Closed form for this sum?
in [Math]
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From: Jon Slaughter on 3 Dec 2009 04:07 mike3 wrote: > On Dec 2, 7:57 pm, "Jon Slaughter" <Jon_Slaugh...(a)Hotmail.com> wrote: > <snip> > > Perhaps I wasn't clear... what I want is it written in explicit form > without > derivatives, just as a sum of terms dependent only on p and/or q and > not on n. > I.e. like how we can expand sums of powers with Faulhaber's formula > (slightly > different form as this sums from 0 to n-1 not 1 to n but still...): > > sum_{k=0...n-1} k^p = 1/(p+1) sum_{k=0...p} C_(p+1, k) B_k n^(p+1-k) > = 1/(p+1) (B_(p+1)(n) - B_(p+1)(0)) > > where B_k are Bernoulli numbers, B_k(x) are Bernoulli polynomials, > and "C" is the binomial coefficient function. You can find a million ways to express the sum. I'm still unsure what you want. If I were to express it as a sum of Bernoulli polynomials then would it satisfy your criteria or would you find something wrong with it? Do you think sum_{k=0...n-1} k^p is more complex than 1/(p+1) sum_{k=0...p} C_(p+1, k) B_k n^(p+1-k)? I believe that sum_{k=0...n-1} k^p is much more intuitive and simpler. I'm not saying the other expression is useless... infact that is the majority of what mathematics is about. Each represetation of an "concept" offers a different perspective and might be the appropriate one for the problem. But I'm unsure of exactly what you want. In fact it seems you want simply a more complex or different expression. The problem is that there are an infitude of these and you have no real criteria what you want. So... sum(k^p*exp(qk)) we can write exp as as sum sum(sum(k^p*(qk)^j/j!)) = sum(q^j/j!*sum(k^(p*j)) now, we could, if we wanted, define a_k(x) = sum(k^(x)) and then we have the infinite sum(a_k(x)*q^j/j!) we could give a name to the sequence a_k(x) such as the mike3 sequence. (note that we actually have an uncountable many sequences express by a_k(x) if x is a complex number) But since you seem to allow p to be only an integer(don't know if you stated this somewhere), we actually know another experssion for a_k(x) when x is restricted to an integer, i.e., Faulhaber's formula... so = sum(q^j/j!*sum(1/(pj+1)*Cr(pj+1,k)*B_k*n^(pj+1-k))) = sum(B_k*n^(1-k)*sum(1/(pj+1)*Cr(pj+1,k)*(q*n^p)^j/j!)) Note that the sums do not depend on n. Of course we have a nasty infinite sum in there. If it wasn't for that 1/(pj+1)*Cr(pj+1,k) term then we would just have sum(B_k*n^(1-k)*e^(q*n^p)) In any case this is an expression involving only the criteria you specified accept is not closed though which is the big problem. http://www.research.att.com/~njas/sequences/?q=1%2C7%2C42%2C210%2C840%2C2520%2C5040&sort=0&fmt=0&language=english&go=Search looking at the inner sum we see that sum(1/(pj+1)*Cr(pj+1,k)*(q*n^p)^j/j!)) seems to be(or can be manipulated to be) 1/k!*sum(T(pj,k-1)*(q*n^p)^j/j!)) which seems to give something like = p*q*exp(q*n^p)*sum(B_k*n^(1-k)*Q_k(p,q,n)) so what I have done is take sum(k^p*exp(qk)) and expressed it in a more complex manner. in this case it is closed form(assuming I make no mistakes) because Q_k are finite polynomials. Note how it is very similar to the original Faulhaber's formula with the extra Q_k per term. We could continue. We could try to write the inner sum in terms of the hypergeometric series and use some of the manipulation techniques. There might be a generating function for Q_k that would further simplify the expression. Of course the expression ultimately is simply sum(QQ_k(n,p,n)*n^(-k)) Which is analogous to Faulhaber's formula with some analogous constants. For Faulhaber's it is the Bernolli numbers(+ some other stuff) but in your case it is something else. If your express is useful we might end up calling them the mike3 polynomials or something like that.
From: Tim Little on 3 Dec 2009 04:26 On 2009-12-03, Jon Slaughter <Jon_Slaughter(a)Hotmail.com> wrote: >> PolyLog[-p, E^q] - E^(q n) LerchPhi[E^q, -p, n] > > This is not a closed-form. "Closed form" is a relative term, not an absolute. It depends upon what functions are considered well-known enough to be included. There are plenty of mathematicians familiar enough with the listed functions for that expression to be considered "closed form". Familiar enough to be included in a packaged computer algebra system such as Mathematica, for example. > You seem to think, at least by the answer you gave, that simple > symbol juggling is enough to produce closed form solutions. If that > is the case then everything has a closed form solution because we > can just make a new symbol for it's non-closed form expression. You could do that - but it's still not considered closed form until it becomes more widespread. - Tim
From: Tim Little on 3 Dec 2009 04:45 On 2009-12-03, Jon Slaughter <Jon_Slaughter(a)Hotmail.com> wrote: > mike3 wrote: >> sum_{k=0...n-1} k^p = 1/(p+1) sum_{k=0...p} C_(p+1, k) B_k n^(p+1-k) >> = 1/(p+1) (B_(p+1)(n) - B_(p+1)(0)) [...] > Do you think sum_{k=0...n-1} k^p is more complex than 1/(p+1) sum_{k=0...p} > C_(p+1, k) B_k n^(p+1-k)? I think you missed the last line there, where it was expressed without an indexed sum. Just 1/(p+1) (B_(p+1)(n) - B_(p+1)(0)) - Tim
From: Jon Slaughter on 3 Dec 2009 05:33 Tim Little wrote: > On 2009-12-03, Jon Slaughter <Jon_Slaughter(a)Hotmail.com> wrote: >> mike3 wrote: >>> sum_{k=0...n-1} k^p = 1/(p+1) sum_{k=0...p} C_(p+1, k) B_k n^(p+1-k) >>> = 1/(p+1) (B_(p+1)(n) - B_(p+1)(0)) > [...] >> Do you think sum_{k=0...n-1} k^p is more complex than 1/(p+1) >> sum_{k=0...p} C_(p+1, k) B_k n^(p+1-k)? > > I think you missed the last line there, where it was expressed without > an indexed sum. Just > > 1/(p+1) (B_(p+1)(n) - B_(p+1)(0)) > > No, those are the polynomials... They are still sums.
From: Jon Slaughter on 3 Dec 2009 05:36
Tim Little wrote: > On 2009-12-03, Jon Slaughter <Jon_Slaughter(a)Hotmail.com> wrote: >>> PolyLog[-p, E^q] - E^(q n) LerchPhi[E^q, -p, n] >> >> This is not a closed-form. > > "Closed form" is a relative term, not an absolute. It depends upon > what functions are considered well-known enough to be included. There > are plenty of mathematicians familiar enough with the listed functions > for that expression to be considered "closed form". > LerchPhi is an infinite series... http://mathworld.wolfram.com/LerchTranscendent.html So you can call it closed if you want... it's up to you. I agree it is relative but you are streching the truth a lot if you think that it is useful in closed forms. > Familiar enough to be included in a packaged computer algebra system > such as Mathematica, for example. > > >> You seem to think, at least by the answer you gave, that simple >> symbol juggling is enough to produce closed form solutions. If that >> is the case then everything has a closed form solution because we >> can just make a new symbol for it's non-closed form expression. > > You could do that - but it's still not considered closed form until > it becomes more widespread. > It's not just about being wide spread. Although I guess if it becomes wide spread enough maybe that means it is relatively simple. Why we might not be able to agree on what is absolutely closed form surely we can agree that there are relative levels? Finite sums of simple things like exponential and cos functions, sqrt's, etc are more "closed" than things involving infinite sums? |