# Is it possible to express this function with a hypergeometric function?

I have this expression.

$$\tag{1} \sum_{i=0}^q\color{red}{(-1)^i\binom{2q+1-i}{i}2^{2q+1-2i}} \frac{1}{2^{N+2q-2i+1}}\binom{N+2q-2i+1}{\frac{N+1}{2}+q-i}$$

I want to express this with the hypergeometric function $${_2F_1}$$, but I don't know if it is possible, so this might just be a wild goose chase.

Some insight:

The red part is the coefficients of the Chebyshev Polynomials of the second kind. And apparently Chebyshev Polynomial can be expressed with $${_2F_1}$$, i.e.,

$$U_n(x) = (n+1)\times {_2F_1}\left(-n,n+2;\frac{3}{2};\frac{1}{2}(1-x)\right).$$

This is why I wondered if (1) can be expressed with $${_2F_1}(m,a;b;f)$$ as well.

This is the first time I am trying to use hypergeometric functions, but from what I gathered $$m=-q$$. Then, I tried to make my expression look like the $${_2F_1}$$ expression in Wikipedia, i.e., $${_2F_1}(-m, a;b;f) = \sum_{i=0}^{m}(-1)^i\binom{m}{i}\frac{(a)_i}{(b)_i}f^i.$$

The only part I managed to do something was the first binomial term, i.e.,

$$\binom{2q+1-i}{i} = \frac{(2q-2i+2)_i}{(q-i)_i}\binom{q}{i}.$$

I tried to do something with the second binomial as well. I used

$$\binom{2z}{z} = \frac{2^{2z}}{\sqrt{\pi}}\frac{\Gamma(z+\frac{1}{2})}{\Gamma(z+1)}.$$

This was kind of nice to get rid of the $$\exp_2(-N-2q+2i-1)$$ term but I couldn't proceed any further.

So, my question is

• Is it possible to express this with $${_2F_1}$$?
• If it is, how should I proceed?

I appreciate any input, as this is literally the first time I started learning about $${_2F_1}$$. And if (1) can be expressed with $${_2F_1}$$, apparently taking the derivative would be neater so this is not just about trying to find a good closed form expression.

I was too focused on $${_2F_1}$$ functions that I missed the very obvious fact that it is not the only hypergeometric function.

This was the first time ever that I had dealings with Gamma Function, double factorials and hypergeometric function. So, I'll lay out all steps, some of which must be too obvious for veteran mathematicians.

Firstly, $$(1)$$ is full of falling factorials, which are not great for hypergeometric functions, so I reordered the sum by substituting $$q-i \rightarrow i$$, i.e.,

$$\tag{2} S =2\times(-1)^q\sum_{i=0}^q(-1)^i \binom{q+1+i}{q-i} 2^{2i} \frac{1}{2^{N+2i+1}} \binom{N+2i+1}{\frac{N+1}{2}+i}$$

Then, I used the binomial identity for $$\binom{2z}{z}$$ that I stated above and reached here.

$$\tag{3} S =\frac{2\times(-1)^q}{\sqrt{\pi}}\sum_{i=0}^q(-1)^i \binom{q+1+i}{q-i} 2^{2i} \frac{\Gamma(\frac{N}{2}+i+1)}{\Gamma(\frac{N+1}{2}+i+1)}$$

Then, I expanded the binomial. It has the only falling factorial in this expression.

$$\tag{4} S = \frac{2\times(-1)^q}{\sqrt{\pi}}\sum_{i=0}^q(-1)^i \frac{(q+i+1)!}{(q-i)!(2i+1)!} 2^{2i} \frac{(\frac{N}{2}+i)!}{(\frac{N+1}{2}+i)!}$$

The get rid of the falling factorial, I multiplied both the numerator and the denominator by $${q!i!}$$.

$$\tag{5} S = \frac{2\times(-1)^q}{q!\sqrt{\pi}}\sum_{i=0}^q(-1)^i\binom{q}{i} \frac{(q+i+1)!i!}{(2i+1)!} 2^{2i} \frac{(\frac{N}{2}+i)!}{(\frac{N}{2}+i+\frac{1}{2})!}$$

It was kind of straightforward except for the $$(2i+1)!$$.

\begin{align} \tag{6} S &= \frac{2(q+1)(q!)(-1)^q}{q!\sqrt{\pi}}\sum_{i=0}^q(-1)^i\binom{q}{i} \frac{(q+2)_i(1)_i}{(2i)!!(2i+1)!!} \frac{(\frac{N}{2}+i)!}{(\frac{N}{2}+i+\frac{1}{2})!}4^{i}\\ \tag{7} &= \frac{2(q+1)(-1)^q}{\sqrt{\pi}}\sum_{i=0}^q(-1)^i\binom{q}{i} \frac{(q+2)_i(1)_i}{2^{2i}i!(i+\frac{1}{2})!} \frac{(\frac{N}{2}+i)!}{(\frac{N}{2}+i+\frac{1}{2})!}4^{i}\\ \tag{8} &=\frac{2(q+1)(-1)^q}{\sqrt{\pi}}\frac{\Gamma(\frac{N}{2}+1)}{\Gamma(\frac{N+3}{2})}\sum_{i=0}^q(-1)^i\binom{q}{i} \frac{(q+2)_i}{(3/2)_i} \frac{(\frac{N}{2}+1)_i}{(\frac{N+3}{2})_i} \end{align}

I am not sure if it can be expressed as $${}_2F_1$$, but I know that it can be expressed as

$$S = \frac{2(q+1)(-1)^q}{\sqrt{\pi}}\frac{\Gamma(\frac{N}{2}+1)}{\Gamma(\frac{N+3}{2})} {}_3F_2(-q,q+2,\frac{N}{2}+1;\frac{3}{2},\frac{N+3}{2};1).$$

During the derivation, I also understood why there is an $$(n+1)$$ factor leading $${}_2F_1$$ for the Chebyshev Polynomial. As a buyer's remorse, I am not sure if a closed-form expression helps with anything.

• The $\binom qi$ will make a $\frac1{(q-i)!}$ with a “$-1$” in front of the $i$. How can one make a“$+1$” in front of the $i$ and apply the infinite sum definition of $_3F_2$? This linked definition does not allow an $(a-i)!$ in the sum Commented Apr 27, 2023 at 21:02
• @TymaGaidash Apparently when one of the nominator arguments of the $_PF_Q$ is negative, it is converted into a finite sum. I was also curious about this and asked this question specifically to understand if it extends to hypergeometric functions other than $_2F_1$ Commented Apr 27, 2023 at 21:06

Here is another variation to derive the generalised hypergeometric function $${_3F_2}$$. We obtain using rising factorials $$(a)_{k}:=a(a+1)\cdots(a+k-1)$$ \begin{align*} \color{blue}{\frac{1}{2^N}}&\color{blue}{\sum_{i=0}^q\underbrace{(-1)^i\binom{2q+1-i}{i}\binom{N+2q-2i+1}{\frac{N+1}{2}+q-i}}_{=:t_i}}\\ &=\frac{1}{2^N}\sum_{i=0}^q t_i\\ &= \frac{1}{2^N}t_0\sum_{i=0}^q\prod_{j=0}^{q-1}\frac{t_{j+1}}{t_j}\\ &=\frac{1}{2^N}\binom{N+2q+1}{\frac{N+1}{2}+q}\sum_{i=0}^q\prod_{j=0}^{i-1}\left((-1)^{j+1}\binom{2q-j}{j+1}\binom{N+2q-2j-1}{\frac{N-1}{2}+q-j}\right.\\ &\qquad\qquad\qquad\cdot\left.(-1)^{-j}\binom{2q+1-j}{j}^{-1}\binom{N+2q-2j+1}{\frac{N+1}{2}+q-j}^{-1}\right)\\ &=\frac{1}{2^N}\binom{N+2q+1}{\frac{N+1}{2}+q}\sum_{i=0}^q\prod_{j=0}^{i-1} \frac{(j-q)\left(j-q-\frac{1}{2}\right)\left(j-q-\frac{N+1}{2}\right)}{(j+1)(j-2q-1)\left(j-q-\frac{N}{2}\right)}\\ &=\frac{1}{2^N}\binom{N+2q+1}{\frac{N+1}{2}+q}\sum_{i=0}^{q}\frac{(-q)_i\left(-q-\frac{1}{2}\right)_i\left(-q-\frac{N+1}{2}\right)_i}{(-2q-1)_i\left(-q-\frac{N}{2}\right)_i}\frac{1}{i!}\\ &\,\,\color{blue}{= \frac{1}{2^N}\binom{N+2q+1}{\frac{N+1}{2}+q}\ {_3F_2} (-q,-q-\frac{1}{2},-q-\frac{N+1}{2};-2q-1,-q-\frac{N}{2};1)}\tag{1} \end{align*}

Hint: Section 4.3 Relations between $${_3F_2}(1)$$ Series in Generalized Hypergeometric Functions might be helpful to relate (1) with the series representation provided by @C.Koca.

• Apparently the result was not unique? Even this is a very valuable information, thank you! Commented May 16, 2023 at 9:50
• @C.Koca: You're welcome. Thanks for accepting my answer. I've added a hint which might be helpful. Commented May 16, 2023 at 10:26
• No problem at all. I composed an answer because I needed it for my thesis. You composed an answer to help other people out. Commented May 16, 2023 at 11:30

Here we follow OPs approach, but do not use Legendres duplication formula \begin{align*} \Gamma(2z)=\frac{2^{2z-1}}{\sqrt{\pi}}\Gamma(z)\Gamma\left(z+\frac{1}{2}\right)\tag{1} \end{align*} from which \begin{align*} \binom{2z}{z} = \frac{2^{2z}}{\sqrt{\pi}}\frac{\Gamma(z+\frac{1}{2})}{\Gamma(z+1)}\tag{2} \end{align*} follows. Instead we simply use \begin{align*} \binom{2z}{z}=\frac{(2z)!}{z!z!} \end{align*} and check which hypergeometric function representation we get this way.

We obtain \begin{align*} \color{blue}{\frac{1}{2^N}} &\color{blue}{\sum_{i=0}^q(-1)^i\binom{2q+1-i}{i}\binom{N+2q-2i+1}{\frac{N+1}{2}+q-i}}\\ &=\frac{(-1)^q}{2^N}\sum_{i=0}^q(-1)^i\binom{q+1+i}{q-i}\binom{N+2i+1}{\frac{N+1}{2}+i}\tag{3.1}\\ &=\frac{(-1)^q}{2^N}\sum_{i=0}^q(-1)^i\frac{(q+1+i)!}{(q-i)!(2i+1)!} \,\frac{(N+2i+1)!}{\left(\frac{N+1}{2}+i\right)!\left(\frac{N+1}{2}+i\right)!}\tag{3.2}\\ &=\frac{(-1)^q}{2^N}\sum_{i=0}^q(-1)^i\frac{(-1)^i(q+1)(-q)_i(q+2)_i}{2^{2i}(1)_i\left(\frac{3}{2}\right)_i}\\ &\qquad\qquad\qquad\qquad\cdot\frac{(N+1)!2^{2i}\left(\frac{N}{2}+1\right)_i\left(\frac{N}{2}+\frac{3}{2}\right)_i} {\left(\frac{N+1}{2}\right)!\left(\frac{N}{2}+\frac{3}{2}\right)_i\left(\frac{N+1}{2}\right)!\left(\frac{N}{2}+\frac{3}{2}\right)_i}\tag{3.3}\\ &\color{blue}{=\frac{(-1)^q(q+1)}{2^N}\binom{N+1}{\frac{N+1}{2}}\ {_3F_2} (-q,q+2,\frac{N}{2}+1;\frac{3}{2},\frac{N}{2}+\frac{3}{2};1)}\tag{3.4} \end{align*}

We observe we get precisely the same representation as OP in his answer by noting that using the duplication formula in the form (2) we have \begin{align*} \color{blue}{\binom{N+1}{\frac{N+1}{2}}=\frac{2^N\Gamma\left(\frac{N}{2}+1\right)} {\sqrt{\pi}\Gamma\left(\frac{N}{2}+\frac{3}{2}\right)}} \end{align*}

Comment:

• In (3.1) we change the order of summation $$i\to q-i$$.

• In (3.2) we write the binomial coefficients using factorials.

• In (3.3) we transform the factorials using rising factorials $$(a)_{k}:=a(a+1)\cdots(a+k-1)$$. We have \begin{align*} (N+2i+1)!&=(N+1)!\prod_{j=0}^{2i-1}\left(N+j+2\right)\\ &=(N+1)!\prod_{j=0}^{i-1}\left(N+2j+2\right)\prod_{k=0}^{i-1}\left(N+2k+3\right)\\ &=(N+1)!2^{2i}\left(\frac{N}{2}+1\right)_{i}\left(\frac{N}{2}+\frac{3}{2}\right)_i\\ \left(\frac{N+1}{2}+i\right)!&=\left(\frac{N+1}{2}\right)!\prod_{j=1}^i\left(\frac{N+1}{2}+j\right)\\ &=\left(\frac{N+1}{2}\right)!\left(\frac{N}{2}+\frac{3}{2}\right)_i\\ (2i+1)!&=\prod_{j=2}^{2i+1}j=\prod_{j=1}^i\left(2j\right)\prod_{j=1}^i(2j+1)\\ &=\prod_{j=0}^{i-1}\left(2j+2\right)\prod_{k=0}^{i=1}(2k+3)\\ &=2^{2i}(1)_i\left(\frac{3}{2}\right)_i\\ \frac{\left(q+1+i\right)!}{\left(q-i\right)!} &=\prod_{j=0}^{i-1}\left(q-i+j+1\right)(q+1)\prod_{k=0}^{i-1}\left(q+k+2\right)\\ &=\prod_{j=0}^{i-1}\left(q-j\right)\left(q+1\right)\left(q+2\right)_i\\ &=(-1)^i\prod_{j=0}^{i-1}(j-q)(q+1)(q+2)_i\\ &=(-1)^i\left(q+1\right)\left(-q\right)_i\left(q+2)\right)_i \end{align*}