Wikipedia:Reference desk/Archives/Mathematics/2012 September 22

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

September 22

a?

problem 20. I saw the answer key but still unable to understand what is going on. I understand the binary expansion part. I get lost in the statement "it follows that the coefficient of x^2012 is equal to 2^0 + 2^1 + 2^5 = 2^6. answer key part 1, answer key part 2. Thanks!Pendragon5 (talk) 00:02, 22 September 2012 (UTC)

A product of sums is the sum of all the possible products; say (a+b)*(c+d)*(e+f): the result will contain every product you can make by picking one of each sum,so for example a*d*e, b*c*e, b*c*f and five others.

In your polynomial, you know which factors supply the power of x part, so the rest must provide the coefficient: showing only the parts that make up the x²⁰¹² term : (x¹⁰²⁴+..)*(x⁵¹²+..)*(x²⁵⁶+..)*(x¹²⁸+..)*(x⁶⁴+..)*(..+ 32)*(x¹⁶+..)*(x⁸+..)*(x⁴+..)*(..+2)*(..+1). And 1*2*32=64=2⁶ Ssscienccce (talk) 02:11, 22 September 2012 (UTC)

Oh okay I think I got it now. Thanks! I also have another question. How can I convert 2012 into binary number without the answer key?Pendragon5 (talk) 19:37, 22 September 2012 (UTC)

Binary_numeral_system#Conversion_to_and_from_other_numeral_systems, first paragraph. — Quondum 21:02, 22 September 2012 (UTC)

For a number like 2012, close to 2048, (=2¹¹), a shorter way is: you know that 2¹¹-1 = 2047 is 11111111111, and 2047-2012=35; So you have to subtract 35 which is expressed as a sum of powers of two: 32 + 2 + 1, so from (32;16;8;4;2;1) you know the positions that have to become zeros to get 2012: 11111011100 Ssscienccce (talk) 08:56, 25 September 2012 (UTC)

Commutability of integration

Is $\int (\int f(r)dr)dv$ the same as $\int (\int f(r)dv)dr$? 71.207.151.227 (talk) 21:04, 22 September 2012 (UTC)

See Fubini's theorem. Sławomir Biały (talk) 21:09, 22 September 2012 (UTC)