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

  3. anonymous
    • 5 years ago
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    Any ideas?

  4. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    Solved: Thanks gamesguru. I also just found out that Wikipedia has the Taylor series listed and my answer was right.

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