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Math Help - Series Representation

  1. #1
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    Series Representation

    How would i find a series representation for ln(x) from the series of ln(1+x)? which is: [(-1)^n * x^n] / n
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by jules027 View Post
    How would i find a series representation for ln(x) from the series of ln(1+x)? which is: [(-1)^n * x^n] / n
    i dont know if this would help..

    \ln x = \ln (1+ (x-1))
    and so \frac{(-1)^n (x-1)^n}{n} = \frac{(1-x)^n}{n}

    haha.. am i ryt?
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by kalagota View Post
    i dont know if this would help..

    \ln x = \ln (1+ (x-1))
    and so \frac{(-1)^n (x-1)^n}{n} = \frac{(1-x)^n}{n}

    haha.. am i ryt?
    \ln(x+1)=\int\frac{dx}{x+1}=\int\sum_{n=0}^{\infty  }(-1)^nx^n=\sum_{n=0}^{\infty}\frac{(-1)^nx^{n+1}}{n+1}


    Now to find ln(x) in terms of this

    Kalgotta was correct the first step is to see that \ln(x)=\ln(1+(x-1))

    So by letting \lambda=x-1 we get

    \ln(1+\lambda)=\sum_{n=0}^{\infty}\frac{(-1)^n\lambda^{n+1}}{n+1}

    Substituting back in what u equals we get

    \ln(x)=\sum_{n=0}^{\infty}\frac{(-1)^n(x-1)^{n+1}}{n+1}


    To verify we see that

    \ln(x)=\int\frac{dx}{x}=\int\sum_{n=0}^{\infty}(-1)^n(x-1)^ndx=\sum_{n=0}^{\infty}\frac{(-1)^n(x-1)^{n+1}}{n+1}

    QED
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