# Math Help - Maclaurin series expansion

1. ## Maclaurin series expansion

I want to calculate the Maclaurin series for $log(1+x^2)$ by using a previous result, that $log(1+x) = x - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \frac{1}{4}x^4 + ...$

The following is not correct, but I don't know why, nor how to do it the right way

let $u=x^2$. Then

$log(1+x^2) = log(1+u) = u - \frac{1}{2}u^2 + \frac{1}{3}u^3 - \frac{1}{4}u^4 + ... = x^2 - \frac{1}{2}x^4 + \frac{1}{3}x^6 - \frac{1}{4}x^8 + ...$

Thanks for any help.

2. Originally Posted by HD09
I want to calculate the Maclaurin series for $log(1+x^2)$ by using a previous result, that $log(1+x) = x - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \frac{1}{4}x^4 + ...$

The following is not correct, but I don't know why, nor how to do it the right way

let $u=x^2$. Then

$log(1+x^2) = log(1+u) = u - \frac{1}{2}u^2 + \frac{1}{3}u^3 - \frac{1}{4}u^4 + ... = x^2 - \frac{1}{2}x^4 + \frac{1}{3}x^6 - \frac{1}{4}x^8 + ...$

Thanks for any help.
Why is that not correct (on [-1,1])?

3. Because my solutions say so and the long-way calculation I did (which turns out to have a mistake in it now I check again) agreed with those instead of this!! Thanks. I was flabbergasted at how this could possibly be incorrect... what a relief.

4. Now you've got me curious! What was the "solution sheet" answer that "the long way" agreed with?

5. It only gave the first two terms but the solutions claim the co-efficient of the second term is 1/4 rather than 1/2