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.

As a follow up of my previous question here is an improved version of the series:

$$\sum_{i=1}^{2NR}{i\cdot \left( \dfrac{1}{1-p} \right)^i}\, \left(1-\dfrac{2R-(2NR-i)\Delta}{2R} \right) $$

where:

  • $p\in [0,1]$
  • $R\in [0,1]$
  • $N$ positive integer
  • $\Delta$ infinitesimal

Notice that for $i=2NR$ the last term of the series, that is actually a probability, is 0 and it is correct.

My question is: it is possible to get a closed form expression?

share|improve this question
add comment

1 Answer

up vote 2 down vote accepted

Note that a little algebra will transform your sum into $$\sum_{i=1}^{2NR}\left( Ki +\frac{\Delta}{2R}i^2\right)\left(\frac{1}{1-p}\right)^i,$$ where $K$ is an easily computed constant.

The answer to the previous post tells you how to find $\sum_1^{2NR} ix^i$. Multiply that sum by $K$. The only remaining task is to find $\sum_1^{2NR}i^2x^i$.

This is done using more or less the same idea as the preceding answer by Marvis. Let $f(x)=1+x+x^2+\cdots+x^n$. There is an explicit closed form for $f(x)$. Note that $$xf'(x)=\sum_{i=0}^n ix^{i}.$$ If $g(x)=xf'(x)$, then by the same trick $$xg'(x)= \sum_{i=0}^n i^2x^{i}.$$

share|improve this answer
    
Thanks: the only problem was exactly how to treat the probability term. :) –  Claudio Fiandrino May 22 '12 at 14:33
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.