Question

How do you write the first four nonzero terms of the taylor series for `ln(1+x^2)`about x=0?

Answer 1

`ln(1+x^2)=x^2-x^4/2+x^6/3-x^8/4+...`

Explanation:

First note:

`d/dy(ln(1+x^2))=(2x)/(1+x^2)`

We can break down the taylor series for this as follows using the well known series (about `x=0`):

`1/(1+x)=1-x+x^2-x^3+...`

=>

`1/(1+x^2)=1-x^2+x^4-x^6+...`

=>

`(2x)/(1+x^2)=2x-2x^3+2x^5-2x^7+...`

We can then integrate this term by term to obtain the required Taylor series:

`ln(1+x^2)=x^2-x^4/2+x^6/3-x^8/4+...`

This approach allows us to skip a lot of tedious steps when determining the series term by term.