# To express $f(x,z)=\sum \limits_{n=0}^\infty \frac{e^{-\alpha n^2 x+\beta n z}}{n!}$ as known functions

$\alpha,\beta >0$ $$f(x,z)=\sum \limits_{n=0}^\infty \frac{e^{-\alpha n^2 x+\beta n z}}{n!}$$

$$\frac{\partial{f(x,z)}}{\partial z}=\beta \sum \limits_{n=1}^\infty \frac{e^{-\alpha n^2 x+\beta n z}}{(n-1)!}$$

$$\frac{\partial{f(x,z)}}{\partial z}|_{z=2 \frac{ \alpha}{\beta} x+ z_1}=\beta \sum \limits_{n=1}^\infty \frac{e^{-\alpha n^2 x+\beta n (2 \frac{ \alpha}{\beta} x+ z_1)}}{(n-1)!}$$

$$\frac{\partial{f(x,z)}}{\partial z}|_{z=2 \frac{ \alpha}{\beta} x+ z_1}=\beta e^{\alpha x+ \beta z_1} \sum \limits_{n=1}^\infty \frac{e^{-\alpha (n-1)^2 x+\beta (n-1) z_1}}{(n-1)!}$$

$$\frac{\partial{f(x,z)}}{\partial z}|_{z=2 \frac{ \alpha}{\beta} x+ z_1}=\beta e^{\alpha x+ \beta z_1} \sum \limits_{n=0}^\infty \frac{e^{-\alpha n^2 x+\beta n z_1}}{n!}$$

$$\frac{\partial{f(x,z)}}{\partial z}|_{z=2 \frac{ \alpha}{\beta} x+ z_1}=\beta e^{\alpha x+ \beta z_1} f(x,z_1)$$

I do not know how to solve this kind differential equations.

Do you know how to solve that?

Can we express the function as known functions such as Jacobi Theta Functions etc?

EDIT:

Another property is:

$$-\alpha\frac{\partial^2{f(x,z)}}{\partial z^2}=\beta^2 \frac{\partial{f(x,z)}}{\partial x}$$

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If $\alpha=0$, things would have been easy. Even the $\beta=0$ case does not seem to admit an easy expression... – J. M. Jul 20 '12 at 7:47
You may assume $\beta=1$. Put $z:=is$ and think of $x$ as "time". Then your $f$ solves the heat equation (look at your edit!) for $s\in {\mathbb R}/(2\pi)$ with initial condition $f(0,i s)=\exp(e^{is})$. Theta functions would result from simpler initial conditions, like a $\delta$-function. – Christian Blatter Jul 20 '12 at 20:10
@ChristianBlatter :If so, Which initial conditions are required to get the function above? Thanks – Mathlover Jul 21 '12 at 11:36
@Mathlover: As I said: Putting $x=0$ ($x$ is time!) in the definition of your $f$ gives $$f(0,z)=\sum_{n=0}^\infty {(e^{\beta z})^n\over n!}=\exp(e^{\beta z})\ .$$ – Christian Blatter Jul 21 '12 at 12:43
@Christian Blatter: But since heat equation can only be solved by using separation of variables to construct kernel form solutions, which will return to the original summation form and no help. – doraemonpaul Jul 21 '12 at 16:32

First note that $\alpha$ and $x$ , $\beta$ and $z$ are both taking the same roles in $f(x,z)=\sum\limits_{n=0}^\infty\dfrac{e^{-\alpha n^2x+\beta nz}}{n!}$ and should be burdensome. So you should simplify your definition to $f(x,z)=\sum\limits_{n=0}^\infty\dfrac{e^{-xn^2+zn}}{n!}$ , where $x\geq0$ .

$f(x,z)=\sum\limits_{n=0}^\infty\dfrac{e^{-xn^2+zn}}{n!}$ can create the following differential equation:

$\dfrac{df(x,z)}{dz}=\sum\limits_{n=0}^\infty\dfrac{ne^{-xn^2+zn}}{n!}=\sum\limits_{n=1}^\infty\dfrac{ne^{-xn^2+zn}}{n!}=\sum\limits_{n=1}^\infty\dfrac{e^{-xn^2+zn}}{(n-1)!}=\sum\limits_{n=0}^\infty\dfrac{e^{-x(n+1)^2+z(n+1)}}{n!}=\sum\limits_{n=0}^\infty\dfrac{e^{-xn^2+(z-2x)n+z-x}}{n!}=e^{z-x}\sum\limits_{n=0}^\infty\dfrac{e^{-xn^2+(z-2x)n}}{n!}=e^{z-x}f(x,z-2x)$

Note that this is a DDE. Unfortunately even the forms of the general solutions of DDEs we still can't know well, we can't know the number of I.C.s should be required. So this approach fails.

$f(x,z)=\sum\limits_{n=0}^\infty\dfrac{e^{-xn^2+zn}}{n!}$ can create the following PDE:

$\dfrac{\partial^2f(x,z)}{\partial z^2}=\sum\limits_{n=0}^\infty\dfrac{n^2e^{-xn^2+zn}}{n!}$

$\dfrac{\partial f(x,z)}{\partial x}=\sum\limits_{n=0}^\infty\dfrac{-n^2e^{-xn^2+zn}}{n!}$

$\therefore\dfrac{\partial f(x,z)}{\partial x}+\dfrac{\partial^2f(x,z)}{\partial z^2}=0$

Note that the form of the general solution of this PDE is as follows:

Let $f(x,z)=X(x)Z(z)$ ,

Then $X'(x)Z(z)+X(x)Z''(z)=0$

$X'(x)Z(z)=-X(x)Z''(z)$

$-\dfrac{X'(x)}{X(x)}=\dfrac{Z''(z)}{Z(z)}=s^2$

$\begin{cases}\dfrac{X'(x)}{X(x)}=-s^2\\Z''(z)-s^2Z(z)=0\end{cases}$

$\begin{cases}X(x)=c_3(s)e^{-xs^2}\\Z(z)=\begin{cases}c_1(s)\sinh zs+c_2(s)\cosh zs&\text{when}~s\neq0\\c_1z+c_2&\text{when}~s=0\end{cases}\end{cases}$

$\therefore f(x,z)=C_1z+C_2+\int_sC_3(s)e^{-xs^2}\sinh zs~ds+\int_sC_4(s)e^{-xs^2}\cosh zs~ds$ or $C_1z+C_2+\sum\limits_sC_3(s)e^{-xs^2}\sinh zs+\sum\limits_sC_4(s)e^{-xs^2}\cosh zs~ds$

It is clear that two I.C.s should be required to solve this PDE uniquely, however from the functional property itself, only $f(0,z)=\sum\limits_{n=0}^\infty\dfrac{e^{zn}}{n!}=\sum\limits_{n=0}^\infty\dfrac{(e^z)^n}{n!}=e^{e^z}$ is trivially known, others for example $f(x,0)=\sum\limits_{n=0}^\infty\dfrac{e^{-xn^2}}{n!}$ cannot be trivially known.

So we still no hope to express $f(x,z)=\sum\limits_{n=0}^\infty\dfrac{e^{-xn^2+zn}}{n!}$ as known functions.

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