Take the 2-minute tour ×
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.

I am trying to work these out: $$\sum\limits_{u=0}^{k}{\left( \begin{matrix} r+u \\ u \\ \end{matrix} \right){{\left( {\alpha} \right)}^{u}}{{\left( {1-\alpha} \right)}^{r+1}}}$$ $$\sum\limits_{u=0}^{k}{\left( \begin{matrix} r+u \\ u \\ \end{matrix} \right){{\left( \alpha \right)}^{u}}{{\left( 1-\alpha \right)}^{r+1}}\sum\limits_{m=u+r+1}^{\infty }{\frac{{{e}^{-\lambda }}{{\lambda }^{m}}}{m!}}}$$

Thanks!

share|improve this question
    
You can have a closed form for the sum in terms of the hypergeometric function. –  Mhenni Benghorbal Feb 16 '13 at 0:49
    
What is $k$? If "sufficiently large" (so that you can live with $k \rightarrow \infty$), then $\binom{-r - 1}{u} = (-1)^u \binom{r + u}{u}$, and you have a good approximation for the first sum if $\lvert \alpha \rvert$ is small. –  vonbrand Feb 16 '13 at 3:17
    
Thanks @ Mhenni Benghorbal. I found that also I can use regularized beta function too. –  Eln Feb 16 '13 at 4:09
    
Thanks @ vonbrand. I can definitely use that for approximation, in some case k could be quite large. For the second integral, I was trying to change the order of summations, see if that helps. –  Eln Feb 16 '13 at 4:23

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.