Maclaurin series expansion

**HD09** · October 9th 2009, 03:51 AM

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.

**Danny** · October 9th 2009, 05:09 AM

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])?

**HD09** · October 9th 2009, 06:24 AM

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.

**HallsofIvy** · October 9th 2009, 06:35 AM

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

**HD09** · October 9th 2009, 07:00 AM

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

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