Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

What is $I_{0.5}(a,b)$ where I is the regularized incomplete beta function?

share|improve this question
    
Could you please give us a reference to the definition of $I$? –  Robin Chapman May 6 '10 at 9:25
    
You can get your answer yourself from mathworld.wolfram.com/IncompleteBetaFunction.html. –  Jacques Carette May 6 '10 at 10:06
    
There's no $I_{0.5}$ there :-( –  Robin Chapman May 6 '10 at 10:40
    
@Neil: I'd join Robin and recommend you to decipher your $I_{0.5}(a,b)$ by giving an integral or series expression. Otherwise it sounds like you are not interested in getting an answer. –  Wadim Zudilin May 6 '10 at 12:07
    
sorry for the late reply. $I$ is defined here: mathworld.wolfram.com/RegularizedBetaFunction.html but there is no expression for $I_0.5$ –  Neil May 7 '10 at 2:10
add comment

1 Answer

up vote 0 down vote accepted

You mean this? http://en.wikipedia.org/wiki/Beta_function $$ \frac{\int_{0}^{\frac{1}{2}} t^{a - 1} (1 - t)^{b - 1} d t}{\int_{0}^{1} t^{a - 1} (1 - t)^{b - 1} d t} = \\ \quad{}\quad{}\frac{\mathrm{hypergeom} \Bigl([a,-b + 1],[1 + a],\frac{1}{2}\Bigr) \Gamma (a + b)}{2^{a} a \Gamma (b) \Gamma (a)} $$ Why do you think there is anything simpler in general?

share|improve this answer
    
Thanks, Gerald, for demystefying the piece! Yes, there is no reduction of the $_2F_1$ hypergeometric series for general $a$ and $b$. But there are other ways to write it hypergeometrically. –  Wadim Zudilin May 6 '10 at 13:35
    
I thought there would be something simpler because when a and b are natural numbers, the function yields a natural number. –  Neil May 7 '10 at 2:11
    
But, thank you for your answer... –  Neil May 7 '10 at 2:11
    
(...yields a rational number) –  Neil May 7 '10 at 7:39
add comment

Your Answer

 
discard

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.