Question

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

Answer 1

`ln abs(1+x^5) = sum_(n=0)^oo (-1)^nx^(5n+5)/(n+1)`

with radius of convergence `R=1`.

Explanation:

Start from the geometric series:

`1/(1-q) = sum_(n=0)^oo q^n`

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

`1/(1+x^5) = sum_(n=0)^oo (-x^5)^n= sum_(n=0)^oo (-1)^nx^(5n)`

Still with `R=1` because `abs(-x^5) <= 1 => abs x <=1`
Multiply by `x^4`:

`x^4/(1+x^5) = sum_(n=0)^oo (-1)^nx^(5n+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/(1+t^5)dx = sum_(n=0)^oo (-1)^n int_0^x t^(5n+4)dt`

`1/5 int_0^x (d(1+t^5))/(1+t^5)dx = sum_(n=0)^oo (-1)^n[ t^(5n+5)/(5n+5)]_0^x`

`1/5 ln abs(1+x^5) = sum_(n=0)^oo (-1)^nx^(5n+5)/(5n+5)`

`ln abs(1+x^5) = sum_(n=0)^oo (-1)^nx^(5n+5)/(n+1)`