2
$\begingroup$

In Closed form solution of a hypergeometric sum. we provided a family of hypergeometric sums that have a closed form solution. Now we would like to investigate another family and come up with corresponding closed form solutions. Let $n \ge 0$ and $m \ge 0$ be integers. We consider a following sum: \begin{equation} {\mathfrak S}^{(m)}_n(x) := \sum\limits_{i=0}^\infty \binom{m \cdot i}{i+n} \cdot x^i \end{equation} Now, by using the duplication formula for the factorial, meaning by using the identity: \begin{equation} (2 i)! = 4^i i! \frac{(i-1/2)!}{(-1/2)!} \end{equation} for $i \ge 0$ and then by using known binomial identities we computed the sum in question for $m=2$. We have: \begin{eqnarray} {\mathfrak S}^{(2)}_n(x) = \frac{\sqrt{\pi} n!}{(n-1/2)!} \left[ \sum\limits_{p=0}^n \sum\limits_{q=0}^n \binom{n}{q} \binom{n-p+q-1}{q-1} \binom{-1/2}{p} (-1)^{n+q} \frac{(4 x)^{p-q}}{(1-4 x)^{p+1/2}}\right.\\ - \sum\limits_{p=0}^{n-1} \binom{n-1/2}{p} (-1)^p \binom{2 n-p-1}{n} \frac{1}{(4 x)^{n-p}} \left.\right] \end{eqnarray} valid for $x \in (0,1/4)$.

Now the obvious question is of course what is the result for arbitrary values of $n$ and $m$?

$\endgroup$
0
$\begingroup$

By no means is this going to be a full answer to this question. However I am going to post it because I believe that this approach can eventually lead to a full solution. Our plan would be to express the sum in question through hypergeometric sums and then use identities governing the later to obtain closed form solutions.

To be specific let us take $m=3$ and $n$ being an odd integer. Then by using elementary properties of Pochhammer symbols and factorials we have: \begin{eqnarray} (3 i)! &=& 3^{3 i} \prod\limits_{\xi=0}^2 (1-\frac{\xi}{3})^{(i)}\\ (2 i-2n-1)! &=& 2^{2(i-n)-1} \cdot \left( \frac{1}{2} - n\right)^{(i)} \cdot \frac{(-n-1/2)!}{(-1/2)!} \cdot 1^{(i)} \cdot \frac{1}{\prod\limits_{\xi=0}^n(i-\xi) }\\ (i+2n+1)!&=& 1^{(i)} \cdot \prod\limits_{\xi=1}^{2 n+1} (i+\xi) \end{eqnarray} Therefore the sum in question reads: \begin{eqnarray} {\mathfrak S}^{(3)}_{2 n+1}(x) = \frac{(-1/2)!}{(-n-1/2)!} \sum\limits_{i=0}^\infty \frac{(2/3)^{(i)} \cdot (1/3)^{(i)} \cdot }{(1/2-n)^{(i)} } \cdot \frac{3^{3 i} x^i}{2^{2(i-n)-1} i!} \cdot \frac{\prod\limits_{\xi=0}^n (i-\xi)}{\prod\limits_{\xi=1}^{2n+1}(i+\xi)} \end{eqnarray} Now we are using the usual "tricks". We have: \begin{eqnarray} \prod\limits_{\xi=0}^n (i-\xi) &=& \left. \frac{d^{n+1}}{d \theta^{n+1}} \theta^i \right|_{\theta=1} \\ \frac{1}{\prod\limits_{\xi=1}^{2n+1}(i+\xi)} &=& \int\limits_0^1 \frac{(1-\xi)^{2 n}}{(2 n)!} \xi^i d\xi \\ \int\limits_0^1 \frac{(1-\xi)^{2 n}}{(2 n)!} \cdot \xi^{n+1} \cdot F_{2,1} \left[ \begin{array}{rr} a & b \\ c \end{array}; \xi x\right] d\xi &=& \frac{(c-1)_{(3n+2)}}{(a-1)_{(3n+2)} (b-1)_{(3n+2)}} \cdot \frac{d^{n+1}}{d x^{n+1}} \cdot \frac{1}{x^{2 n+1}} \left\{ F_{2,1} \left[ \begin{array}{rr} a-3n-2 & b-3n-2 \\c-3n-2\end{array};x\right] - \sum\limits_{l=0}^{3 n+1} \frac{(a-3n-2)^{(l)} (b-3n-2)^{(l)}}{(c-3n-2)^{(l)}} \frac{x^l}{l!}\right\} \end{eqnarray} Inserting those into the equation before them gives the following: \begin{eqnarray} &&{\mathfrak S}^{(3)}_{2 n+1}(x) = \frac{1}{2} \frac{(-1/2)!}{(-n-1/2)!} \cdot \frac{(2/3)^{(n+1)} (1/3)^{(n+1)} }{(1/2-n)^{(n+1)} } \cdot (27 x)^{n+1} \cdot \\ &&\int\limits_0^1 \frac{(1-\xi)^{2 n}}{(2n)!} \cdot \xi^{n+1} F_{2,1}\left[ \begin{array}{rr} 5/3+n & 4/3+n \\ 3/2 \end{array}; \frac{27}{4} x \xi\right] d\xi \\ &&=\frac{1}{2} \frac{(-1/2)!}{(-n-1/2)!} \cdot \frac{(2/3)^{(n+1)} (1/3)^{(n+1)} (-1/2-3n)^{(3n+2)}}{(1/2-n)^{(n+1)} (-1/3-2n)^{(3n+2)} (-2/3-2 n)^{(3n+2)} } \cdot (27 x)^{n+1} \cdot \\ &&\sum\limits_{p=0}^{n+1} \binom{n+1}{p} \frac{(-1)^{n+1-p}}{(27/4 x)^{3n+2-p}} \frac{(3n+1-p)!}{(2n)!} \left( \frac{(-1/3-2 n)^{(p)} (-2/3-2 n)^{(p)}}{(-1/2-3 n)^{(p)}} F_{2,1}\left[ \begin{array}{rr} -1/3-2n+p& -2/3-2 n+p \\ -1/2-3 n+p\end{array};\frac{27}{4} x\right]-\sum\limits_{l=p}^{3n+1} \frac{(-1/3-2 n)^{(l)} (-2/3-2 n)^{(l)}}{(-1/2-3 n)^{(l)}} \cdot \frac{(27/4 x)^{l-p}}{(l-p)!}\right) \end{eqnarray} Herewith we are done,i.e we have accomplished our goal. There are only hypergeometric functions in the right hand side. Now we will demonstrate that we can actually evaluate those quantities in closed form.

Let us take $n=0$ and $p=1$ first. Then we have: \begin{eqnarray} &&F_{2,1}\left[ \begin{array}{rr} -1/3-2n+p& -2/3-2 n+p \\ -1/2-3 n+p\end{array};\frac{27}{4} x\right] = F_{2,1}\left[ \begin{array}{rr} 2/3& 1/3 \\ 1/2\end{array};\frac{27}{4} x\right] =\\ &&\left(1- \frac{27}{4} x\right)^{-1/2} \cdot F_{2,1} \left[ \begin{array}{rr} 1/6 & -1/6 \\ 1/2\end{array};\frac{27}{4} x\right] =\\ && \left(1- \frac{27}{4} x\right)^{-1/2} \cdot \cos \left( \frac{1}{6} \cdot \arccos\left( 1-\frac{27}{2} x\right)\right)=\\ &&\frac{\cos\left( \frac{1}{3} \cdot \arcsin \left( \frac{3 \sqrt{3}}{2} \sqrt{x}\right)\right)}{\sqrt{1-\frac{27}{4} x}} \end{eqnarray} where in the first line we used the Euler transformation of the hypergeometric sum and in the second line we used https://en.wikipedia.org/wiki/Hypergeometric_function#Other_points .

Now let us take $n=0$ and $p=0$. Then we have: \begin{eqnarray} &&F_{2,1}\left[ \begin{array}{rr} -1/3-2n+p& -2/3-2 n+p \\ -1/2-3 n+p\end{array};\frac{27}{4} x\right] = F_{2,1}\left[ \begin{array}{rr} -1/3& -2/3 \\ -1/2\end{array};\frac{27}{4} x\right] =\\ &&1- 3 \int\limits_0^1 F_{2,1}\left[ \begin{array}{rr} 2/3& 1/3 \\ 1/2\end{array};\frac{27}{4} \xi\right] d\xi = 1- 3 \int\limits_0^x \frac{\cos \left( \frac{1}{3} \cdot \arcsin\left( \frac{3 \sqrt{3}}{2} \sqrt{\xi}\right)\right)}{\sqrt{1- \frac{27}{4} \xi}}d\xi =\\ && 1-4/9 \int\limits_0^{27/4 x} \frac{\cos\left(\frac{1}{3} \arcsin\left( \sqrt{\xi} \right)\right)}{\sqrt{1-\xi}}d\xi = \\ && \frac{1}{2} \left( \sqrt{4-27 x} \cos\left( \frac{1}{3} \arcsin\left( \frac{3 \sqrt{3} \sqrt{x}}{2}\right)\right) + \sqrt{3 x} \sin\left( \frac{1}{3} \arcsin\left( \frac{3 \sqrt{3} \sqrt{x}}{2}\right)\right)\right) \end{eqnarray}

This concludes the case of $n=1$. We believe that this approach can be generalized to arbitrary values of $n$.

$\endgroup$
0
$\begingroup$

Let us denote the sum in question as: \begin{equation} {\mathfrak S}^{(m)}_j(w) := \sum\limits_{i=0}^\infty \binom{m\cdot i}{i+j} \cdot w^i \end{equation} We provide an answer for arbitrary $m\ge 2$ and $n=0$. In https://www.researchgate.net/profile/Larry_Glasser/publication/252738212_Hypergeometric_functions_and_the_trinomial_equation/links/5454c36a0cf2bccc490c41ba.pdf the author proves an interesting result.

Let $|w| < (m-1)^{m-1}/m^m$. Out of the solutions of the following transcendental equation: \begin{equation} 1-x+w \cdot x^m=0 \end{equation} we choose the one the one that is closest to unity.

Then we have: \begin{equation} x=1+ \sum\limits_{i=1}^\infty \binom{m \cdot i}{i} \cdot \frac{w^i}{((m-1)i+1)} \end{equation} Now by substituting for $\tilde{w}:=w^{1/(m-1)}$ then by multiplying both sides of the above by $\tilde{w}$ and differentiating with respect to $\tilde{w}$ we readily get the following identity: \begin{equation} \sum\limits_{i=0}^\infty \binom{m \cdot i}{i} \cdot w^i = \frac{x\cdot\left(1-w \cdot x^{m-1} \right)}{1-m \cdot w \cdot x^{m-1}}=\frac{x}{(1-m) x+m} \end{equation} Now, let us take arbitary value of $j\ge 0$. Then clearly we have: \begin{eqnarray} \binom{m \cdot i}{i+j}&=& \binom{m \cdot i}{i} \cdot \frac{((m-1) \cdot i)_{(j)}}{(i+1)^{(j)}}\\ &=&\binom{m \cdot i}{i} \cdot\left[(m-1)^j + \sum\limits_{l=1}^j \binom{l \cdot m+j-l-1}{j} j \binom{j-1}{l-1} \cdot \frac{(-1)^{j-l+1}}{(i+l)}\right] \end{eqnarray} where in the second line we decomposed the ratio of Pochhammer factors into partial fractions with respect to the variable $i$.

Now since $1/(i+l) = \int\limits_0^1 \theta^{i+l-1} d\theta$ we have: \begin{eqnarray} {\mathfrak S}^{(m)}_j(w)&=& (m-1)^j \cdot \frac{x}{(1-m) x+m} +\sum\limits_{l=1}^j \binom{l \cdot m+j-l-1}{j} j \binom{j-1}{l-1} \cdot (-1)^{j-l+1}\cdot \\ &&\int\limits_1^x \left( \frac{\xi-1}{w \xi^m}\right)^{l-1} \cdot \frac{1}{w \xi^{m}} d\xi\\ &=& (m-1)^j \cdot \frac{x}{(1-m) x+m} +\sum\limits_{l=1}^j \binom{l \cdot m+j-l-1}{j} j \binom{j-1}{l-1} \cdot \frac{(-1)^{j-l+1}}{w^l} \\ &&\int\limits_0^{x-1} \frac{u^{l-1}}{(1+u)^{l m}} du\\ \end{eqnarray} Therefore the final result reads: \begin{eqnarray} {\mathfrak S}^{(m)}_j(w) &=& \frac{x}{((1-m)x+m)(x-1)^j}+\sum\limits_{l=1}^j (-1)^{j-l+1} \binom{l \cdot m+j-l-1}{j-l} \cdot \frac{1}{w^{l}} \end{eqnarray}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.