Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider the classical hypergeometric function $F(5/4,3/4; 2, z)$ for $z\in (0,1]$. Is this bounded by some real number (independent of $z$)?

I'm aware of Euler's formula:

$$F(5/4,3/4; 2, z) = \frac{1}{\Gamma(5/4)\Gamma(3/4)}\int_0^1 t^{-1/4} (1-t)^1/4 (1-tz)^{-5/4} dt.$$

The best I can do using this formula is $$F(5/4, 3/4; 2, z) \leq \frac{4}{ 3\Gamma(5/4) \Gamma(3/4)}\frac{1}{1-z}. $$ This is not good though, because it blows up as $z$ tends to $1$.

Any suggestions?

Maybe it is easy to show that $F$ is strictly increasing on $(0,1]$ and continuous on $\mathbf{R}$. Then, we simply need to estimate $F(5/4,3/4;2,1)$.

share|cite|improve this question
«uniform boundedness» is a property of sets of functions. A set consisting of one function is uniformly bounded iff that function is bounded, so one never uses the «uniform» in that situation. – Mariano Suárez-Alvarez Mar 10 '12 at 18:17
I didn't use "uniform" did I? Or are you referring to the "independent of $z$" within parentheses? – seporhau Mar 10 '12 at 18:22
Well, you wrote the title :) – Mariano Suárez-Alvarez Mar 10 '12 at 18:35
haha. you're right. :) – seporhau Mar 10 '12 at 19:09
up vote 0 down vote accepted

It is defined for $z \in [-1,1)$, but it diverges to $+\infty$ as $z \to 1^-$.


All terms in the power series are nonnegative, so the function is increasing on $[0,1)$. Actually it is increasing on $[-1,1)$. I get numerically $F(5/4,3/4;2;1/4) \approx 1.1409$ as the bound for $[0,1/4]$.

share|cite|improve this answer
Ok thnx. What if we stick to $z\in [0,1/4]$. Is $F$ strictly increasing on this interval? If yes, then I can try to approximate its value at $1/4$ to get an upper bound for $F$ on this interval. – seporhau Mar 10 '12 at 18:02
That's all I need. Thanks. – seporhau Mar 10 '12 at 19:54

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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