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Math Help - Help with calculating a summation

  1. #1
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    Help with calculating a summation

    Hi there,

    I have a trigonometrical-series type of solution as follows:

    C=C_{0}-\frac{4C_{0}}{\pi}\sum\frac{(-1)^n}{2n+1}\exp(\frac{-D(2n+1)^2\pi^2t}{4l^2}\cos(\frac{(2n+1)\pi\times(x  )}{2l})

    I am having trouble calculating the sum part of the expression. Let's assume

    \frac{Dt}{l^2}=1 and x = 0 and divide through by C_{0}, then I get the sum part to be:

    \sum\frac{(-1)^n}{2n+1}\exp(\frac{-(2n+1)^2\pi^2}{4})\times1

    The reference I am using gives the overall answer to be:

    \frac{C}{C_{0}}=1 - 0.1080 = 0.8920

    but I am getting much smaller answers. Could anyone suggest what I might be doing wrong?

    Any help would be much appreciated
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  2. #2
    Super Member
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    Athens, OH, USA
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    Re: Help with calculating a summation

    Your sum has no limits, so I'm assuming an infinite series starting at n = 0. Notice this is an alternating series with the magnitude of the terms decreasing to 0 as n goes to infinity. So the series converges and you know the value of a partial sum with m term differs from the value of the series by at most the magnitude of the (m+1)st term. With your calculator you find that the magnitude of the 3nd term is of the order of E-28. So just the sum of the 1st two terms is very close to the true value. I find the answer you gave to be correct.
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  3. #3
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    Re: Help with calculating a summation

    Thanks johng!

    I just reached the same conclusion - you beat me to it! I had initally started at n = 1 and seeing as the n = 0 part contributes the most, my answer was much too small.

    I have run into another similar problem, perhaps you could help me with that too? The sum is:



    we can let be a constant. The limits in this case are n = 1 to infinity (I haven't mastered that in latex yet sorry...)

    Essentially, I am getting answers that tend to oscillate for different values of r (ranging from 0 to 1).

    I have tried to break down each part of the sum to see how it contributes, but I am not having much luck.

    Again, any light thrown on this would be very much appreciated!
    Last edited by hannah89; January 3rd 2013 at 07:36 PM.
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