prose. As I've stated, we don't know which of the many different possible causes of that is the real one, but the fact that he cannot write readable English is not in doubt. When you pretend it is a realistic possibility for him to write an article by himself and send it to a journal, you're implicitly mocking his lack of written skills and setting him up to be the laughingstock of some journal editor somewhere. That is not helpful; it is just cruel. –Henning Makholm (talk) 14:43, 10 April 2011 (UTC)[reply]

I can assure you that I was not mocking him. I'm sorry that you have misinterpreted my intentions. The very fact that we don't know anything about the OP means that we cannot judge if the OP is or is not capable of writing a publishable article. The only thing to do in such cases it to assume good faith and to answer by providing a truthful response, i.e. it's damn hard but if you want to then you should do this…. Fly by Night on Tour (talk) 17:08, 10 April 2011 (UTC)[reply]

Can we continue this on a talk page rather than the referenece desk.--Salix (talk): 20:41, 10 April 2011 (UTC)[reply]

Salix, you may continue wherever you choose; I'd finished. Fly by Night on Tour (talk) 21:23, 10 April 2011 (UTC)[reply]

April 9

Limit

Does anyone know how to prove x^n tends to 0 for fixed x in [0,1)? Somehow I can't recall how to do the simplest of things. Money is tight (talk) 15:10, 9 April 2011 (UTC)[reply]

x^{n}

is a decreasing sequence bounded below, so it has a limit. The limit L must be a fixed point of

L\mapsto xL

, so L=0. Sławomir Biały (talk) 15:20, 9 April 2011 (UTC)[reply]

Thanks, that's a very clever proof. Money is tight (talk) 04:21, 10 April 2011 (UTC)[reply]

April 10

Differentiation problem

I am differentiating

${\frac {1}{1+e^{-a(x+b)}}}$

I use the chain rule to get

${\frac {-1}{(1+e^{-a(x+b)})^{2}}}\times -ae^{-a(x+b)}={\frac {ae^{-a(x+b)}}{(1+e^{-a(x+b)})^{2}}}$

But the actual answer is ${\frac {ae^{a(x+b)}}{(1+e^{a(x+b)})^{2}}}$

What happened to the negative signs in the exponents? —Preceding unsigned comment added by 130.102.158.15 (talk) 03:55, 10 April 2011 (UTC)[reply]

I did this quickly mentally so may be wrong, but I think the two expressions are the same. To see this multiply both numerator and denominator by

(e^{a(x+b)})^{2}

. Zunaid 06:58, 10 April 2011 (UTC)[reply]

Correct. Trivial algebra shows that the two are the same. Michael Hardy (talk) 16:27, 10 April 2011 (UTC)[reply]

Set Theory

Is there a ~~surjection~~ bijection - in an arithmetical language - from the set of natural numbers on the set of prime numbers?

If there is - then: how would it look like (in an arithmetical language)?
If there isn't - then: wouldn't the continuum hypothesis be refuted by the set of prime numbers, on which there is no first-order ~~surjection~~ bijection from the set of natural numbers, and ~~from which there is~~ no first-order ~~surjection on~~ bijection from the set of real numbers? Note that when Paul Cohen proved that the continuum hypothesis does not derive from ZF, he proved that for first order mappings only.

HOOTmag (talk) 06:51, 10 April 2011 (UTC)[reply]

The primes are a subset of the natural numbers. So let

f:N\to P

be the map defined by f(x)=x when x is prime, or f(x)=2 when x is not prime. I'm not sure if that will satisfy your requirement of "arithmetical language", but surely the condition that x be prime is expressible in an "arithmetical language"? Staecker (talk) 11:31, 10 April 2011 (UTC)[reply]

Yes, thank you, but I was wrong with the "surjection"; It should have been a "bijection". I've just fixed that (after reading your response), by striking out the "surjection" and by replacing it by "bijection". See above.

I'm still looking for an answer for my question in its new version, see above.

HOOTmag (talk) 12:43, 10 April 2011 (UTC)[reply]

The obvious bijection is the function that sends n to the nth smallest prime. I'm not sure what you mean by "in an arithmetical language", but this function is first-order definable in the natural numbers considered as a structure for the language with symbols for plus and times, for example. Algebraist 15:19, 10 April 2011 (UTC)[reply]

Is the term "nth smallest prime" first order definable? I'm really eager to know how you define the bijection. HOOTmag (talk) 19:32, 10 April 2011 (UTC)[reply]

Yes. Moreover, the function enumerating the primes is computable (in fact primitive recursive) so is simply (Delta 1 at most) definable in Peano arithmetic. To see it's primative recursive is fairly straightforward as bounded existentials and minimization is, and a fine computable bound for the next prime after n is 2n. As for what you are claiming, it's not quite true. Any finite set, for example, has a different cardinality than the naturals and reals, so there are no bijections. A counterexample to the continuum hypothesis would be a set A such that there is no surjection from the naturals to A and there is no injection from A to the reals. Under the assumption of AC, this is the same as saying the cardinality of A is strictly between the cardinality of the naturals and reals. Also, to not be misleading, what Trovatore said below is correct. This being first order definable doesn't really matter too much. There are continuum many functions from N to N, but only countably many definable ones in any countable language.Wgunther (talk) 20:20, 10 April 2011 (UTC)[reply]

Yes, and you can check that the original version of my question was really about no surjection from the set of natural numbers on A, and no surjection from A on the set of real numbers; I was wrong when I finally struck out the surjection and replaced it by bijection. Anyways, I think you made a mistake, in your statement about an injection from A to the reals, instead of a surjection from A to the reals (or an injection from the reals to A). HOOTmag (talk) 21:02, 10 April 2011 (UTC)[reply]

It was Euclid who first (as far as anyone knows, apparently) proved that there actually are infinitely many primes. See Euclid's theorem. That is essential to this problem. Michael Hardy (talk) 18:23, 10 April 2011 (UTC)[reply]

I don't see how this would bear upon the continuum hypothesis. The continuum hypothesis is false if there is a cardinality between that of the natural numbers and that of the reals. But the primes are a subset of the naturals, so their cardinality is not more than that of the naturals. Michael Hardy (talk) 18:25, 10 April 2011 (UTC)[reply]

"Between" - yes, and that's what I meant by bijection, which is a one-to-one correspondence between sets.

"The cardinality of the primes is not more that that of the reals": yes, but cardibality is defined by the existence of a bijection between all the sets belonging to the same cardinality. Now, notice that when Paul Cohen proved that the continuum hypothesis does not derive from ZF, he proved that for first order bijections only, so he meant that there may be an (unconstructible) set from which there is no first-order bijection to the set of natural numbers, and no first-order bijection to the set of real numbers. Note that Paul Cohen has never promised that there is no bijections at all, e.g bijections in higher orders! That's why I would like to know whether there is a first order bijection between the set of natural numbers and the set of prime numbers, because if there isn't - then the set of prime numbers refutes the continuum hypothesis (at least as far as first order languages are concerned), which is a stronger result than Cohen's, who didn't refute the continuum hypothesis but rather only proved (by referring to first order bijections) that the continuum hypothesis does not derive from ZF...

HOOTmag (talk) 19:32, 10 April 2011 (UTC)[reply]

No, sorry, I think you misunderstand the result. There is a "first-order" involved here, but it's not about the bijections themselves.

What Cohen proved is, there is no proof of CH from ZFC, in first-order logic. (The second-order version of ZFC, interpreted in full second-order logic, either implies or refutes CH, we just don't know which one.)

But the bijections involved are not restricted to first-order definable ones. (I assume that by "first-order bijection" you actually mean "first-order definable bijection"; your literal term "first-order bijections" simply doesn't mean anything in standard terminology.) --Trovatore (talk) 19:37, 10 April 2011 (UTC)[reply]

Are those mappings restricted to finitely definable ones? For my question to make sense, please check any randomal real number, i.e. any infinite set S of randomal natural numbers; Such a set is not constructible, i.e. it can't be obtained by using ZFC axioms only; However, it can be defined by an infinite first order formula: {x|(x=a_1) V (x=a_2) V ... (x=a_n) V ...}, i.e. by an illegitimate formula in ZFC. Further, there is no (finitely definable) surjection from the set of natural numbers on S, and no surjection from S on the set of real numbers, so may S be an (unconstructible) counterexample to the continuum hypothesis (assuming that the very existence of S is consistent in ZFC)? HOOTmag (talk) 20:54, 10 April 2011 (UTC)[reply]

My response below, timestamped 21:10, 10 April 2011 (UTC), addresses this point. --Trovatore (talk) 21:17, 10 April 2011 (UTC)[reply]

My new question was about whether "those mappings [are] restricted to finitely definable ones", so I understand now that your answer is simply: "no", isn't it? HOOTmag (talk) 21:28, 10 April 2011 (UTC)[reply]

There certainly exists an injection from the set of primes to the set of natural numbers (the inclusion map). Therefore, by definition, the cardinality of the set of primes is no greater than the cardinality of the set of natural numbers.

I can give you many examples of sets which have neither a bijection to the set of natural numbers nor a bijection to the set of real numbers: the empty set, any finite set, the set of all functions from R to R, etc. In order for the continuum hypothesis to be false, there would need to exist a set S such that there exist injections from the set of natural numbers into S and from S into the set of real numbers, but there do not exist injections from the set of real numbers into S or from S into the set of natural numbers. Since the set of prime numbers plainly fails the last of these criteria, it cannot possibly be a counterexample to the continuum hypothesis. —Bkell (talk) 19:42, 10 April 2011 (UTC)[reply]

Yes, and you can check that the original version of my question was really about no surjection from the set of natural numbers on S, and no surjection from S on the set of real numbers. I was wrong when I finally struck out the surjection and replaced it by bijection.

Anyways, how about a randomal real number, i.e an infinite set S of randomal natural numbers? Such a set is not constructible, i.e. it can't be obtained by using ZFC axioms only; However, it can be defined by an infinite first order formula: {x|(x=a_1) V (x=a_2) V ... (x=a_n) V ...}, i.e. by an illegitimate formula in ZFC. Further, there is no (finitely definable) injection from S to the set of natural numbers, and no injection from the set of real numbers to S, so may S be an (unconstructible) counterexample to the continuum hypothesis (assuming that the very existence of S is consistent in ZFC)?

HOOTmag (talk) 20:54, 10 April 2011 (UTC)[reply]

It's frustrating to discuss these things with you, because you keep saying things that sort of almost make sense, but not quite, and I have trouble pinpointing the exact thing that you're missing.

For the moment, let's just be clear on what I tried to point out before: The statement of CH has nothing to do with whether the various maps (bijections or surjections as you like) are definable or not. The result that says CH is not provable from ZFC, is talking about CH full stop, not about some version of CH restricted to definable maps. Where the "first-order" part comes in is not in regards to the meaning of CH, but only in regards to the methods of proof that are being considered.

So for example, one way of stating CH is there is a bijection between the real numbers and the countable ordinals. Now, it's tempting to think Cohen proved there could not be any definable such bijection, because if there were, you might think that you could just give the definition, and your proof of CH is complete.

But that's wrong. There "could" be a definable such bijection, for some value of "could". For example, if V=HOD and CH holds, then there is a definable bijection between the reals and the countable ordinals, and ZFC does not refute this situation. So you can give a definition that may define a bijection between the reals and the countable ordinals. What you can't prove, in ZFC alone, is that the definition works. --Trovatore (talk) 21:10, 10 April 2011 (UTC)[reply]

My new question was about whether "those mappings [are] restricted to finitely definable ones", so I understand now that your answer is simply: "no", isn't it? HOOTmag (talk) 21:26, 10 April 2011 (UTC)[reply]

Yes, the answer to that question is indeed "no". There could (in some sense of "could") be a non-definable counterexample to CH, but it cannot possibly be a set of natural numbers. --Trovatore (talk) 21:45, 10 April 2011 (UTC)[reply]

Is "randomal" a technical term? Randomal redirects to a city in Armenia, and Google is not helpful either. –Henning Makholm (talk) 22:39, 10 April 2011 (UTC)[reply]

AFAIK, not a technical term. My best guess in what was meant by was "randomal real" was a non-constructible real (in the L sense). Wgunther (talk) 22:57, 10 April 2011 (UTC)[reply]

Very easy maths question

Hi guys. This is probably the easiest question that has ever been asked on this forum but please bear in mind that I'm very stupid and I can't seem to find an answer anywhere.

I have an equation ½v² = g x h. I want to find the value of v but I'm really struggling to know what to do with the 1/2 and square, ie. to isolate the v - do they cancel each other out? If I times by v then I assume I get rid of the 1/2 but does that leave me with v³ or not?

I know this is embarrassingly easy but I have no intuitive maths skills at all.

Thanks

Pantscat (talk) —Preceding undated comment added 12:08, 10 April 2011 (UTC).[reply]

x^y is a common computer notation for x^y.

Your equation can be written ½ × v^2 = g × h. You can isolate v^2 by multiplying both sides by 2, since 2 × ½ = 1. Once you have isolated v^2, you can take the square root on both sides to isolate v. Consider whether v can be both positive and negative in your problem. For example, v^2 = 9 would have two integer solutions: v = 3, and v = -3. PrimeHunter (talk) 13:01, 10 April 2011 (UTC)[reply]

Thanks Prime Hunter. So does that leaves me with v = square root of 2gh? If so, does that mean I times the product of g x h by 2? The alternative is that I multiply 2g by 2h, which gives a totally different answer.

Yours dazed and confused Pantscat (talk) —Preceding undated comment added 13:35, 10 April 2011 (UTC).[reply]

Yes, your final answer is plus or minus the square root of 2gh. This can be interpreted in two ways, either 2(gh), which is the way that you describe, multiplying the product gh by 2, or (2g)h, which takes the product 2g and multiplies it by h (note: NOT by 2h). These should give the same answer. --COVIZAPIBETEFOKY (talk) 14:19, 10 April 2011 (UTC)[reply]

If you multiplied 2g by 2h, you would get 4gh. -- Meni Rosenfeld (talk) 15:56, 10 April 2011 (UTC)[reply]

If I understand the question correctly, the you would like to solve

{\frac {1}{2}}v^{2}=gh\,,

with respect to v. Since both sides of the equation are equal, it follows that if we do the same thing to both sides of the equation then they will also be equal. So, first of all, multiply both sides by 2, this gives v² = 2gh. Next, we take the square root of both sides of the equation, not forgetting the ±, this gives

v=\pm {\sqrt {2gh}}\,.

The ± appears because, for example, 5² = 25 and (−5)² = 25. It follows that the square root of 25 is either +5 or −5.

So, whenever we use a square root, we need to add a ± to account for that fact. Fly by Night on Tour (talk) 18:30, 10 April 2011 (UTC)[reply]

Thanks, guys. All makes sense now. Pantscat (talk) —Preceding undated comment added 19:07, 10 April 2011 (UTC).[reply]

Map from z-plane to w-plane

If you consider the map from the z-plane to the w-plane of $C_{r}$ , the circle centred at the origin, radius r in the z-plane, given by $w=f(C_{r})$ where $f(z)=(z-a)^{n}$ for some complex constant a and some integer n, how can you analytically determine the values of r at which the curve in the w-plane has a cusp (perhaps more explicitly, the values of r such that the curve has an extra loop for just greater than r than it has for just less than r). Perhaps the general case is hard to treat; if so, can an answer for n=3, 4 be given? Thanks. asyndeton talk 20:52, 10 April 2011 (UTC)[reply]

Intuitively, there'll be a cusp when your circle passes through an odd-degree root of the derivative of f. You'll probably want to consider roots of even degree too, though there will be no visible cusp in that case. –Henning Makholm (talk) 22:33, 10 April 2011 (UTC)[reply]

Thanks for that, it's very useful but apparently the use of the word 'cusp' in my question was misleading. Essentially, I want to know how to analytically determine at which values of r an extra loop will evolve. asyndeton talk 22:55, 10 April 2011 (UTC)[reply]

Then ignore whether the degree of the root is odd or even. (That's for the sense of "loop" I think would be most appropriate. You're welcome to suggest a more precise definition for what you're looking for). –Henning Makholm (talk) 23:14, 10 April 2011 (UTC)[reply]

I'm unsure how to describe it mathematically. Perhaps a second attempt would be: how do you analytically determine the values of r for which some part of the curve crosses over itself (inducing a loop, by my definition)? asyndeton talk 23:35, 10 April 2011 (UTC)[reply]

You know that "interesting" things happen to

f(C_{r})

when

C_{r}

passes through a root of

f(z)=0

. If

f(z)=(z-a)^{n}

then the only root of

f(z)=0

is at

z=a

, where there is a root of degree n. So pick values for a and n and try plotting

f(C_{r})

for two values of r, one just less than |a|, and one just greater than |a|. Do this for a few different values of n to see how the behaviour depends on the order of the root. Gandalf61 (talk) 08:50, 11 April 2011 (UTC)[reply]

integral

can someone tell me how to write on wikipedia the definite integral sign? i need both top and bottom limits, the middle function, and dx at the end thnx —Preceding unsigned comment added by 129.94.130.220 (talk) 22:51, 10 April 2011 (UTC)[reply]

Something like

\int _{-\pi }^{\pi -h}f(x)\,dx

. Or, if you prefer an upright d,

\int _{-\pi }^{\pi -h}f(x)\,\mathrm {d} x

April 11

Why are the imaginary parts of some nontrivial Riemann zeta zeroes so close to integers?

If you look at the list of the first few nontrivial Riemann zeta zeroes of the form 1/2 + i x for x > 0 (see below), you see quite a few that are close to integers. E.g. values for x that are approximately within 1/100 of an integer should occur on average once in every 50 entries, however in the first 10 entries we already see 2 of them. In the later entries we don't see many of these values so close to integers, so the total number of the values in the table containing 151 entries within 1/100 of an integer isn't that large (there are 3 of them, the probability of getting more than 2 is approximately 0.08). However, the entries close to integers are still so close that you wouldn't have expected to see them at all in the table. So, x = 48.005150881 or an x closer to an integer should occur with probability of approximately 1/100 (probability for this to occur once or more within the first ten entries is about 1/10 ), x = 146.000982487 or a x closer to an integer should occur with probability 1/500 (probability for this to occur once or more in the table is approximately 1/10). So, it looks as if most entries are random, but you have some that are anomalously close to integers.

Table:

    14.134725142
    21.022039639
    25.010857580
    30.424876126
    32.935061588
    37.586178159
    40.918719012
    43.327073281
    48.005150881
    49.773832478
    52.970321478
    56.446247697
    59.347044003
    60.831778525
    65.112544048
    67.079810529
    69.546401711
    72.067157674
    75.704690699
    77.144840069
    79.337375020
    82.910380854
    84.735492981
    87.425274613
    88.809111208
    92.491899271
    94.651344041
    95.870634228
    98.831194218
   101.317851006
   103.725538040
   105.446623052
   107.168611184
   111.029535543
   111.874659177
   114.320220915
   116.226680321
   118.790782866
   121.370125002
   122.946829294
   124.256818554
   127.516683880
   129.578704200
   131.087688531
   133.497737203
   134.756509753
   138.116042055
   139.736208952
   141.123707404
   143.111845808
   146.000982487

Count Iblis (talk) 01:09, 11 April 2011 (UTC)[reply]

It does seem slightly more than would be expected:

                  VALUES       VALUES        VALUES 
                  WITHIN       WITHIN        WITHIN
                  0.10         0.03          0.01
                  ---------    --------      ------
PERCENT FOUND     13/51=25%    6/51=12%      2/51=4%   
PERCENT EXPECTED        20%          6%           2%

However, presumably you stopped where you did to maximize this effect. If we stopped one sooner, the results would be:

                  VALUES       VALUES        VALUES 
                  WITHIN       WITHIN        WITHIN
                  0.10         0.03          0.01
                  ---------    --------      ------
PERCENT FOUND     12/50=24%    5/50=10%      1/50=2%   
PERCENT EXPECTED        20%          6%           2%

This is close enough to just be a coincidence. StuRat (talk) 04:19, 11 April 2011 (UTC)[reply]

Number sequence

In relation to the above sequence; If a have a random number generating function which produces a number from a set of 'interesting numbers' with probability p, and an 'uninteresting number' with probability 1-p, and analyse the produced sequence by deliberately selecting from the total sequence a continuous string that starts and finishes with one of the interesting numbers, what is the expected increase in the expectation of 'interesting numbers' in my string compared to a randomly selected string?