Simple question

Discussion in 'Math' started by boks, Dec 13, 2008.

  1. boks

    Thread Starter Active Member

    Oct 10, 2008
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    Can somebody explain why

    \frac{-1}{(2 \pi )(2+in)}[e^{- \pi (2 + in)} - e^{\pi (2 + in)}]

    can be written

    \frac{(-1)^n}{2+in} \frac{e^{2 \pi} - e^{-2 \pi}}{2 \pi}

    where n is an integer?
     
  2. studiot

    AAC Fanatic!

    Nov 9, 2007
    5,005
    513
    Rewrite the numerator as

    {[exp(2π) . exp(inπ)] - [exp(-2π) . exp(-inπ)]}

    Noting that (Eulers theorem)
    exp (iπ) = cos (π) + isin (π) = 1
    exp(-iπ) = cos (π) - isin (π) = 1

    and substituting

    The (-1)^{n}

    comes from putting n into Eulers formula, the result is alternately +1 and -1

    Sorry the π (pi) symbols are almost indistinguishable from the n.
     
  3. boks

    Thread Starter Active Member

    Oct 10, 2008
    218
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    Great, thanks!
     
  4. boks

    Thread Starter Active Member

    Oct 10, 2008
    218
    0
    cos \pi = -1
     
  5. studiot

    AAC Fanatic!

    Nov 9, 2007
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    Damn spell checker again! I'll try not to scribble so quickly in future.
    you're right, it = -1 but that does work out.
     
  6. boks

    Thread Starter Active Member

    Oct 10, 2008
    218
    0
    I still don't understand why it's

    \frac{(-1)^n}{2+in} \frac{e^{2 \pi} - e^{-2 \pi}}{2 \pi}

    and not

    \frac{-(-1)^n}{2+in} \frac{e^{2 \pi} - e^{-2 \pi}}{2 \pi}
     
  7. studiot

    AAC Fanatic!

    Nov 9, 2007
    5,005
    513
    Don't forget I rearranged your orignal expression by taking the -1 insde the bracket. It is part of the numerator.

    The last term in the third line should read exp minus (i n pi)

    [​IMG]
     
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