## zmudz one year ago 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 $$P(x-(1/x)) = x^5 - (1/x)^5.$$

1. beginnersmind

/cc

2. myininaya

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

3. myininaya

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

4. thomas5267

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

5. myininaya

$P(x-\frac{1}{x})=x^5-\frac{1}{x^5} \\ \text{ Let } z=x-\frac{1}{x} \\ zx=x^2-1 \\ x^2-zx-1=0 \\ x=\frac{ z \pm \sqrt{z^2+4}}{2} \\ P(z)=(\frac{z+\sqrt{z^2+4}}{2})^5-(\frac{2}{z+\sqrt{z^2+4}})^5$ 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

6. thomas5267

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

Find more explanations on OpenStudy