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| {{DISPLAYTITLE:Proof that 22/7 exceeds {{pi}}}}
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| {{Pi box}}
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| [[mathematical proof|Proofs]] of the famous mathematical result that the [[rational number]] 22/7 is greater than [[pi|<span class=texhtml>π</span>]] (pi) date back to antiquity. One of these proofs, more recently developed but requiring only elementary techniques from calculus, has attracted attention in modern mathematics due to its [[mathematical elegance]] and its connections to the theory of [[diophantine approximations]]. Stephen Lucas calls this proof, “One of the more beautiful results related to approximating <span class=texhtml>π</span>”.<ref name="Lucas2005">{{Citation
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| | last = Lucas
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| | first = Stephen
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| | title = Integral proofs that 355/113 > {{pi}}
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| | journal = Australian Mathematical Society Gazette
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| | volume = 32
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| | issue = 4
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| | pages = 263–266
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| | year = 2005
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| | url = http://www.austms.org.au/Publ/Gazette/2005/Sep05/Lucas.pdf
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| | mr = 2176249
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| | zbl = 1181.11077}}</ref>
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| Julian Havil ends a discussion of [[continued fraction]] approximations of <span class=texhtml>π</span> with the result, describing it as “impossible to resist mentioning” in that context.<ref>{{Citation
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| | last = Havil
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| | first = Julian
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| | title = Gamma. Exploring Euler's Constant
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| | place = Princeton, NJ
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| | publisher = Princeton University Press
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| | year = 2003
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| | page = 96
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| | mr = 1968276
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| | zbl = 1023.11001
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| | isbn = 0-691-09983-9}}</ref>
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| The purpose of the proof is not primarily to convince its readers that 22/7 {{nowrap|(or 3<sup>1</sup>⁄<sub>7</sub>)}} is indeed bigger than {{pi}}; systematic methods of computing the value of {{pi}} exist.
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| If one knows that {{pi}} is approximately 3.14159, then it trivially follows that {{pi}} < 22/7, which is approximately 3.142857. But it takes much less work to show that {{pi}} < 22/7 by the method used in this proof than to show that {{pi}} is approximately 3.14159.
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| ==Background==
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| 22/7 is a widely used [[Diophantine approximation]] of {{pi}}. It is a convergent in the simple [[continued fraction]] expansion of {{pi}}. It is greater than {{pi}}, as can be readily seen in the [[decimal expansion]]s of these values:
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| :<math>\begin{align}
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| \frac{22}{7} & = 3. \overline{142\,857}, \\
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| \pi\, & = 3.141 \,592\,65\ldots
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| \end{align}</math>
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| The approximation has been known since antiquity. [[Archimedes]] wrote the first known proof that 22/7 is an overestimate in the 3rd century BCE, although he may not have been the first to use that approximation. His proof proceeds by showing that 22/7 is greater than the ratio of the [[perimeter]] of a [[circumscribe]]d [[regular polygon]] with 96 sides to the diameter of the circle. Another rational approximation of {{pi}} that is far more accurate is [[355/113]].
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| == The proof ==
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| The proof can be expressed very succinctly:
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| :<math> 0 < \int_0^1 \frac{x^4(1-x)^4}{1+x^2} \, dx = \frac{22}{7} - \pi. </math>
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| Therefore 22/7 > {{pi}}.
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| The evaluation of this integral was the first problem in the 1968 [[William Lowell Putnam Mathematical Competition|Putnam Competition]].<ref>{{Citation
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| | last = Alexanderson
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| | first = Gerald L.
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| | last2 = Klosinski
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| | first2 = Leonard F.
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| | last3 = Larson
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| | first3 = Loren C. (editors)
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| | title = The William Lowell Putnam Mathematical Competition: Problems and Solutions: 1965–1984
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| | place = Washington, D.C.
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| | publisher = The Mathematical Association of America
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| | year = 1985
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| | url = http://books.google.com/books?id=HNLRgSGZrWMC&pg=PA9&dq=December-7-1968+Putnam+Mathematical-Competition&ei=DZCfR4iRJJu4sgPRu-CwCg&sig=u4-SIYyVtV3rbwN-p56c42BGUKw
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| | isbn = 0-88385-463-5
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| | zbl = 0584.00003}}</ref>
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| It is easier than most Putnam Competition problems, but the competition often features seemingly obscure problems that turn out to refer to something very familiar. This integral also has been used in the entrance examinations for the [[Indian Institutes of Technology]].<ref>[http://entrance.icbse.com/iit-jee/solutions/2010/fiitjee-1.pdf 2010 IIT Joint Entrance Exam], question 38 on page 15 of the mathematics section.</ref>
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| ==Details of evaluation of the integral==
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| That the [[integral]] is positive follows from the fact that the [[integrand]] is a quotient whose numerator and denominator are both [[non-negative]], being sums or products of powers of non-negative [[real numbers]]. Since the integrand is positive, the integral from 0 to 1 is positive because the lower limit of integration is less than the upper limit of integration.
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| It remains to show that the integral in fact evaluates to the desired quantity:
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| : <math>
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| \begin{align}
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| 0 & < \int_0^1\frac{x^4(1-x)^4}{1+x^2}\,dx \\[8pt]
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| & = \int_0^1\frac{x^4-4x^5+6x^6-4x^7+x^8}{1+x^2}\,dx\quad\text{(expansion of terms in the numerator)} \\[8pt]
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| & = \int_0^1 \left(x^6-4x^5+5x^4-4x^2+4-\frac{4}{1+x^2}\right) \,dx \\
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| & {} \qquad \text{(polynomial long division)} \\[8pt]
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| & = \left.\left(\frac{x^7}{7}-\frac{2x^6}{3}+ x^5- \frac{4x^3}{3}+4x-4\arctan{x}\right)\,\right|_0^1 \quad \text{(definite integration)} \\[6pt]
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| & = \frac{1}{7}-\frac{2}{3}+1-\frac{4}{3}+4-\pi\quad (\text{since }\arctan(1) = \pi/4 \text{ and } \arctan(0) = 0) \\[8pt]
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| & = \frac{22}{7}-\pi. \quad \text{(addition)}
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| \end{align}
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| </math>
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| (See [[polynomial long division]].)
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| ==Quick upper and lower bounds==
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| In {{harvtxt|Dalzell|1944}}, it is pointed out that if 1 is substituted for {{math|''x''}} in the denominator, one gets a lower bound on the integral, and if 0 is substituted for {{math|''x''}} in the denominator, one gets an upper bound:<ref>{{citation|last=Dalzell|first=D. P.|year=1944|title=On 22/7|journal=Journal of the London Mathematical Society|volume=19|pages=133–134|mr=0013425|zbl=0060.15306}}.</ref>
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| :<math>\frac{1}{1260} = \int_0^1\frac{x^4 (1-x)^4}{2}\,dx < \int_0^1\frac{x^4 (1-x)^4}{1+x^2}\,dx < \int_0^1\frac{x^4 (1-x)^4}{1}\,dx = {1 \over 630}.</math> | |
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| Thus we have
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| :<math>{22 \over 7} - {1 \over 630} < \pi < {22 \over 7} - {1 \over 1260},</math>
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| hence 3.1412 < {{pi}} < 3.1421 in decimal expansion. The bounds deviate by less than 0.015% from {{pi}}. See also {{harvtxt|Dalzell|1971}}.<ref>{{citation|last=Dalzell|first=D. P.|year=1971|title=On 22/7 and 355/113|journal=Eureka; the Archimedeans' Journal|volume=34|pages=10–13|issn=0071-2248}}.</ref>
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| ==Proof that 355/113 exceeds {{pi}}==
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| As discussed in {{harvtxt|Lucas|2005}}, the well-known Diophantine approximation and far better upper estimate [[355/113]] for {{pi}} follows from the relation
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| :<math>0<\int_0^1\frac{x^8(1-x)^8(25+816x^2)}{3164(1+x^2)}\,dx=\frac{355}{113}-\pi.</math> | |
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| Note that
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| :<math>\frac{355}{113}= 3.141\,592\,92\ldots,</math>
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| where the first six digits after the period agree with those of {{pi}}. Substituting 1 for {{math|''x''}} in the denominator, we get the lower bound
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| :<math>\int_0^1\frac{x^8(1-x)^8(25+816x^2)}{6328}\,dx =\frac{911}{5\,261\,111\,856} = 0.000\,000\,173\ldots,</math>
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| substituting 0 for {{math|''x''}} in the denominator, we get twice this value as an upper bound, hence
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| :<math>\frac{355}{113}-\frac{911}{2\,630\,555\,928}<\pi<\frac{355}{113}-\frac{911}{5\,261\,111\,856}\,.</math>
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| In decimal expansion, this means
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| '''3.141''' '''592''' 57 < {{pi}} < '''3.141''' '''592''' 74, where the bold digits of the lower and upper bound are those of {{pi}}.
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| ==Extensions==
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| The above ideas can be generalized to get better approximations of {{pi}}, see also {{harvtxt|Backhouse|1995}}<ref>{{citation|last=Backhouse|first=Nigel|date=July 1995|title=Note 79.36, Pancake functions and approximations to {{pi}}|journal=The Mathematical Gazette|volume=79|issue=485|pages=371–374|jstor=3618318}}</ref> and {{harvtxt|Lucas|2005}} (in both references, however, no calculations are given). For explicit calculations, consider, for every integer {{math|''n'' ≥ 1}},
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| :<math>
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| \frac1{2^{2n-1}}\int_0^1 x^{4n}(1-x)^{4n}\,dx
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| <\frac1{2^{2n-2}}\int_0^1\frac{x^{4n}(1-x)^{4n}}{1+x^2}\,dx
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| <\frac1{2^{2n-2}}\int_0^1 x^{4n}(1-x)^{4n}\,dx,
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| </math>
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| where the middle integral evaluates to
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| :<math>\begin{align}
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| &\frac1{2^{2n-2}}\int_0^1\frac{x^{4n}(1-x)^{4n}}{1+x^2}\,dx\\
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| &\qquad=\sum_{j=0}^{2n-1}\frac{(-1)^j}{2^{2n-j-2}(8n-j-1)\binom{8n-j-2}{4n+j}}
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| +(-1)^n\biggl(\pi-4\sum_{j=0}^{3n-1}\frac{(-1)^j}{2j+1}\biggr)
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| \end{align}</math>
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| involving {{pi}}. The last sum also appears in [[Leibniz formula for pi|Leibniz' formula for {{pi}}]]. The correction term and [[error bound]] is given by
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| :<math>\begin{align}\frac1{2^{2n-1}}\int_0^1 x^{4n}(1-x)^{4n}\,dx
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| &=\frac{1}{2^{2n-1}(8n+1)\binom{8n}{4n}}\\
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| &\sim\frac{\sqrt{\pi n}}{2^{10n-2}(8n+1)},
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| \end{align}</math>
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| where the approximation (the tilde means that the quotient of both sides tends to one for large {{math|''n''}}) of the [[central binomial coefficient]] follows from [[Stirling's formula]] and shows the fast convergence of the integrals to {{pi}}.
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| <div style="clear:both;width:95%;" class="NavFrame">
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| <div class="NavHead" style="background-color:#FFFAF0; text-align:left; font-size:larger;">Calculation of these integrals</div>
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| <div class="NavContent" style="text-align:left;display:none;">
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| For all integers {{math|''k'' ≥ 0}} and {{math|''l'' ≥ 2}} we have
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| :<math>\begin{align}
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| x^k(1-x)^l&=(1-2x+x^2)x^k(1-x)^{l-2}\\
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| &=(1+x^2)\,x^k(1-x)^{l-2}-2x^{k+1}(1-x)^{l-2}. | |
| \end{align}</math>
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| Applying this formula recursively {{math|2''n''}} times yields
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| :<math>x^{4n}(1-x)^{4n}
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| =(1+x^2)\sum_{j=0}^{2n-1}(-2)^jx^{4n+j}(1-x)^{4n-2(j+1)}+(-2)^{2n}x^{6n}.</math>
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| Furthermore,
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| :<math>\begin{align}
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| x^{6n}-(-1)^{3n}
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| &=\sum_{j=1}^{3n}(-1)^{3n-j}x^{2j}-\sum_{j=0}^{3n-1}(-1)^{3n-j}x^{2j}\\ | |
| &=\sum_{j=0}^{3n-1}\bigl((-1)^{3n-(j+1)}x^{2(j+1)}-(-1)^{3n-j}x^{2j}\bigr)\\
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| &=-(1+x^2)\sum_{j=0}^{3n-1}(-1)^{3n-j}x^{2j},\\
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| \end{align}</math>
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| where the first equality holds, because the terms for {{math|1 ≤ ''j'' ≤ 3''n'' − 1}} cancel, and the second equality arises from the index shift {{math|''j'' → ''j'' + 1}} in the first sum.
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| Application of these two results gives
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| :<math>\begin{align}\frac{x^{4n}(1-x)^{4n}}{2^{2n-2}(1+x^2)}
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| &=\sum_{j=0}^{2n-1}\frac{(-1)^j}{2^{2n-j-2}}x^{4n+j}(1-x)^{4n-2j-2}\\
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| &\qquad{}-4\sum_{j=0}^{3n-1}(-1)^{3n-j}x^{2j}+(-1)^{3n}\frac4{1+x^2}.\qquad(*)
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| \end{align}</math>
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| For integers {{math|''k'', ''l'' ≥ 0}}, using [[integration by parts]] {{math|''l''}} times, we obtain
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| :<math>\begin{align}
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| \int_0^1x^k(1-x)^l\,dx
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| &=\frac l{k+1}\int_0^1x^{k+1}(1-x)^{l-1}\,dx\\
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| &=\cdots\\
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| &=\frac l{k+1} \frac{l-1}{k+2}\cdots\frac1{k+l}\int_0^1x^{k+l}\,dx\\
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| &=\frac{1}{(k+l+1)\binom{k+l}{k}}.\qquad(**)
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| \end{align}</math>
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| Setting {{math|''k'' {{=}} ''l'' {{=}} 4''n''}}, we obtain
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| :<math>\int_0^1 x^{4n}(1-x)^{4n}\,dx
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| =\frac{1}{(8n+1)\binom{8n}{4n}}.</math>
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| Integrating (*) from 0 to 1 using (**) and {{math|arctan(1) {{=}} π/4}}, we get the claimed equation involving {{pi}}.
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| </div>
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| </div>
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| The results for {{math|''n'' {{=}} 1}} are given above. For {{math|''n'' {{=}} 2}} we get
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| :<math>\frac14\int_0^1\frac{x^8(1-x)^8}{1+x^2}\,dx=\pi -\frac{47\,171}{15\,015}</math>
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| and
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| :<math>\frac18\int_0^1 x^8(1-x)^8\,dx=\frac1{1\,750\,320},</math>
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| hence '''3.141''' '''592''' 31 < {{pi}} < '''3.141''' '''592''' 89, where the bold digits of the lower and upper bound are those of {{pi}}. Similarly for {{math|''n'' {{=}} 3}},
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| :<math>\frac1{16}\int_0^1\frac{x^{12}(1-x)^{12}}{1+x^2}\,dx= \frac{431\,302\,721}{137\,287\,920}-\pi</math>
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| with correction term and error bound
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| :<math>\frac1{32}\int_0^1 x^{12}(1-x)^{12}\,dx=\frac1{2\,163\,324\,800},</math>
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| hence '''3.141''' '''592''' '''653''' 40 < {{pi}} < '''3.141''' '''592''' '''653''' 87. The next step for {{math|''n'' {{=}} 4}} is
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| :<math>\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}</math>
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| with
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| :<math>\frac1{128}\int_0^1 x^{16}(1-x)^{16}\,dx=\frac1{2\,538\,963\,567\,360},</math>
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| which gives '''3.141''' '''592''' '''653''' '''589''' 55 < {{pi}} < '''3.141''' '''592''' '''653''' '''589''' 96.
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| ==See also==
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| *[[Approximations of π|Approximations of {{pi}}]]
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| *[[Chronology of computation of π|Chronology of computation of {{pi}}]]
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| *[[Proof that π is irrational|Proof that {{pi}} is irrational]]
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| *[[Lindemann–Weierstrass theorem]] (proof that {{pi}} is transcendental)
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| *[[List of topics related to π|List of topics related to {{pi}}]]
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| ==References==
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| {{reflist|2}}
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| ==External links==
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| * [http://web.archive.org/web/20070730150749/http://www.kalva.demon.co.uk/putnam/putn68.html The problems of the 1968 Putnam competition], with this proof listed as question A1.
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| * [http://www.cecm.sfu.ca/%7Ejborwein/pi-slides.pdf The Life of Pi] by Jonathan Borwein—see page 5 for this integral.
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| {{DEFAULTSORT:Proof That 22 7 Exceeds}}
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| [[Category:Pi]]
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| [[Category:Article proofs]]
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