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