# Thread: Solving an equation for a proof

1. ## 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**):

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?

2. ## Re: Solving an equation for a proof

you are integrating $\displaystyle \int ~f(g)~dg$ ?

3. ## Re: Solving an equation for a proof

Originally Posted by romsek
you are integrating $\displaystyle \int ~f(g)~dg$ ?
Yes,

4. ## Re: Solving an equation for a proof

Originally Posted by sjaffry
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**):

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?