Question

Derive the maclaurin series for ln (1+x^2)

Answer 1

If my memory serves me well the MacLaurin series is:
\[\sum_{n=0}^{\infty}\frac{f^n(0)x^n}{n!}\]

Answer 2

Please can you solve it out for me?
thanks!!!

Answer 3

http://www.wolframalpha.com/input/?i=maclaurin+series+for+ln(1%2Bx^2)
I honestly don't remember how to do it. I thought you could take a derivative and then re-integrate it. But I can't get it to work out quite right.

Answer 4

Start by taking a bunch of derivatives, and evaluating at 0 (I used wolfram alpha to get the higher order derivatives rather than trudging through quotient and chain rules). What you will get is a series of numbers:

0, 0, 2, 0, -12, 0, 240, 0, -10080, 0, 725760, 0, ...

Assuming that a factorial is involved, try to find a pattern (you could skip this step if you have trouble finding the pattern):
2nd der: 2 = 2!/1
4th der: 12 = 4!/2
6th der: 240 = 6!/3
8th der: 10080 = 8!/4
10th der: 725760 = 10!/5
for n even n-th der = n!/(n/2)

Now, plug in and simplify. The factorials cancel, and you are left with
\[x^2 - \frac{1}{2}x^4+\frac{1}{3}x^6-\frac{1}{4}x^8 + ...\]

Spot the pattern?? Write the series!
\[\sum_{n=1}^{\infty}\frac{ (-1)^{n+1}}{n}x^{2n}\]

Answer 5

Thanks!!!