Question

How do you find the Maclaurin series of `f(x)=ln(1+x^2)` ?

Answer 1

Plug in `x^2` for every x in the Maclaurin series for `f(x)=ln(1+x)`

Explanation:

So, you have the Maclaurin series for `f(x)=ln(1+x)` defined as:

`f(x)=ln(1+x) = x-x^2/2+x^3/3-... =sum_(k=1)^n(((-1)^(k+1)x^k) / k)`

The question is essentially asking you to plug in `x^2` for every x here, so:

`(x^2)^1-(x^2)^2/2+(x^2)^3/3-(x^2)^4/4+...`
which gives... `x^2 - x^4/2+x^6/3-x^8/4+... = sum_(k=1)^n(((-1)^(k+1)(x^2)^k) / k)`