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Thread: Algebra

  1. #1
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    Algebra

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

    Show that for $\displaystyle \lambda>0$

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

    where

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

    $\displaystyle x! = 1 , x=0$
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  2. #2
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    Re: Algebra

    You can re-write the expression as

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

    If you can show
    $\displaystyle \sum_{j=0}^{\infty }\frac{e^{-\lambda}\lambda^{j}}{j!}$ 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 $\displaystyle e^{-\lambda}$ is constant and that the remaining is the Taylor Series expression of $\displaystyle e^{\lambda}$
    Last edited by MacstersUndead; Mar 16th 2013 at 09:29 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

    $\displaystyle \sum_{j=0}^{\infty }\frac{e^{-\lambda}\lambda^{j}}{j!}$ 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 $\displaystyle \frac{i}{i!}=\frac{i}{(i)(i-1)...(1)}=\frac{\not i}{(\not{i})(i-1)...(1)}=\frac{1}{(i-1)!}$
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