$$e^x=\sum \limits\_{k=0}^\infty \frac{x^k}{k!}$$


We can rewrite the equation as
$$e^x=\sum \limits\_{k=0}^\infty \frac{x^k}{ \Gamma(k+1)} \tag{1}$$


because $x!=\Gamma(x+1)$ where $x$ is non-negative integer.


$\Gamma(x)$ ([Gamma function](https://en.wikipedia.org/wiki/Gamma_function)) also has undefined property of any of the integers {0, −1, −2,... . } at those points; it is a meromorphic function with simple poles at the non-positive integers.


$m$ is non-positive, $$\frac{1}{\Gamma(m)}=0 \tag{2}$$  

Let's combine the property $(1)$ with $(2)$ and the formula $(1)$ can be extended for $x \neq 0$


$$e^x=\sum \limits\_{k=-\infty}^\infty \frac{x^k}{ \Gamma(k+1)} \tag{3}$$


If we change sum variable $k'=k+n$ where n is an integer . We get
$$e^x=\sum \limits\_{k'=-\infty}^\infty \frac{x^{k'+n}}{ \Gamma(k'+n+1)} \tag{4}$$


$n \in Z $ {...,-2,-1,0,1,2,...}


I wanted to find out if the equation (4) holds true if we select $n \in C $ for $x \neq 0$ ?


If we select , $n=\frac{1}{2}$


$$\Gamma(1/2)=\sqrt{\pi}$$
$$f(x)=\sqrt{x}\sum \limits\_{k=-\infty}^\infty \frac{x^{k}}{ \Gamma(k+3/2)} \tag{5}$$


I have noticed $f(x)$ has the same derivative result as $e^x$.


$$f'(x)=f(x) \tag{6}$$


Proof:
------


$$f(x)=\sqrt{x}\sum \limits\_{k=-\infty}^\infty \frac{x^{k}}{ \Gamma(k+3/2)} \tag{7}$$


$$f(x)=\frac{2\sqrt{x}}{\sqrt{\pi}}(1+\frac{2x}{3}+\frac{2^2x^2}{3.5}+\frac{2^3x^3}{3.5.7}+....)+\frac{1}{\sqrt{\pi x}}(1-\frac{1}{2x}+\frac{3}{2^2x^2}-\frac{3.5}{2^3x^3}+.... ) \tag{8}$$


$$f'(x)=\frac{1}{\sqrt{\pi x}}+\frac{2\sqrt{x}}{\sqrt{\pi}}(1+\frac{2x}{3}+\frac{2^2x^2}{3.5}+\frac{2^3x^3}{3.5.7}+....)-\frac{1}{\sqrt{\pi x}}(\frac{1}{2x}-\frac{3}{2^2x^2}+\frac{3.5}{2^3x^3}-.... ) \tag{9}$$


$$f'(x)=\frac{2\sqrt{x}}{\sqrt{\pi}}(1+\frac{2x}{3}+\frac{2^2x^2}{3.5}+\frac{2^3x^3}{3.5.7}+....)+\frac{1}{\sqrt{\pi x}}-\frac{1}{\sqrt{\pi x}}(\frac{1}{2x}-\frac{3}{2^2x^2}+\frac{3.5}{2^3x^3}-.... ) \tag{10}$$


$$f'(x)=\frac{2\sqrt{x}}{\sqrt{\pi}}(1+\frac{2x}{3}+\frac{2^2x^2}{3.5}+\frac{2^3x^3}{3.5.7}+....)+\frac{1}{\sqrt{\pi x}}(1-\frac{1}{2x}+\frac{3}{2^2x^2}-\frac{3.5}{2^3x^3}+.... ) \tag{11}$$


Thus
$$f'(x)=f(x) \tag{12}$$


More generally,


$$f\_n(x)=\sum \limits\_{k=-\infty}^\infty \frac{x^{k+n}}{ \Gamma(k+n+1)} \tag{13}$$
where $n \in C $ for $x \neq 0$ ?


The function $f\_n(x)$ has same differential property $f'\_n(x)=f\_n(x)$ with $e^x$.


Please advice how to prove or disprove that $f\_n(x)=e^x$ where $n \in C $ for $x \neq 0$ ?


Thanks a lot



Investigation of $\sum \limits_{k=-\infty}^\infty \frac{x^{k+n}}{ \Gamma(k+n+1)}$ where $n \in C$?
Set $a:=1+n\in \mathbb C$ and assume $a\notin\mathbb Z$; we use the variable $z:=1/x$. The question has already been answered in the comments: the power series $$S(z):=\sum\_{k=0}^\infty \frac{z^k}{\Gamma(a-k)}$$ is divergent (null radius of convergence). Yet a more research-level angle can be reached with the help of Borel-Laplace sumation. 


The formal power series $S(z)$ is Gevrey of class $1$, and its Borel transform is $$\hat S(z):=\sum\_{k=0}^\infty\frac{z^k}{\Gamma(a+1-k)k!}=\sum\_{k=0}^\infty\frac{z^k}{\Gamma(a+1-k)k!},$$ which is convergent on the disc of radius $1$. Indeed the reflection formula and Stirling equivalent yield $$\Gamma(a-k)k!\sim\_{k\to\infty}\frac{(-1)^k\sin(\pi a)}{\pi(k-a)^a}.$$ The only singularity of this function on the boundary is located at $z=-1$. One can show that $\hat S$ can be analytically continued along any line $\theta\mathbb R\_{\geq0}$ with $|\theta|=1$ as long as $\theta\neq-1$ with exponential growth of order at most $1$ near $\infty$, so that the Laplace transform $$S\_\theta(z):=\theta\int\_0^\infty\exp(-t/z)S(\theta t)\mathrm{d}t$$ defines a holomorphic function on any sector $A\_\theta:=\{0<|z| : |\arg(z/\theta)|<\delta\}$ of aperture $2\delta<\pi$. Its asymptotic expansion at $0$ coincides with $S$.


Standard terminology says that $\theta$ is a **Stokes direction** for $S$ if $S\_\theta$ cannot be continued to a sector of aperture strictly greater than $\pi$, and that $S$ is **sumable in the sense of Borel-Laplace** if there exists only finitely many Stokes directions. 


In the present situation one can show that the only Stokes direction is $\theta=-1$, and that for $\theta:=1$ the sectorial sum $S\_\theta$ is actually analytically continuable to a (self-intersecting) sector of aperture $3\pi$. Therefore we can fix $\theta:=1$ in the following, and consider the multivalued function $S\_1$ on the sector $A\_1$ for $\delta:=3\pi/2$. 


The nice thing about Borel-Laplace summable function of Gevrey class is that every algebro-differential operation can be carried out formally and is valid too for the sum $S\_1$. Therefore the improper computations performed at a formal level by the OP hold true for the function $S\_1$. Also the function is canonical, in the sense that the bigger-than-$\pi$ aperture of $A\_1$ ensures the uniqueness of $S\_1$ among all sectorial functions with asymptotic expansion $S$ at $0$. It is called the **Borel-Laplace sum of $S$ on $A\_1$**. 


As a matter of consequence we do have for $a:=3/2$ and $1/x\in A\_1$ the relationship $$\exp x=\frac{e}{f(1)+S\_1(1)}\sqrt{x}~\left(f(x)+S\_1(1/x)\right)$$ where $f$ is the entire function $f(x):=\sum\_{k=1}^\infty\frac{x^k}{\Gamma(k+3/2)}$.




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To conclude, a few words about numerics. It is very likely that numerics bolster the OP's claim, as the divergence of the series happens «far away». For instance one can compare the behavior of two series: $$a(-100):=\sum\_k (-100)^k/k!$$ and $$b(-100):=\sum\_kk!/(-100)^k.$$ The partial sums of order $N=100$ are given respectively by $5,34\times10^{41}$ and $0,99019$, while for $N=300$ they are closer to their true limit $8,153\times10^{-16}\simeq \exp (-100)$ and $2,29\times 10^{14}\simeq\infty$. Notice that in the case of $b(-100)$ the value is very close to $$-100\int\_0^{-0,01}\frac{\exp(100+1/t}{t}\mathrm{d}t.$$ This integral is not random, and appears when solving the differential equation $$x^2y'(x)=y(x)-x$$ by quadrature on $]-\infty,0[$ (over which there is a single solution tending to $0$ at $0$). The formal power series $-b(x)/100$ is solution of this equation (and the method of variation of the constant provides the Borel-Laplace sectorial sum), which seems to indicate that performing the «best term summation» yields something very close to the actual solution. This was known from a long time (actually Euler used the trick to propose a sum to Wallis hypergeometric series $$\sum(-1)^nn!\simeq-0,5963$$), and used in actual calculations (astronomy, quantum mechanics…) Borel-Laplace sumation of Gevrey series explains why this is so and validates the process. For details see works by Balser, Malgrange, Ramis, Sibuya…


I do like the following quote by Poincaré, summing it all up:



> 
> Il y a entre les géomètres et les astronomes une sorte de malentendu au sujet de la signification du mot convergence. Les géomètres préoccupés de la parfaite rigueur et souvent trop indifférents à la longueur de calculs inextricables dont
>  ils conçoivent la possibilite, sans songer à les entreprendre effectivement, disent qu’une série est convergente quand la somme des termes tend vers une limite déterminée, quand même les premiers termes diminueraient très lentement. Les astronomes, au contraire, ont coutume de dire qu’une série converge quand les 20 premiers termes, par exemple, diminuent très rapidement, quand même les termes suivants devraient croître indéfiniment.
> 
> 
> 


which roughly translates as



> 
> There exists a misunderstanding between geometers and astronomers as to what a convergent series should be. Geometers, busy with perfect rigour and often indifferent to the intractable length of calculations they conceive, without thinking of actually carrying them out, say that a series converges when the sum of terms tends to a well-determined limit, even though the first terms would decrease very slowly. Astronomers, on the contrary, usually say a series is convergent when the 20 first terms, say, rapidly decrease, even though the following terms should grow without bound.
> 
> 
> 


