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Math Help - representation of a power series

  1. #1
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    representation of a power series

    express the following as a power series:

    \int \frac{ln(1-t)}{t} dt

    so far I have

      \mbox{let} \ g(t) = ln(1-t) \rightarrow g(t) = \int g'(t) dt = \int -\frac{1}{1-t} \ dt= \int - \sum_{n=0}^{\infty} t^n \ dt

     - \sum_{n=0}^{\infty} \frac{t^{n-1}}{n} = - \sum_{n=1}^{\infty} \frac{t^{n}}{n}

    now putting it back into the original:

     \int - \sum_{n=1}^{\infty} \frac{t^n}{n} \cdot \frac{1}{t} \ dt = \int -\frac{1}{n} \sum_{n=1}^{\infty} t^{n-1} \ dt =  -\frac{1}{n} \int \sum_{n=1}^{\infty} t^{n-1} \ dt

    -\frac{1}{n} \int \sum_{n=1}^{\infty} t^{n-1} \ dt = -\frac{1}{n} \sum_{n=1}^{\infty} \frac{t^{n-2}}{n-1} = -\sum_{n=1}^{\infty} \frac{t^{n-2}}{n^2-n}

    is this correct?
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by lllll View Post
    express the following as a power series:

    \int \frac{ln(1-t)}{t} dt

    so far I have

     \mbox{let} \ g(t) = ln(1-t) \rightarrow g(t) = \int g'(t) dt = \int -\frac{1}{1-t} \ dt= \int - \sum_{n=0}^{\infty} t^n \ dt

     - \sum_{n=0}^{\infty} \frac{t^{n-1}}{n} = - \sum_{n=1}^{\infty} \frac{t^{n}}{n}

    now putting it back into the original:

     \int - \sum_{n=1}^{\infty} \frac{t^n}{n} \cdot \frac{1}{t} \ dt = \int -\frac{1}{n} \sum_{n=1}^{\infty} t^{n-1} \ dt = -\frac{1}{n} \int \sum_{n=1}^{\infty} t^{n-1} \ dt

    -\frac{1}{n} \int \sum_{n=1}^{\infty} t^{n-1} \ dt = -\frac{1}{n} \sum_{n=1}^{\infty} \frac{t^{n-2}}{n-1} = -\sum_{n=1}^{\infty} \frac{t^{n-2}}{n^2-n}

    is this correct?
    Yes as far as I can see it is correct good job
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Wait

    Quote Originally Posted by lllll View Post
    express the following as a power series:

    \int \frac{ln(1-t)}{t} dt

    so far I have

     \mbox{let} \ g(t) = ln(1-t) \rightarrow g(t) = \int g'(t) dt = \int -\frac{1}{1-t} \ dt= \int - \sum_{n=0}^{\infty} t^n \ dt

     - \sum_{n=0}^{\infty} \frac{t^{n-1}}{n} = - \sum_{n=1}^{\infty} \frac{t^{n}}{n}

    now putting it back into the original:

     \int - \sum_{n=1}^{\infty} \frac{t^n}{n} \cdot \frac{1}{t} \ dt = \int -\frac{1}{n} \sum_{n=1}^{\infty} t^{n-1} \ dt = -\frac{1}{n} \int \sum_{n=1}^{\infty} t^{n-1} \ dt

    -\frac{1}{n} \int \sum_{n=1}^{\infty} t^{n-1} \ dt = -\frac{1}{n} \sum_{n=1}^{\infty} \frac{t^{n-2}}{n-1} = -\sum_{n=1}^{\infty} \frac{t^{n-2}}{n^2-n}

    is this correct?
    I do see one problem \int{t^{n-1}}dt=\frac{t^n}{n}
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  4. #4
    Math Engineering Student
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    Quote Originally Posted by lllll View Post
     - \sum_{n=0}^{\infty} \frac{t^{n-1}}{n} = - \sum_{n=1}^{\infty} \frac{t^{n}}{n}
    Careful here.

    The thing is, you should have memorized the power series representation for \ln(1-x), that is -\sum_{k\,=\,1}^\infty\frac{x^k}k for |x|<1. From here you can easily find \frac{\ln(1-x)}x as a series, then interchange sum by integral and you're done.

    There's no much here, simple calculations, it's easy.
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