How do you find a power series representation for $\frac{1 + x}{1 - x}$ $(1+x)/(1-x)$ and what is the radius of convergence?

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How do you find a power series representation for $\frac{1 + x}{1 - x}$ $(1+x)/(1-x)$ and what is the radius of convergence?

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Answer 1 · 2015-10-01T21:18:17

Simplify $\frac{1 + x}{1 - x}$ $(1+x)/(1-x)$ to express in terms of the power series for $\frac{1}{1 - x}$ $1/(1-x)$ and find:

$\frac{1 + x}{1 - x} = 1 + \infty \sum n = 1 2 x^{n}$ $(1+x)/(1-x) = 1 + sum_(n=1)^oo 2x^n$

with radius of convergence $1$ $1$

Explanation:

$\frac{1 + x}{1 - x} = \frac{2 - (1 - x)}{1 - x} = \frac{2}{1 - x} - 1$ $(1+x)/(1-x) = (2-(1-x))/(1-x) = 2/(1-x) - 1$

$\frac{1}{1 - x} = \infty \sum n = 0 x^{n}$ $1/(1-x) = sum_(n=0)^oo x^n$ with radius of convergence $1$ $1$

To see that for yourself, notice that:

$(1 - x) \infty \sum n = 0 x^{n}$ $(1-x) sum_(n=0)^oo x^n$

$= \infty \sum n = 0 x^{n} - x \infty \sum n = 0 x^{n}$ $=sum_(n=0)^oo x^n - x sum_(n=0)^oo x^n$

$= \infty \sum n = 0 x_{n} - \infty \sum n = 1 x^{n}$ $=sum_(n=0)^oo x_n - sum_(n=1)^oo x^n$

$= 1$ $=1$

.. if the sums converge.

Now the sum $\infty \sum n = 0 x^{n}$ $sum_(n=0)^oo x^n$ is a geometric series with common ratio $x$ $x$ . So it converges when $| x | < 1$ $abs(x) < 1$ .

So:

$\frac{1 + x}{1 - x} = 1 + \infty \sum n = 1 2 x^{n}$ $(1+x)/(1-x) = 1 + sum_(n=1)^oo 2x^n$

with radius of convergence $1$ $1$

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How do you find a power series representation for $\frac{1 + x}{1 - x}$ $(1+x)/(1-x)$ and what is the radius of convergence?

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Explanation:

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How do you find a power series representation for $\frac{1 + x}{1 - x}$ $(1+x)/(1-x)$ and what is the radius of convergence?