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I am at my wit's end trying to show the following lower bound:

$${}_2F_1[-m, -m; -(m+l); z]\geq (1-z)^m,\tag{1}$$

where ${}_2F_1[a,b;c;z]$ is the Gauss hypergeometric function, $m,l=0,1,2,\ldots$, and $0<z<1$. Numerical experiments seem to confirm that this bound holds, but I cannot figure out how to prove it. Any help?

What I tried

When argument $c$ in ${}_2F_1[a,b;c;z]$ is a negative integer ${}_2F_1[a,b;c;z]$ is usually undefined, however, per discussion on DLMF, here we can express the LHS as

$${}_2F_1[-m, -m; -(m+l); z]=\sum_{n=0}^m (-1)^n\binom{m}{n}\frac{(-m)_n}{(-(m+l))_n}z^n=\sum_{n=0}^m \binom{m}{n}\frac{\binom{m}{n}}{\binom{m+l}{n}}(-z)^n,$$ where $(-m)_n=\left\{\begin{array}{rl}\frac{(-1)^nm!}{(m-n)!},&0\leq n\leq m \\ 0, &n>m\end{array}\right.$ is the Pochhammer's symbol for $m$ and $n$ nonnegative integers.

To prove the bound above, I originally attempted to show that

$$\binom{m}{n}\frac{\binom{m}{n}}{\binom{m+l}{n}}z^n-\binom{m}{n+1}\frac{\binom{m}{n+1}}{\binom{m+l}{n+1}}z^{n+1}\geq\binom{m}{n}z^n-\binom{m}{n+1}z^{n+1}\tag{2}$$

for $m,l,n=0,1,2,\ldots$ and $0<z<1$.
Edit: @VarunVejalla pointed to a counterexample $m,l,n=2,2,1$ in a comment, showing that (2) doesn't hold. However, this doesn't rule out the truth of (1).

Any ideas how to prove (1), or a counterexample to it, are appreciated!

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  • $\begingroup$ For $m, l, n = 2, 2, 1$, it doesn't seem that the last inequality holds up. $\endgroup$ Sep 26, 2020 at 21:59
  • $\begingroup$ @VarunVejalla You are absolutely right -- thank you for pointing this out! The path through showing the last inequality is out and I will modify my question accordingly. I still think that the lower bound on ${}_2F_1$ that I'm trying to prove holds, but it might a very difficult problem. $\endgroup$
    – M.B.M.
    Sep 26, 2020 at 22:16
  • $\begingroup$ $(1)$ appears to be true for $l=0$, so maybe try proof by induction? $\endgroup$
    – Andy Walls
    Sep 26, 2020 at 23:14
  • $\begingroup$ @M.B.M. Just curious: in what problem/context did this hypergeometric function arise? $\endgroup$
    – Andy Walls
    Sep 27, 2020 at 23:10
  • $\begingroup$ @AndyWalls This was originally for a large network science proof. However, later I saw a mistake in formulation, correcting which obliterated the ${}_2F_1$-functions. But I kept thinking about the neat lower bound I saw while playing with the incorrect formulation in Mathematica... hence this question. $\endgroup$
    – M.B.M.
    Sep 27, 2020 at 23:22

1 Answer 1

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Proof by induction:

Inequality $(1)$ holds for $l = 0$

$$\begin{align*}{}_2F_1[-m, -m; -(m+0); z]&=\sum_{n=0}^m (-1)^n\binom{m}{n}\frac{(-m)_n}{(-(m+0))_n}z^n\\ \\ &=\sum_{n=0}^m \binom{m}{n}\left(-z\right)^n\\ \\ &=(1-z)^m \quad \text{(binomial theorem)}\\ \\ \therefore \;{}_2F_1[-m, -m; -(m+0); z]&\ge (1-z)^m \\ \end{align*}$$

Assume inequality $(1)$ holds for $l-1$, namely:

$${}_2F_1[-m, -m; -(m+l-1); z]\ge (1-z)^m$$

So now

$$\begin{align*} {}_2F_1[-m, -m; -(m+l); z] &=\sum_{n=0}^m (-1)^n\binom{m}{n}\frac{(-m)_n}{(-(m+l))_n}z^n\\ \\ &= \sum_{n=0}^m (-1)^n\binom{m}{n}\frac{(-m)_n}{\left(\frac{(-1)^n(m+l)!}{(m+l-n)!}\right)}z^n\\ \\ &= \sum_{n=0}^m (-1)^n\binom{m}{n}\frac{(-m)_n}{\frac{(-1)^n(m+l-1)!}{(m+l-1-n)!}\cdot\frac{m+l}{m+l-n}}z^n\\ \\ &= \sum_{n=0}^m \dfrac{m+l-n}{m+l}(-1)^n\binom{m}{n}\frac{(-m)_n}{(-(m+l-1))_n}z^n\\ \\ &= {}_2F_1[-m, -m; -(m+l-1); z] \\ &\quad + \dfrac{1}{m+l}\sum_{n=0}^m n(-1)^{n+1}\binom{m}{n}\frac{(-m)_n}{(-(m+l-1))_n}z^n \\ \\ &= {}_2F_1[-m, -m; -(m+l-1); z] \\ &\quad - \dfrac{z}{m+l}\sum_{n=1}^m \frac{(-m)_n(-m)_n}{(-(m+l-1))_n}\dfrac{z^{n-1}}{(n-1)!} \\ \\ &= {}_2F_1[-m, -m; -(m+l-1); z] \\ &\quad - \dfrac{z}{m+l}\sum_{k=0}^{m-1} \frac{(-m)_{k+1}(-m)_{k+1}}{(-(m+l-1))_{k+1}}\dfrac{z^{k}}{k!} \\ \\ &= {}_2F_1[-m, -m; -(m+l-1); z] \\ &\quad + \dfrac{m^2 z}{(m+l)(m+l-1)}\sum_{k=0}^{m-1} \frac{(-(m-1))_{k}(-(m-1))_{k}}{(-(m-1+l-1))_{k}}\dfrac{z^{k}}{k!} \\ \\ &= {}_2F_1[-m, -m; -(m+l-1); z] \\ &\quad + \dfrac{m^2 z}{(m+l)(m+l-1)}{}_2F_1[-(m-1), -(m-1); -(m-1+l-1); z] \\ \\ {}_2F_1[-m, -m; -(m+l); z] &\ge (1-z)^m + \dfrac{m^2 z}{(m+l)(m+l-1)}(1-z)^{m-1}\\ \\ \therefore \; {}_2F_1[-m, -m; -(m+l); z] &\ge (1-z)^m\\ \end{align*}$$

Whew!

Please check all the steps. I make no guarantees that the above is free from errors. :)

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  • $\begingroup$ I think that you almost got it! To complete the proof, note that one can write the sum in your last line is: $\dfrac{1}{m+l}\sum_{n=0}^m n(-1)^{n+1}\binom{m}{n}\frac{(-m)_n}{(-(m+l-1))_n}z^n=\dfrac{m^2}{(m+l)(m+l-1)}\sum_{n=0}^m (-1)^{n-1}\binom{m-1}{n-1}\frac{(-(m-1))_{n-1}}{(-(m+l-2))_{n-1}}zz^{n-1}=\dfrac{zm^2}{(m+l)(m+l-1)}\sum_{n=0}^{m-1} (-1)^{n}\binom{m-1}{n}\frac{(-(m-1))_{n}}{(-(m+l-2))_{n}}z^{n}=\dfrac{zm^2}{(m+l)(m+l-1)}{}_2F_1[-(m-1),-(m-1);-(m+l-2);z]\geq\dfrac{zm^2}{(m+l)(m+l-1)}(1-z)^{m-1}\geq 0$, which yields the proof! If this checks out, feel free to update and I'll accept. $\endgroup$
    – M.B.M.
    Sep 27, 2020 at 19:57
  • $\begingroup$ Also, there appears to be extraneous $(-1)^n$ in the denominator in the fraction of your second and third equality. You get rid of it in the fourth equality. $\endgroup$
    – M.B.M.
    Sep 27, 2020 at 20:06
  • $\begingroup$ @M.B.M. Yeah, I just had independently worked out that last sum exactly as you have written in your comment. I came here to update the answer. :) $\endgroup$
    – Andy Walls
    Sep 27, 2020 at 22:27
  • $\begingroup$ @M.B.M. The $(-1)^n$ in that denominator fraction came from the expansion(?) of the Pochhammer of a negative quantity. I think it's correct. $\endgroup$
    – Andy Walls
    Sep 27, 2020 at 22:30

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