Question

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

Answer 1

To express \( x^5 - \frac{1}{x^5} \) as a polynomial in \( x - \frac{1}{x} \), we will first define a variable \( y = x - \frac{1}{x} \).

Using the identity for the powers of \( x - \frac{1}{x} \), we can derive a few key terms to help us find \( p(y) \).

1. First, we compute \( x^2 + \frac{1}{x^2} \):
\[
\left(x - \frac{1}{x}\right)^2 = x^2 - 2 + \frac{1}{x^2} \implies y^2 = x^2 - 2 + \frac{1}{x^2},
\]
Thus,
\[
x^2 + \frac{1}{x^2} = y^2 + 2.
\]

2. Next, we compute \( x^3 + \frac{1}{x^3} \):
\[
x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x}\right)\left(x^2 + \frac{1}{x^2}\right) - \left(x + \frac{1}{x}\right).
\]
We know that \( x + \frac{1}{x} = y + 2 \). Now using \( x^2 + \frac{1}{x^2} = y^2 + 2 \):
\[
x^3 + \frac{1}{x^3} = (y + 2)(y^2 + 2) - (y + 2).
\]
Simplifying this:
\[
= (y + 2)(y^2 + 2) - (y + 2) = (y + 2)(y^2 + 2 - 1) = (y + 2)(y^2 + 1) = y^3 + 3y + 2.
\]

3. We can compute \( x^4 + \frac{1}{x^4} \) now using \( (x^2 + \frac{1}{x^2})^2 - 2 \):
\[
x^4 + \frac{1}{x^4} = (y^2 + 2)^2 - 2 = y^4 + 4y^2 + 4 - 2 = y^4 + 4y^2 + 2.
\]

4. Finally, we compute \( x^5 + \frac{1}{x^5} \):
\[
x^5 + \frac{1}{x^5} = (x + \frac{1}{x})(x^4 + \frac{1}{x^4}) - (x^3 + \frac{1}{x^3}) = (y + 2)(y^4 + 4y^2 + 2) - (y^3 + 3y + 2).
\]
Expanding this, we need to compute \( (y + 2)(y^4 + 4y^2 + 2) \):
\[
= y^5 + 4y^3 + 2y + 2y^4 + 8y^2 + 4.
\]
Therefore,
\[
x^5 + \frac{1}{x^5} = y^5 + 4y^3 + 2y + 2y^4 + 8y^2 + 4 - (y^3 + 3y + 2) = y^5 + 3y^3 + 5y + 2.
\]

Now, we note that:
\[
x^5 - \frac{1}{x^5} = (x^5 + \frac{1}{x^5}) - 2.
\]
Thus, we have:
\[
x^5 - \frac{1}{x^5} = y^5 + 3y^3 + 5y + 2 - 2 = y^5 + 3y^3 + 5y.
\]

Therefore, \( p(y) \) such that \( p(x - \frac{1}{x}) = x^5 - \frac{1}{x^5} \) is:
\[
p(y) = y^5 + 3y^3 + 5y.
\]