Question

find a power series representation for the function.

$f(x)=\frac{x}{(1 + 7x)^2}$

$f(x)=\sum_{n = 0}^{\infty}(\square x)$

determine the radius of convergence, $r$.

$r = 1/7$

Explanation:

Step1: Recall the power - series formula

We know that \(\frac{1}{1 - t}=\sum_{n = 0}^{\infty}t^{n}\), for \(|t|\lt1\). First, rewrite \(f(x)\) in a form related to the geometric series. Let \(t=-7x\), then \(\frac{1}{1 + 7x}=\frac{1}{1-(-7x)}=\sum_{n = 0}^{\infty}(-7x)^{n}=\sum_{n = 0}^{\infty}(- 1)^{n}7^{n}x^{n}\), for \(|7x|\lt1\) (i.e., \(|x|\lt\frac{1}{7}\)).

Step2: Differentiate the geometric series

Differentiate \(\frac{1}{1 + 7x}=\sum_{n = 0}^{\infty}(-1)^{n}7^{n}x^{n}\) with respect to \(x\). Using the formula \((\sum_{n = 0}^{\infty}a_{n}x^{n})^\prime=\sum_{n = 1}^{\infty}na_{n}x^{n - 1}\), we have \(\frac{-7}{(1 + 7x)^{2}}=\sum_{n = 1}^{\infty}(-1)^{n}7^{n}nx^{n - 1}\).

Step3: Manipulate the differentiated series

Multiply both sides of \(\frac{-7}{(1 + 7x)^{2}}=\sum_{n = 1}^{\infty}(-1)^{n}7^{n}nx^{n - 1}\) by \(-\frac{x}{7}\).

$$
LATEXBLOCK0
$$

Let \(m=n + 1\), then \(n=m - 1\). So \(\frac{x}{(1+7x)^{2}}=\sum_{m = 1}^{\infty}(-1)^{m - 1}m7^{m - 1}x^{m}=\sum_{n = 1}^{\infty}(-1)^{n - 1}n7^{n - 1}x^{n}\)

Answer:

\(f(x)=\sum_{n = 1}^{\infty}(-1)^{n - 1}n7^{n - 1}x^{n}\)