Talk:Bernoulli polynomials

message: ::Definite integrals
::::\int_0^1 B_n(t) B_m(t)\,dt = 
(-1)^{n-1} \frac{m! n!}{(m+n)!} B_{n+m}
\quad \mbox { for } m,n \ge 1 
::::\int_0^1 E_n(t) E_m(t)\,dt = 
(-1)^{n} 4 (2^{m+n+2}-1)\frac{m! n!}{(m+n+2)!} B_{n+m+2}
Should it be (−1)n + m instead of (−1)n − 1?  Obviously the integrals must depend in a symmetrical way on the two variables n and m. Michael Hardy 19:18, 7 September 2006 (UTC)
replies: text: Nope, but either m or n will do. If one is odd and the other even, the right-hand side is zero. So the sign only matters when they are both odd or both even, and in that case it is sufficient to look at one of them. Fredrik Johansson 19:51, 7 September 2006 (UTC)
thread_title: Error in integrals?

message: I expect the second formula in "differences and derivatives" to be a misprint : replace by En(x+1) + En(x) = 2*x^n  ? [as a consequence of {(e^D + 1)/2} En = x^n ].
One should tell where the Fourier expansion of Euler polynomials is valid : on [0,1] (open interval for n=1). The formula for Bn is more elegant than for En. Write the latter as : ...*[sum over all odd integers of exp(Pi i kx)/k^(n+1)] ?
About the 3d formula in "Integrals", one could explain : by Parseval formula applied to the Fourier expansions.
Periodic Bernoulli also appear naturally for p-adic l values. I would add :
The Fourier expansion given for the Bernoulli polynomial Bn is in fact a Fourier expansion for Pn. If Pn is viewed as a function on the circle group R/Z, the multiplication theorem can be written as Pn(x) = m^(n-1) [sum over y such that my=x of Pn(y)]Kibour (talk) 20:27, 12 December 2008 (UTC)
thread_title:  misprint ? suggestions. 

message: In the basic defining formula ("explicite formula") the index k is running along the set of bernoulli-numbers and oppositely to the powers of x. Because the binomial coefficients are a symmetric set of numbers for one row of the binomial matrix it seems to be arbitray, in which direction I use the index. So this formula seems to have no problems.      
However, later there is the generalization to fractional orders of the bernoulli-polynomials. But for fractional orders the set of nonzero binomial coefficients is infinite and thus not symmetric. Then the order of indexing is relevant - and then the given indexing in the formula is wrong. If we consider fractional orders, then we have to replace the bernoulli-numbers by zeta-values of fractional arguments, and the index for the binomial coefficients and the index for the zeta-arguments must run in the same direction, so it should always run in the opposite direction for the bernoulli-numbers.   
The direction of indexes is "correct" in this sense in the section of the alternate formula ("another explicte formula"): there the index of the binomial coefficients run with the exponents of x, which is correct also for the case of fractional orders.
Note: I've just adapted the first formula by changing the index at the binomial coefficients.
--Gotti 11:37, 19 August 2011 (UTC)
thread_title:  unfortunate indexing in the basic formula 

message: does that fourier series true for any value of x? ISHANBULLS (talk) 13:43, 20 March 2019 (UTC)
thread_title:  Fourier series 

message: :The following is a closed discussion of a requested move. Please do not modify it. Subsequent comments should be made in a new section on the talk page. Editors desiring to contest the closing decision should consider a move review after discussing it on the closer's talk page. No further edits should be made to this discussion.
The result of the move request was: not moved.  – robertsky (talk) 15:49, 10 January 2024 (UTC)
replies: text: Bernoulli polynomials →  – The article Bernoulli polynomials is almost as much about the Euler polynomials as it is about the Bernoulli polynomials, so it appears appropriate to rename the article.  --Lambiam 09:14, 24 December 2023 (UTC)
thread_title:  Requested move 24 December 2023 

message: In the section "Maximum and minimum", I am pretty sure that 2n! has to be read as 2\,n! and not as (2n)!. However, in that case, setting n=1 leads to  \frac 12=M_1<\frac{2\cdot 1!}{(2\pi)^1}=\frac 1\pi, i.e., \pi<2. (Same problem applies to the bound for m_1). Could someone please clarify? Hagman (talk) 12:54, 10 March 2024 (UTC)
thread_title:  Wrong maximum?