Monday

December 22, 2014

December 22, 2014

Posted by **Count Iblis** on Monday, July 23, 2007 at 2:42am.

p(x) = (1 + a1 x)(1 + a2 x)(1 + a3 x)

Take the logarithm:

Log[p(x)] =

Log(1 + a1 x) + Log(1 + a2 x)

+ Log(1 + a3 x)

Expand in powers of x by using that:

Log(1 + x) = x - x^2/2 +x^3/3 - x^4/4 + ...

Then, if we denote the sum of the n-th powers of the ai by Sn, we find:

Log[p(x)] = S1 - S2/2 x^2 + S3/3x^3 - S4/4 x^4 + ...

If we now exponentiate this we should get p(x) back:

Exp{Log[p(x)]} = p(x)

We can calculate

Exp[S1 - S2/2 x^2 + S3/3x^3 - S4/4 x^4 + ...]

using the series expansion:

Exp[X] = 1 + X + X^2/2! + X^3/3! + X^4/4! + ...

If we put in here X = S1 - S2/2 x^2 + S3/3x^3 - S4/4 x^4 + ...]

the result must be p(x). Snce p(x) is a third degree polynomial, this means that

all powers of x^4 and higher must vanish.

If you work out the coefficient of x^4 and equate it to zero you get the equation:

S4 = S1^4/6 - S1^2 S2 + S2^2/2 + 4/3 S1 S3

So, if S1 = 4, S2 = 10, and S3 = 22, then S4 must be 50.

the sum of three numbers is 4, the sum of their squares is 10 and the sum of their cubes is 22. what is the sum of their fourth powers?

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