Proofs of the mathematical result that the rational number 22 / 7 is greater than π (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 π ".[1]
Julian Havil ends a discussion of continued fraction approximations of π with the result, describing it as "impossible to resist mentioning" in that context.[2]
This is not a perfect 22/7 circle, because 22/7 is not a perfect representation of pi.
The purpose of the proof is not primarily to convince its readers that 22 / 7 (or 3+ 1 / 7 ) is indeed bigger than π ; systematic methods of computing the value of π exist. If one knows that π is approximately 3.14159, then it trivially follows that π < 22 / 7 , which is approximately 3.142857. But it takes much less work to show that π < 22 / 7 by the method used in this proof than to show that π is approximately 3.14159.
Background
edit
22 / 7 is a widely used Diophantine approximation of π . It is a convergent in the simple continued fraction expansion of π . It is greater than π , as can be readily seen in the decimal expansions of these values:
22
7
=
3.
142
857
¯
,
π
=
3.141
592
65
…
{\displaystyle {\begin{aligned}{\frac {22}{7}}&=3.{\overline {142\,857}},\\\pi \,&=3.141\,592\,65\ldots \end{aligned}}}
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 regular polygon with 96 sides to the diameter of a circle it circumscribes.[note 1]
The proof can be expressed very succinctly:
0
<
∫
0
1
x
4
(
1
−
x
)
4
1
+
x
2
d
x
=
22
7
−
π
.
{\displaystyle 0<\int _{0}^{1}{\frac {x^{4}\left(1-x\right)^{4}}{1+x^{2}}}\,dx={\frac {22}{7}}-\pi .}
Therefore, 22 / 7 > π .
The evaluation of this integral was the first problem in the 1968 Putnam Competition .[4]
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 has also been used in the entrance examinations for the Indian Institutes of Technology .[5]
Details of evaluation of the integral
edit
That the integral is positive follows from the fact that the integrand is non-negative ; the denominator is positive and the numerator is a product of nonnegative numbers. One can also easily check that the integrand is strictly positive for at least one point in the range of integration, say at 1 / 2 . Since the integrand is continuous at that point and nonnegative elsewhere, the integral from 0 to 1 must be strictly positive.
It remains to show that the integral in fact evaluates to the desired quantity:
0
<
∫
0
1
x
4
(
1
−
x
)
4
1
+
x
2
d
x
=
∫
0
1
x
4
−
4
x
5
+
6
x
6
−
4
x
7
+
x
8
1
+
x
2
d
x
expansion of terms in the numerator
=
∫
0
1
(
x
6
−
4
x
5
+
5
x
4
−
4
x
2
+
4
−
4
1
+
x
2
)
d
x
using polynomial long division
=
(
x
7
7
−
2
x
6
3
+
x
5
−
4
x
3
3
+
4
x
−
4
arctan
x
)
|
0
1
definite integration
=
1
7
−
2
3
+
1
−
4
3
+
4
−
π
with
arctan
(
1
)
=
π
4
and
arctan
(
0
)
=
0
=
22
7
−
π
.
addition
{\displaystyle {\begin{aligned}0&<\int _{0}^{1}{\frac {x^{4}\left(1-x\right)^{4}}{1+x^{2}}}\,dx\\[8pt]&=\int _{0}^{1}{\frac {x^{4}-4x^{5}+6x^{6}-4x^{7}+x^{8}}{1+x^{2}}}\,dx&{\text{expansion of terms in the numerator}}\\[8pt]&=\int _{0}^{1}\left(x^{6}-4x^{5}+5x^{4}-4x^{2}+4-{\frac {4}{1+x^{2}}}\right)\,dx&{\text{ using polynomial long division}}&\\[8pt]&=\left.\left({\frac {x^{7}}{7}}-{\frac {2x^{6}}{3}}+x^{5}-{\frac {4x^{3}}{3}}+4x-4\arctan {x}\right)\,\right|_{0}^{1}&{\text{definite integration}}\\[6pt]&={\frac {1}{7}}-{\frac {2}{3}}+1-{\frac {4}{3}}+4-\pi \quad &{\text{with }}\arctan(1)={\frac {\pi }{4}}{\text{ and }}\arctan(0)=0\\[8pt]&={\frac {22}{7}}-\pi .&{\text{addition}}\end{aligned}}}
(See polynomial long division .)
Quick upper and lower bounds
edit
In Dalzell (1944) , it is pointed out that if 1 is substituted for x in the denominator, one gets a lower bound on the integral, and if 0 is substituted for x in the denominator, one gets an upper bound:[6]
1
1260
=
∫
0
1
x
4
(
1
−
x
)
4
2
d
x
<
∫
0
1
x
4
(
1
−
x
)
4
1
+
x
2
d
x
<
∫
0
1
x
4
(
1
−
x
)
4
1
d
x
=
1
630
.
{\displaystyle {\frac {1}{1260}}=\int _{0}^{1}{\frac {x^{4}\left(1-x\right)^{4}}{2}}\,dx<\int _{0}^{1}{\frac {x^{4}\left(1-x\right)^{4}}{1+x^{2}}}\,dx<\int _{0}^{1}{\frac {x^{4}\left(1-x\right)^{4}}{1}}\,dx={1 \over 630}.}
Thus we have
22
7
−
1
630
<
π
<
22
7
−
1
1260
,
{\displaystyle {\frac {22}{7}}-{\frac {1}{630}}<\pi <{\frac {22}{7}}-{\frac {1}{1260}},}
hence 3.1412 < π < 3.1421 in decimal expansion. The bounds deviate by less than 0.015% from π . See also Dalzell (1971) .[7]
Proof that 355/113 exceeds π
edit
As discussed in Lucas (2005) , the well-known Diophantine approximation and far better upper estimate 355 / 113 for π follows from the relation
0
<
∫
0
1
x
8
(
1
−
x
)
8
(
25
+
816
x
2
)
3164
(
1
+
x
2
)
d
x
=
355
113
−
π
.
{\displaystyle 0<\int _{0}^{1}{\frac {x^{8}\left(1-x\right)^{8}\left(25+816x^{2}\right)}{3164\left(1+x^{2}\right)}}\,dx={\frac {355}{113}}-\pi .}
355
113
=
3.141
592
92
…
,
{\displaystyle {\frac {355}{113}}=3.141\,592\,92\ldots ,}
where the first six digits after the decimal point agree with those of π . Substituting 1 for x in the denominator, we get the lower bound
∫
0
1
x
8
(
1
−
x
)
8
(
25
+
816
x
2
)
6328
d
x
=
911
5
261
111
856
=
0.000
000
173
…
,
{\displaystyle \int _{0}^{1}{\frac {x^{8}\left(1-x\right)^{8}\left(25+816x^{2}\right)}{6328}}\,dx={\frac {911}{5\,261\,111\,856}}=0.000\,000\,173\ldots ,}
substituting 0 for x in the denominator, we get twice this value as an upper bound, hence
355
113
−
911
2
630
555
928
<
π
<
355
113
−
911
5
261
111
856
.
{\displaystyle {\frac {355}{113}}-{\frac {911}{2\,630\,555\,928}}<\pi <{\frac {355}{113}}-{\frac {911}{5\,261\,111\,856}}\,.}
In decimal expansion, this means 3.141 592 57 < π < 3.141 592 74 , where the bold digits of the lower and upper bound are those of π .
Extensions
edit
The above ideas can be generalized to get better approximations of π ; see also Backhouse (1995) [8] and Lucas (2005) (in both references, however, no calculations are given). For explicit calculations, consider, for every integer n ≥ 1 ,
1
2
2
n
−
1
∫
0
1
x
4
n
(
1
−
x
)
4
n
d
x
<
1
2
2
n
−
2
∫
0
1
x
4
n
(
1
−
x
)
4
n
1
+
x
2
d
x
<
1
2
2
n
−
2
∫
0
1
x
4
n
(
1
−
x
)
4
n
d
x
,
{\displaystyle {\frac {1}{2^{2n-1}}}\int _{0}^{1}x^{4n}(1-x)^{4n}\,dx<{\frac {1}{2^{2n-2}}}\int _{0}^{1}{\frac {x^{4n}(1-x)^{4n}}{1+x^{2}}}\,dx<{\frac {1}{2^{2n-2}}}\int _{0}^{1}x^{4n}(1-x)^{4n}\,dx,}
where the middle integral evaluates to
1
2
2
n
−
2
∫
0
1
x
4
n
(
1
−
x
)
4
n
1
+
x
2
d
x
=
∑
j
=
0
2
n
−
1
(
−
1
)
j
2
2
n
−
j
−
2
(
8
n
−
j
−
1
)
(
8
n
−
j
−
2
4
n
+
j
)
+
(
−
1
)
n
(
π
−
4
∑
j
=
0
3
n
−
1
(
−
1
)
j
2
j
+
1
)
{\displaystyle {\begin{aligned}{\frac {1}{2^{2n-2}}}&\int _{0}^{1}{\frac {x^{4n}(1-x)^{4n}}{1+x^{2}}}\,dx\\[6pt]={}&\sum _{j=0}^{2n-1}{\frac {(-1)^{j}}{2^{2n-j-2}(8n-j-1){\binom {8n-j-2}{4n+j}}}}+(-1)^{n}\left(\pi -4\sum _{j=0}^{3n-1}{\frac {(-1)^{j}}{2j+1}}\right)\end{aligned}}}
involving π . The last sum also appears in Leibniz' formula for π . The correction term and error bound is given by
1
2
2
n
−
1
∫
0
1
x
4
n
(
1
−
x
)
4
n
d
x
=
1
2
2
n
−
1
(
8
n
+
1
)
(
8
n
4
n
)
∼
π
n
2
10
n
−
2
(
8
n
+
1
)
,
{\displaystyle {\begin{aligned}{\frac {1}{2^{2n-1}}}\int _{0}^{1}x^{4n}(1-x)^{4n}\,dx&={\frac {1}{2^{2n-1}(8n+1){\binom {8n}{4n}}}}\\[6pt]&\sim {\frac {\sqrt {\pi n}}{2^{10n-2}(8n+1)}},\end{aligned}}}
where the approximation (the tilde means that the quotient of both sides tends to one for large n ) of the central binomial coefficient follows from Stirling's formula and shows the fast convergence of the integrals to π .
Calculation of these integrals: For all integers k ≥ 0 and ℓ ≥ 2 we have
x
k
(
1
−
x
)
ℓ
=
(
1
−
2
x
+
x
2
)
x
k
(
1
−
x
)
ℓ
−
2
=
(
1
+
x
2
)
x
k
(
1
−
x
)
ℓ
−
2
−
2
x
k
+
1
(
1
−
x
)
ℓ
−
2
.
{\displaystyle {\begin{aligned}x^{k}(1-x)^{\ell }&=(1-2x+x^{2})x^{k}(1-x)^{\ell -2}\\[6pt]&=(1+x^{2})\,x^{k}(1-x)^{\ell -2}-2x^{k+1}(1-x)^{\ell -2}.\end{aligned}}}
Applying this formula recursively 2n times yields
x
4
n
(
1
−
x
)
4
n
=
(
1
+
x
2
)
∑
j
=
0
2
n
−
1
(
−
2
)
j
x
4
n
+
j
(
1
−
x
)
4
n
−
2
(
j
+
1
)
+
(
−
2
)
2
n
x
6
n
.
{\displaystyle x^{4n}(1-x)^{4n}=\left(1+x^{2}\right)\sum _{j=0}^{2n-1}(-2)^{j}x^{4n+j}(1-x)^{4n-2(j+1)}+(-2)^{2n}x^{6n}.}
Furthermore,
x
6
n
−
(
−
1
)
3
n
=
∑
j
=
1
3
n
(
−
1
)
3
n
−
j
x
2
j
−
∑
j
=
0
3
n
−
1
(
−
1
)
3
n
−
j
x
2
j
=
∑
j
=
0
3
n
−
1
(
(
−
1
)
3
n
−
(
j
+
1
)
x
2
(
j
+
1
)
−
(
−
1
)
3
n
−
j
x
2
j
)
=
−
(
1
+
x
2
)
∑
j
=
0
3
n
−
1
(
−
1
)
3
n
−
j
x
2
j
,
{\displaystyle {\begin{aligned}x^{6n}-(-1)^{3n}&=\sum _{j=1}^{3n}(-1)^{3n-j}x^{2j}-\sum _{j=0}^{3n-1}(-1)^{3n-j}x^{2j}\\[6pt]&=\sum _{j=0}^{3n-1}\left((-1)^{3n-(j+1)}x^{2(j+1)}-(-1)^{3n-j}x^{2j}\right)\\[6pt]&=-(1+x^{2})\sum _{j=0}^{3n-1}(-1)^{3n-j}x^{2j},\end{aligned}}}
where the first equality holds, because the terms for 1 ≤ j ≤ 3n – 1 cancel, and the second equality arises from the index shift j → j + 1 in the first sum.
Application of these two results gives
x
4
n
(
1
−
x
)
4
n
2
2
n
−
2
(
1
+
x
2
)
=
∑
j
=
0
2
n
−
1
(
−
1
)
j
2
2
n
−
j
−
2
x
4
n
+
j
(
1
−
x
)
4
n
−
2
j
−
2
−
4
∑
j
=
0
3
n
−
1
(
−
1
)
3
n
−
j
x
2
j
+
(
−
1
)
3
n
4
1
+
x
2
.
(
1
)
{\displaystyle {\begin{aligned}{\frac {x^{4n}(1-x)^{4n}}{2^{2n-2}(1+x^{2})}}=\sum _{j=0}^{2n-1}&{\frac {(-1)^{j}}{2^{2n-j-2}}}x^{4n+j}(1-x)^{4n-2j-2}\\[6pt]&{}-4\sum _{j=0}^{3n-1}(-1)^{3n-j}x^{2j}+(-1)^{3n}{\frac {4}{1+x^{2}}}.\qquad (1)\end{aligned}}}
For integers k , ℓ ≥ 0 , using integration by parts ℓ times, we obtain
∫
0
1
x
k
(
1
−
x
)
ℓ
d
x
=
ℓ
k
+
1
∫
0
1
x
k
+
1
(
1
−
x
)
ℓ
−
1
d
x
⋮
=
ℓ
k
+
1
ℓ
−
1
k
+
2
⋯
1
k
+
ℓ
∫
0
1
x
k
+
ℓ
d
x
=
1
(
k
+
ℓ
+
1
)
(
k
+
ℓ
k
)
.
(
2
)
{\displaystyle {\begin{aligned}\int _{0}^{1}x^{k}(1-x)^{\ell }\,dx&={\frac {\ell }{k+1}}\int _{0}^{1}x^{k+1}(1-x)^{\ell -1}\,dx\\[6pt]&\,\,\,\vdots \\[6pt]&={\frac {\ell }{k+1}}{\frac {\ell -1}{k+2}}\cdots {\frac {1}{k+\ell }}\int _{0}^{1}x^{k+\ell }\,dx\\[6pt]&={\frac {1}{(k+\ell +1){\binom {k+\ell }{k}}}}.\qquad (2)\end{aligned}}}
Setting k = ℓ = 4n , we obtain
∫
0
1
x
4
n
(
1
−
x
)
4
n
d
x
=
1
(
8
n
+
1
)
(
8
n
4
n
)
.
{\displaystyle \int _{0}^{1}x^{4n}(1-x)^{4n}\,dx={\frac {1}{(8n+1){\binom {8n}{4n}}}}.}
Integrating equation (1) from 0 to 1 using equation (2) and arctan(1) = π / 4 , we get the claimed equation involving π .
The results for n = 1 are given above. For n = 2 we get
1
4
∫
0
1
x
8
(
1
−
x
)
8
1
+
x
2
d
x
=
π
−
47
171
15
015
{\displaystyle {\frac {1}{4}}\int _{0}^{1}{\frac {x^{8}(1-x)^{8}}{1+x^{2}}}\,dx=\pi -{\frac {47\,171}{15\,015}}}
and
1
8
∫
0
1
x
8
(
1
−
x
)
8
d
x
=
1
1
750
320
,
{\displaystyle {\frac {1}{8}}\int _{0}^{1}x^{8}(1-x)^{8}\,dx={\frac {1}{1\,750\,320}},}
hence 3.141 592 31 < π < 3.141 592 89 , where the bold digits of the lower and upper bound are those of π . Similarly for n = 3 ,
1
16
∫
0
1
x
12
(
1
−
x
)
12
1
+
x
2
d
x
=
431
302
721
137
287
920
−
π
{\displaystyle {\frac {1}{16}}\int _{0}^{1}{\frac {x^{12}\left(1-x\right)^{12}}{1+x^{2}}}\,dx={\frac {431\,302\,721}{137\,287\,920}}-\pi }
with correction term and error bound
1
32
∫
0
1
x
12
(
1
−
x
)
12
d
x
=
1
2
163
324
800
,
{\displaystyle {\frac {1}{32}}\int _{0}^{1}x^{12}(1-x)^{12}\,dx={\frac {1}{2\,163\,324\,800}},}
hence 3.141 592 653 40 < π < 3.141 592 653 87 . The next step for n = 4 is
1
64
∫
0
1
x
16
(
1
−
x
)
16
1
+
x
2
d
x
=
π
−
741
269
838
109
235
953
517
800
{\displaystyle {\frac {1}{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
1
128
∫
0
1
x
16
(
1
−
x
)
16
d
x
=
1
2
538
963
567
360
,
{\displaystyle {\frac {1}{128}}\int _{0}^{1}x^{16}(1-x)^{16}\,dx={\frac {1}{2\,538\,963\,567\,360}},}
which gives 3.141 592 653 589 55 < π < 3.141 592 653 589 96 .
See also
edit
^ Proposition 3: The ratio of the circumference of any circle to its diameter is less than 3+ 1 / 7 but greater than 3+ 10 / 71 .[3]
Citations
edit
^ Lucas, Stephen (2005), "Integral proofs that 355/113 > π " (PDF) , Australian Mathematical Society Gazette , 32 (4): 263–266, MR 2176249 , Zbl 1181.11077
^ Havil, Julian (2003), Gamma. Exploring Euler's Constant , Princeton, NJ: Princeton University Press, p. 96, ISBN 0-691-09983-9 , MR 1968276 , Zbl 1023.11001
^ Archimedes (2002) [1897], "Measurement of a circle", in Heath, T.L. (ed.), The Works of Archimedes , Dover Publications, pp. 93–96, ISBN 0-486-42084-1
^ Alexanderson, Gerald L. ; Klosinski, Leonard F.; Larson, Loren C., eds. (1985), The William Lowell Putnam Mathematical Competition: Problems and Solutions: 1965–1984 , Washington, DC: The Mathematical Association of America, ISBN 0-88385-463-5 , Zbl 0584.00003
^ 2010 IIT Joint Entrance Exam , question 41 on page 12 of the mathematics section.
^ Dalzell, D. P. (1944), "On 22/7", Journal of the London Mathematical Society , 19 (75 Part 3): 133–134, doi :10.1112/jlms/19.75_part_3.133 , MR 0013425 , Zbl 0060.15306 .
^ Dalzell, D. P. (1971), "On 22/7 and 355/113", Eureka; the Archimedeans' Journal , 34 : 10–13, ISSN 0071-2248 .
^ Backhouse, Nigel (July 1995), "Note 79.36, Pancake functions and approximations to π ", The Mathematical Gazette , 79 (485): 371–374, doi :10.2307/3618318 , JSTOR 3618318 , S2CID 126397479
External links
edit