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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 $F(5/4,3/4;2,1)$.

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«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
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1 Answer

up vote 0 down vote accepted

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

added

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 $[0,1/4]$.

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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
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