A curious power series

Question

I would like to find a closed form for the following power series: $$f(x;r) = \sum_{n=1}^\infty \frac{x^{nr}}{\Gamma(nr)},$$ where $r$ is a positive integer. Until now, I've tried to let Mathematica compute it and I looked on the net but found nothing similar... Any help will be very appreciated.

Answer 1

Except for the very first terms, it seems to me that there is no closed form.

What I obtained is $$f(x;1)=e^x x$$ $$f(x;2)=x \sinh (x)$$ $$f(x;3)=\frac{1}{3} e^{-x/2} x \left(e^{3 x/2}-2 \sin \left(\frac{1}{6} \left(3 \sqrt{3} x+\pi \right)\right)\right)$$ $$f(x;4)=-\frac{1}{2} x (\sin (x)-\sinh (x))$$ For $r>4$ appear generalized hypergeometric functions but some of them can simplify $$f(x;5)=\frac{1}{5} \left(e^x-\sqrt[5]{-1} e^{-\sqrt[5]{-1} x}+(-1)^{2/5} e^{(-1)^{2/5} x}-(-1)^{3/5} e^{-(-1)^{3/5} x}+(-1)^{4/5} e^{(-1)^{4/5} x}\right) x$$ $$f(x;6)=\frac{1}{3} x \left(\sinh (x)+\cos \left(\frac{\sqrt{3} x}{2}\right) \sinh \left(\frac{x}{2}\right)-\sqrt{3} \sin \left(\frac{\sqrt{3} x}{2}\right) \cosh \left(\frac{x}{2}\right)\right)$$ I give up with nightmares !

Answer 2

Since $f(x,r) = x \sum \limits _{n=1} ^\infty \frac {x^ {nr-1}} {(nr-1)!}$, for fixed $r \in \Bbb N \setminus \{0\}$ let $g(x) = \sum \limits _{n=1} ^\infty \frac {x^ {nr-1}} {(nr-1)!}$. Convince yourself that after deriving $r$ times you get $g^{(r)} = g$. But this is an easy differential equation, as it has constant coefficients! If $\varepsilon$ is a primitive root of unity of order $r$, the general solution to your differential equation is $\sum \limits _{k=0} ^{r-1} c_k \space \Bbb e ^{\varepsilon ^k x}$. In order to find the coefficients $c_k$ note that for $0 \le k \le r-2$ you have $g^{(k)} (0) = 0$ and $g^{(r-1)} (0) = 1$. This means

$$\begin{eqnarray*} c_0 + c_2 + \dots + c_{r-1} = 0 \\ c_0 + \varepsilon \space c_1 + \varepsilon ^2 \space c_2 + \dots + \varepsilon ^{r-1} \space c_{r-1} = 0 \\ c_0 + (\varepsilon ^2) \space c_1 + (\varepsilon ^2) ^2 \space c_2 + \dots + (\varepsilon ^2) ^{r-1} \space c_{r-1} = 0 \\ \dots \\ c_0 + (\varepsilon ^{r-2}) \space c_1 + (\varepsilon ^{r-2}) ^2 \space c_2 + \dots + (\varepsilon ^{r-2}) ^{r-1} \space c_{r-1} = 0 \\ c_0 + (\varepsilon ^{r-1}) \space c_1 + (\varepsilon ^{r-1}) ^2 \space c_2 + \dots + (\varepsilon ^{r-1}) ^{r-1} \space c_{r-1} = 1 . \end{eqnarray*}$$

The determinant of this system is a particular form of a Vandermonde determinant easily seen to be nonzero, so the system has a unique solution. I do not know whether the minors associated to the unknowns are easily computable (they exhibit a lot of symmetry).

In any case, $f(x,r) = x \sum \limits _{k=0} ^{r-1} c_k \space \Bbb e ^{\varepsilon ^k x}$.

Answer 3

Following the idea of Alex.M, and exploiting the symmetry of the system of equations, we get closed expressions for the $c_k$. Indeed, the matrix involved in the system is a particular form of a Vandermonde matrix, which is a Discrete Fourier Transform matrix. This matrix is easily invertible and thus we get the $c_k$.

More precisely, we want to compute the following series: $$\sum_{n=1}^\infty \frac{x^{nr}}{\Gamma(nr)} = x \sum_{n=1}^\infty \frac{x^{nr-1}}{(nr-1)!}.$$ Then, let us define: $$g(x;r) = \sum_{n=1}^\infty \frac{x^{nr-1}}{(nr-1)!}, \quad f(x;r) = \sum_{n=1}^\infty \frac{x^{nr}}{\Gamma(nr)} = x g(x;r).$$ We have then: $$g(x;r) = \frac{x^{r-1}}{(r-1)!} + \frac{x^{2r-1}}{(2r-1)!}+ \dots $$ Then, we have: $$ \frac{\partial }{\partial x^k} g(x;r) = \sum_{n=1}^\infty \frac{x^{r-1-k}}{(r-1-k)!},\quad \forall k=1,\dots,r.$$ Then, in particular, we have: $$g^{(r)}(x;r) = g(x;r). $$ The characteristic polynomial of this simple differential equation is then $z^r -1=0$. Let then $z$ be a primary root of $1$, that is, without loss of generality, $z = e^{\frac{ 2\pi i }{r}}.$ Then, the general solution of the this differential equation is given by $\sum_{k=0}^{r-1} c_k e^{z^k}.$ For $k=0,\dots, r-2$, we have $g^{(k)}(0;r) = 0$ and $g^{(r-1)}(0,r) = 1.$ Then, identifying the terms, we have the following $r\times r$ system of equations: $$ \left\{ \begin{array}{l} c_0 + c_1 + c_2 + \dots + c_{r-1} = 0 \\ c_0 + c_1 z + c_2 z^2 + \dots + c_{r-1} z^{r-1} = 0\\ \qquad \vdots \\ c_0 + c_1 z^{r-2} + c_2 (z^2)^{r-2} + \dots + c_{r-1}(z^{r-1})^{r-2} = 0\\ c_0 + c_1 z^{r-1} + c_2 (z^2)^{r-1} + \dots + c_{r-1}(z^{r-1})^{r-1} = 1. \end{array}\right. , $$ or in matrix form, $\mathbf{F} \mathbf{c} = (0,\dots,0,1)$, where $\mathbf{F}$ is the following matrix: $$\mathbf{F} = \begin{pmatrix} 1 & 1 & 1 & \cdots & 1 \\ 1 & z^1 & z^2 & \cdots & z^{r-1} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & z^{r-1} & (z^2)^{r-1} & \dots & (z^{r-1})^{r-1} \end{pmatrix}, $$ and $\mathbf{c}=(c_0,c_1,\dots, c_{r-1})^t$. Since $z$ is a unitary root of the unity, this particular matrix is a discrete Fourier transform matrix. We will prove that $\frac{1}{\sqrt{r}}\mathbf{F}$ is unitary, that is: $$ [\mathbf{F}]_k \cdot [\mathbf{F}]^*_l = r \delta_{kl},$$ where $[\mathbf{F}]_k$ denote the $k-$th column of the matrix $\mathbf{F}$. Then, we have: $$[\mathbf{F}]_k \cdot [\mathbf{F}]^*_l = \sum_{m=0}^{r-1} e^{\frac{2\pi i}{r} km } e^{\frac{2\pi i}{r} (-l)m } = r \delta_{kl}. $$ Indeed, if $k=l$, the result is trivial. If $k\neq l$, we have: $$\sum_{m=0}^{r-1} e^{\frac{2\pi i}{r} km } e^{\frac{2\pi i}{r} (-l)m } = \sum_{m=0}^{r-1} (e^{\frac{2\pi i}{r}(k-l)})^m = \frac{(e^{\frac{2\pi i}{r}(k-l)})^r - 1 }{e^{\frac{2\pi i}{r}(k-l)}- 1 } = 0,$$ since $(e^{\frac{2\pi i}{r}(k-l)})^r = 1$. Then, we can write the inverse of $\mathbf{F}$ as: $$\mathbf{F}^{-1} = \frac{1}{r}\mathbf{F}^* =\frac{1}{r} \begin{pmatrix} 1 & 1 & 1 & \cdots & 1 \\ 1 & \overline{z}^1 & \overline{z}^2 & \cdots & \overline{z}^{r-1} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & \overline{z}^{r-1} & (\overline{z}^2)^{r-1} & \dots & (\overline{z}^{r-1})^{r-1} \end{pmatrix}. $$ Then, we have: $$[\mathbf{F}^{-1}]_r =\frac{1}{r} \begin{pmatrix} 1 \\ \overline{z}^{r-1}\\ \vdots \\ (\overline{z}^{(r-1)(r-1)} \end{pmatrix}. $$ Then, we get: $$ \begin{pmatrix} c_0\\ \vdots \\ c_{r-1} \end{pmatrix} = \mathbf{F}^{-1} \begin{pmatrix} 0\\ \vdots \\ 0\\ 1 \end{pmatrix} = [\mathbf{F}^{-1}]_r. $$ Then, we can conclude: $$c_k =\frac{1}{r} \overline{z}^{k(r-1)}, k=0,1,\dots, r-1.$$ We can finally conclude that $$f(x;r) = x g(x;r) = \frac{x}{r}\sum_{k=0}^{r-1} \overline{z}^{k(r-1)}e^{z^k x}.$$ Most of this answer is due to the edit of Alex M., thanks a lot.