# How do you find the maclaurin series expansion of f(x) = ln abs(1+x^5)?

Jul 31, 2018

$\ln \left\mid 1 + {x}^{5} \right\mid = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{5 n + 5} / \left(n + 1\right)$

with radius of convergence $R = 1$.

#### Explanation:

Start from the geometric series:

$\frac{1}{1 - q} = {\sum}_{n = 0}^{\infty} {q}^{n}$

having radius on convergence $R = 1$.
Let now $q = - {x}^{5}$:

$\frac{1}{1 + {x}^{5}} = {\sum}_{n = 0}^{\infty} {\left(- {x}^{5}\right)}^{n} = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{5 n}$

Still with $R = 1$ because $\left\mid - {x}^{5} \right\mid \le 1 \implies \left\mid x \right\mid \le 1$
Multiply by ${x}^{4}$:

${x}^{4} / \left(1 + {x}^{5}\right) = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{5 n + 4}$

In the interior of the interval of convergence we can integrate term by term obtaining a series with at least the same radius of convergence:

${\int}_{0}^{x} {t}^{4} / \left(1 + {t}^{5}\right) \mathrm{dx} = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {\int}_{0}^{x} {t}^{5 n + 4} \mathrm{dt}$

$\frac{1}{5} {\int}_{0}^{x} \frac{d \left(1 + {t}^{5}\right)}{1 + {t}^{5}} \mathrm{dx} = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {\left[{t}^{5 n + 5} / \left(5 n + 5\right)\right]}_{0}^{x}$

$\frac{1}{5} \ln \left\mid 1 + {x}^{5} \right\mid = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{5 n + 5} / \left(5 n + 5\right)$

$\ln \left\mid 1 + {x}^{5} \right\mid = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{5 n + 5} / \left(n + 1\right)$