# How do you use the Maclaurin series for f(x) = ln abs((1+x)/(1-x))?

Apr 15, 2017

$\ln \left\mid \frac{1 + x}{1 - x} \right\mid = 2 {\sum}_{n = 0}^{\infty} {x}^{2 n + 1} / \left(2 n + 1\right)$

for $\left\mid x \right\mid < 1$

#### Explanation:

Differentiate the function:

$\frac{d}{\mathrm{dx}} \ln \left\mid \frac{1 + x}{1 - x} \right\mid = \frac{d}{\mathrm{dx}} \ln \left\mid 1 + x \right\mid - \frac{d}{\mathrm{dx}} \ln \left\mid 1 - x \right\mid = \frac{1}{1 + x} + \frac{1}{1 - x} = \frac{1 - x + 1 + x}{\left(1 - x\right) \left(1 + x\right)} = \frac{2}{1 - {x}^{2}}$

Now note that:

$\frac{1}{1 - {x}^{2}}$

is the sum a geometric series with ratio ${x}^{2}$, so we have:

$\frac{2}{1 - {x}^{2}} = 2 {\sum}_{n = 0}^{\infty} {\left({x}^{2}\right)}^{n} = 2 {\sum}_{n = 0}^{\infty} {x}^{2 n}$

for $\left\mid x \right\mid < 1$.

We can now integrate term by term:

$\ln \left\mid \frac{1 + x}{1 - x} \right\mid = 2 {\int}_{0}^{x} \frac{\mathrm{dt}}{1 - {t}^{2}} = 2 {\sum}_{n = 0}^{\infty} {\int}_{0}^{x} {t}^{2 n} \mathrm{dt} = 2 {\sum}_{n = 0}^{\infty} {\left[{t}^{2 n + 1} / \left(2 n + 1\right)\right]}_{0}^{x}$

So, always for $\left\mid x \right\mid < 1$:

$\ln \left\mid \frac{1 + x}{1 - x} \right\mid = 2 {\sum}_{n = 0}^{\infty} {x}^{2 n + 1} / \left(2 n + 1\right)$