Is there a closed form for the sequences mentioned in the body of this question?

Question

Is there any hope for closed forms for expressions like: $$\sum_{k=0}^{\infty}\frac{x^k}{(3k+i)!}\text{ and/or }\sum_{k=0}^{\infty}\frac{kx^k}{(3k+i)!}$$where $i\in\{0,1,2\}$?

I am interested because I am trying to find an answer to this question.

Thank you in advance and sorry if this is a duplicate.

Answer 1

Hint:

$$ \sum\limits_{k = 0}^\infty {{{x^{\,k} } \over {\left( {3k + i} \right)!}}} = x^{-i/3} \sum\limits_{k = 0}^\infty {{{\left( {x^{1/3} } \right)^{\,3k + i} } \over {\left( {3k + i} \right)!}}} $$

and the last sum is a tri-section of $\exp(x^{1/3})$

So for instance $$ \sum\limits_{k = 0}^\infty {{{z^{\,3k} } \over {\left( {3k} \right)!}}} = {1 \over 3}\left( {e^{\,z} + 2e^{\, - z/2} \cos {{\sqrt 3 z} \over 2}} \right) $$ and have a look at the above link for the espressions for the other values of $i$.

And let me add that the $3$ tri-sections share the property of $\cosh, \, \sinh$ that the derivative of one equals the other with lower $i$, in cycle, as it is easily seen.

Answer 2

Considering $$f_i=\sum_{k=0}^{\infty}\frac{x^k}{(3k+i)!}\qquad \text{and} \qquad g_i=\sum_{k=0}^{\infty}\frac{kx^k}{(3k+i)!}$$ a CAS gives $$f_i=\frac{1}{i!}\,\, _1F_3\left(1;\frac{i+1}{3},\frac{i+2}{3},\frac{i+3}{3};\frac{x}{27}\right)$$ $$g_i=\frac{x}{(i+3)!}\,\, _1F_3\left(2;\frac{i+4}{3},\frac{i+5}{3},\frac{i+6}{3};\frac{x}{27}\right)$$ Only for the specific cases you asked for $(i=0,1,2)$, we can write the results in a nice form, defining $$F_i=3 e^{\frac{t}{2 \sqrt{3}}}\left(\frac{t}{\sqrt3}\right)^{i-1}\,f_i-e^{\frac{\sqrt{3} }{2}t}\qquad \text{where} \qquad \color{red}{t=\sqrt{3} \sqrt[3]{x}}$$ $$F_0=2 \cos \left(\frac{t}{2}\right)\qquad F_1=-2 \sin \left(\frac{\pi}{6} -\frac{t}{2}\right)\qquad F_2=-2 \sin \left(\frac{\pi}{6} +\frac{t}{2}\right)$$

It seems that for $f_i$ there is no problem for the expansion for any $i$. For $g_i$, it does not seem to be the same story at all except for $i=0$ (in such a case $g_0=x f_0'$).

$$g_0=\frac{t}{9 \sqrt{3}}e^{\frac{t}{\sqrt{3}}}\left(1-2 e^{-\frac{\sqrt{3}}{2}t} \sin \left(\frac{\pi}{6} +\frac{t}{2} \right)\right)$$

Answer 3

Let $$f_i(x)=\sum_{k=0}^{\infty}\frac{x^k}{(3k+i)!}$$

For $i\in \{0,1,2\}$.

First we manipulate the series for $f_1(x)$: $$f_1(x)=\sum_{k=0}^{\infty}\frac{x^k}{(3k+1)!}\implies f_1(x^3)=\sum_{k=0}^{\infty}\frac{x^{3k}}{(3k+1)!}\implies xf_1(x^3)=\sum_{k=0}^{\infty}\frac{x^{3k+1}}{(3k+1)!}$$

Taking the derivatives of both sides: $$\frac{d\left(xf_1(x^3)\right)}{dx}=\frac{d\left(\sum_{k=0}^{\infty}\frac{x^{3k+1}}{(3k+1)!}\right)}{dx}=\sum_{k=0}^{\infty}\frac{d\left(\frac{x^{3k+1}}{(3k+1)!}\right)}{dx}=\sum_{k=0}^{\infty}\frac{(3k+1)x^{3k}}{(3k+1)!}=\sum_{k=0}^{\infty}\frac{x^{3k}}{(3k)!}=f_0(x^3)$$

Performing similar manipulations on $f_2(x)$ yields $$\frac{d^2\left(x^2f_2(x^3)\right)}{dx^2}=f_0(x^3)$$

Lastly, if you consider the fact that each number is either congruent to either $0,1,$ or $2$ $\operatorname{mod} 3$, you get that $$e^x=\sum_{j=0}^{\infty}\frac{x^j}{j!}=\sum_{k=0}^{\infty}\frac{x^{3k}}{(3k)!}+\sum_{k=0}^{\infty}\frac{x^{3k+1}}{(3k+1)!}+\sum_{k=0}^{\infty}\frac{x^{3k+2}}{(3k+2)!}=f_0(x^3)+xf_1(x^3)+x^2f_2(x^3)$$

So $$e^x=f_0(x^3)+xf_1(x^3)+x^2f_2(x^3)$$

If we differentiate both sides wrt to $x$ twice we get $$\frac{d^2\left(e^x\right)}{dx^2}=\frac{d^2\left(f_0(x^3)\right)}{dx^2}+\frac{d^2\left(xf_1(x^3)\right)}{dx^2}+\frac{d^2\left(x^2f_2(x^3)\right)}{dx^2}$$

Using what we have already shown and the fact that $e^x$ is fixed under differentiation, we get $$e^x=\left(f_0(x^3)\right)''+\left(f_0(x^3)\right)'+f_0(x^3)$$

If we substitute $g(x)=f(x^3)$, we get $$e^x=g''(x)+g'(x)+g(x)$$ which is a simple ODE.

You can easily check that substituting $\frac{e^x}{3}$ for $g(x)$ will satisfy the above equation.

The general solution for the ODE is then given by $\frac{e^x}{3}+h(x)$ where $h(x)$ is the general solution for $$0=h''(x)+h'(x)+h(x)$$

From the basic theory of ODEs, we may find the genera $h(x)$ using the characteristic equation of the $h(x)$ ODE.

The roots are $$\frac{-1\pm i\sqrt{3}}{2}$$

Using this, the general solution for $g(x)$ becomes $\frac{e^x}{3}+Ae^{\frac{-1+ i\sqrt{3}}{2}x}+Be^{\frac{-1- i\sqrt{3}}{2}x}$ with $A$ and $B$ being arbitrary constants.

$f_0(x^3)$ is a special case $g(x)$ so by plugging in values of $x$ where $f_0(x^3)$ (or its first derivative) may be easily evaluated, we can solve for $A$ and $B$, and so solve for $f_0(x^3)$ and hence $f_0(x)$. Then using the relationships $f_1(x)$ and $f_2(x)$ have to $f_0(x)$, they may be solved for too.

Note: If there's anything wrong or confusing in my answer, feel free to edit it (and comment please) or just comment