HP49-50G plotting of sum(x=0 to 0, 1, 2,...10, of x!/(2x)!)

[Thomas Klemm]

(11-06-2023 12:25 AM)Thomas Klemm Wrote:  For Sum[(k!)^2/(2k)!, k] the result is:

\[
\begin{align}
\sum_{k=0}^\infty \frac{(k!)^2}{(2 k)!}
= \frac{2}{27} \left(18 + \sqrt{3} \pi \right)
\approx 1.73639985871872
\end{align}
\]

I was not able to find a closed solution for the associated generating function using WolframAlpha.
But after trying some values for \(x\) I could guess it:

\[
\begin{align}
\sum_{k=0}^{\infty} \frac{(k!)^2}{(2k)!} \; x^k
&= \frac{1 + \frac{\frac{\sqrt{x}}{2}}{\sqrt{1 - \frac{x}{4}}} \sin^{-1}\left( \frac{\sqrt{x}}{2}\right)}{1 - \frac{x}{4}}
\end{align}
\]

Examples

\(x = 1\)

\[
\begin{align}
\sum_{k=0}^{\infty} \frac{(k!)^2}{(2k)!}
&= \frac{2}{27} \left(18 + \sqrt{3} \pi \right) \\
&\approx 1.736399858718715077909795 \\
\end{align}
\]

\(x = 2\)

\[
\begin{align}
\sum_{k=0}^{\infty} \frac{(k!)^2}{(2k)!} \; 2^k
&= \frac{4 + \pi}{2} \\
&\approx 3.570796326794896619231322 \\
\end{align}
\]

\(x = 3\)

\[
\begin{align}
\sum_{k=0}^{\infty} \frac{(k!)^2}{(2k)!} \; 3^k
&= \frac{4}{3} \left(3 + \sqrt{3} \pi \right) \\
&\approx 11.25519745693687140237631 \\
\end{align}
\]

Differential Equation

I wondered how to find a differential equation based on the coefficients of the sequence.

Substitution

The formula above suggests the following substitution:

\[
\begin{align}
u &= \frac{\sqrt{x}}{2} \\
u^2 &= \frac{x}{4} \\
x &= 4u^2
\end{align}
\]

We can also rewrite the coefficients using Double Factorials:

\[
\begin{align}
y(x)
&= \sum_{k=0}^{\infty} \frac{(k!)^2}{(2k)!} \; x^k \\
\\
y(u)
&= \sum_{k=0}^{\infty} \frac{(k!)^2}{(2k)!} \; 2^{2k} u^{2k} \\
\\
&= \sum_{k=0}^{\infty} \frac{((2k)!!)^2}{(2k)!} \; u^{2k} \\
\\
&= \sum_{k=0}^{\infty} \frac{(2k)!!}{(2k-1)!!} \; u^{2k} \\
\\
&= \sum_{k=0}^{\infty} a_{k} \; u^{2k} \\
\end{align}
\]

Example

Here's an example of the transformation of the coefficient \(a_k\) for \(k = 4\):

\[
\begin{align}
\frac{(k!)^2}{(2k)!} \; 2^{2k}
&= \frac{(1 \cdot 2 \cdot 3 \cdot 4)^2}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8} \; 2^8 \\
\\
&= \frac{(1 \cdot 2 \cdot 3 \cdot 4)^2}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8} \; \left( 2^4 \right)^2 \\
\\
&= \frac{(1 \cdot 2 \cdot 3 \cdot 4 \cdot \; 2^4)^2}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8} \\
\\
&= \frac{(2 \cdot 4 \cdot 6 \cdot 8)^2}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8} \\
\\
&= \frac{2 \cdot 4 \cdot 6 \cdot 8}{1 \cdot 3 \cdot 5 \cdot 7} \\
\\
&= \frac{8!!}{7!!} \\
\end{align}
\]

Difference of Coefficients

Based on this definition the difference \(a_{k} - a_{k-1}\) can be calculated:

\[
\begin{align}
a_k
&= \frac{(2k)!!}{(2k-1)!!} \\
\\
a_k - a_{k-1}
&= \frac{(2k)!!}{(2k-1)!!} - \frac{(2k-2))!!}{(2k-3)!!} \\
\\
&= \frac{(2k-2)!!}{(2k-3)!!} \frac{2k}{2k-1} - \frac{(2k-2))!!}{(2k-3)!!} \\
\\
&= \frac{(2k-2)!!}{(2k-3)!!} \cdot \left( \frac{2k}{2k-1} - 1 \right) \\
\\
&= a_{k-1} \cdot \frac{1}{2k-1}
\end{align}
\]

Example

\[
\left \{ a_k \mid k = 0, 1, 2, 3, \cdots \right \} =
\left \{
1, \;
\frac{2}{1}, \;
\frac{2 \cdot 4}{1 \cdot 3}, \;
\frac{2 \cdot 4 \cdot 6}{1 \cdot 3 \cdot 5}, \;
\frac{2 \cdot 4 \cdot 6 \cdot 8}{1 \cdot 3 \cdot 5 \cdot 7}, \;
\frac{2 \cdot 4 \cdot 6 \cdot 8 \cdot 10}{1 \cdot 3 \cdot 5 \cdot 7 \cdot 9}, \;
\cdots
\right \}
\]

For \(k = 4\) we get:

\[
\begin{align}
a_4 - a_3
&= \frac{2 \cdot 4 \cdot 6 \cdot 8}{1 \cdot 3 \cdot 5 \cdot 7} - \frac{2 \cdot 4 \cdot 6}{1 \cdot 3 \cdot 5} \\
\\
&= \frac{2 \cdot 4 \cdot 6}{1 \cdot 3 \cdot 5} \cdot \left(\frac{8}{7} - 1 \right) \\
\\
&= \frac{2 \cdot 4 \cdot 6}{1 \cdot 3 \cdot 5} \cdot \frac{1}{7} \\
\\
&= a_3 \cdot \frac{1}{7} \\
\end{align}
\]

This allows us to find a differential equation for \(y(u)\):

\[
\begin{align}
y(u) \cdot \left( 1 - u^2 \right)
&= \sum_{k=0}^{\infty} a_{k} \; u^{2k} \cdot \left( 1 - u^2 \right) \\
\\
&= \sum_{k=0}^{\infty} a_{k} \; u^{2k} - \sum_{k=0}^{\infty} a_{k} \; u^{2k+2} \\
\\
&= \sum_{k=0}^{\infty} a_{k} \; u^{2k} - \sum_{k=1}^{\infty} a_{k-1} \; u^{2k} \\
\\
&= 1 + \sum_{k=1}^{\infty} \left( a_{k} - a_{k-1} \right) \; u^{2k} \\
\\
&= 1 + \sum_{k=1}^{\infty} a_{k-1} \cdot \frac{1}{2k-1} \; u^{2k} \\
\\
\frac{y(u) \cdot \left( 1 - u^2 \right)}{u} &= \frac{1}{u} + \sum_{k=1}^{\infty} a_{k-1} \cdot \frac{1}{2k-1} \; u^{2k-1} \\
\\
\frac{\mathrm{d}}{\mathrm{d} u} \; y(u) \cdot (\frac{1}{u} - u) &= - \frac{1}{u^2} + \sum_{k=1}^{\infty} a_{k-1} \; u^{2k-2} \\
\\
\frac{\mathrm{d}}{\mathrm{d} u} \; y(u) \cdot (\frac{1}{u} - u) &= - \frac{1}{u^2} + \sum_{k=0}^{\infty} a_{k} \; u^{2k} \\
\\
\frac{\mathrm{d}}{\mathrm{d} u} \; y(u) \cdot (\frac{1}{u} - u) &= - \frac{1}{u^2} + y(u) \\
\end{align}
\]

Solution

I'm a bit lazy, so I just post the step-by-step solution that WolframAlpha gave me:




To calculate \(\mu(u)\) use:




Back Substitution

What's left is using the initial condition \(y(0) = 1\) to deduce that \(c_1 = 0\) and replace \(u\) by:

\[
\begin{align}
u &= \frac{\sqrt{x}}{2} \\
\end{align}
\]