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Math Help - Integration of a Power Series Representation of a Function

  1. #1
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    Integration of a Power Series Representation of a Function

    Hello!

    I am trying to express ln(6) as a series \sum{c_n} with rational numbers c_n

    Before, I showed that the power series represenation of
    f(x)=\frac{1}{3+2x}
    was
    \sum_{n=0}^{\infty}\frac{(-1)^n2^n}{3^{n+1}}*x^n

    (http://www.mathhelpforum.com/math-he...tml#post401824)

    \integrate both, and we get

    F(x) =\int\frac{1}{3+2x}dx=ln(3+2x)
    with power series representation
    \sum_{n=0}^{\infty}\frac{(-1)^n2^n}{3^{n+1}}*\frac{x^{n+1}}{n+1} + c
    where c is a constant of integration,

    also, we can find the constant of integration c by letting x=0
    ln (3+2*0) = \sum_{n=0}^{\infty}\frac{(-1)^n2^n}{3^{n+1}}*\frac{0^{n+1}}{n+1} + c
    gives us c= ln 3

    so, we have now  ln6=ln(3)-ln(3/6)

    if we let ln(3/6)=ln(3+2x), we get x=-5/4 (which is in the radius of convergence R=3/2)

    I can't seem to put it all together, however, I still can't seem to answet the question, all I have is

    ln(6)= ln(3)-ln(3/6)
    ln(6)= ln(3) - (the series evaluated at x=-5/4) + ln(3)
    where the 2nd ln(3) is the constant of integration

    the series evaluated at x=-5/4 is \sum_{n=0}^{\infty}\frac{(-1)^n2^n}{3^{n+1}}*\frac{(-5/4)^{n+1}}{n+1}

    solving for ln(6) gives us ln(6) = ln (3) + c + \sum_{n=0}^{\infty}\frac{(-1)^n2^n}{3^{n+1}}*\frac{(-5/4)^{n+1}}{n+1}
    but c= ln 3, so...

    ln(6) = 2ln(3) +\sum_{n=0}^{\infty}\frac{(-1)^n2^n}{3^{n+1}}*\frac{(-5/4)^{n+1}}{n+1}

    yikes! so basically, my question is
    is the above expression a suitable answer to the original question,
    "I am trying to express ln(6) as a series \sum{c_n} with rational numbers c_n"??

    sorry if this is not totally clear, I will try and clarify if this doesn't make sense!

    Any help is greatly appriciated!
    Sorry for being long-winded here!
    Thank you!
    Last edited by matt.qmar; November 11th 2009 at 06:14 PM.
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  2. #2
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    whoops! cleaned up the latex a bit, should make more sense now, sorry!
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  3. #3
    MHF Contributor Drexel28's Avatar
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    How about this? A commonly known fact is that for |x|<1

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

    So \ln(6)=-\ln\left(\frac{1}{6}\right)=-\ln\left(1-\frac{5}{6}\right)=\sum_{n=1}^{\infty}\frac{5^n}{n  \cdot 6^n}
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