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Math Help - Algebra

  1. #1
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    Algebra

    Calling on all Math Whizzes... Can anyone please help...

    Show that for \lambda>0

    \sum_{i=1}^{\infty }i  \times \frac{e^{-\lambda}\lambda^{i}}{i!}  = \lambda

    where

    x!=\prod_{j=0}^{x=1}(x-j)=x \times (x-1)\times ... \times 1, x=1,2,3....
    and

    x! = 1                                                                            , x=0
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  2. #2
    Senior Member MacstersUndead's Avatar
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    Re: Algebra

    You can re-write the expression as

    \sum_{i=1}^{\infty }\frac{e^{-\lambda}\lambda^{i}}{(i-1)!} by factorial property
    \lambda \sum_{i=1}^{\infty }\frac{e^{-\lambda}\lambda^{i-1}}{(i-1)!} factor out lambda not dependent on i
    \lambda \sum_{j=0}^{\infty }\frac{e^{-\lambda}\lambda^{j}}{j!} Let j=i-1 change of index.

    If you can show
    \sum_{j=0}^{\infty }\frac{e^{-\lambda}\lambda^{j}}{j!} is equal to 1, then you are done. This sum looks extremely familiar to me, might be from statistics. I'll help you find a solution for this part if you need it online.

    EDIT://

    Note that e^{-\lambda} is constant and that the remaining is the Taylor Series expression of e^{\lambda}
    Last edited by MacstersUndead; March 16th 2013 at 10:29 PM.
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  3. #3
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    Re: Algebra

    Quote Originally Posted by MacstersUndead View Post
    You can re-write the expression as

    \sum_{j=0}^{\infty }\frac{e^{-\lambda}\lambda^{j}}{j!} is equal to 1, then you are done. This sum looks extremely familiar to me, might be from statistics. I'll help you find a solution for this part if you need it online.
    Yes please, I would appreciate it so much. Also, I haven't done factorials ever, so can you please expand on the first line?
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  4. #4
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    Re: Algebra

    MacstersUndead's edit is the solution. The first line comes from the fact that \frac{i}{i!}=\frac{i}{(i)(i-1)...(1)}=\frac{\not i}{(\not{i})(i-1)...(1)}=\frac{1}{(i-1)!}
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