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Thread: Solving an equation for a proof

  1. #1
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    Solving an equation for a proof

    Hi all,

    I am reading a research paper and trying to understand an equation. I have tried it (I guess more or less) 10 times.
    But the result I am getting is completely different what author has achieved.

    $$ f(g) = e^{-K-g} \sum_{k=0}^\infty \frac{(Kg)^k}{(k!)^2} $$

    Now I do not understand and could not solve beyond this. The author says that the CDF can be simplified by applying 2.321.2 from the following attached image (reference: **Table of Integrals, Series, and Products**):

    Solving an equation for a proof-11111.png

    The result (which is a CDF from $0$ to $g$) that author has yield is :


    $$ F_{f(g)}(g) = e^{-K}\sum_{k=0}^\infty \frac{K^{k}}{k!} \Bigg\{e^{-g}\sum_{m=0}^k(-1)^{2m+1}m! \binom{m}{k} x^{k-m} + k!\Bigg\}$$

    I really cannot understand how he came up with this.
    My result yields the following

    $$F_{f(g)}(g) = e^{-K}\sum^\infty_{k=0} \sum^k_{m=0} \frac{K^k}{k!} {-e^{-g}g^{k-m}(e^{-g} - 1)}$$

    Any idea?
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  2. #2
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    Re: Solving an equation for a proof

    you are integrating $\displaystyle \int ~f(g)~dg$ ?
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  3. #3
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    Re: Solving an equation for a proof

    Quote Originally Posted by romsek View Post
    you are integrating $\displaystyle \int ~f(g)~dg$ ?
    Yes,
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  4. #4
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    Re: Solving an equation for a proof

    Quote Originally Posted by sjaffry View Post
    Hi all,

    I am reading a research paper and trying to understand an equation. I have tried it (I guess more or less) 10 times.
    But the result I am getting is completely different what author has achieved.

    $$ f(g) = e^{-K-g} \sum_{k=0}^\infty \frac{(Kg)^k}{(k!)^2} $$
    Write this as $f(g)= e^{-K}\sum_{k=0}^\infty e^{-g}\frac{(Kg)^k}{(k!)^2}$
    and integrate the sum term by term using integration of parts for each term.

    Now I do not understand and could not solve beyond this. The author says that the CDF can be simplified by applying 2.321.2 from the following attached image (reference: **Table of Integrals, Series, and Products**):

    Click image for larger version. 

Name:	11111.png 
Views:	6 
Size:	37.1 KB 
ID:	38463

    The result (which is a CDF from $0$ to $g$) that author has yield is :


    $$ F_{f(g)}(g) = e^{-K}\sum_{k=0}^\infty \frac{K^{k}}{k!} \Bigg\{e^{-g}\sum_{m=0}^k(-1)^{2m+1}m! \binom{m}{k} x^{k-m} + k!\Bigg\}$$

    I really cannot understand how he came up with this.
    My result yields the following

    $$F_{f(g)}(g) = e^{-K}\sum^\infty_{k=0} \sum^k_{m=0} \frac{K^k}{k!} {-e^{-g}g^{k-m}(e^{-g} - 1)}$$

    Any idea?
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