# How do you find the taylor series for ln(1+x^2)?

Jan 14, 2018

$\ln \left(1 + {x}^{2}\right) = {\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n + 1} / n {x}^{2 n} = {x}^{2} - {x}^{4} / 2 + {x}^{6} / 3 - {x}^{8} / 4. . .$

Convergent when $| x | < 1$

#### Explanation:

We start by working out a taylor series for $\ln \left(1 + x\right)$. I will be expanding around $0$, so it will be a Maclaurin series.

The general formula for a Maclaurin series is:
f(x)=sum_(n=0)^oof^n(0)/(n!)x^n

This means we need to work out the nth derivative of $\ln \left(1 + x\right)$. Let's start by taking some derivatives and see what values they produce at $x = 0$:

$f \left(x\right) = \ln \left(1 + x\right) \to 0$

$f ' \left(x\right) = \frac{1}{1 + x} \to 1$

$f ' ' \left(x\right) = - \frac{1}{1 + x} ^ 2 \to - 1$

${f}^{3} \left(x\right) = \frac{2}{1 + x} ^ 3 \to 2$

${f}^{4} \left(x\right) = - \frac{6}{x + 1} ^ 4 \to - 3$

${f}^{5} \left(x\right) = \frac{24}{x + 1} ^ 5 \to 24$
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f^n(x)=(-1)^(n+1)((n-1)!)/(x+1)^(n)-> (-1)^(n+1)(n-1)!

We can now plug this into the Maclaurin expansion (note that we ignore the first term, since it is $0$):
ln(1+x)=sum_(n=1)^oo((-1)^(n+1)(n-1)!)/(n!)x^n

This simplifies to:
$\ln \left(1 + x\right) = {\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n + 1} / n {x}^{n}$

Now that we have a series for $\ln \left(1 + x\right)$, we can replace all the $x$'s with ${x}^{2}$ to get a series for $\ln \left(1 + {x}^{2}\right)$:
$\ln \left(1 + {x}^{2}\right) = {\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n + 1} / n {\left({x}^{2}\right)}^{n} =$

$= {\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n + 1} / n {x}^{2 n} = {x}^{2} - {x}^{4} / 2 + {x}^{6} / 3 - {x}^{8} / 4. . .$

which is the series we were looking for.

To test the convergence, we can use the ratio test:
${\lim}_{n \to \infty} - \frac{{x}^{2 \left(n + 1\right)} / \left(n + 1\right)}{{x}^{2 n} / n} = {\lim}_{n \to \infty} - \frac{{x}^{2} \cancel{{x}^{2 n}}}{n + 1} \cdot \frac{n}{\cancel{{x}^{2 n}}} =$

$= {\lim}_{n \to \infty} - \frac{n {x}^{2}}{n + 1} = - {x}^{2}$

So, we need to look at when $| - {x}^{2} | < 1$:
$| - {x}^{2} | < 1$

${x}^{2} < 1$

$\sqrt{{x}^{2}} < \sqrt{1}$

$| x | < 1$

which means the series is convergent when $| x | < 1$