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How can I get the sum inside the exponential?

  1. Jun 6, 2012 #1
    1. The problem statement, all variables and given/known data
    Show if true:
    [tex]\sum_{i=1}^{n-1}e^{2 \pi ift}=\frac{e^{2 \pi ifn}-1}{e^{2 \pi if}-1}=e^{\pi if(n-1)}\frac{e^{\pi ifn}-e^{-\pi i f n}}{e^{ \pi if}-e^{-\pi if}}\\\\\\\\[/tex]

    2. Relevant equations

    I'm really stuck here, just looking for a suggestion as to what equation to use...
    Can I take the log? How can I get the sum inside the exponential?

    3. The attempt at a solution
     
    Last edited: Jun 6, 2012
  2. jcsd
  3. Jun 6, 2012 #2

    vela

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    Re: Euler

    Hint: Geometric series.
     
  4. Jun 6, 2012 #3

    D H

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    Re: Euler

    This does not look right, no matter how I read your sum. The expression is incorrect if the sum is over i; you should have a t somewhere on the right hand side. The expression is also incorrect if the sum is over t rather than i.

    [tex]\sum_{i=1}^{n-1} e^{2 \pi i f t} =
    \frac{e^{2\pi f n t} - e^{2\pi f t}} {e^{2\pi f t} - 1}
    = e^{2\pi f t} \frac{e^{2\pi f (n-1) t} - 1} {e^{2\pi f t} - 1}[/tex]

    and

    [tex]\sum_{t=1}^{n-1} e^{2 \pi i f t} =
    \frac{e^{2\pi i f n} - e^{2\pi i f}} {e^{2\pi i f} - 1}
    = e^{2\pi i f} \frac{e^{2\pi i f (n-1)} - 1} {e^{2\pi i f} - 1}[/tex]

    Since you mentioned Euler in the title, I suspect it is the latter that you are supposed to prove.

    Or perhaps it is

    [tex]\sum_{t=0}^{n-1} e^{2 \pi i f t} =
    \frac{e^{2\pi i f n} - 1} {e^{2\pi i f} - 1}[/tex]
     
  5. Jun 6, 2012 #4
    Re: Euler

    Hi,

    You are correct my mistake.


    [tex]\sum_{t=0}^{n-1}e^{2 \pi ift}=\frac{e^{2 \pi ifn}-1}{e^{2 \pi if}-1}=e^{\pi if(n-1)}\frac{e^{\pi ifn}-e^{-\pi i f n}}{e^{ \pi if}-e^{-\pi if}}\\\\\\\\[/tex]
     
    Last edited: Jun 6, 2012
  6. Jun 6, 2012 #5
    Re: Euler

    Using Geometric Series I get...

    [tex] \frac{1-e^{2 \pi ifn}}{1-e^{2 \pi if}} =\frac{e^{2 \pi ifn}-1}{e^{2 \pi if}-1}[/tex]

    Do I want to split up the pieces in the numerator? or Factor?
     
    Last edited: Jun 6, 2012
  7. Jun 13, 2012 #6
    Re: Euler

    You have:

    [itex]\frac{e^{2nx}-1}{e^{2x}-1}=\frac{e^{nx}}{e^{x}}\;\frac{e^{nx}-e^{-nx}}{e^x-e^{-x}}[/itex]

    and the result follows simplifying the first factor in the right had side of the equality.
     
  8. Jun 13, 2012 #7
    Re: Euler

    How is this even related to Euler's formula, since i is an index? But, as vela said, geometric series makes it obvious here. But first you really have to simplify that mess. Replace 2πft with something else, like ##A##, or even ##x## since it isn't used so far.
     
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