# Ratio of hypergeometric function

Given $$a>b>0$$, is there any upper bound of the following ratio of hypergeometric function? $$\frac{_2F_1(a,1-b;a+1;x)}{_2F_1(a,1-b;a+1;y)}$$ for $$1>x>y>0$$ ideally in the form like some powers of $$x/y$$

## 1 Answer

If $$b$$ is not an integer, this ratio is bounded by a constant.

Indeed, Remark 1.2.4 pp 34-35 of "From Gauss to Painlevé" by Iwasaki-Kimura-Shimomura-Yoshida can be applied to your case, in which $$\gamma-\alpha-\beta=b$$. It yields that the function $$f={}_2F_1(\alpha,\beta;\gamma,\cdot)$$ you consider is continuous on [0,1]. the exact value at $$1$$ is given on page 73 of the same book at as $$\frac{\Gamma(a+1)\Gamma(b)}{\Gamma(1)\Gamma(a+b)}$$, in particular it is not $$0$$. As f(0)=1, using the local uniqueness of solutions in $$(0,1)$$, one may conclude the existence of $$K>0$$ such that $$f(x)>K$$, for every $$x\in[0,1]$$.

Then the function $$F:[0,1]\times [0,1]\rightarrow \mathbb R, (x,y)\mapsto f(x)/f(y)$$ is continuous on a compact, whence the conclusion.

Edit: Actually, the derivative $$f'$$ is $$\left(\frac{a+1}{a(1-b)}\right){}_2F_1(\alpha+1,\beta+1;\gamma+1,\cdot)$$ and must have the sign of $$\frac{a+1}{a(1-b)}$$. So an explicit upper bound can be given:

• if $$1-b>0$$ then $$F$$ is bounded by $$f(1)/f(0)=\frac{\Gamma(a+1)\Gamma(b)}{\Gamma(1)\Gamma(a+b)}$$,

• if $$1-b<0$$ then $$F$$ is bounded by $$f(0)/f(1)=\left(\frac{\Gamma(a+1)\Gamma(b)}{\Gamma(1)\Gamma(a+b)}\right)^{-1}$$.

• If you read the details in the mentioned book, I think you can remove the condition of b integral, replacing the remark by the preceding Lemma. For the quantitative part, if $b=1$, then $f$ is constant, so the optimal bound is $1$. – gcousin May 7 '19 at 3:40