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.

Sorry for this uninteresting question but hopefully someone can provide some help.

Is there a way to simplify the following expression?

$$\binom{n}{m}\sum_{\nu=0}^{n-m}(-1)^{\nu}\binom{n-m}{v}\left(\frac{n-m-v}{n}\right)^{r}\displaystyle \frac{m}{m+v}$$

$$-\binom{n}{m}\sum_{\nu=0}^{n-m}(-1)^{\nu}\binom{n-m}{v}\left(\frac{n-m-v}{n}\right)^{r}$$

This is a hint for a problem, but I don't know how to proceed.

UPDATE: Would it bother you if I add the original problem?. Maybe some context is needed in order to solve this problem by using this hint.

share|improve this question
1  
Perhaps, you can add a new question with the original problem and link that to this one. You might get an answer which does not use this hint at all... –  Aryabhata Dec 28 '10 at 1:31
    
@Moron: Ok, I'll do that. –  Robert Smith Dec 28 '10 at 2:34
add comment

2 Answers 2

up vote 1 down vote accepted

I doubt it, the case $m=0$ would give a simpler formula for Stirling numbers of the second kind.

$$n!\ S(r,n) = \sum_{v = 0}^{n} (-1)^{v} {n \choose v} (n-v)^{r}$$

Perhaps you can write your expression in terms of these numbers...

share|improve this answer
    
Yes, I tried that but the first expression is not really a $S(r,n)$ because the coefficient $\frac{m}{m+v}$ changes with the sum. –  Robert Smith Dec 27 '10 at 21:53
    
@Rob: I said the case $m=0$, which leaves us with the second expression. Even in special case of $m=0$, you get $S(r,n)$, so with $m \neq 0$, it will certainly be more difficult! –  Aryabhata Dec 27 '10 at 21:57
    
@Rob: In case you were responding to the last sentence of the answer...I would suggest looking that the proofs of the identity with $S(n,r)$ I gave. Perhaps you can modify one of those to cater to your $m/(m+v)$. –  Aryabhata Dec 27 '10 at 22:31
    
Uhm, yes, I was looking for an identity but didn't find any. I suppose this shouldn't be relying in some obscure identity (if there is any which fits the requirements). –  Robert Smith Dec 27 '10 at 23:03
add comment

At a quick glance, it looks like there are some common factors to the two sums and that factoring is likely to help in simplifying. Also, since the summations have the same indexing and range, they could be combined into a single summation.

share|improve this answer
    
It looks like you can use $\frac{m}{m+v}-1=\frac{-v}{m+v}$ to combine the terms –  Ross Millikan Dec 27 '10 at 21:23
    
@Ross: That's a good idea, but what can I do with term with $\frac{v}{m+v}$? Supposedly, I should be able to find a single expression. –  Robert Smith Dec 27 '10 at 21:55
    
That gets you to a single sum. You can take out the v against the $\binom{n-m}{v}$ but that doesn't seem to help. If I could go farther, I would have posted an answer instead of a comment. –  Ross Millikan Dec 27 '10 at 22:12
    
@Ross: Right. It could have been helpful, but not for this particular problem. –  Robert Smith Dec 27 '10 at 22:51
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.