Talk:John Wallis

message: See here for an article on his contribution to breaking French codes and a Times challenge to rebreak them - since Wallis failed to leave a description of the method he used. Malick78 (talk) 07:47, 11 April 2009 (UTC)
thread_title: Times challenge on his code-breaking

message: To the authors:  Thank you for a very educational article on John Wallis.  I want to point out that your expression of Wallis's product is probably going to be misleading because of the numerator not quite seeming to keep up with the denominator.  Note that if you express your representation of the product as (2/3)^2 (4/5)^2 (6/7)^2 ..., every term will be less than 1, resulting in a product that is less than 1, which pi/2 is not.   In fact, this product approaches zero if it is continued indefinitely.  I do not know how Wallis himself expressed the product.  However, I have seen it expressed as something like this
:\frac{\pi}{2} = \frac{2 \cdot 2 \cdot 4 \cdot 4 \cdot 6 \cdot 6 \cdots}{1 \cdot 3 \cdot 3 \cdot 5 \cdot 5 \cdot 7 \cdots}.
This significantly differs from your expression only in that it avoids squaring the last odd integer in the denominator.  I'm sure that it will be helpful to some readers to align the numerator and denominator in the expression so that one of each of the two even integers in the numerator hovers directly over one odd integer that is one greater than itself and one that is one less.
:It's not just a matter of alignment, but of ill-defined limits. Obviously one cannot form the diverging products of numerator and denominator and then divide the two infinities. So it is essential that divisions are performed first, and the product is one of (infinitely many) fractions. I've just fixed that. Marc van Leeuwen (talk) 16:24, 14 April 2010 (UTC)
thread_title:  Moved anonymous comments from article 

message: It seems to me some things have gone wrong in the integral calculus section.
*"he was unacquainted with the binomial theorem, he could not effect the quadrature of the circle" seems nonsensical. Since Wallis was capable of integrating (1-x^2)^n for n=0,1,2,3 he was most likely in particular aware of the binomial theorem for integer exponents (at least small ones), which was well known around the time (and Pascal had just published his Traité du triangle arithmétique in 1653). What Wallis did not know however was the generalized binomial formula for fractional exponents (notably for 1/2). And all this is not about the quadrature of the circle, just about computing its area, 
*"Wallis argued, we have in fact a series 1,  , ,..." But those numbers were from a different problem (x-x^2)^n rather than (1-x^2)^n
*The text seems to be in part copied literally (except for some transcription errors) from the Online Encyclopedia (although that text breaks off with "[ <-- This needs a lot of work! ] " It is not so clear what has gone on here (and I dont have "A Short Account of the History of Mathematics by W. W. Rouse Ball (4th Edition, 1908)" which the latter text claims as (public domain) source), but it makes me uncomfortable (and the current WP article acknowledges nothing).
Marc van Leeuwen (talk) 16:24, 14 April 2010 (UTC)
replies: text: The Rouse Ball material will be public domain, because of the date. These days we ask that direct copying should be acknowledged under a heading in the External links section (I use the semi-colon indent to make such headings - there is no need for a subsection). But, from the point of view of the article, it would be better to report what W. W. Rouse Ball said: e.g. "in Rouse Ball's opinion, Wallis used an interpolation argument because the binomial theorem was not available to him". Charles Matthews (talk) 18:23, 14 April 2010 (UTC)
thread_title:  Making sense of the Integral calculus section 

message: Wallis was elected to a fellowship at Queens' College, Cambridge in 1644, which he however had to resign following his marriage on 14 March 1645 to Susanna Glyde
Why did he have to resign? Was it a rule surviving from Catholic traditions that fellows have to be celibate? Or was it due to purely technical difficulties having to do e.g. with moving to live elsewhere?
Top.Squark (talk) 12:35, 10 July 2010 (UTC)
replies: text: It was a rule that lasted until about 1870 - only the Master of an Oxbridge college could marry. Charles Matthews (talk) 12:41, 10 July 2010 (UTC)
replies: text: I think it will benefit the article if you add a few words about this rule and a citation Top.Squark (talk) 13:32, 10 July 2010 (UTC)
thread_title:  Resigned fellowship because of Marriage? 

message: Thus, as the ordinate of the circle y = \sqrt{1 - x^2} is the geometrical mean between the ordinates of the curves y = (1 - x^2)^0 and y = (1 - x^2)^1, it might be supposed that, as an approximation, the area of the semicircle \int_{0}^{1} \sqrt{1 - x^2}\, dx which is \begin{matrix} \frac{1}{4} \end{matrix} \pi might be taken as the geometrical mean between the values of 
:\int_{0}^{1} (1 - x^2)^0 \, dx\text{ and }\int_{0}^{1} (1 - x^2)^1 \, dx
that is, 1 and \begin{matrix} \frac{2}{3} \end{matrix}; this is equivalent to taking 4 \sqrt{\begin{matrix} \frac{2}{3} \end{matrix}} or 3.26... as the value of π.
So far so good.
But, Wallis argued, we have in fact a series 1, \begin{matrix} \frac{1}{6} \end{matrix}, \begin{matrix} \frac{1}{30} \end{matrix}, \begin{matrix} \frac{1}{140} \end{matrix},... and therefore the term interpolated between 1 and \begin{matrix} \frac{1}{6} \end{matrix} ought to be chosen so as to obey the law of this series
Unclear. Where does this series come from? Top.Squark (talk) 13:43, 10 July 2010 (UTC)
replies: text: The sequence is of the quadratures (i.e. definite integrals) mentioned a couple of paragraphs above. Charles Matthews (talk) 21:08, 10 July 2010 (UTC)
replies: text: But there we have the integral of (x - x^2)^p whereas here we have the integral of (1 - x^2)^p. There is some confusion here between computing the area of a quarter circle with radius 1 and the area of a half-circle with radius 1/2. Top.Squark (talk) 12:53, 23 July 2010 (UTC)
thread_title:  Derivation of the Wallis product 

message: He [van Heuraet] supposes the curve to be referred to rectangular axes; if this be so, and if (x, y) be the coordinates of any point on it, and n be the length of the normal...
Unclear. What is the "length of the normal"? The normal is a vector orthogonal to the curve, that is, orthogonal to its tangent. It is usually taken either of unit or of arbitrary length.
Top.Squark (talk) 13:47, 10 July 2010 (UTC)
replies: text: I believe this will refer to the Descartes method, which is basically to find the circle touching the curve (so resting on the tangent), which comes down to saying that "the" normal would be the radius vector from the point to the centre of that circle. Rouse Ball talks about this.. I think that because this page is about the original book of Van H, and describes his method as coming from the Descartes normal method and Hudde's rules. This all should probably be in arc length rather than this article. Charles Matthews (talk) 21:26, 10 July 2010 (UTC)
replies: text: So, the "length of the normal" is the inverse curvature of the curve? Top.Squark (talk) 12:57, 23 July 2010 (UTC)
replies: text: That's what it looks like. The Frenet formula approach is probably implied, in fact, which would have been familiar mathematics for Rouse Ball's readers: the normal coming from the second derivative? Charles Matthews (talk) 13:48, 23 July 2010 (UTC)
replies: text: Doesn't seem to make sense. For example ds : dx = n : y can't hold since for a straight line n is infinite. Also, I don't understand the relevance of W. W. Rouse Ball who was only born about 200 years later. Top.Squark (talk) 14:44, 24 July 2010 (UTC)
thread_title:  Van Heuraët's method 

message: He [Wallis] is usually credited with the proof of the Pythagorean theorem using similar triangles. However, Thabit Ibn Qurra (AD 901), an Arab mathematician, had produced a generalization of the Pythagorean theorem applicable to all triangles 6 centuries earlier. It is a reasonable conjecture that Wallis was aware of Thabit's work
Does it mean that Wallis' proof was similar / based on Thabit's proof?
Top.Squark (talk) 14:00, 10 July 2010 (UTC)
thread_title:  Wallis and Thabit Ibn Qurra 

message: In the introduction to the article, John Wallis is credited with introducing the symbol ∞  for infinity, but there is no further mention of this in the body of the article. I found that he defined the symbol for infinity in his Treatise on the Conic Sections, which was published in 1656. In his paper, he wrote : "(let the altitude of each one of these be an infinitely small part, \frac{1}{\infty }of the whole altitude, and let the symbol ∞ denote Infinity)". I have added this to the body of the article under the Analytical Geometry section.  —Preceding unsigned comment added by Tmancarella635 (talk • contribs) 20:12, 5 December 2010 (UTC)
replies: text: Very interesting.  It may be worth elaborating further in this direction. Tkuvho (talk) 05:38, 6 December 2010 (UTC)
thread_title:  Infinity 

message: Same question with the last few entries, just wondering if they are all true and genuine Mental Calculators, that's all?  — Preceding unsigned comment added by Iamnofool6 (talk • contribs) 00:35, 12 February 2014 (UTC)
thread_title:  Are all the new additions to mental calculators actually mental calculators ? 

message: Concerning John Wallis' article and the fact it may need a cleanup to meet Wikipedia's quality standards, I would like to say that: as proved by Amir Alexander, Wallis wasn't only and mainly an "experimental" mathematician, but he was also a man of his times (an ecclesiastic and a politician member of parliament) during a dark period for the sixteenth century science. His appointment as professor of mathematics was mainly political: at that point his knowledge of mathematics was definitively insufficient to hold the position. You have to keep this into consideration dear wiki friends. Mgvongoeden (talk) 13:32, 29 December 2015 (UTC)
thread_title:  Concerning John Wallis' article 

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thread_title:  External links modified 

message: The "clergyman and" and the category that I added () were based on other parts of the article.  According to the references, part of the article was derived from , which says that he took orders (see Holy Orders), so please don't remove "clergyman and."  If you don't think the Westminster category should be there, please also remove its mention from the text.  208.95.51.38 (talk) 13:06, 27 October 2017 (UTC)
replies: text: The addition of the descriptor clergyman is trivial and irrelevant, and is not appropriate for the lede which is meant to summarize the  major features of the article. Of course Wallis took Holy Orders, as all Fellows of colleges were required to do. There is no evidence for notability for his work as a clergyman (as there is, for example, in the case of Jeremy Taylor). Wallis is famous for his work as a mathematician. I have transferred this thread from my talk page to the Wallis talk page. Please see the note at the top of my own talk page. Xxanthippe (talk) 21:35, 27 October 2017 (UTC)
replies: text: How is it trivia?  He wouldn't have been summoned to Parliament's assembly to define the doctrine and government of the Church if he were just an average cleric.  They didn't summon mathematicians to Westminster, except for men like Wallis who also happened to mathematicians as well as divines.  208.95.51.38 (talk) 15:42, 30 October 2017 (UTC)
replies: text:  Hi, I'm just a bystander in this debate, but let me contribute another point of view: the lead (or lede) at the top of the article is supposed to summarize the article below. (You can see more at Wikipedia:Manual of Style/Lead section) So, if you think that he was notable as a cleric, then you should first add it to the article with appropriate references (e.g. under the "Contributions" section), and then the first sentence of the article can serve as an adequate summary of the article. Regards, Shreevatsa (talk) 02:41, 31 October 2017 (UTC)
thread_title:  John Wallis 

message: "John Wallis is often known to be a modern-day Newton..."
This is a rather odd thing to say given that he was a contemporary of Isaac Newton. Cocaineninja (talk) 19:05, 5 December 2018 (UTC)
replies: text: Quite right and also not in the source. I've replaced it. --Bill Cherowitzo  (talk) 19:49, 5 December 2018 (UTC)
replies: text: There has recently been some poor quality editing of this article which will have to be assessed. Xxanthippe (talk) 21:33, 5 December 2018 (UTC)
thread_title:  A modern-day Newton? 

message: Hi @Xxanthippe, re: adding the category "British Latinists"; he doesn't seem to be listed as an "English Latinist", although he is categorised within "English logicians". Am I missing something? On the other hand, there isn't a category for "English Latinists", it seems, which isn't good, as he's pre-British really. Jim Killock (talk) 22:32, 26 March 2023 (UTC)
replies: text: What is the definition of a "Latinist"? Xxanthippe (talk) 23:37, 26 March 2023 (UTC)
thread_title:  Latinist