Talk:Proof that 22/7 exceeds π

message: There are plenty of pages that involve math on Wikipedia.  Why is this one getting so much attention?  There is no rule that all encyclopedia articles must be accessible to all people.
"The following argument will be readily understood by persons with no knowledge of mathematics beyond first-year calculus. "
Well, what about those of us who don't even have that? It's all just funny looking symbols to me :) Adam Bishop 03:25, 29 Nov 2003 (UTC)
replies: text: Well, "simple" is a subjective term. I don't think its possible to give a simpler proof than this, because even defining π involves calculus (the length of a curve is the limit of a sum, an integral).  -- Arvindn 03:52, 29 Nov 2003 (UTC)
replies: text: The only part I had trouble with was that... I'm not sure what its called... curly thing with the 1 and 0 Nat2 23:04, 2 February 2006 (UTC)
replies: text: So, in other words, you didn't understand the proof at all!
I suspect that any reasonable proof of the same proposition that avoids knowledge of calculus would be more complicated.  As for calculus being involved in defining π, if someone who knows no calculus asks me what π is, I would not hesitate to answer that it is the ratio of a circle's circumference to its diameter; multiply the diameter by π to get the circumference. Michael Hardy 21:26, 29 Nov 2003 (UTC)

text: Do you know a (formal) definition of circumference that doesn't involve integration? If not, what I said still holds, doesn't it? -- Arvindn 03:27, 30 Nov 2003 (UTC)
replies: text: It can be formally defined in terms of the limit of circumferences of polyhedra. This does need the notion of a limit, but not the fully machinery of integral calculus. However, since the formalization of the limit was done precisely to facilitate the development of calculus, the distinction is perhaps not a strong one in the history of formalism. However, in the history of Pi the earliest estimates of Pi make informal use of limits, but not of integration. (BTW, thanks for the name change from A very elementary...) ~ Jeff 18:33, 1 Dec 2003 (UTC)
thread_title:  Required calculus 

message: The point of this whole page is the funny integral. That is no "simple" proof for the fact that 22/7 exceeds pi, but rather a mathematical joke; the page is relevant and informative, but the caption is misleading at best; this is a page about a putnam competition problem and some subsequent analysis on a particular definite integral. Funny, but miscaptioned. This is not THE proof for that thing; THE proof is indeed naively using the calculator so readily available to anyone who has gotten this far... --sigs
: The article is of interest quite independently of the Putnam Competition.  Of course, the brute-force calculation yields no understanding of this simple argument, and the brute-force calculation is far more complicated than this argument. Michael Hardy 02:50, 20 February 2007 (UTC)
thread_title:  The title is misleading 

message: :0<\int_0^1\frac{x^4(1-x)^4}{1+x^2}\,dx=\frac{22}{7}-\pi.
/ \
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That's not very simple!
::It's a routine calculus problem.  One expects secondary-school students to do such problems in a couple of minutes.  Sheesh.  Michael Hardy 22:46, 14 Mar 2005 (UTC)
replies: text: No, one expects calculus students to do such a problem in a couple minutes, but certainly not your average secondary-school student. I think your time at MIT has warped your sense of reality. Revolver 03:03, 9 September 2005 (UTC)
replies: text: I don't think so.  I never said one expects this of all high-school students.  But at typical high schools, there are some students who take calculus, and are expected to evaluate integrals.  Some of those integrals are challenging, but for this one you just plod through the steps mechanically and you've got it. Michael Hardy 03:40, 2 April 2006 (UTC)
thread_title:  0 < integral of [x^4(1-x)^4/(1+x^2)] from 0 to 1 = 22/7-pi 

message: Although many people know this numerical value of π from school, far fewer know how to compute it.
:I have no idea what this sentence is supposed to mean, or how it is supposed to relate to the rest of the introduction. Revolver 03:10, 9 September 2005 (UTC)
thread_title: statement

message: Does anyone have any idea what's going on behind this result?  I will clarify what I mean, and if anyone knows an answer to my question, it might belong in the article.  Those who don't know calculus might not get much out of what I have to ask - sorry about that.  I certainly don't advocate making the article more confusing, so if that's the only possible result of this, let's drop it.  At this stage though, it can't hurt to ask the question, which follows without further ado:
The integrand expands to a polynomial minus 4*arctan(x).  That polynomial is p(x) = x6 - 4x5 + 5x4 - 4x2 + 4.  Have a look at a graph of that integrand - it's rather symmetrical.  That prettiness leads me to believe that p(x) is some kind of canonical approximation of the function 4*arctan(x), and not just a polynomial that happens to equal 22/7 at x = 1.  It's not a Taylor approximation, or it would be closest at one point and diverge on either side.  What is it, and is there a reason based on numerical approximation theory that one would be certain it's an over-approximation, and that its error should have such a pretty factorization? -GTBacchus(talk) 06:26, 28 November 2005 (UTC)
replies: text: GTBacchus has clarified on my talk page that it is p(x) - 4/(1+x) that is symmetrical — in the range [0,1]. Also, p(x) could only approximate 4/(1+x2), not 4*arctan(x). -- Paddu 18:31, 18 March 2006 (UTC)
thread_title: Question

message: There are, obviously, many different proofs for this result. While this page was loading up, I was fully expecting to see the Archimedean proof given here; it is obviously the most historic and arguably the most notable proof. I think the article would gain from having that proof reproduced here more fully. The proof presented is certainly notable, it a proof I have seen before (and indeed, I am sure far more people have seen the details of this proof than Archimedes') but I have only seen it given as a piece of "fun" maths, an exceptionally elegant method that is presented for instructive purposes on the topics of calculus and polynomial long division, with the interesting historical backdrop that 22/7 was a classical approximation to Pi. Archimedes' proof, on the other hand, was a serious (and relatively important) piece of mathematical research, so it seems a little unbalanced to cover it with a one sentence summary! On a slightly different note, it would be really nice to know who discovered, or at least first published, the particular integral given here; it would add a lot to the context of the article. TheGrappler 13:58, 19 March 2006 (UTC)
thread_title: Details of Archimedes' Proof

message: Does anyone object if I move this article to 22/7 (number)?  The article is really about the number; the proof that it exceeds pi is interesting & notable because 22/7 is an important approximation of pi.  It was suggested on the AfD debate that the article be renamed, and this seems like the best suggestion to me. Mangojuice 15:48, 20 March 2006 (UTC)
replies: text: Absolutely should be moved! This page is not about the proof but rather about a definite integral involved in a putnam competition; imho you should move this under Putnam competition 1968, first assignment or similar and link to it from 22/7 (Number). —The preceding unsigned comment was added by 138.246.7.59 (talk) 21:01, 14 February 2007 (UTC)
replies: text:  It's not about a Putnam problem.  This argument is of interest quite independently of its appearance in the Putnam.  See the AfD discussion.  Obviously the argument has great charm, and that has nothing to do with the Putnam. Michael Hardy 03:53, 15 February 2007 (UTC)

text: I object.  The article is about the proof, not the number.  I can't make heads or tails of your reason for disagreeing.  We might as well argue that the Pythagorean theorem article  is really about right triangles so should be moved to right triangle.  --C S (Talk) 01:02, 21 March 2006 (UTC)

text: I think the best is to keep the main part of the article (proof part), and merge some of the introduction as a stub at 22/7 (number).  Certainly a lot more could be written about 22/7 and its historical importance.  But the rest of the article should be here.--C S (Talk) 01:06, 21 March 2006 (UTC)
replies: text: No, a proof that 85/27 would probably not be as interesting.  85/27 is actually further from π than 22/7 despite the fact that the denominator, 27, is bigger than the other denominator, 7.  You can't come closer to π than 22/7 with any rational number until the denominator is actually more than 100.  I think the fact that 22/7 is a convergent in the continued fraction is relevant here. Michael Hardy 04:02, 30 March 2006 (UTC)

text: I would have to see the proof that 85/27 exceeds pi in order to deem it interesting or not.  Much of the interest of this proof is because of certain mathematical aspects of the proof.  --C S (Talk) 01:02, 21 March 2006 (UTC)
replies: text: OK. We start with the observation that 0<\frac{1189}{189}+\int_0^1\frac{x^4(1-x)^4}{1+x^2}\,dx :) . LambiamTalk 22:00, 22 March 2006 (UTC)
replies: text: \mbox{That should be }\frac{1}{189}\mbox{ instead of }\frac{1189}{189} \mbox{.} -- Paddu 08:10, 1 April 2006 (UTC)
thread_title: Move suggestion

message: This article is not primarily about that number.  It's about the fact that this proof is though-provoking and causes one to suspect some deeper and enlightening fact is a large iceberg whose tip is this argument. Michael Hardy 04:07, 30 March 2006 (UTC)
thread_title: Don't move to "22/7"

message: It would be useful if Diophantine approximation explained why this gadget works, which it doesn't now; in fact, it's rather stubby. (That is the title, the present link from the article is a redirect.) Septentrionalis 15:36, 16 June 2006 (UTC)
thread_title: Diophantine approximation

message: In the 'details' section, shouldn't those equal signs be less than signs? --Carifio24 21:48, 22 July 2006 (UTC)
replies: text: Certainly not.  Why would you say that? Michael Hardy 22:32, 22 July 2006 (UTC)
replies: text: Don't mind me, I didn't really read the whole thing right...wasn't totally awake. Now I've made myself like a fool, great  --Carifio24 00:33, 23 July 2006 (UTC)
thread_title: Equal signs

message: You all know what brought me here, but that isn't what I want to talk about. What I want to say is simply "thanks". Because even though I had seen the proof before, I found it nowhere as succinctly and yet as clearly stated as here. Shinobu (talk) 16:01, 16 December 2007 (UTC)
thread_title:  Thanks, people 

message: I hate to nitpick, but I'm so busy these days I don't have time to make substantial contributions. But I was thinking, would this not more properly be titled "Proof that 22 sevenths exceeds π"? Maybe I'm splitting hairs. --Cheeser1 (talk) 05:12, 17 December 2007 (UTC)
replies: text: I disagree.  I haven't recently looked at Wikipedia:Manual of Style (dates and numbers), but there is a difference between "every number except one" and "every number except 1".  When writing about the number 1, rather than using it in such a sentence as "There is one excecption", I think it should be written as a digit, and this seems to be a somewhat similar case. Michael Hardy (talk) 05:20, 17 December 2007 (UTC)
replies: text: I completely understand your point about "all but one" vs "all but 1" - I don't know that the same could be said here (how could one confuse "sevenths" with say "7ths" or "over 7" - not to mention "over seven"). It's just always been my experience that unless ambiguity is possible, one says "nths" whenever n is a reasonably small integer (as opposed to "over n"). It's not so much whether or not we spell it out, but whether we say "over 7." I am aware that I'm certainly being a bit picky, but I'm just curious about it. I presume a / is not possible in a title, is it? --Cheeser1 (talk) 20:14, 19 December 2007 (UTC)
replies: text: I don't know if a slash is a good idea, but it is certainly possible per WP:Subpages#Slashes in article titles (main namespace).  I don't have any strong opinions on the title, but I would prefer twenty-two sevenths or 22/7 or 22 over 7 rather than 22 sevenths.  I find "22 sevenths" hard to read, but I cannot reliably distinguish 2+b from 2+5 from s+5, so feel free to ignore. JackSchmidt (talk) 20:30, 19 December 2007 (UTC)

text: Fractions are normally spelled out; use the fraction form if they occur in a percentage or with an abbreviated unit (⅛ mm, but never an eighth of a mm) or they are mixed with whole numerals.
Looks to me like it should be spelled out. However, it may also make sense to write 22/7. I am uncertain of what we'd want, although my preference would be 22/7 now that I've realized that a / is acceptable in articlespace titles. --Cheeser1 (talk) 20:59, 19 December 2007 (UTC)

text: However: This is an article title, and those are not as flexible in some respects as the text within the article.  In particular, slashes in article names are used to indicate subpages. Michael Hardy (talk) 22:09, 19 December 2007 (UTC)

text: That's actually not a problem, according to this policy that User:JackSchmidt pointed us to. I remember seeing people repeat "mainspace subpages do not exist" and this appears to be the case. So I suggest renaming it to 22/7 - anyone agree/disagree? --Cheeser1 (talk) 22:28, 19 December 2007 (UTC)
replies: text: I've added a clause to the "Exceptions" in the MOS on numbers, saying: "Numbers in mathematical formulas are never spelled out (3 < π < 22/7, not three < π < 22 sevenths)." (That's what I like about Wikipedia: if you think the rules as formulated get in the way, you can not only ignore but even change them.) It can be plausibly argued that in the context here 22/7 is a formula. Count this as agree.  --Lambiam 09:30, 20 December 2007 (UTC)
thread_title:  Possible rename? 

message: The section Proof that 22/7 exceeds π#Extensions is quite intriguing, particulay these subsections:
The above ideas can be generalized to get better approximations of π. More precisely, for every integer n ≥ 1,
:
\frac1{2^{2n-1}}\int_0^1 x^{4n}(1-x)^{4n}\,dx
<\frac1{2^{2n-2}}\int_0^1\frac{x^{4n}(1-x)^{4n}}{1+x^2}\,dx
<\frac1{2^{2n-2}}\int_0^1 x^{4n}(1-x)^{4n}\,dx,
where the middle integral evaluates to
:\begin{align}
&\frac1{2^{2n-2}}\int_0^1\frac{x^{4n}(1-x)^{4n}}{1+x^2}\,dx\\
&\qquad=\sum_{j=0}^{2n-1}\frac{(-1)^j}{2^{2n-j-2}(8n-j-1)\binom{8n-j-2}{4n+j}}
+(-1)^n\biggl(\pi-4\sum_{j=0}^{3n-1}\frac{(-1)^j}{2j+1}\biggr)
\end{align}
involving π. The last sum also appears in Leibniz' formula for π. The correction term and error bound is given by
:\begin{align}\frac1{2^{2n-1}}\int_0^1 x^{4n}(1-x)^{4n}\,dx
&=\frac{1}{2^{2n-1}(8n+1)\binom{8n}{4n}}\end{align}
.............
The results for n = 1 are given above. For n = 2 we get
:\frac14\int_0^1\frac{x^8(1-x)^8}{1+x^2}\,dx=\pi -\frac{47\,171}{15\,015}
and
:\frac18\int_0^1 x^8(1-x)^8\,dx=\frac1{1750320},
hence  3.14159231 < π < 3.14159289, where the bold digits of the lower and upper bound are those of π. Similarly for n = 3,
:\frac1{16}\int_0^1\frac{x^{12}(1-x)^{12}}{1+x^2}\,dx= \frac{431\,302\,721}{137\,287\,920}-\pi
with correction term and error bound
:\frac1{32}\int_0^1 x^{12}(1-x)^{12}\,dx=\frac1{2\,163\,324\,800},
hence 3.14159265340 < π < 3.14159265387. The next step for n = 4 is
:\frac1{64}\int_0^1\frac{x^{16}(1-x)^{16}}{1+x^2}\,dx= \pi-\frac{741\,269\,838\,109}{235\,953\,517\,800}
with
:\frac1{128}\int_0^1 x^{16}(1-x)^{16}\,dx=\frac1{2\,538\,963\,567\,360},
which gives 3.14159265358955 < π < 3.14159265358996.
Now if, instead of n = 1,2,3,4,..., similar formulas are developed with n=2,4,6,8 - something like:
The above ideas can be generalized to get better approximations of π. More precisely, for every even integer n ≥ 2,
:
\frac1{2^{n-1}}\int_0^1 x^{2n}(1-x)^{2n}\,dx
<\frac1{2^{n-2}}\int_0^1\frac{x^{2n}(1-x)^{2n}}{1+x^2}\,dx
<\frac1{2^{n-2}}\int_0^1 x^{2n}(1-x)^{2n}\,dx,
where the middle integral evaluates to
:\begin{align}
&\frac1{2^{n-2}}\int_0^1\frac{x^{2n}(1-x)^{2n}}{1+x^2}\,dx\\
&\qquad=\sum_{j=0}^{n-1}\frac{(-1)^j}{2^{n-j-2}(4n-j-1)\binom{4n-j-2}{2n+j}}
+(-1)^{n/2}\biggl(\pi-4\sum_{j=0}^{(3n-1)/2}\frac{(-1)^j}{2j+1}\biggr)
\end{align}
involving π. The last sum also appears in Leibniz' formula for π. The correction term and error bound is given by
:\begin{align}\frac1{2^{n-1}}\int_0^1 x^{2n}(1-x)^{2n}\,dx
&=\frac{1}{2^{n-1}(4n+1)\binom{4n}{2n}}\end{align}
.............
The results for n = 2 are given above. For n = 4 we get
:\frac14\int_0^1\frac{x^8(1-x)^8}{1+x^2}\,dx=\pi -\frac{47\,171}{15\,015}
and
:\frac18\int_0^1 x^8(1-x)^8\,dx=\frac1{1750320},
hence  3.14159231 < π < 3.14159289, where the bold digits of the lower and upper bound are those of π. Similarly for n = 6,
:\frac1{16}\int_0^1\frac{x^{12}(1-x)^{12}}{1+x^2}\,dx= \frac{431\,302\,721}{137\,287\,920}-\pi
with correction term and error bound
:\frac1{32}\int_0^1 x^{12}(1-x)^{12}\,dx=\frac1{2\,163\,324\,800},
hence 3.14159265340 < π < 3.14159265387. The next step for n = 8 is
:\frac1{64}\int_0^1\frac{x^{16}(1-x)^{16}}{1+x^2}\,dx= \pi-\frac{741\,269\,838\,109}{235\,953\,517\,800}
with
:\frac1{128}\int_0^1 x^{16}(1-x)^{16}\,dx=\frac1{2\,538\,963\,567\,360},
which gives 3.14159265358955 < π < 3.14159265358996.
then similar relationships involving n = 1,3,5,7,..., will give relationships involving the natural logarithm of 2 (ln 2 - 2/3, 38,429/55,440 - ln 2, ln 2 - 1,290,876,029/1,862,340,480, and 356,281,790,621/514,005,972,480 - ln 2) instead of pi. I shall show that info (verifyable on TI-89 and similar calculators) on this page in the near future.  --Glenn L (talk) 09:32, 4 December 2009 (UTC)
thread_title:  Extensions section with even n? 

message: Is there an article that similarly proves that the value of the natural logarithm of 2 exceeds 2/3 ? It appears that the main article's proof that 22/7 > π can easily be used for this proof:
===Basic idea===
The basic idea behind the proof can be expressed very succinctly:
:0<\int_0^1\frac{x^2(1-x)^2}{1+x^2}\,dx = \log(2) - \frac{2}{3}.
Therefore log (2) > 2⁄3 .
===Details===
:
0\,	<\int_0^1\frac{x^2(1-x)^2}{1+x^2}\,dx              	
   	=\int_0^1\frac{x^2-2x^3+x^4}{1+x^2}\,dx            	(expanded terms in numerator)
   	=\int_0^1 \left(x^2-2x+\frac{2x}{1+x^2}\right) \,dx	(performed polynomial long division)
   	_0^1                                               	(definite integration)
   	=\frac{1}{3}-1+\log(2)\                            	(use log(1) = 0)
   	=\log(2)-\frac{2}{3}.                              	(addition)
-- Glenn L (talk) 18:30, 24 April 2010 (UTC)
thread_title:  Proof that natural log (2) exceeds 2/3 

message: :
0\,	<\int_0^1\frac{x^2(1-x)^2}{1+x^2}\,dx              	
   	=\int_0^1\frac{x^2-2x^3+x^4}{1+x^2}\,dx            	(expanded terms in numerator)
   	=\int_0^1 \left(x^2-2x+\frac{2x}{1+x^2}\right) \,dx	(performed polynomial long division)
   	_0^1                                               	(definite integration)
   	=\frac{1}{3}-1+\log(2)\                            	(use log(1) = 0)
   	=\log(2)-\frac{2}{3}.                              	(addition)
-- Glenn L (talk) 18:30, 24 April 2010 (UTC)
thread_title: Details

message: People really do know how to miss the point completely.  This edit is one of the most extreme cases of stupidly missing the point that ever happened, but I'm finding out this is not an isolated case.  Googling the title of this article shows a yahoo answers page about it.  A huge number of people asserted that "just" (sic!) knowing that π = 3.14159265...... is enough to show that 22/7 > π.  And so it is, but you have to do a lot of work in order to "just" know that.  A lot more work than needs to get done in this article.  It is appalling that so many people are so gullible.  If all the books say that π = 3.14159265......, then they think it's the self-evident innerant word of God.  Appalling.
: 
Should we say something more than we've said in this article to correct that mentality? Michael Hardy (talk) 03:55, 24 December 2010 (UTC)
thread_title:  Missing the point 

message: There is a move discussion in progress on Talk:Proof that π is irrational which affects this page. Please participate on that page and not in this talk page section. Thank you.  —RM bot 17:46, 24 April 2011 (UTC)
thread_title: Move discussion in progress

message: There is a move discussion in progress on Talk:Pi which affects this page. Please participate on that page and not in this talk page section. Thank you.  —RM bot 13:15, 2 June 2011 (UTC)
thread_title: Move discussion in progress

message: In a recent posting to mathoverflow, Harvard professor Noam Elkies wrote about the proof that this article is about:
: 
Is this something one could fairly cite in this article?  Would it help satisfy those who have strangely claimed they see no elegance in this proof (not that those people are active now, but they've made themselves memorable)? Michael Hardy (talk) 20:24, 13 June 2011 (UTC)
thread_title:  "striking" 

message: In the section currently:
:\begin{align}
\pi\,        & \approx 3.141\,592\,65\ldots
\end{align}
Shouldn't it be like this:
:\begin{align}
\pi\,        & = 3.141 \,592\,65\ldots
\end{align}
or this:
:\begin{align}
\pi\,        & \approx 3.141\,592\,65
\end{align}
What's going on in here?  — Preceding unsigned comment added by Th4n3r (talk • contribs) 17:43, 14 March 2013 (UTC)
thread_title:  Background section 

message: Why we use  as approximation of π for primary school? Exessia (talk) 06:03, 30 June 2019 (UTC)
replies: text: Yep, Exessia (talk)? Secondary school may use  or  instead of  for more precise approximations of π. Exessia (talk) 06:07, 30 June 2019 (UTC)
thread_title:  Why we use 22/7 as approximation of pi for primary school?