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Thread: Sum = Integral

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    Sum = Integral

    I have to prove that $\displaystyle \sum_{x=0}^{k-1}\frac{(\lambda w)^{x}e^{- \lambda w}}{x!}=\int_{\lambda w}^{\infty }\frac{z^{k-1}e^{-z}}{(k-1)!}dz$

    How would I go about doing that?
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  2. #2
    Grand Panjandrum
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    Re: Sum = Integral

    Quote Originally Posted by BrownianMan View Post
    I have to prove that $\displaystyle \sum_{x=0}^{k-1}\frac{(\lambda w)^{x}e^{- \lambda w}}{x!}=\int_{\lambda w}^{\infty }\frac{z^{k-1}e^{-z}}{(k-1)!}dz$

    How would I go about doing that?
    Induction on k

    CB
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    Re: Sum = Integral

    How do I start?
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  4. #4
    Grand Panjandrum
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    Re: Sum = Integral

    Quote Originally Posted by BrownianMan View Post
    How do I start?
    You prove a base case, for this show that the formula is true when $\displaystyle k=1$ that is show that:

    $\displaystyle \frac{(\lambda w)^0e^{-\lambda w}}{0!}=e^{-\lambda w}=\int_{\lambda w}^{\infty} \frac{z^0 e^{-z}}{0!}\; dz=\int_{\lambda w}^{\infty} e^{-z}\; dz$

    That is show:

    $\displaystyle e^{-\lambda w}=\int_{\lambda w}^{\infty} e^{-z}\; dz$.

    Next assume it true for some $\displaystyle k$, then prove it true for $\displaystyle k+1$. This will involve integration by parts, start by using the induction assumption:

    $\displaystyle \sum_{x=0}^{(k+1)-1}\frac{(\lambda w)^xe^{-\lambda w}}{x!}=\sum_{x=0}^{k-1}\frac{(\lambda w)^xe^{-\lambda w}}{x!}+\frac{(\lambda w)^ke^{-\lambda w}}{k!}\\ \\ \\ \phantom{xxxxxxxxx} =\int_{\lambda w}^{\infty} \frac{z^{k-1} e^{-z}}{(k-1)!}\; dz+\frac{(\lambda w)^ke^{-\lambda w}}{k!}$

    Now show using integration by parts on the integral on the right that:

    $\displaystyle \int_{\lambda w}^{\infty} \frac{z^{k-1} e^{-z}}{(k-1)!}\; dz+\frac{(\lambda w)^ke^{-\lambda w}}{k!}=\int_{\lambda w}^{\infty} \frac{z^{(k+1)-1} e^{-z}}{((k+1)-1)!}\; dz$

    CB
    Last edited by CaptainBlack; Sep 14th 2011 at 06:22 AM.
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    Re: Sum = Integral

    Thanks so much!
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