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Math Help - power series representation #2

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    Senior Member mollymcf2009's Avatar
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    power series representation #2

    Evaluate the indefinite integral as a series:

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

    Here is what I did:

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

    \frac{1}{t} \sum^{\infty}_{n=0}  (t)^n

    So, what do I do now? Since I differentiated to get \frac{1}{1-t}, do I just need to integrate it now and then multiply \frac{1}{t} back in? Did I miss anything ?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mollymcf2009 View Post
    Evaluate the indefinite integral as a series:

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

    Here is what I did:

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

    \frac{1}{t} \sum^{\infty}_{n=0}  (t)^n

    So, what do I do now? Since I differentiated to get \frac{1}{1-t}, do I just need to integrate it now and then multiply \frac{1}{t} back in? Did I miss anything ?
    and exactly why would you factor a function of t out of a dt-integral? you can only factor out constants (with respect to the variable of integration) remember?

    Hint: you know the power series for the given log function, just write that in. then divide it through by t. then integrate term by term. well, that's not really a hint, that's how to do the problem

    by the way, your power series for the log is incorrect, you wrote the power series for 1/(1 - t) ...which is NOT the integral of ln(1 - t), by the way
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  3. #3
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by Jhevon View Post
    and exactly why would you factor a function of t out of a dt-integral? you can only factor out constants (with respect to the variable of integration) remember?

    Hint: you know the power series for the given log function, just write that in. then divide it through by t. then integrate term by term. well, that's not really a hint, that's how to do the problem

    by the way, your power series for the log is incorrect, you wrote the power series for 1/(1 - t) ...which is NOT the integral of ln(1 - t), by the way
    Ok, so what I would have for my power series is:
    <br />
\sum^{\infty}_{n=0} \int \frac{(t)^n}{t} dt ?

    Is that what you said for me to do?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mollymcf2009 View Post
    Ok, so what I would have for my power series is:
    <br />
\sum^{\infty}_{n=0} \int \frac{(t)^n}{t} dt ?

    Is that what you said for me to do?
    question: what does the power series for ln(1 - t) look like? did you use that here?
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