Source and context of $\frac{22}{7} - \pi = \int_0^1 (x-x^2)^4 dx/(1+x^2)$? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T21:12:51Z http://mathoverflow.net/feeds/question/67384 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/67384/source-and-context-of-frac227-pi-int-01-x-x24-dx-1x2 Source and context of $\frac{22}{7} - \pi = \int_0^1 (x-x^2)^4 dx/(1+x^2)$? Noam D. Elkies 2011-06-09T21:43:41Z 2012-01-24T22:23:42Z <p>Possibly the most striking proof of Archimedes's inequality $\pi &lt; 22/7$ is an integral formula for the difference: $$\frac{22}{7} - \pi = \int_0^1 (x-x^2)^4 \frac{dx}{1+x^2},$$ where the integrand is manifestly positive. This formula is "well-known" but its origin remains somewhat mysterious. I ask:</p> <blockquote> <p>Who discovered this integral, and in what context?</p> </blockquote> <p>The earliest reference I know of is Problem A-1 on the 29th Putnam Exam (1968). According to J.H.McKay's report in the <em>American Math. Monthly</em> (Vol.76 (1969) #8, 909-915), the Questions Committee consisted of N.D.Kazarinoff, Leo Moser, and Albert Wilansky. Is one of them the discoverer, and if so which one?</p> <p>The printed solution, both in the <em>Monthly</em> article and in the book by Klosinski, Alexanderson, and Larson, says only "The standard approach, from elementary calculus, applies. By division, rewrite the integrand as a polynomial plus a rational function with numerator of degree less than 2. The solution follows easily." But surely there's more to be said, because this integral is a minor miracle of mathematics:</p> <p>$\bullet$ Not only is the integrand manifestly positive, but it is always small: $x-x^2 \in [0,1/4]$ for $x \in [0,1]$, and the denominator $1+x^2$ is at least 1, so $(x-x^2)^4/(x^2+1) &lt; 1/4^4 = 1/256$. A better upper bound on the integral is $\int_0^1 (x-x^2)^4 dx$, which comes to $1/630$ either by direct expansion or by recognizing the Beta integral $B(5,5)=4!^2/9!$. Hence $\frac{22}{7} - \pi &lt; 1/630$, which also yields Archimedes's lower bound $\pi > 3\frac{10}{71}$.</p> <p>$\bullet$ The "standard approach" explains how to evaluate the integral, but not why the answer is so simple. When we expand $$\frac{(x-x^2)^4}{1+x^2} = x^6 - 4x^5 + 5x^4 - 4x^2 + 4 - \frac4{x^2+1},$$ the coefficient of $x/(x^2+1)$ vanishes, so there's no $\log 2$ term in the integral. [This much I can understand: the numerator $(x-x^2)^4$ takes the same value $(1\pm i)^4 = -4$ at both roots of the denominator $x^2+1$.] When we integrate the polynomial part, we might expect to combine fractions with denominators of 2, 3, 4, 5, 6, and 7, obtaining a complicated rational number. But only 7 appears: there's no $x$ or $x^3$ term; the $x^4$ coefficient 5 kills the denominator of 5; and the terms $-4x^5-4x^2$ might have contributed denominators of 6 and 3 combine to yield the integer $-1$.</p> <p>Compare this with the next such integrals $$\int_0^1 (x-x^2)^6 \frac{dx}{1+x^2} = \frac{38429}{13860} - 4 \log 2$$ and $$\int_0^1 (x-x^2)^8 \frac{dx}{1+x^2} = 4\pi - \frac{188684}{15015},$$ which yield better but much more complicated approximations to $\log 2$ and $\pi$...</p> <p>This suggests a refinement of the "in what context" part of the question:</p> <blockquote> <p>Does that integral for $(22/7)-\pi$ generalize to give further approximations to $\pi$ (or $\log 2$ or similar constants) that are useful for the study of Diophantine properties of $\pi$ (or $\log 2$ etc.)?</p> </blockquote> http://mathoverflow.net/questions/67384/source-and-context-of-frac227-pi-int-01-x-x24-dx-1x2/67388#67388 Answer by Michael Renardy for Source and context of $\frac{22}{7} - \pi = \int_0^1 (x-x^2)^4 dx/(1+x^2)$? Michael Renardy 2011-06-09T22:29:56Z 2012-01-24T22:23:42Z <p>Wikipedia quotes the following source: D.P. Dalzell, On 22/7, <em>J. London Math. Soc.</em> <strong>19</strong> (1944), 133–134, and the <a href="http://www.ams.org/mathscinet-getitem?mr=13425" rel="nofollow">MathSciNet review</a> says "By the use of integral calculus the author establishes the inequalities ${\textstyle\frac{22}{7}}-{\textstyle\frac 1{1260}}>\pi>{\textstyle\frac{22}{7}}-{\textstyle\frac 1{630}}$. He then proceeds to develop a series $\pi={\textstyle\frac{22}{7}}+\sum_{n-1}^\infty a_n$, where the $a_n$'s are less in magnitude than the terms of a geometric series of ratio ${\textstyle\frac 1{1024}}$." The latter link is cited by <a href="http://dx.doi.org/10.4169/193009709X469922" rel="nofollow">S.K. Lucas, Approximations to $\pi$ derived from integrals with nonnegative integrands, <em>Amer. Math. Monthly</em> <strong>116</strong> (2009), no. 2, 166–172</a>, whose abstract is as follows: "An intriguing definite integral due to Dalzell equals $22/7 - \pi$ where the integrand is nonnegative, and can be used to derive an infinite series for $\pi$. Here we extend Dalzell's results in two ways. First we look at a new family of integrals leading to series for π that converge arbitrarily fast. Then we show how integrals with nonnegative integrands can be found that equal $z - \pi$ or $\pi - z$ for any real $z$." The paper can be downloaded from the author's page <a href="http://educ.jmu.edu/~lucassk/Papers/more%20on%20pi.pdf" rel="nofollow">here</a> and it contains some further historical account from the 1960s.</p> http://mathoverflow.net/questions/67384/source-and-context-of-frac227-pi-int-01-x-x24-dx-1x2/67395#67395 Answer by Andrey Rekalo for Source and context of $\frac{22}{7} - \pi = \int_0^1 (x-x^2)^4 dx/(1+x^2)$? Andrey Rekalo 2011-06-09T23:28:45Z 2011-06-09T23:28:45Z <p>Jonathan M. Borwein, David H. Bailey and Roland Girgensohn discuss this and related formulae for $\pi$ in their book <em>"Experimentation in Mathematics"</em> (see Section 1.1, p. 3). They claim that</p> <blockquote> <p>The integral was apparently shown by Kurt Mahler to his students in the mid-1960s, and it had appeared in a mathematical examination at the University of Sydney in November, 1960.</p> </blockquote> <p>They mention also a paper by Beuker who further developed the method of integral representations to obtain the irrationality estimate $$\left|\pi-\frac{p}{q}\right|\geq\frac{1}{q^{21.04...}},$$<br> which holds true for all integers $p$, $q$ with sufficiently large $q$. The exponent $21.04...$ is rather far from being optimal. </p> <p><a href="http://carma.newcastle.edu.au/~jb616/Preprints/Books/MbyE/expbook-II.pdf" rel="nofollow">A draft</a> of the book is freely available on J.M. Borwein's home page.</p> http://mathoverflow.net/questions/67384/source-and-context-of-frac227-pi-int-01-x-x24-dx-1x2/67399#67399 Answer by Gerry Myerson for Source and context of $\frac{22}{7} - \pi = \int_0^1 (x-x^2)^4 dx/(1+x^2)$? Gerry Myerson 2011-06-10T01:17:44Z 2011-06-10T01:17:44Z <p>Generalizations are discussed in S. K. Lucas, Integral approximations to $\pi$ with nonnegative integrands, Amer Math Monthly 116 (2009) 166-172. If you don't have access to the Monthly, you can find a preprint on Lucas' website. Lucas agrees with Wikipedia in citing Dalzell. You might also want to see Lucas' earlier paper, Integral proofs that $355/113\gt\pi$, Gazette Aust. Math. Soc. 32 (2005) 263-266.</p> http://mathoverflow.net/questions/67384/source-and-context-of-frac227-pi-int-01-x-x24-dx-1x2/67410#67410 Answer by Geoff Robinson for Source and context of $\frac{22}{7} - \pi = \int_0^1 (x-x^2)^4 dx/(1+x^2)$? Geoff Robinson 2011-06-10T07:55:31Z 2011-06-10T08:04:18Z <p>It may be covered by the articles referred to in the earlier answers, but if you integrate $$\frac{(x-x^2)^{8k+4}}{1+x^2}$$ over the unit interval (for $k$ a non-negative integer), and rewrite <code>$(x-x^2)^{8k+4}$</code> as <code>$x^{8k+4}(1+x^{2} -2x)^{4k+2}$</code>, then rewrite <code>$2^{4k+2}x^{12k+6}$</code> as <code>$2^{4k+2}(x^{12k+6} +1) -2^{4k+2}$</code>, you can see that you get a rational approximation to $2^{4k} \pi$ with an error less than $4^{-(8k+4)}$, where the rational approximation is the integral of a polynomial with integer coefficients over the unit interval. However the denominator is not usually so straightforward.</p>