in the following exercises, given that 1/(1 - x) = ∑(n = 0 to ∞) x^n with convergence in (-1, 1), find the…

Explanation:

Step1: Rewrite the denominator

First, factorize the denominator $5 - 4x+x^{2}=(x - 1)(x - 5)$. Then, use partial - fraction decomposition. Let $\frac{x^{2}}{5 - 4x+x^{2}}=\frac{x^{2}}{(x - 1)(x - 5)}=\frac{A}{x - 1}+\frac{B}{x - 5}$. Cross - multiply: $x^{2}=A(x - 5)+B(x - 1)$. Let $x = 1$, then $1=A(1 - 5)+B(0)\Rightarrow A=-\frac{1}{4}$. Let $x = 5$, then $25=A(0)+B(5 - 1)\Rightarrow B=\frac{25}{4}$. So $\frac{x^{2}}{5 - 4x+x^{2}}=-\frac{1}{4(x - 1)}+\frac{25}{4(x - 5)}$.

Step2: Rewrite in terms of $(x - 2)$

We know that $\frac{1}{x - 1}=\frac{1}{(x - 2)+1}= \sum_{n = 0}^{\infty}(-1)^{n}(x - 2)^{n}$ for $|x - 2|\lt1$, and $\frac{1}{x - 5}=\frac{1}{(x - 2)-3}=-\frac{1}{3}\frac{1}{1-\frac{x - 2}{3}}=-\frac{1}{3}\sum_{n = 0}^{\infty}(\frac{x - 2}{3})^{n}$ for $|\frac{x - 2}{3}|\lt1$.

Step3: Combine the series

[ \begin{align*} \frac{x^{2}}{5 - 4x+x^{2}}&=-\frac{1}{4}\sum_{n = 0}^{\infty}(-1)^{n}(x - 2)^{n}+\frac{25}{4}\left(-\frac{1}{3}\sum_{n = 0}^{\infty}(\frac{x - 2}{3})^{n}\right)\ &=\sum_{n = 0}^{\infty}\left[\frac{(-1)^{n + 1}}{4}-\frac{25}{12}\left(\frac{1}{3}\right)^{n}\right](x - 2)^{n} \end{align*} ]

Step4: Find the interval of convergence

For $\sum_{n = 0}^{\infty}(-1)^{n}(x - 2)^{n}$, the interval of convergence is $|x - 2|\lt1$, i.e., $1\lt x\lt3$. For $\sum_{n = 0}^{\infty}(\frac{x - 2}{3})^{n}$, the interval of convergence is $|\frac{x - 2}{3}|\lt1$, i.e., $-1\lt x\lt5$. The intersection of these two intervals gives the interval of convergence of the power - series of $f(x)$ as $1\lt x\lt3$.

Answer:

The power series of $f(x)$ is $\sum_{n = 0}^{\infty}\left[\frac{(-1)^{n + 1}}{4}-\frac{25}{12}\left(\frac{1}{3}\right)^{n}\right](x - 2)^{n}$ and the interval of convergence is $(1,3)$.