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Math Help - [SOLVED] McLaurin series for ln(...)

  1. #1
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    [SOLVED] McLaurin series for ln(...)

    Need McLaurin series of <br />
\ln{\left(\frac{1+2x}{1-3x+2x^2}\right)}<br />

    Tried to make it like \ln\left({1+x}\right) and found <br />
\ln\left(1+\frac{5x-2x^2}{1-3x+2x^2}\right)<br />

    So, can i put that \;\left(\frac{5x-2x^2}{1-3x+2x^2}\right) into \sum\limits_{n= 1}^\infty\;\frac{{\left(-1\right)}^{n-1}}{n}x^n

    if i can not, how can i solve that problem?
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  2. #2
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    Quote Originally Posted by Lafexlos View Post
    Need McLaurin series of <br />
\ln{\left(\frac{1+2x}{1-3x+2x^2}\right)}<br />

    Tried to make it like \ln\left({1+x}\right) and found <br />
\ln\left(1+\frac{5x-2x^2}{1-3x+2x^2}\right)<br />

    So, can i put that \;\left(\frac{5x-2x^2}{1-3x+2x^2}\right) into \sum\limits_{n= 1}^\infty\;\frac{{\left(-1\right)}^{n-1}}{n}x^n

    if i can not, how can i solve that problem?
    I'd start by using the log rule ln(A/B) = ln(A) - ln(B).
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  3. #3
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    That looks much better.
    Thx for help. =)
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