From: spellucci@mathematik.tu-darmstadt.de (Peter Spellucci) Newsgroups: sci.math.research Subject: Re: bernoulli numbers Date: 26 Nov 1998 09:58:58 GMT In article , Sebastien Blot writes: |> |> I'd like to know if the sums of integer can always be factorised ? |> |> for example :sum(i)=n(n+1)/2 |> sum(i^2)=n(n+1)(2n+1)/6 |> sum(i^3)=(n^2(n+1)^2)/4 |> |> |> are there any theorems about this ?!!! |> yes. let B_k(x) be the k-th Bernoulli-polynomial. then \sum_{i=1}^n i^{k-1} = (1/k)(B_k(n+1)-B_k(0)) B_k(x) has the explicit form B_k(x) = \sum_{i=0}^k ( k choose i) B_i(0)x^{k-i} with B_i(0) the Bernoulli numbers, which in turn are defined by the series expansion x/(exp(x)-1) = \sum_{i=0}^\infty B_i(0)/(i!)x^i hope this helps peter ============================================================================== From: "Achava Nakhash, the Loving Snake" Newsgroups: sci.math Subject: Re: finite-series formulas Date: Fri, 25 Dec 1998 09:48:00 -0800 Steve Leman wrote: > I had a question regarding sumations. > > Sum[k=0 to n] of k = (1/2)n(1+n) > > is fairly easy to derive. However for Sum of k^2, Sum k^3, etc... I am a > little lost. I was able to get other closed forms, but they all relied > on the belief that the resulting form will be a polynomial with a degree > one greater than what I am summing. > > Sum k = degree 2 poly > Sum k^2 = degree 3 poly > etc... > > I hope I have been clear enough in what I am trying to accomplish. > > Thanks, > Steve Leman Well, here it is, Christmas morning, and I find myself at work. Still no books in front of me to remind me of basic definitions like that of the Bernoulli numbers, so let's see what I can come up with to continue the discussion from yesterday. Remember that e^0x = 1 + 0(x^1/1!) + 0(x^2/2!) + 0(x^3/3!) e^x = 1^0(x^0/0!) + 1^1(x^1/1!) + 1^2(x^2/2!) + 1^3(x^3/3!) + ... so that e^2x = 2^0(x^0/0!) + 2^1(x^1/1!) + 2^2(x^2/2!) + 2^3(x^3/3!) + ... e^3x = 3^0(x^0/0!) + 3^1(x^1/1!) + 3^2(x^2/2!) + 3^3(x^3/3!) + ... etc. Now we form the sum 1 + e^x + e^2x + ... + e^(n-1)x = (e^nx - 1) / (e^x - 1) since it is a geometric progression. Now looking at the above table and adding like powers of x (we will be equating coefficients later, a step whose justification I will skip, but it is pretty easy), so it is time to settle on some notation. In particular, let S(n, k) = 1^k + 2^k + ... + (n-1)^k for k > 0, and S(n, 0) = n. Then (e^nx - 1) / (e^x - 1) = S(n, 0)(x^1/1!) + S(n, 1)(x^2/2!) + S(n, 2)(x^2/2!) + ..., so if only we could easily represent the function on the left as a power series independently of our knowledge of the various S(n, k), then we have an easy method of determining the formulas for the S(n, k). If only life were that easy. Fortunately there is something that will have to do. Notice that x/(e^x - 1) is actually an analytic function at 0 (meaning effectively that it has a power series development around 0, and we define the Bernoulli numbers B(n) via x/(e^x - 1) = B(0)/0! + B(1)(x^1/1!) + B(2)(x^2/2!) + B(3)(x^3)/3! + ... I hope. They should come out to B(0) = 1, B(1) = -1/2, B(2) = 1/6, again I am relying on very flawed memory, so this should be checked. My definition should be right up to a plus/minus ambibuity. That being done, multiply this power series by the power series for (e^nx - 1) / x which you already know and voila!, out pop the formulas for all of the S(n, k) as polynomials of the correct degree and whose coefficients are closely related to the Bernoulli numbers. It looks like I like before about every other coeffecient being 0, since B(1) is not equal to 0. For every odd number n except 1, B(n) is in fact 0, so I was almost right. I realize that you still have a little work to do from here, but it should be more fun that way. Also, notice that I summed everything from 1 to n-1 and you summed from 1 to n. This is an easy difference to account for. Finally, all of this kind of stuff and lots more very interesting mathematics, for instance about the Bernoulli numbers and their various congruence properties, can be found in many standard books on Number Theory. They will be advanced books usually, but this material is not advanced, so you if you are not up to the more advanced part, you just have hunt for the stuff about Bernoulli numbers and sums of powers, for instance in the book by Borevich and Shafarevich. Hope this is of interest, Achava