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Exponential series

  1. Nov 24, 2005 #1
    This should be a proof of the fact that exp(x)*exp(y)=exp(x+y). Have a look at it:
    Now, I understand everything fairly well, except for one step: what is the operation to get from the 4th line to the 5th?
    Help will be appreciated very much.
    Best regards...Cliowa
  2. jcsd
  3. Nov 24, 2005 #2


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    There's a typo, the 5th line should be:

    [tex]=\sum_{n=0}^{\infty}\frac{\left(\sum_{k=0}^n\left (\begin{array}{*{1}{c}}n\\k\end{array}\right)x^ky^ {n-k}\right)}{n!}[/tex]

    They've just changed the order of summation.
  4. Nov 25, 2005 #3


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    A little help from Rudin

    I'm not sure the formula used to evaluate the product of the infinite series [going from line # 3 -> #4] is correct (though it may well be correct), Rudin gives the formula for the Cauchy product of series as it is presented in this link.

    A quick version is:

    Suppose [itex] \sum_{n=0}^{\infty} a_n[/itex] and [itex] \sum_{n=0}^{\infty} b_n[/itex] converge absolutely. Then

    [tex]\left( \sum_{n=0}^{\infty} a_n\right) \left( \sum_{n=0}^{\infty} b_n\right) = \sum_{n=0}^{\infty} \sum_{k=0}^{n} a_{k}b_{n-k}[/tex] also converges absolutely.

    Alternately, look here, under the heading A Variant.
    Last edited: Nov 25, 2005
  5. Nov 25, 2005 #4
    @shmoe: Thanks alot, you really opened my eyes. Thank you very much. It seems clear now.

    @benorin: I do think going from line 3 to 4 is alright, because all you do is you write n!/(k!*n-k)!) as the binomial coefficient of n and k, which works fine. Thanks anyway, the link was quite useful.

    Best regards...Cliowa
  6. Nov 25, 2005 #5


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    Lines 1 to 5 are essentially deriving the Cauchy product for these two series (with some algebraic simplification mixed in and not mentioning where absolute convergence is used). cliowa, you might want try to applying the form benorin gave to line 1 directly.
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