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

Aug 10, 2015

$\ln \left(1 + {x}^{2}\right) = {x}^{2} - {x}^{4} / 2 + {x}^{6} / 3 - {x}^{8} / 4 + \ldots$

#### Explanation:

First note:

$\frac{d}{\mathrm{dy}} \left(\ln \left(1 + {x}^{2}\right)\right) = \frac{2 x}{1 + {x}^{2}}$

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

$\frac{1}{1 + x} = 1 - x + {x}^{2} - {x}^{3} + \ldots$

=>

$\frac{1}{1 + {x}^{2}} = 1 - {x}^{2} + {x}^{4} - {x}^{6} + \ldots$

=>

$\frac{2 x}{1 + {x}^{2}} = 2 x - 2 {x}^{3} + 2 {x}^{5} - 2 {x}^{7} + \ldots$

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

$\ln \left(1 + {x}^{2}\right) = {x}^{2} - {x}^{4} / 2 + {x}^{6} / 3 - {x}^{8} / 4 + \ldots$

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