Talk:Faulhaber's formula

Umbral section

In the 'Umbral' section it looks as if the factor 1 / (p+1) has been omitted twice.(Crackling 22:55, 13 March 2007 (UTC))[reply]

First sum fomula: no need for "special case"

In this article, in the first formula, there is no need for a special case for p=1: because if ${(-1)^{p}}$ be inserted, all is well - since all odd Bernoulli numbers (excepting B_1) vanish. (There should be no other consequences - as far as I can see.) Hair Commodore 16:32, 10 April 2007 (UTC)[reply]

I have changed it now. Eric Kvaalen (talk) 06:37, 14 April 2010 (UTC)[reply]

about faulhaber observation in the last of this article

There is an error in faulhaber's polynomials. This article says that : 1^{11}+ 2^{11} + 3^{11} + \cdots + n^{11} = {32a^6 - 64a^5 + 68a^4 -40a^3 + 5a^2 \over 6} But It's : 1^{11}+ 2^{11} + 3^{11} + \cdots + n^{11} = {32a^6 - 64a^5 + 68a^4 -40a^3 + 10a^2 \over 6} Thank you. —Preceding unsigned comment added by 63.243.163.199 (talk • contribs)

Setting n=1 makes it clear that there was an error, and I have changed it according to what you say. However, Knuth's paper does say 5n^2. I don't know whose mistake it is. Eric Kvaalen (talk) 06:37, 14 April 2010 (UTC)[reply]

A letter from Knuth informs me that in the paper version of his article the mistake is not there. So probably Faulhaber had it right and it was just a misprint in the preprint that was later corrected. Eric Kvaalen (talk) 09:16, 14 May 2010 (UTC)[reply]

Upper limit of sum

Where is the mistake? Here on in the Bernoulli_number article? The upper limits differ by 1.--MathFacts (talk) 19:58, 24 April 2009 (UTC)[reply]

Both articles are right. This one says:

The formula says

\sum _{k=1}^{n}k^{p}={1 \over p+1}\sum _{j=0}^{p}{p+1 \choose j}B_{j}n^{p+1-j}\qquad \left({\mbox{with }}B_{1}={1 \over 2}{\mbox{ rather than }}-{1 \over 2}\right)

(the index j runs only up to p, not up to p + 1).

The other one says:

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

This article takes B₁ to be +1/2; the other has −1/2. That's why they don't contradict each other. Look at the polynomials listed in this article, and you see that in each case if you change the coefficient of the second-highest-degree term from plus to minus, the effect is exactly the same as that of deleting the last term on the left side of the identity. Michael Hardy (talk) 21:27, 24 April 2009 (UTC)[reply]

did or did not know?

The article says Faulhaber did not know the formula in this form while MathWorld claims the opposite:

In a 1631 edition of Academiae Algebrae, J. Faulhaber published the general formula for...

What information is correct? Did Faulhaber know (and publish) the general formula? Maxal (talk) 04:10, 30 April 2009 (UTC)[reply]

I seem to recall that he knew the first couple of dozen or so cases. I'll see if I can find that. Michael Hardy (talk) 20:12, 30 April 2009 (UTC)[reply]

No, Faulhaber did definitely not know this formula. It is absurd to call this formula Faulhaber's formula. Mathworld is, again, unreliable and cannnot be used a primary source. Wirkstoff (talk) 12:26, 4 July 2009 (UTC)[reply]

Move?

Since the formula is not actually due to Faulhaber, should we move the article to "Bernoulli's formula"? Eric Kvaalen (talk) 06:37, 14 April 2010 (UTC)[reply]

The Wikipedia article title policy says that articles should be listed under their common names: "[Use] names and terms commonly used in reliable sources, and so likely to be recognized...Articles are normally titled using the most common English-language name of the subject of the article." This holds even when that name is erroneous or misattributed. For example, Pell's equation is under that title because the equation is called "Pell's equation", even though Pell had nothing to do with it, the equation was solved by Brouncker, and the solution was attributed to Pell entirely by mistake. Similarly Zorn's lemma is not listed as "Kuratowski's lemma", even though Kuratowski unquestionably published it first.

Unless it appears that the term "Bernoulli's formula" is more common than "Faulhaber's formula", the article should be titled "Faulhaber's formula", perhaps with "Bernoulli's formula" as a redirect to it. —Dominus (talk)

NPOV

Lots of things in math aren't named after the person to actually discover them. We don't need to remind the reader at every opportunity, and certainly don't need the commentary. Twin Bird (talk) 22:05, 21 October 2010 (UTC)[reply]

Comments on new sections? ("Proof" and "Alternate expression")

I have more material I would like to add to this page, but I don't want to do it without gauging whether it is in the spirit of Wikipedia, since I'm aware that proof-like material is not quite encyclopaedic.

I believe what I've written so far is explicative rather than merely demonstrative, since the proof is in a historic style, illuminating the relation between coefficients that led to the formulation of the Bernoulli numbers, and the transformation between the two expressions explains why there are two conventions for B_1.

What I would like to add next is a generalisation to non-natural powers which shows the relationship between Faulhauber's formula and the Hurwitz (and Riemann) zeta function, and further demonstrates why the alternate expression is more natural (since the more conventional one is not a Taylor series when p is non-natural). — Preceding unsigned comment added by Haklo (talk • contribs) 21:25, 13 June 2012 (UTC)[reply]

A new Theorem?

Please see at point 2 of Discussion in the following article: http://upload.wikimedia.org/wikipedia/commons/6/6c/Pubblicazione_english.pdf

--Ancora Luciano (talk) 15:47, 11 August 2013 (UTC)[reply]

That observation is a special case of the Stolz–Cesàro_theorem, for

a_{n+1}-a_{n}=(n+1)^{m}

, and

b_{n+1}-b_{n}=(n+1)^{m+1}-n^{m+1}

. It is interesting though, isn't it?

Haklo (talk) 02:06, 12 August 2013 (UTC)[reply]

Performing calculations with Excel, I encounter these other amazing results:

     n
lim (Σ_n n³)/(Σ_n n). n² = 1/2     
n→∞  1
     n
lim (Σ_n n⁵)/(Σ_n n). n⁴ = 1/3     
n→∞  1
     n
lim (Σ_n n⁷)/(Σ_n n). n⁶ = 1/4     
n→∞  1
     n
lim (Σ_n n⁹)/(Σ_n n). n⁸ = 1/5     
n→∞  1

which, by induction, can be generalized in a formula. Note that the denominators of the results are the positions of the exponents in the numerator in the sequence of odd numbers. Also this is a special case? To see the genesis of this, go to: http://en.wikipedia.org/wiki/Talk:Summation#Sum_of_the_first_.22n.22_cubes_-_even_cubes_-_odd_cubes_.28geometrical_proofs.29

--Ancora Luciano (talk) 15:49, 12 August 2013 (UTC)[reply]