difficult series

[Feynman]

Good morning ,
I have to calculate these series , Help me

\displaystyle\sum_{k=1}^{i-1}C_{i}^{k}(tk)^{k-2}k(i-k)^{i-k+1}
Thanks

[Wong]

Are you sure it is C_{i}^{k}? When k is smaller than i, I think C_{i}^{k} is 0?

[Feynman]

YA I'm sure
so???????????

[Zurtex]

I don't really understand the maths involved here, but I think the point was if C_{i}^{k} = 0 then your sum is 0 + 0 + 0 + ... + 0.

[Feynman]

u write it is C_{k}^{i}

[Feynman]

So how calculate these serieS?

[Feynman]

Please help me how we calculate these SERIES??

[Wong]

Maybe you should try to use Maple for this particular problem? Some guys told me that Maple can handle this kind of symbolic computations.

[Feynman]

Good morning
I'm must proof the calculation step of this series and not the result from maple!!!!

[Wong]

Now, your series contains a variable t. Surely if the sum makes sense, it should sum to a polynomial. So, the problem is what do you want to prove? The problem makes sense if the proposition is like "prove that the series sums to a polynomial in t with coefficients of the form..." or like "prove that the series sums to the nth derivative of ... function".

So what exactly is the proposition?

[Feynman]

I need the steps calculation of these series!!

[Wong]

sorry then....maybe I can't help, as I am unclear about what you want to prove. The expression is a polynomial in t. It may be "simplified" into any form that you may deem suitable.

[Feynman]

Can u calculate me this series?

[Feynman]

So Wong can u help me?

[Wong]

Sorry, maybe I can't help....

[Zurtex]

Have you tried expanding it out and trying the first few terms? I'm just doing that now.

But it does look complex, I get the 1st 2 terms as being (i/t)(i - 1)^i and i(i-1)(i-2)^{i-1}

Perhaps looking at this wrong, it may be more easy for trying different values of i.

[pnaj]

Write out, on paper, the first 5 terms, the kth term and the last two terms. See if you can find a pattern. For example, the first term MIGHT pair with the last term to equal zero, the second might pair with the second to last, etc.

If that doesn't work, try induction on k, see how far you get.

At least have a go!

EDIT: Zurtex ... I didn't notice your post, I'm just pretty much repeating what you said.

[Zurtex]

Edit: It's all wrong, sorry.

[Zurtex]

Edit: Sorry I tried reducing the seris in a way that was just plain wrong. However it does occur to me that you can take out i!/t^2 as a common factor.

[Feynman]

Zurtex
THe i is not a complex it is the index!!!!!!!!!!!!!!!!!!!!

[Zurtex]

Zurtex
THe i is not a complex it is the index!!!!!!!!!!!!!!!!!!!!
I know, however it is just a constant and thus is common to the whole sum, I'm trying to help you please do not talk to me like i'm an idoit. I've been working on this a bit more and I don't see how it can be reduced.

Whatever value of i you choose you get a polynomial in t, where the coefficient of each power of t is dependent on both on the power t and the value of i. Such that the final form is:

\sum_{k=1}^{i-1} a_{k,i}t^{k-2}

Well the sum you have is:

\sum_{k=1}^{i-1} \frac{i!k^k(i-k)^{i+1}}{(i-k)!k!k(i-k)^k}t^{k-2}

This is of the same form :uhh:. But I would like to point out you can take i!/t^2 out of it like so:

\frac{i!}{t^2}\sum_{k=1}^{i-1} \frac{k^k(i-k)^{i+1}}{(i-k)!k!k(i-k)^k}t^k

[Feynman]

Excuse Me Zurtex But How we can continue ?

[Zurtex]

Excuse Me Zurtex But How we can continue ?
Sorry I don't understand your question, but choose a value for i and work it out.

[Feynman]

Zurtex , I need the exact value of my series for every i
So what should i do?

[Zurtex]

Well I wrote a little program and this is what I got:

For i = 2: 2t^{-1}

For i = 3: 24t^{-1}+6

For i = 4: 324t^{-1}+96+36t

For i = 5: 5120t^{-1}+1620+720t+320t^2

For i = 6: 93750t^{-1}+30720+14580t+7680t^2+3750t^3

For i = 7: 1959552t^{-1}+656250+322560t+181440t^2+105000t^3+54432t^4

I can give you up to 12 if you want, if you can spot a pattern and prove it inductively good luck.

Edit: I'm not 100% confident in my calculations, I'll check them over later.

[Feynman]

So Zurtex I need the exact value of these series

[Zurtex]

So Zurtex I need the exact value of these series
Sorry, your going to have to explain yourself better than that. What do you mean by "exact value"?

[matt grime]

I suspect that he means something like:

consider the series
\sum_{r=0}^n \binom{n}{r}t^r

then its "exact value" is

(1+t)^n

it might help, feynman, if you told us where these inexplicable sums come from, because that might help us decide what the generating functions are.
You could also stop being quite so insistently demanding, Zurtex is doing you a favour here, and these are far from elementary or obvious sums.

[Feynman]

Hello Everybody
My complex sum come from resolving this problem:
I have to proove that the exact solution of Smoluchowski coagulation equation wich is
\ds \frac{\partial c}{\partial
t}(i,t)=\frac{1}{2}\sum_{j=1}^{i-1}K(i-j,j)c(i-j,t)c(j,t)-\sum_{j=1}^{\infty}
K(i,j)c(i,t)c(j,t)

With this condition :
c(1,0):=1 \quad et \quad \forall i>1,\quad c(i,0):=0.
And K(i,j)=i+j
Is e^{-t} B(1-e^{-t},x)
and B(t,x)=\frac{(tx)^{x-1}e^{-tx}}{x!}

[Zurtex]

Excuse me I need your help ,
So My real name is Toufic and you?
You are supposed to politely request help on a forum not demand it. Think what you are typing and imagine how someone would read it, so far you have been generally rude to people who you want to spend their free time to help you.

[pnaj]

Feynmann,

Let us try to solve this problem ... when you say 'us', do you mean it?

You've told us what the problem is ... now show us how you have have tried to solve it and what went wrong.

pnaj.

P.S. I don't think you're going to get help unless you are polite. That might mean learning a little more English.

[Feynman]

so no body need to help me please?

[Zurtex]

so no body need to help me please?
No one can help unless you show us what you have done, I don't think you will find anyone on this forum who will just solve a problem for you as that doesn't help.

[Feynman]

Zurtex I help evrybody on this forum so>?.............

[HallsofIvy]

So you should know better! What have you attempted on this problem so far?

[Feynman]

ok thak you fot all

[Feynman]

So.................. ?????????
Who know how to calculate this dificult series??????????
I nead to solve it.........
soooooooooo????????
any one help .