Question

Given that \(x^n - (1/x^n)\) is expressible as a polynomial in \(x - (1/x)\) with real coefficients only if \(n\) is an odd positive integer, find \(P(z)\) so that \(P(x-(1/x)) = x^5 - (1/x)^5.\)

Answer 1

/cc

Answer 2

For some reason I first thought is to factor x^5-(1/x)^5

Answer 3

hmm I don't think I like that either
I will continue to think

Answer 4

If that helps:
\[
x^5\left(x^5-\frac{1}{x^5}\right)=x^{10}-1
\]

Answer 5

\[P(x-\frac{1}{x})=x^5-\frac{1}{x^5} \\ \text{ Let } z=x-\frac{1}{x} \\ zx=x^2-1 \\ x^2-zx-1=0 \\ x=\frac{ z \pm \sqrt{z^2+4}}{2} \\ P(z)=(\frac{z+\sqrt{z^2+4}}{2})^5-(\frac{2}{z+\sqrt{z^2+4}})^5\]
I don't know which choice in x I should have made
and P doesn't look like a polynomial but maybe it can be fix up I guess

might be worth looking at what @thomas5267 said

Answer 6

\[
P(x)=x^5+ax^4+bx^3+cx^2+dx+e\\
\left(x-\frac{1}{x}\right)^5=x^5-5x^3+10x-10x^{-1}+5x^{-3}-x^{-5}\\
\left(x-\frac{1}{x}\right)^4=x^4-4x^2+6-4x^{-2}+x^{-4}\\
\left(x-\frac{1}{x}\right)^3=x^3-3x+3x^{-1}-x^{-3}\\
\left(x-\frac{1}{x}\right)^2=x^2-2+x^{-2}\\
x-\frac{1}{x}\\
\left(x-\frac{1}{x}\right)^5+a\left(x-\frac{1}{x}\right)^4+b\left(x-\frac{1}{x}\right)^3\\+c\left(x-\frac{1}{x}\right)^2+d\left(x-\frac{1}{x}\right)+e\\
=x^5+x^{-5}\\
a=0\\
b=5\\
c=0\\
d=5\\
e=0
\]
Inelegant, but it works.