# Cauchy product on exponential-looking power series

Original posting by dioxen here: Double summation including power and factorial

I am finding some trouble in computing the following sum:

$$\sum_{k=0}^\infty \frac{x^k}{k!}\;\sum_{m=0}^k\frac {y^m}{m!}$$

Could you please provide a result?

After a failed attempt at getting to the binomial expansion, I tried to play around with it a little bit. I came up with two ideas of a solution, but can't see it through. Both involve applying the Cauchy product. $$\Bigg(\sum_{n=0}^\infty a_n \Bigg) \Bigg(\sum_{n=0}^{\infty} b_n \Bigg) = \sum_{n=0}^{\infty} c_n \space \text{ where } c_n = \sum_{k=0}^n a_k b_{n-k} \space \space \space \space \space \space \space \space \space \space \space \space (1)$$

1: In this problem, we cound treat $$c_n= \frac{x^n}{n!}\;\sum_{k=0}^n\frac {y^k}{k!}=\frac{1}{n!}\sum_{k=0}^n\frac {(xy)^{n-k}}{(n-k)!}x^{k}=\frac{1}{n!}\sum_{k=0}^n a_k b_{ n-k}$$

So if we're thinking within the framework of the first two summations in (1), our $a_n$ and $b_n$ are as follows: $a_n=x^n$ and $b_n = \frac {(xy)^{n}}{n!}\\$

our only problem is the we're left with the $\frac{1}{n!}$ in front of the summation. If we thought up some way to get rid of it, we'd have a nice geometric series and $e^{xy}$. But I can't think of a way...any leads here?

2: Alternatively, we could leave the $\frac{x^n}{n!}$ outside and treat it as a number (because it is, before we choose to sum it up over $n$). $$c_n= \frac{x^n}{n!}\;\sum_{k=0}^n\frac {y^k}{k!} 1^{n-k}=\frac{x^n}{n!}\sum_{k=0}^n a_k b_{ n- k}$$ which gives us $a_n=\frac{y^n}{n!}$ and $b_n=1$. Could we do something about $\frac{x^n}{n!}$?

$$\Bigg(\sum_{n=0}^\infty \frac{x^n}{n!} \Bigg) \Bigg(\sum_{n=0}^{\infty} \frac{x^n}{n!} \frac{y^n}{n!} \Bigg)$$ Does this even make sense? Because otherwise I can't think of a way this sum would be computable.

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I don't get it. Why don't you write $e^x$ instead of the power-series for it? –  deufeufeu May 26 '13 at 9:48
@deufeufeu Because the two sums are dependent. The second one is FINITE! –  shimee May 26 '13 at 12:48
ok I'm sorry. I misread the initial sum. –  deufeufeu May 26 '13 at 13:07
I tried Laplace transform of it over $y$. Then I(Mathematica) could do the summation. I ended up with $\frac{s e^x-e^{x/s}}{(s-1) s}$ which I couldn't calculate the Inverse Laplace transformation of it. However, I can write that as an Integral, If that is good enough I will turn it to an answer. –  Ali May 29 '13 at 18:45
@Ali sure, I'd be happy to see it (if you could also include some intermediate steps) –  shimee May 29 '13 at 20:02

This function probably doesn't have a closed form, but here's how you can handle it. Define $$f(x,y) = \sum_{0\leq m\leq k} \frac{x^k y^m}{k! m!},$$ and simplify the inner sum using $$\sum_{0\leq m\leq k} \frac{y^m}{m!} = e^y \frac{k!-\gamma(1+k,y)}{k!} = e^y\left( 1 - \frac1{k!}\gamma(1+k,y)\right),$$ using equation (9) here: http://mathworld.wolfram.com/IncompleteGammaFunction.html. After that, we have $$f(x,y) = e^{x+y} - e^y \sum_{k\geq0} \frac{x^k}{k!^2} \frac{y^{1+k}}{k+1} F\left(\begin{array}{c}k+1\\k+2\end{array}\middle| -y\right) = e^{x+y} - e^y h(x,y),$$ where $F$ is a hypergeometric function appearing in (7) on that page. Now the sum $h(x,y)$ can be simplified using the definition of a hypergeometric to get $$h(x,y) = \sum_{k,l\geq0} \frac{x^k y^{1+k+l} (-1)^l}{k!^2 (k+1+l) l!}.$$ If we differentiate once w.r.t. $y$, the sum separates: $$\partial_y h(x,y) = \sum_{k,l\geq0} \frac{x^k y^k}{k!^2} \sum_{l\geq0}\frac{(-y)^l}{l!} = I_0(2\sqrt{xy})e^{-y},$$ where $I_0$ is a modified Bessel function.

Finally using $h(x,0)=0$, we can integrate $\partial_y h$ w.r.t. $y$ over $[0,y]$ to find that $$f(x,y) = e^{x+y} - e^y \int_0^y e^{-y}I_0(2\sqrt{xy})\,dy.$$

It is very likely that that integral has no simple or closed form, as it doesn't appear in Gradshteyn and Ryzhik, but I could be wrong. This form is perhaps somewhat simpler than the above single sum with an incomplete gamma function.

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We start by Laplace Transformation of the inner sum argument, this way we will get rid of the factorial in the denominator and end up with a simple series:

$$\mathcal{L}\left(\frac{y^m}{m!} \right)=s^{-(m+1)}$$

Where $\mathcal{L}$, stands for Laplace transform from $y$ to $s$(something tells me, this will deeply assume y is positive). Now we can do the inner summation easily:

$$\sum_{m=0}^k{s^{-(m+1)}}=\frac{s^{-k-1} \left(s^{k+1}-1\right)}{s-1}=\frac{1-s^{-k-1}}{s-1}$$

Now putting this back into the initial sum, we will have:

$$\sum_{k=0}^\infty \frac{x^k}{k!} \frac{1-s^{-k-1}}{s-1}=\frac{1}{s-1} \sum_{k=0}^\infty\left(\frac{x^k}{k!}-\frac{(\frac{x}{s})^k}{s\ k!} \right)$$

These sums are just the expansion of $e^z$, so we have:

$$\frac{e^x}{s-1}-\frac{e^\frac{x}{s}}{s(s-1)}=\frac{e^x}{s-1}-\frac{e^{x/s}}{s-1}+\frac{e^{x/s}}{s}$$

Now, we have to calculate the inverse Laplace transform from $s$ to $y$. I could produce explicit expressions for all but the middle term(which I will write as an integral)

$$\sum_{k=0}^\infty \frac{x^k}{k!}\ \sum_{m=0}^k\frac {y^m}{m!}=\mathcal{L}^{-1}\left(\frac{e^x}{s-1}-\frac{e^{x/s}}{s-1}+\frac{e^{x/s}}{s} \right) \\ =e^{x+y}+J_0 (2\sqrt{xy})-\frac{1}{2 \pi i} \lim_{T\to\infty}\int_{ \gamma - i T}^{ \gamma + i T} e^{sy} \frac{e^{x/s}}{s-1}\,ds$$ where $\gamma$ is a real number so that the contour path of integration is in the region of convergence of $\frac{e^{x/s}}{s-1}$.

This is how far I could come. Hope it's a useful result for you.

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(+1) very nice technique. –  Mhenni Benghorbal May 31 '13 at 19:54

Sadly, I believe there is no closed form for your double summation. The closest form I can get is $$\begin{eqnarray}\sum_{k=0}^\infty \frac{x^k}{k!}\;\sum_{m=0}^k\frac {y^m}{m!} &=& \sum_{k=0}^\infty \frac{x^k}{k!} \; {}_0F_1\left(\left.\begin{array}{c} - \\ k+1 \end{array} \; \right| xy \right) \\ &=& \frac 1{\pi}\sum_{k=0}^\infty \left(\frac xy \right)^{\frac k2} \; \int_0^{\pi} e^{2 \sqrt {xy} \cos(t)} \cos (kt) dt \end{eqnarray}$$

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What is the expression $_0 F_1 ()$? Hypergeometric series? –  shimee May 30 '13 at 13:41