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vestirme4

Answered question

2021-10-29

Find a power series representation for the function and determine the radius of convergence.

f (x) = \frac{x}{{(1 + 4 x)}^{2}}

Answer & Explanation

Faiza Fuller

Skilled2021-10-30Added 108 answers

Step 1
We will start with the power series of $\frac{1}{1 - x}$
We know that
$\frac{1}{1 - x} = \sum_{n = 0}^{\infty} x^{n}$
Step 1 : Replace x with $- 4 x$
$\frac{1}{1 - (- 4 x)} = \sum_{n = 0}^{\infty} {(- 4 x)}^{n}$
$\frac{1}{1 + 4 x} = \sum_{n = 0}^{\infty} {(- 4)}^{n} x^{n}$
Step 2
Differentiate both sides
$\frac{d}{d x} [\frac{1}{1 + 4 x}] = \frac{d}{d x} [\sum_{n = 0}^{\infty} {(- 4)}^{n} x^{n}]$
$- \frac{4}{{(1 + 4 x)}^{2}} = \sum_{n = 0}^{\infty} {(- 4)}^{n} \frac{d [x^{n}]}{d x}$
$- \frac{4}{{(1 + 4 x)}^{2}} = \sum_{n = 1}^{\infty} {(- 4)}^{n} (n) x^{n - 1}$
Not that the sum in RHS is starting from $n = 1$ , because the first term died during differentiation
Divide both sides by $- 4$ , To get
$\frac{1}{{(1 + 4 x)}^{2}} = \sum_{n = 1}^{\infty} {(- 4)}^{n - 1} (n) x^{n - 1}$
$\frac{1}{{(1 + 4 x)}^{2}} = \sum_{n = 1}^{\infty} {(- 1)}^{n - 1} \cdot 4^{n - 1} \cdot (n) x^{n - 1}$
Step 3 : Multiply both sides by x
$\frac{x}{{(1 + 4 x)}^{2}} = \sum_{n = 1}^{\infty} {(- 1)}^{n - 1} {(4)}^{n - 1} (n) x^{n}$
If you want you can start the sum from $n = 0$ , but if you do that, you should replace n with $n + 1$ in the sum

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