Talk:Bernoulli number

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No neutral point of view - removed paragraphs

There is no place for such remarks:

So Pavlyk could have saved his employer Wolfram a lot of resources had he used free and open source software. Pavlyk wrote: "The numerator begins with 571642756... and ends with ...597039303." This result is false as a quick check with the asymptotic formula in the next section reveals. Interestingly, none of the math-professors in the newsgroup seem to have noticed.

The following removed paragraphs are personal opinions of Peter Luschny, see an essay of Luschny The Bernoulli Confusion. This guy (is he an expert on number theory or a graduate mathematician at all? I don't think so!) rigidly claims, against the consensus of experts, to change the definition of Bernoulli numbers so that B(1) = 1/2. For both cases, B(1) = 1/2 or B(1) = -1/2, there are advantages and disadvantages. There is no favored viewpoint! One can start with the summation of powers up to n-1 (simple) or the Riemann zeta function on the negative x-axis (advanced).

This convention is looked at as an unhappy choice by many mathematicians as it is not in accordance with the viewpoint suggested by the Riemann zeta function. Therefore more and more modern writers depart from it. John H. Conway for example uses the convention B_n = B_n(1) in his books. He remarked that this convention also makes "certain formulas more aesthetically pleasing to our eyes".

and is often preferred by number theorists

In conclusion: The Bernoulli numbers B_n admit a variety of different representations. Since not all of these agree with one another conventions are introduced. Which among them should be adopted may depend not only on mathematical insight. But mathematical insight, simplicity and aesthetics suggest that the convention supporting the concordance is the best one in almost all of the cases.

Indeed for many mathematicians there is no valid alternative to the above concordance as they agree with G. H. Hardy: "The mathematician's patterns, like the painter's or the poet's must be beautiful; the ideas like the colours or the words, must fit together in a harmonious way. Beauty is the first test: there is no permanent place in the world for ugly mathematics."

Cross-references with no clear referents

There are a couple of points in the article where a promise is made to explain something, but this promise is not kept.

1. "Knuth's in-depth study of Faulhaber's formula concludes (the nonstandard notation on the LHS is explained further on)"

2. "to the modern form (more on different conventions in the next paragraph)"

In case (1) it is not clear where the nonstandard notation is explained, and in case (2) the different conventions are not covered in the next paragraph.

This kind of internal cross-reference does not work in Wikipedia as it is not robust against editing. Explanations need to be in the same paragraph or in footnotes. 08:23, 30 July 2022 (UTC) Gdr 08:23, 30 July 2022 (UTC)

I have now deleted both of these remarks. It is now 2026 and apparently the original author is either gone or unable to fix his own wording. Wikipedia is an electronic text and everything needs to be explicit with hyperlinks, footnotes or concrete references. Usage as "see below" or "see above" should be avoided. Danko Georgiev (talk) 13:34, 2 January 2026 (UTC)

Should the re-typeset image of Bernoulli's book be replaced by a scan of the original?

A scan of Bernoulli's book compared to the re-typeset version currently in use.

It seems to me that a high-quality scan of the original Ars Conjectandi would be preferable to the (not entirely accurately) retypeset version currently on this page. The Smithsonian made a nice scan and hosted it on the internet archive at https://archive.org/details/jacobibernoulli00bern/page/97

In particular, the beige color seems significantly better than the bright yellow, the scan gives a better idea of what the original notation actually looked like, and there are no typos accidentally introduced by the modern typesetter.

Should we replace this? If so, I'm not sure if it would be best to crop like I have done here, or just to include the full page. We don't have too many images near this section of the article, so it would even be possible to stack two full pages vertically there (possibly with the margins cropped)

Thoughts? –jacobolus (t) 11:08, 26 February 2023 (UTC)

No! If someone wants to watch ugly scanned bitmap images, he can use any of the public domain scans in the Internet Archive. The page color can be white if one chooses black-and-white PDF, https://archive.org/details/bub_gb_kz9nvk99EWoC/page/n140/mode/1up

To me personally, any coloring of the page is tormenting the brain, so I would vote for the yellow color to be replaced by white color. In addition, the original notation of nn instead of n² is also ridiculous and better be modernized because it does not change the meaning. In contrast, the original error in the term

-{\tfrac {1}{12}}n^{2}

should be kept because it is a meaningful historic fact. Danko Georgiev (talk) 09:38, 2 January 2026 (UTC)

The point is to show the original appearance in context, not to "modernize" it; a "modern" version is redundant with the entire rest of the article. –jacobolus (t) 12:01, 2 January 2026 (UTC)

Also, there's nothing "ugly" about scans of old books. –jacobolus (t) 12:01, 2 January 2026 (UTC)

Nothing "ugly"?? These are bitmap images with enormous size, contain smudges, blobs, transparent page with lines visible on the opposite side of the page, and so on. Also, the shrift is old and barely recognizable until you perform a Google search to discover that "f" in old Latin texts is actually meant to stand for "s", etc. You improve the quality of reading, by keeping historical accuracy (i.e., you do not change the intended meaning), and if someone wants to enjoy the "beauty" of the scan texts, he is free to do this at archive.org. Danko Georgiev (talk) 12:13, 2 January 2026 (UTC)

The text is in Latin. Nobody who can even read it is going to be confused by the long s. Also, your scare quotes, double question marks, bold, etc. are unnecessary here. To be honest I find all of your objections to be missing the point of the illustration to the point of absurdity, and the sarcasm (or whatever tone you are going for with those) makes it still less convincing. –jacobolus (t) 12:37, 2 January 2026 (UTC)

Well, I could not read the Latin and got confused by the long "s". The OCR text recognition at https://archive.org/details/bub_gb_kz9nvk99EWoC/page/n140/mode/1up also "got confused" and when you copy-paste the Latin text you get "f" instead of "s". So, your superiority over me is now officially established and I am one of the nobodies. Thank you for your opinion, but I will not accept it. Danko Georgiev (talk) 13:06, 2 January 2026 (UTC)

Why does it matter if you are temporarily slightly confused about the spelling of a few words that you are unable to read either way? This does suggest a decent idea though: the image description page should contain an English translation. –jacobolus (t) 00:51, 3 January 2026 (UTC)

The original PNG image is .tex generated, contains typesetting errors that do not correspond to the original Latin book, and also Wikipedia itself recommends that the PNG image is replaced with .tex and then DELETED from Wikipedia.

Here, I have created a .tex box that can easily be edited to have different background color (p.s. normal box template has colors, but does not float; after switching to float box, there is no longer background color settings, just gray color ?!)

... Atque sic porrò ad altiores gradatim potestates pergere, levique negotio sequentem adornare laterculum licet:

Summæ Potestatum
 $\textstyle \int n={\frac {1}{2}}nn+{\frac {1}{2}}n$ 
 $\textstyle \int nn={\frac {1}{3}}n^{3}+{\frac {1}{2}}nn+{\frac {1}{6}}n$ 
 $\textstyle \int n^{3}={\frac {1}{4}}n^{4}+{\frac {1}{2}}n^{3}+{\frac {1}{4}}nn$ 
 $\textstyle \int n^{4}={\frac {1}{5}}n^{5}+{\frac {1}{2}}n^{4}+{\frac {1}{3}}n^{3}-{\frac {1}{30}}n$ 
 $\textstyle \int n^{5}={\frac {1}{6}}n^{6}+{\frac {1}{2}}n^{5}+{\frac {5}{12}}n^{4}-{\frac {1}{12}}nn$ 
 $\textstyle \int n^{6}={\frac {1}{7}}n^{7}+{\frac {1}{2}}n^{6}+{\frac {1}{2}}n^{5}-{\frac {1}{6}}n^{3}+{\frac {1}{42}}n$ 
 $\textstyle \int n^{7}={\frac {1}{8}}n^{8}+{\frac {1}{2}}n^{7}+{\frac {7}{12}}n^{6}-{\frac {7}{24}}n^{4}+{\frac {1}{12}}nn$ 
 $\textstyle \int n^{8}={\frac {1}{9}}n^{9}+{\frac {1}{2}}n^{8}+{\frac {2}{3}}n^{7}-{\frac {7}{15}}n^{5}+{\frac {2}{9}}n^{3}-{\frac {1}{30}}n$ 
 $\textstyle \int n^{9}={\frac {1}{10}}n^{10}+{\frac {1}{2}}n^{9}+{\frac {3}{4}}n^{8}-{\frac {7}{10}}n^{6}+{\frac {1}{2}}n^{4}-{\frac {1}{12}}nn$ 
 $\textstyle \int n^{10}={\frac {1}{11}}n^{11}+{\frac {1}{2}}n^{10}+{\frac {5}{6}}n^{9}-1n^{7}+1n^{5}-{\frac {1}{2}}n^{3}+{\frac {5}{66}}n$ 
Quin imò qui legem progressionis inibi attentius inspexerit, eundem etiam continuare poterit absque his ratiociniorum ambagibus: Sumtâ enim c pro potestatis cujuslibet exponente, fit summa omnium  $n^{c}$  seu
 $\textstyle \int n^{c}={\frac {1}{c+1}}n^{c+1}+{\frac {1}{2}}n^{c}+{\frac {c}{2}}An^{c-1}+{\frac {c\cdot c-1\cdot c-2}{2\cdot 3\cdot 4}}Bn^{c-3}$ 
 $\qquad +{\frac {c\cdot c-1\cdot c-2\cdot c-3\cdot c-4}{2\cdot 3\cdot 4\cdot 5\cdot 6}}Cn^{c-5}$ 
 $\qquad +{\frac {c\cdot c-1\cdot c-2\cdot c-3\cdot c-4\cdot c-5\cdot c-6}{2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7\cdot 8}}Dn^{c-7}\ldots$ 
& ita deinceps, exponentem potestatis ipsius  $n$  continuè minuendo binario, quosque perveniatur ad  $n$  vel  $nn$ . Literæ capitales  $A$ ,  $B$ ,  $C$ ,  $D$  &c. ordine denotant coëfficientes ultimorum terminorum pro  $\textstyle \int nn$ ,  $\textstyle \int n^{4}$ ,  $\textstyle \int n^{6}$ ,  $\textstyle \int n^{8}$  &c. nempe
 $\textstyle A={\frac {1}{6}},B=-{\frac {1}{30}},C={\frac {1}{42}},D=-{\frac {1}{30}}.$  

 From Jacob Bernoulli's Ars Conjectandi, 1713^[a]

Danko Georgiev (talk) 11:47, 2 January 2026 (UTC)

I agree with jacobolus, having the original scan looks better because it shows what Bernoulli actually wrote originally. That is the most historically accurate you could get. There are many subtleties in the original scan that haven't been replicated in the text version, like the "equal signs" he used and those extra asterisks, and the spacing. —lightbulbMEOW!!! (meow) 09:47, 4 January 2026 (UTC)

I also agree with jacobolus. A good-quality scan brings more encyclopedic value to the article, illustrating how mathematics was communicated in the past and what a document from that era looks like when it survives into ours. The re-typeset version with the bright yellow background is an awkward half-and-half thing that is not really useful: why modernize the notation and not translate the Latin? It is neither historically accurate nor helpful to the typical modern reader. Who benefits? (Also, the long s is not an obscure thing. It's probably one of the first things anybody who looks at old documents will discover, being all over the Bill of Rights, the First Folio, etc., etc. There was even a joke in Mystery Science Theater 3000 about it.) Stepwise Continuous Dysfunction (talk) 21:42, 4 January 2026 (UTC)

The two generating functions only differ by t.

Really?
First, if one divides ${\frac {te^{t}}{e^{t}-1}}$ by ${\frac {t}{e^{t}-1}}$ the result is $e^{t}$ , so the two generating functions differ by a factor of $e^{t}$ and not a factor of $t$ .
Second, if one actually reads the text, it is clear that the correct wording should be that the two generating functions differ by the "sign of t" and not "by t".
Sometimes, if you do not know how to use plain English, it is better to leave the mathematics as it is without trying to explain what you do not understand. Danko Georgiev (talk) 09:47, 2 January 2026 (UTC)

The original claim is correct as is. The difference is

{\frac {te^{t}}{e^{t}-1}}-{\frac {t}{e^{t}-1}}={\frac {t(e^{t}-1)}{e^{t}-1}}=t

.—lightbulbMEOW!!! (meow) 13:18, 6 January 2026 (UTC)

No, the original claim is NOT correct as it is. It says they "differ" and NOT "the arithmetic difference is". The English should be used unambiguously. The reader may not possess telepathic skills to read the mind of an incompetent writer. And by the way, the arithmetic difference can have a positive or a negative sign depending on the order in which you do the subtraction. Danko Georgiev (talk) 16:48, 6 January 2026 (UTC)

I don't know if English is your first language, but when something says "A and B only differ by X", it always means that the difference between A and B is X. I don't see any other way you could interpret that.

Technically the difference should be

|t|

since difference should always be positive, but that's being too pedantic. The reader just needs to know the difference between the two generating functions is

t

. They'll understand what it means, no need to overcomplicate it. —lightbulbMEOW!!! (meow) 10:59, 9 January 2026 (UTC)

The original text was saying that "The two generating functions differ by t", which was wrong:

{\frac {t}{e^{t}-1}}-{\frac {te^{t}}{e^{t}-1}}=-t

.

When you edit encyclopedia, you should pay attention to "details" such as "order of appearance" in the main text, etc. p.s. "They'll understand what it means" is what an incompetent writer will expect from the reader. Well, guess what? I was the reader and did not understand it! If the purpose is to blame the reader, and not the author, then so be it. I do not accept your claim that the verb "differ" implies that one should be performing subtraction and then take absolute values! I have nothing more to say. Danko Georgiev (talk) 11:57, 9 January 2026 (UTC)

Notes

[a]
Translation of the text:
" ... And if [one were] to proceed onward step by step to higher powers, one may furnish, with little difficulty, the following list:
Sums of powers
$\textstyle \int n=\sum _{k=1}^{n}k={\frac {1}{2}}n^{2}+{\frac {1}{2}}n$
⋮
$\textstyle \int n^{10}=\sum _{k=1}^{n}k^{10}={\frac {1}{11}}n^{11}+{\frac {1}{2}}n^{10}+{\frac {5}{6}}n^{9}-1n^{7}+1n^{5}-{\frac {1}{2}}n^{3}+{\frac {5}{66}}n$
Indeed [if] one will have examined diligently the law of arithmetic progression there, one will also be able to continue the same without these circuitous computations: For [if] $\textstyle c$ is taken as the exponent of any power, the sum of all $\textstyle n^{c}$ is produced or
$\textstyle \int n^{c}=\sum _{k=1}^{n}k^{c}={\frac {1}{c+1}}n^{c+1}+{\frac {1}{2}}n^{c}+{\frac {c}{2}}An^{c-1}+{\frac {c(c-1)(c-2)}{2\cdot 3\cdot 4}}Bn^{c-3}+{\frac {c(c-1)(c-2)(c-3)(c-4)}{2\cdot 3\cdot 4\cdot 5\cdot 6}}Cn^{c-5}+{\frac {c(c-1)(c-2)(c-3)(c-4)(c-5)(c-6)}{2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7\cdot 8}}Dn^{c-7}+\cdots$
and so forth, the exponent of its power $n$ continually diminishing by 2 until it arrives at $n$ or $n^{2}$ . The capital letters $\textstyle A,B,C,D,$ etc. denote in order the coefficients of the last terms for $\textstyle \int n^{2},\int n^{4},\int n^{6},\int n^{8}$ , etc. namely
$\textstyle A={\frac {1}{6}},B=-{\frac {1}{30}},C={\frac {1}{42}},D=-{\frac {1}{30}}$ .

Talk:Bernoulli number

No neutral point of view - removed paragraphs

Cross-references with no clear referents

Should the re-typeset image of Bernoulli's book be replaced by a scan of the original?

The two generating functions only differ by t.

Notes

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