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From: mike3 on 2 Dec 2009 15:00
Hi.

Is there a closed form for

sum_{k=0...n-1} k^p exp(qk)

with p, q nonnegative integers?

Like how Faulhaber's formula gives a closed form for sum_{k=0...n-1}
k^p? I tried the "summation by parts" method but it gives a nasty
looking recurrence formula that I would not be at all sure how to
reduce to a "closed form".
From: Jon Slaughter on 2 Dec 2009 15:37
mike3 wrote:
> Hi.
>
> Is there a closed form for
>
> sum_{k=0...n-1} k^p exp(qk)
>
> with p, q nonnegative integers?
>
> Like how Faulhaber's formula gives a closed form for sum_{k=0...n-1}
> k^p? I tried the "summation by parts" method but it gives a nasty
> looking recurrence formula that I would not be at all sure how to
> reduce to a "closed form".


it's the same as

d^p/dq^p sum(exp(qk))

since

d^p/dq^p(exp(qk)) = k^p exp(qk)

sum(exp(qk)) is known.


From: master1729 on 2 Dec 2009 05:53
mike 3 wrote :

> Hi.
>
> Is there a closed form for
>
> sum_{k=0...n-1} k^p exp(qk)
>
> with p, q nonnegative integers?
>
> Like how Faulhaber's formula gives a closed form for
> sum_{k=0...n-1}
> k^p? I tried the "summation by parts" method but it
> gives a nasty
> looking recurrence formula that I would not be at all
> sure how to
> reduce to a "closed form".

ever heard of polylogaritms ?

also known as the Jonquière's function ?

regards

tommy1729
From: mike3 on 2 Dec 2009 19:22
On Dec 2, 1:37 pm, "Jon Slaughter" <Jon_Slaugh...(a)Hotmail.com> wrote:
> mike3 wrote:
> > Hi.
>
> > Is there a closed form for
>
> > sum_{k=0...n-1} k^p exp(qk)
>
> > with p, q nonnegative integers?
>
> > Like how Faulhaber's formula gives a closed form for sum_{k=0...n-1}
> > k^p? I tried the "summation by parts" method but it gives a nasty
> > looking recurrence formula that I would not be at all sure how to
> > reduce to a "closed form".
>
> it's the same as
>
> d^p/dq^p sum(exp(qk))
>
> since
>
> d^p/dq^p(exp(qk)) = k^p exp(qk)
>
> sum(exp(qk)) is known.

So since

sum_{k=0...n-1} exp(qk) = (exp(qn) - 1)/(exp(q) - 1)

then we have (for q > 0 -- for q = 0 we use Faulhaber's formula)

sum_{k=0...n-1} k^p exp(qk) = d^p/dx^p (exp(xn) - 1)/(exp(x) - 1)
eval'd at x=q.

But how do you expand that derivative for arbitrary orders p?
Wouldn't you need Faa di Bruno's formula to do it? Doesn't that
mean the expression grows extraordinarily quickly in length as
p goes up (grows up like the partition function(!)), or is there
some way it simplifies?
From: Jon Slaughter on 2 Dec 2009 21:57
mike3 wrote:
> On Dec 2, 1:37 pm, "Jon Slaughter" <Jon_Slaugh...(a)Hotmail.com> wrote:
>> mike3 wrote:
>>> Hi.
>>
>>> Is there a closed form for
>>
>>> sum_{k=0...n-1} k^p exp(qk)
>>
>>> with p, q nonnegative integers?
>>
>>> Like how Faulhaber's formula gives a closed form for sum_{k=0...n-1}
>>> k^p? I tried the "summation by parts" method but it gives a nasty
>>> looking recurrence formula that I would not be at all sure how to
>>> reduce to a "closed form".
>>
>> it's the same as
>>
>> d^p/dq^p sum(exp(qk))
>>
>> since
>>
>> d^p/dq^p(exp(qk)) = k^p exp(qk)
>>
>> sum(exp(qk)) is known.
>
> So since
>
> sum_{k=0...n-1} exp(qk) = (exp(qn) - 1)/(exp(q) - 1)
>
> then we have (for q > 0 -- for q = 0 we use Faulhaber's formula)
>
> sum_{k=0...n-1} k^p exp(qk) = d^p/dx^p (exp(xn) - 1)/(exp(x) - 1)
> eval'd at x=q.
>
> But how do you expand that derivative for arbitrary orders p?
> Wouldn't you need Faa di Bruno's formula to do it? Doesn't that
> mean the expression grows extraordinarily quickly in length as
> p goes up (grows up like the partition function(!)), or is there
> some way it simplifies?

Closed form simply means a finite number of computable operations that may
include some standard non-closes quantities such as the trig functions.


Your original sum is in closed form already. In fact, your sum is exactly
the exansion of what I gave.

For example, suppose you were to expand several of the derivatives. You
would find that they are exactly sum_{k=0...n-1} k^p exp(qk).

It may be that there are other possible expressions. The derivatives may
offer a different point of few for some other simplification.

Your question is kinda like asking,

"What is the closed form for 9/3x^2/x"

Well, you have 18/6*x, (9x + x)/3 - 3x, and some fourier series and fourier
transform and many other representations... Which one do you want?


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