+63 Given that \( x^n - \frac1{x^n}\) is expressible as a polynomial in \( x - \frac1x\)with real coefficients only if n is an odd positive integer, find P(z) so that \(P\left(x-\frac1x\right) = x^5 - \frac1{x^5}\)
+842 We are given the equation \( p(x - \frac{1}{x}) = x^5 - \frac{1}{x^5} \), where \( p(z) \) is a polynomial in terms of \( z = x - \frac{1}{x} \). Our goal is to express \( x^5 - \frac{1}{x^5} \) as a polynomial in \( z \).
### **Solution By Steps**
*Step 1: Express \( x^5 - \frac{1}{x^5} \) in terms of powers of \( x - \frac{1}{x} \)*
We start by recalling some trigonometric identities and symmetries for powers of \( x \) and their reciprocals.
Let \( z = x - \frac{1}{x} \). We first compute the first few powers of \( z \):
\[
z = x - \frac{1}{x}
\]
Square both sides to compute \( z^2 \):
\[
z^2 = \left(x - \frac{1}{x}\right)^2 = x^2 - 2 + \frac{1}{x^2} = x^2 + \frac{1}{x^2} - 2
\]
Thus,
\[
x^2 + \frac{1}{x^2} = z^2 + 2
\]
Next, let's compute \( x^3 - \frac{1}{x^3} \):
\[
x^3 - \frac{1}{x^3} = \left(x - \frac{1}{x}\right)\left(x^2 + \frac{1}{x^2}\right) = z \cdot (z^2 + 2) = z^3 + 2z
\]
Now, let's compute \( x^5 - \frac{1}{x^5} \) by multiplying \( x^2 + \frac{1}{x^2} \) and \( x^3 - \frac{1}{x^3} \):
\[
x^5 - \frac{1}{x^5} = \left(x^3 - \frac{1}{x^3}\right)\left(x^2 + \frac{1}{x^2}\right) = (z^3 + 2z)(z^2 + 2)
\]
\[
= z^5 + 2z^3 + 2z^3 + 4z = z^5 + 4z^3 + 4z
\]
Thus, we have expressed \( x^5 - \frac{1}{x^5} \) as:
\[
x^5 - \frac{1}{x^5} = z^5 + 4z^3 + 4z
\]
*Step 2: Write the polynomial \( p(z) \)*
From the above expression, we can see that the polynomial \( p(z) \) that satisfies \( p(x - \frac{1}{x}) = x^5 - \frac{1}{x^5} \) is:
\[
p(z) = z^5 + 4z^3 + 4z
\]
### **Final Answer**
The polynomial \( p(z) \) is:
\[
p(z) = z^5 + 4z^3 + 4z
\]
+63 Ahh dang, it seems its incorrect. Though it gave a hint, Define \(P_n\left(x-\frac1x\right)=x^n-\frac1{x^n}\) for odd n and seek a recursive relationship between \(P_{2k+1}\) and lower order polynomials.
