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've asked similar question, but this is much more complicated, I think)

What is the sum of $\sum\limits_{i=1}^{n}i^k p^i$?

Interpretation (why is it important ) :
$f(k,n,p)=\sum\limits_{i=1}^{n}i^k p^i$
if n goes to plus infinity, then $f(1,n,1/2)$ is average length or series of heads while tipping symmetric coin, and $f(2,n,1/2)-f(1,n,1/2)^2$ is the variance of that length, so for another k we get k-th moment

share|improve this question
    
These quantities are well-studied when $p = 1,$ and there are inductive formulae for them. –  Geoff Robinson Aug 8 '12 at 10:36

2 Answers 2

up vote 4 down vote accepted

Well, starting again with $f(p)=\sum_{i=1}^n p^i=\frac {p(p^n-1)}{p-1}$

Consider the theta operator $\Theta=p\frac d{dp}$ then your answer is $\Theta^k(f(p))$
(compute the derivative, multiply by $p$, repeat $k$ times).

share|improve this answer

We use the follwing denotions

Recall that each power of a number can be expressed in terms of its lower factorials $$ i^k=\sum\limits_{r=0}^k S(k, r)(i)_r $$ So we can write $$ \begin{align} f(k,n,p)=\sum\limits_{i=1}^n i^k p^i&= \sum\limits_{i=1}^n \left(\sum\limits_{r=0}^k S(k, r)(i)_r\right) p^i\\ &=\sum\limits_{i=1}^n \sum\limits_{r=0}^k S(k, r)(i)_r p^i \\ &=\sum\limits_{i=1}^n \sum\limits_{r=0}^k S(k, r) p^r \frac{d^r}{dp^r}(p^i)\\ &=\sum\limits_{r=0}^k \sum\limits_{i=1}^n S(k, r) p^r \frac{d^r}{dp^r}(p^i)\\ &=\sum\limits_{r=0}^k S(k, r) p^r\frac{d^r}{dp^r}\left(\sum\limits_{i=1}^n p^i\right)\\ &=\sum\limits_{r=0}^k S(k, r) p^r\frac{d^r}{dp^r}\left(\frac{p^{n+1}-p}{p-1}\right) \end{align} $$ For each fixed $k$ one may compute this sum.

share|improve this answer

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.