Talk:Bernoulli number

From Wikipedia, the free encyclopedia

1 Inconsistency
2 Big O notation
3 Relationship of Bernouilli numbers to Riemann zeta function
4 Kowa Seki
5 Recursive Definition
6 Asymptotic expansion
7 Another set of identities

[edit] Inconsistency

Inconsistency here: Bernoulli numbers as defined on Bernouilli number page are alternately negative and positive.

But Taylor series for Tan(x) and Cot(x) use Benouilli numbers that are all positive. Formula for Taylor series should use absolute value |Bn| and not Bn.

Or Bernouilli numbers should be defined as all positive.

There is no inconsistency at all. The article says:

The Bernoulli numbers also appear in the Taylor series expansion of the tangent...

and that is exactly correct; they appear in the Taylor series expansion. The Taylor series for tan(x) is:

$\tan x = \sum^{\infin}_{n=1} \frac{B_{2n} (-4)^n (1-4^n)}{(2n)!} x^{2n-1}$

in which

B 2 n

denotes the 2nth Bernoulli number; so the Bernoulli numbers do indeed appear in the Taylor series expansion. The multiplication by (-4)ⁿ ensures that all the coefficients are positive. I hope this clears things up for you. -- Dominus 12:40, 29 July 2005 (UTC)

Yes it does, thanks.

Another question:

$\sum_{k=0}^{m-1} k^n = 0^n + 1^n + 2^n + \cdots + {(m-1)}^n$

for various fixed values of n. The closed forms are always polynomials in m of degree n + 1

But term of highest degree appears to be $(m - 1) n$ which has a degree of 'n'.

there is a sum of m terms involved here. If you take in the sum of powers only those summands say with k > m/2− 1 then the cut-off sum will be already bigger than

$(m/2) (m/2)^n ={1\over {2^{n+1}}} m^{n+1} \ .$

Therefore the order of magnitude (for fixed n) is asymptotically

O (m n + 1)

and not

O (m n)

(the exact order is ${1\over{n+1}}\,m^{n+1}+O(m^{n})).$ .

[edit] Big O notation

Thanks for the timely revert, Dmharvey. I mistakenly thought you were using the omega notation in its historical sense, equivalent to big O. Elroch 22:21, 10 February 2006 (UTC)

[edit] Relationship of Bernouilli numbers to Riemann zeta function

I decided to change the language used to describe the relationship of the Bernouilli numbers to the Riemann zeta function, which grated with me as it stood. As I understand it, two sequences are the same "up to a factor" if one is a constant multiple of the other, and describing one sequence as "essentially" another sequence was wooly language at best. Elroch 22:29, 10 February 2006 (UTC)

[edit] Kowa Seki

A recent edit removed the assertion that the Bernoulli numbers were first studied by Bernoulli, and instead attributed them to the great Japanese mathematician Seki "in 1683", and asserting that Bernoulli did not study them until "the 18th Century". The implication here is that Bernoulli was greatly anticipated by Seki. But Bernoulli (1654-1705) and Seki (1642-1708) were contemporaries, and without two dates, I am reluctant to believe any claims of priority.

If anyone has any real information, I would be glad to hear it. Meantime, I am going to change the article again to note Seki's discovery. -- Dominus 01:15, 28 March 2006 (UTC)

[edit] Recursive Definition

I could be mistaken, but I don't see the recursive definition as being recursive. Maybe the 0 on the right hand side should be B_m+1? Psellus 23:19, 7 July 2006

It's recursive but not phrased in an explicitly recursive manner. For example, try substituting m = 3 and then solve for B₃. If you like you can rearrange the equation to show the recursion more explicitly, but it's quite elegant the way it's written currently. Dmharvey 23:48, 7 July 2006 (UTC)

I was afraid it would turn out to be something like this. OK, I will look at it harder. Thanks very much. Psellus 23:54, 7 July 2006 (UTC)

[edit] Asymptotic expansion

I made a check about that formula, it is ok but not especially good. The ordinary formula which is B(n) = 2*n!/Pi^n/2^n (shown just above) is much better : for n=1000 the approximation is good for the first 300 digits(!) and that one is good to something like 20 digits. I don't see the point of showing that formula. Maybe is it new but by far less efficient than the usual one. In my opinion that formula should be removed and we could maybe put a reference to that guy that found it. The litterature on bernoulli numbers is quite large and I am sure many authors looked at these pages.

Plouffe 08:29, 18 February 2007 (UTC)

It is apparently the ordinary asymptotic formula with a Stirling-like approximation for the factorial. Perhaps it could be useful for computing Bernoulli numbers with double (16-digit) precision? Nevertheless, it seems redundant to mention in this article since anyone who'd be interested in an n-digit approximation for large Bernoulli numbers could simply take the original asymptotic formula and substitute an n-digit factorial approximation of his own choice. Maybe the section could be reworded to say something to this effect? By the way, the article should also mention how to compute Bernoulli numbers exactly with the zeta function and the von Staudt-Clausen theorem. Fredrik Johansson 19:50, 18 February 2007 (UTC)

[edit] Another set of identities

Hi -

I've got another set of identities based on an recursive definition, which I found recently, and which is not much known.

However, scanning some internet resources, I found the basic idea also mentioned in Zhi Wei Sun's article about "courious results concerning Bernoulli and Euler-polynomials", where he cites this relation according to von Ettinghaus in the early 19'th century.

This recursive definition relates some basic number-theoretic sequences in a very simple scheme. Would it be appropriate to link to this article of mine?

GeneralizedBernoulliRecursion.pdf

--Gotti 21:08, 18 February 2007 (UTC)

Retrieved from "http://en.wikipedia.org../../../b/e/r/Talk%7EBernoulli_number_5f7c.html"