Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The following integral,

$$ \int_0^1 \frac{x^4(1-x)^4}{x^2 + 1} \mathrm{d}x = \frac{22}{7} - \pi $$

is clearly positive, which proves that $\pi < 22/7$.

Is there a similar integral which proves $\pi > 333/106$?

share|improve this question
    
Hey, how do you know that it is clearly positive? is there something about the integral that makes it positive? –  Tyler Hilton Aug 10 '10 at 20:01
12  
If $f(x)>0$ then $$\int\limits_{a}^{0} f(x)>0$$ –  anonymous Aug 10 '10 at 20:06
5  
@Affan, you're integrating a fraction of even powers: it can't be negative. –  Andrea Ambu Aug 11 '10 at 7:13
1  
You can use $\pi = \int_0^1 \frac{4}{1+x^2}$. Now the power series of $\frac{4}{1+x^2}$ is alternating, thus stopping after odd/even numbers of terms gives under and overevaluations. Thus, for all $a< \pi <b$ you can find $m, n$ so that $a< \int_0^1 P_m(x) dx < \pi < \int_0^1 P_n(x) < b$, where $P_n$ denotes a Taylor Polynomial. –  N. S. Aug 3 '12 at 13:25
add comment

3 Answers 3

up vote 61 down vote accepted

This integral would do the job:

$$\int_0^1 \frac{x^5(1-x)^6(197+462x^2)}{530(1+x^2)}= \pi -\frac{333}{106}$$

  • Also you can refer to S.K. Lucas Integral proofs that $355/113 > \pi$, Gazette, Aust. Math. Soc. 32 (2005), 263-266.

  • This is the link. (Thanks to lhf for pointing out.)

share|improve this answer
1  
Great! I had not expected anyone to answer this so quickly! –  anon Aug 9 '10 at 21:10
6  
How does one come up with that integral, may I ask? –  ShreevatsaR Aug 9 '10 at 21:33
5  
math.jmu.edu/~lucassk/Papers/more%20on%20pi.pdf Please read this article –  anonymous Aug 9 '10 at 21:38
2  
The link above is broken. The current one is educ.jmu.edu/~lucassk/Papers/more%20on%20pi.pdf –  lhf Jun 10 '11 at 19:05
    
@lhf: Thanks for the link. –  user9413 Jun 10 '11 at 19:09
add comment

Although this is not exactly an answer to the question, it seems sufficiently related to mention: there are some direct generalizations, given on the Wikipedia page about this integral. For instance, $$0 < \frac14\int_0^1\frac{x^8(1-x)^8}{1+x^2}\ dx=\pi -\frac{47171}{15015}$$

In general, $$\frac1{2^{2n-1}}\int_0^1 x^{4n}(1-x)^{4n}\ dx <\frac1{2^{2n-2}}\int_0^1\frac{x^{4n}(1-x)^{4n}}{1+x^2}\ dx <\frac1{2^{2n-2}}\int_0^1 x^{4n}(1-x)^{4n}\ dx$$

which for $n=1$ (the integral in the question) gives slightly better bounds than just $\pi < 22/7$: $$ \frac{1}{1260} < \frac{22}{7} - \pi < \frac{1}{630}$$

share|improve this answer
1  
Thank you! Doesn't that imply that pi is irrational? –  anon Aug 9 '10 at 21:17
2  
@muad: Without checking the asymptotics of the respective numerators and denominators, I'm not sure. It doesn't follow simply from the fact that the left and right expressions go to 0: for instance 1 is rational, but you could still find sequences of rational numbers $x_n, y_n, z_n$ such that $x_n$ and $y_n$ go to 0, but $x_n < 1 - z_n < y_n$ for all $n$. –  ShreevatsaR Aug 9 '10 at 21:31
1  
@anon For this to work, you'd have to show that the rational approximants converged suitably fast. See Dirichlet's irrationality test. –  A Walker Oct 19 '11 at 6:23
add comment

In the beginning of 2009 I was posting re similar issue at several sites, namely, at sci.math.symbolic, www.math.utexas.edu, etc.

To repeat: In Paper 1 Lucas found, by brute-force search using Maple programming, several different variants of integral identities which relate each of several first Pi convergents (described in terms of OEIS sequences as A002485(n)/A002486(n)) to Pi.

Further, in my above-mentioned postings, I conjectured the following identity below, which represents a generalization of Stephen Lucas' experimentally obtained identities between Pi and its convergents:

$$(-1)^n\cdot(\pi - \text{A002485}(n)/\text{A002486}(n))$$

$$=(|i|\cdot2^j)^{-1} \int_0^1 \big(x^l(1-x)^m(k+(i+k)x^2)\big)/(1+x^2)\; dx$$

where integer n = 0,1,2,3,... serves as index for terms in OEIS A002485(n) and A002486(n), and {i, j, k, l, m} are some integers (to be found experimentally or otherwise), which are probably some functions of n.

The "interesting" (I think) part of my generalization conjecture is that "i" is present in both:

denominator of the coefficient in front of the integral and in the body of the integral itself

For example for $\frac{22}{7}$

$$\frac{22}{7} - \pi = \int_{0}^{1}\frac{x^4(1-x)^4}{1+x^2}\,\mathrm{d}x$$

with $n=3, i=-1, j=0, k=1, l=4, m=4$ - with regards to my above suggested generalization.

In Maple notation

i:=-1; j:=0; k:=1; l:=4; m:=4;Int(x^l*(1-x)^m*(k+(k+i)x^2)/((1+x^2)(abs(i)*2^j)),x= 0...1)

yields 22/7 - Pi

It also works for found by Lucas

http://www.math.jmu.edu/~lucassk/Papers/more%20on%20pi.pdf

formula for $\frac{333}{106}$

$$\pi - \frac{333}{106} = \frac{1}{530}\int_{0}^{1}\frac{x^5(1-x)^6(197+462x^2)}{1+x^2}\,\mathrm{d}x$$

with $n=4, i=265, j=1, k=197, l=5, m=6$ -with regards to my above suggested generalization.

In Maple notation i:=265; j:=1; k:=197; l:=5; m:=6;Int(x^l*(1-x)^m*(k+(k+i)x^2)/((1+x^2)(abs(i)*2^j)),x= 0...1)

yields Pi - 333/106

And it works for Lucas's formula for $\frac{355}{113}$

$$\frac{355}{113} - \pi = \frac{1}{3164}\int_{0}^{1}\frac{(x^8(1-x)^8(25+816x^2)}{(1+x^2)}$$

with $n=5, i=791, j=2, k=25, l=8, m=8$ -with regards to my above suggested generalization.

In Maple notation

i:=791; j:=2; k:=25; l:=8; m:=8;Int(x^m*(1-x)^l*(k+(k+i)x^2)/((1+x^2)(abs(i)*2^j)),x= 0...1)

yields 355/113 - Pi

And it works as well for Lucas's formula for $\frac{103993}{33102}$

$$\pi - \frac{103993}{33102} = \frac{1}{755216}\int_{0}^{1}\frac{x^{14}(1-x)^{12}(124360+77159x^2)}{1+x^2}\,\mathrm{d}x$$

with $n=6, i= -47201, j=4, k=124360, l=14, m=12$ -with regards to my above suggested generalization.

In Maple notation

i:=-47201; j:=4; k:=124360; l:=14; m:=12;Int(x^l*(1-x)^m*(k+(k+i)x^2)/((1+x^2)(abs(i)*2^j)),x= 0...1)

yields Pi - 103993/33102

And also it works Lucas's formula for $\frac{104348}{33215}$

$$\frac{104348}{33215} - \pi = \frac{1}{38544}\int_{0}^{1}\frac{x^{12}(1-x)^{12}(1349-1060x^2)}{1+x^2}\,\mathrm{d}x$$

with $n=7, i= -2409, j=4, k=1349, l=12, m=12$ - with regards to my above suggested generalization.

In Maple notation

i:=-2409; j:=4; k:=1349; l:=12; m:=12;Int(x^l*(1-x)^m*(k+(k+i)x^2)/((1+x^2)(abs(i)*2^j)),x= 0...1)

yields 104348/33215 - Pi

And it works as well for $\frac{618669248999119}{196928538206400}$

$$\frac{618669248999119}{196928538206400} - \pi = \frac{1}{755216}\int_{0}^{1}\frac{x^{14}(1-x)^{12}(77159+124360x^2)}{1+x^2}\,\mathrm{d}x$$

with $n=6, i= 47201, j=4, k=77159, l=14, m=12$ -with regards to my above suggested generalization.

In Maple notation

i:=47201; j:=4; k:=77159; l:=14; m:=12;Int(x^l*(1-x)^m*(k+(k+i)x^2)/((1+x^2)(abs(i)*2^j)),x= 0...1)

618669248999119/196928538206400 - Pi

I do not have computer math resources (Mathematica, Maple, etc.) to experimentally prove or disprove it for all larger n (but see my comment below).

Best Regards, Alexander R. Povolotsky

share|improve this answer
    
One also could check and see that my above generalization formula also applies to identities obtained by Jaume Oliver Lafont, described in the "Following Lucas (2009)" section in oeis.org/wiki/User:Jaume_Oliver_Lafont/… –  Alex Apr 8 '12 at 21:20
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.