Maclaurin Series for ln(1+x^2)/x

[9mmv2]

Homework Statement

Find the Maclaurin series for : [ln(1+x^2)]/x

Homework Equations

f(x) = \sum f^{n}(0)/n! * x^{n}

f(x) = f(0) + f'(0)(x) + f''(0)(x)^2/2! + ...

The Attempt at a Solution

I got stuck right away, as how do I determine f(0) when you can't divide by 0? Similarly, for f'(0), f''(0)?

 

[lurflurf]

Find the Maclaurin series for [ln(1+x^2)]/x by
Finding the Maclaurin series for ln(1+x^2)
then the Maclaurin series for [ln(1+x^2)]/x is
(the Maclaurin series for [ln(1+x^2)])/x

 

[SammyS]

Method 1: Find the Maclaurin series for ln(1+x^2), then divide that by x.

Method 2: Use the fact that lim_x→0 f(x) = 0, so set f(0) = 0.

If you need to justify this, define g(x) such that g(0) = 0, g(x) = f(x) for all non-zero x. Clearly, g(x) and f(x) have the same Maclaurin series.
​