Express Hypergeometric series as polynomial

Question

How can I prove the following identity between a hypergeometric series and a polynomial (it has been changed meanwhile) $$ \frac{(2k-2)!}{(k-1)!}{}_2F_1(1-2h,1-k,2-2k;2)=(-4)^{k-1}\prod_{j=1}^{k-1}(h-j)=(-4)^{k-1}\frac{(h)_k}{h}? $$ If it can help, they both equal the sum $$ -\frac{1}{4h}\sum_{j=1}^{\min(k,2h)}(-2)^{j}\frac{(2k-j-1)!(2h)!}{(j-1)!(k-j)!(2h-j)!}.$$ Also an advice on what tools can be helpful for this goal can be enough. If needed, I can provide some additional context.
Thank you in advance for any help.

Answer 1

Here we prove OPs identity starting with the summation formula. We show the following is valid for $1\leq k\leq h$: \begin{align*} \color{blue}{-\frac{1}{4h}\sum_{j=1}^{k}(-2)^{j}\frac{(2k-j-1)!(2h)!}{(j-1)!(k-j)!(2h-j)!}=(-4)^{k-1}\prod_{j=1}^{k-1}(h-j)}\tag{1} \end{align*} The product on the right-hand side of (1) indicates the range $k\leq h$, since otherwise the product is zero. So we take $k$ as upper limit of the sum.

The proof that follows is a three-step approach.

• At first we transform (1) into a more convenient binomial identity.

• Then we use generating functions to further transform the identity.

• Finally we apply some bit of magic called Lagrange inversion.

Binomial identity:

We start with the left-hand side of (1) and obtain \begin{align*} \color{blue}{-\frac{1}{4h}}&\color{blue}{\sum_{j=1}^{k}(-2)^{j}\frac{(2k-j-1)!(2h)!}{(j-1)!(k-j)!(2h-j)!}}\\ &=-\frac{1}{2}\sum_{j=1}^{k}(-2)^j\binom{2h-1}{j-1}\frac{(2k-j-1)!}{(k-j)!}\tag{2.1}\\ &=-\frac{1}{2}(k-1)!\sum_{j=1}^{k}(-2)^j\binom{2h-1}{j-1}\binom{2k-j-1}{k-j}\tag{2.2}\\ &\,\,\color{blue}{=(k-1)!\sum_{j=0}^{k-1}(-2)^j\binom{2h-1}{j}\binom{2k-j-2}{k-j-1}}\tag{2.3} \end{align*}

Comment:

• In (2.1) we cancel the factor $2h$ and write factorials as binomial coefficient.

• In (2.2) we also write the other factorials as binomial coefficient.

• In (2.3) we shift the index to start with $j=0$.

The product at the right-hand side of (1) can be written as \begin{align*} \color{blue}{(-4)^{k-1}\prod_{j=1}^{k-1}(h-j)}&=(-4)^{k-1}\frac{(k-1)!}{(h-k)!}\\ &\,\,\color{blue}{=(-4)^{k-1}(k-1)!\binom{h-1}{k-1}}\tag{2.4} \end{align*}

We conclude from (2.3) and (2.4) that proving the identity (1) is equivalent with proving \begin{align*} \sum_{j=0}^{k-1}(-2)^j\binom{2h-1}{j}\binom{2k-j-2}{k-j-1}=(-4)^{k-1}\binom{h-1}{k-1}\qquad 1\leq k\leq h\tag{3} \end{align*}

We get a slightly more convenient representation by shifting $h\to h+1$ and $k\to k+1$ by one. We will therefore show \begin{align*} \color{blue}{\sum_{j=0}^{k}\binom{2h+1}{j}\binom{2k-j}{k-j}(-2)^j=(-4)^{k}\binom{h}{k}\qquad 0\leq k\leq h}\tag{4} \end{align*}

Generating functions:

In the following we use the coefficient of operator $[z^k]$ to denote the coefficient of $z^k$ in a series. This way we can write for instance \begin{align*} \binom{n}{k}=[z^k](1+z)^n\tag{5.1} \end{align*}

We obtain from (4) \begin{align*} \color{blue}{\sum_{j=0}^{k}}&\color{blue}{\binom{2h+1}{j}\binom{2k-j}{k-j}(-2)^j}\\ &=(-1)^k\sum_{j=0}^k\binom{2h+1}{j}\binom{-k-1}{k-j}2^j\tag{5.2}\\ &=(-2)^k\sum_{j=0}^{k}\binom{2h+1}{k-j}\binom{-k-1}{j}\frac{1}{2^j}\tag{5.3}\\ &=(-2)^k\sum_{j=0}^{k}\binom{-k-1}{j}[z^{k-j}](1+z)^{2h+1}\frac{1}{2^j}\tag{5.4}\\ &=(-2)^k[z^k](1+z)^{2h+1}\sum_{j\geq 0}\binom{-k-1}{j}\left(\frac{z}{2}\right)^j\tag{5.5}\\ &\,\,\color{blue}{=(-2)^k[z^k]\frac{(1+z)^{2h+1}}{\left(1+\frac{z}{2}\right)^{k+1}}}\tag{5.6} \end{align*}

Comment:

• In (5.2) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

• In (5.3) we change the order of summation $j\to k-j$.

• In (5.4) we apply the coefficient of operator according to (5.1).

• In (5.5) we do some rearrangements and use the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$. We also extend the upper limit of the series to $\infty$ which has no effect since powers of $z$ greater than $k$ are skipped thanks to the $[z^k]$ operator.

• In (5.6) we apply the binomial series expansion.

Lagrange inversion:

Now it's time for a bit of magic. We follow the paper Lagrange Inversion: when and how by R. Sprugnoli, consider the generating function \begin{align*} F(u)=\sum_{k=0}^{\infty}a_ku^k =\sum_{k=0}^\infty[z^k]A(z)\phi(z)^ku^k \end{align*} and apply formula (G6) from the paper. This formula tells us that if there are functions $A(z)$ and $\phi(z)$, so that the coefficient $a_k$ admits a representation \begin{align*} a_k=[z^k]A(z)\phi(z)^k \end{align*} then the following is valid: \begin{align*} F(u)&=\sum_{k=0}^{\infty}[z^k]A(z)\phi(z)^ku^k\tag{6.1}\\ &=\left.\frac{A(z)}{1-u\phi^{\prime}(z)}\right|_{z=u\phi(z)}\tag{6.2} \end{align*}

We obtain from (5.6) and (6.1) with $A(z)=\frac{(1+z)^{2h+1}}{1+\frac{z}{2}}$ and $\phi(z)=\left(1+\frac{z}{2}\right)^{-1}$ \begin{align*} \color{blue}{F(u)}&\color{blue}{=\sum_{k=0}^\infty [z^k]\frac{(1+z)^{2h+1}}{1+\frac{z}{2}}\left(1+\frac{z}{2}\right)^{-k} u^n}\tag{6.3}\\ &=\frac{(1+z)^{2h+1}}{1+\frac{z}{2}}\cdot\left.\frac{1}{1-u\frac{d}{dz}\left(\left(1+\frac{z}{2}\right)^{-1}\right)}\right|_{z=u\frac{1}{1+\frac{z}{2}}}\tag{6.4}\\ &=\left.\frac{(1+z)^{2h+1}}{1+\frac{z}{2}} \cdot\frac{1}{1+\frac{2z\left(1+\frac{z}{2}\right)}{(2+z)^2}}\right|_{z=u\frac{1}{1+\frac{z}{2}}}\tag{6.5}\\ &=\left.(1+z)^{2h}\right|_{z=u\frac{1}{1+\frac{z}{2}}}\tag{6.6}\\ &=\left(1+(-1+\sqrt{1+2u})\right)^{2h}\tag{6.7}\\ &\,\,\color{blue}{=(1+2u)^h}\tag{6.8} \end{align*}

Comment:

• In (6.3) we arrange the expression according to (6.1).

• In (6.4) we can now apply (6.2).

• In (6.5) we substitute $u=z\left(1+\frac{z}{2}\right)$, do the derivation and some simplifications.

• In (6.6) we do some more simplifications.

• In (6.7) we consider again $u=z\left(1+\frac{z}{2}\right)$ noting that $z=z(u)$ is a quadratic equation in $z$ resulting in \begin{align*} z_{1,2}=-1\pm\sqrt{1+2u} \end{align*} We take the solution $-1+\sqrt{1+2u}$ which is admissible since it can be expanded as a power series.

Finally, combininig (5.6) and (6.8) we obtain \begin{align*} \color{blue}{\sum_{j=0}^{k}}&\color{blue}{\binom{2h+1}{j}\binom{2k-j}{k-j}(-2)^j}\\ &=(-2)^k[z^k]\frac{(1+z)^{2h+1}}{\left(1+\frac{z}{2}\right)^{k+1}}\\ &=(-2)^k[u^k](1+2u)^h\\ &=(-2)^k\binom{h}{k}2^k\\ &\,\,\color{blue}{=(-4)^k\binom{h}{k}} \end{align*} and the claim (4) and so also the claim (1) follow.

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Addendum: We close the circle and derive also from OPs summation formula the hypergeometric function.

We obtain \begin{align*} \color{blue}{-\frac{1}{4h}}&\color{blue}{\sum_{j=1}^{k}(-2)^{j}\frac{(2k-j-1)!(2h)!}{(j-1)!(k-j)!(2h-j)!}}\\ &=(k-1)!\sum_{j=0}^{k-1}\underbrace{(-2)^j\binom{2h-1}{j}\binom{2k-j-2}{k-j-1}}_{=:t_j}\tag{$\to\ $(2.3)}\\ &=(k-1)!t_0\sum_{j=0}^{k-1}\prod_{q=0}^{j-1}\frac{t_{q+1}}{t_q}\\ &=\frac{(2k-2)!}{(k-1)!}\sum_{j=0}^{k-1}\prod_{q=0}^{j-1}(-2)^{q+1}\binom{2h-1}{q+1} \binom{2k-q-3}{k-q-2}\\ &\qquad\qquad\qquad\quad\cdot(-2)^{-q}\binom{2h-1}{q}^{-1}\binom{2k-q-2}{k-q-1}^{-1}\\ &=\frac{(2k-2)!}{(k-1)!}\sum_{j=0}^{k-1}2^j\prod_{q=0}^{j-1}\frac{(-2h+q+1)(-k+q+1)}{(q+1)(-2k+q+2)}\\ &=\frac{(2k-2)!}{(k-1)!}\sum_{j=0}^{k-1}\frac{(1-2h)_j(1-k)_j}{(2-2k)_j}\,\frac{2^j}{j!}\\ &\,\,\color{blue}{=\frac{(2k-2)!}{(k-1)!}{}_2F_1(1-2h,1-k,2-2k;2)} \end{align*} and OPs claim follows.

Answer 2

I'd like to use Mathematica

Clear["Global`*"];
expr = Hypergeometric2F1[1 - 2 h, 1 - k, 2 - 2 k, 
     2]*(2 k - 2)!/(k - 1)! - (-4)^(k - 1)*
    Product[(h - j), {j, 1, k - 1}];

Table[FullSimplify@expr, {k, 2, 100}]

$$ \{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0\} $$

Answer 3

The confusion you and I were discussing in comments comes down to the question of the meaning of ${}_2F_1(a,b,c;z)$ for certain kinds of parameters. In particular: if $c$ is not in $\mathbb{Z}_{\leq 0}$, then a hypergeometric series is unquestionably defined as a formal series, and converges in some appropriate neighborhood of $0$. If $c$ does belong to $\mathbb{Z}_{\leq 0}$ and neither $a$ nor $b$ is a nonpositive integer closer to $0$ than $c$ is, then ${}_2F_1(a,b,c;z)$ is unambiguously undefined (as either a formal or analytic object). When all of $a, b, c$ are nonpositive integers and $|c| \geq |a|, |b|$, then ${}_2F_1(a,b,c;z)$ is again unambiguously defined to be the polynomial $$ {}_2F_1(a,b,c;z) = \sum_{n = 0}^{\min(|a|, |b|)} \frac{(a)_n (b)_n}{(c)_n} \frac{z^n}{n!}. $$

However, when $a, c$ are nonpositive integers with $|a| \leq |c|$ but $b \not \in \mathbb{Z}_{\leq 0}$, there is a potential ambiguity in what the hypergeometric sum means: does it mean $\sum_{n = 0}^{|a|} \frac{(a)_n (b)_n}{(c)_n} \frac{z^n}{n!}$, or does it include also terms for $n > |c|$ with an appropriate cancellation between the $0$ in the numerator and the $0$ in the denominator? I am not sure what is "normal" in this area; my guess is that the "normal" thing to do would be to have the series continue, but that the identity you've asked about is valid (for $k \in \mathbb{Z}_{>0}$) under the definition $$ {}_2F_1(1-2h,1-k,2-2k;2)= \sum_{n = 0}^{k - 1} \frac{(1 - 2h)_n (1 - k)_n}{(2 - 2k)_n} \cdot \frac{2^n}{n!}, $$ which is transparently a polynomial in $h$ of degree $k - 1$. When $h$ is an integer with $0 < h < k$, we're in the case where both $a, b$ are nonpositive integers larger than $|c|$, everything is well behaved, and the fact you observed (that Pfaff says the sum is $0$) agrees with the problem: in this case, the right side also evaluates to $0$. (When $h$ is not an integer, your computation is not valid; when $h$ is an integer outside of that interval, I suspect the discrepancy comes down to going very carefully through the precise statement of the domain on which the Pfaff transformation is valid (especially places where people write things like "by analytic continuation" without going into details).) If you accept this Pfaff-based calculation for $h$ an integer between $1$ and $k - 1$, you're essentially done: you know you have a polynomial of degree $k - 1$, you've found its $k - 1$ roots, and so then you just need to evaluate at one additional point to get the leading coefficient; I recommend $h = \frac{1}{2}$ for this purpose.

Answer 4

By way of enrichment of the answer by @epi163sqrt observe that

$$(-1)^k 2^k [z^k] \frac{(1+z)^{2h+1}}{(1+z/2)^{k+1}} = (-1)^k 2^k \; \underset{z}{\mathrm{res}} \; \frac{1}{z^{k+1}} \frac{(1+z)^{2h+1}}{(1+z/2)^{k+1}}.$$

Now put $z(1+z/2) = w$ so that $z = -1 + \sqrt{1+2w}$ (taking the branch that maps zero to zero) and $dz = \frac{1}{\sqrt{1+2w}} \; dw$ to obtain

$$(-1)^k 2^k \; \underset{w}{\mathrm{res}} \; \frac{1}{w^{k+1}} \sqrt{1+2w}^{2h+1} \frac{1}{\sqrt{1+2w}} \\ = (-1)^k 2^k \; \underset{w}{\mathrm{res}} \; \frac{1}{w^{k+1}} (1+2w)^h = (-1)^k 2^k 2^k {h\choose k} = (-1)^k 2^{2k} {h\choose k}.$$