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## anonymous 5 years ago I am working on the taylor series portion of the course and had a question:

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1. anonymous

Is the taylor series representation of ln (x+1) = $\sum_{0}^{\infty}((-1)^{n+1}X ^{n})/n$ I worked this out on my own and wanted to know if this is correct. Thanks!

2. anonymous

Correction I guess you have to start the series at 1 because of 0 being undefined.

3. anonymous

Any ideas?

4. anonymous

two ways i can think to approach this. one is to calculate the nth derivative of ln[1+x], and find $(-1)^{n+1}/(1+x)^n$ this value at x=0 is simply -(-1)^n or (-1)^(n+1) then proceed with taylor's formula. else use the geometric series representation of closed form (assumes abs[x]<1)$\frac{1}{1+x}=\sum_{n=0}^{\infty}(-x)^n$ then integrate both sides with respect to x

5. anonymous

Solved: Thanks gamesguru. I also just found out that Wikipedia has the Taylor series listed and my answer was right.

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