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September 22

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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)[reply]

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)[reply]

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)[reply]

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

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)[reply]

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)[reply]

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