Sign up ×
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.

Could anyone try to prove that the below conjectured formula is valid for relating $\pi$ with ALL of its convergents - those, which are described in OEIS via A002485(n)/A002486(n) ?

$$(-1)^n\cdot(\pi - \text{A002485}(n)/\text{A002486}(n))$$ $$=(|i|\cdot2^j)^{-1} \int_0^1 \big(x^l(1-x)^m(k+(i+k)x^2)\big)/(1+x^2)\; dx$$

where integer $n = 0,1,2,3,...$ serves as the 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

At this time it could be shown that the formula under question is applicable for some first few convergents (of the A002485(n)/A002486(n) type)

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

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)

yields 618669248999119/196928538206400 - Pi

This question relates to my answer given in Is there an integral that proves $\pi > 333/106$?

share|cite|improve this question
Alot more people would be inclined to try to help if you use latex. This is almost unreadable. – Winther Jul 9 '14 at 5:13
@Winther - done – Alex Jul 9 '14 at 5:21
Also posted at – Joel Reyes Noche Jul 14 '14 at 0:43

1 Answer 1

up vote 2 down vote accepted

Not giving the solution, but some ideas that could lead to it.

If this formula works, it will be because the integral can be decomposed on $P(x) + \frac{1}{1+x^2}$. It's easy to integrate a polynomial (and will give you a rational number), and $1 \over {1+x^2}$ will give you arctan and as such $\pi \over 4$.

So, taking the problem in reverse, you can look at the expansion of arctan between 0 and $\pi \over 4$. You can then convert that polynomial in a continued fraction that will give you the approximations you're looking at.

This reference could help (p10)

I'll try to have a closer look later.

EDIT: Ok, so after peeking through Mathoverflow as well, here is the idea.

Starting from the very general following integral: $I_n = \int_0^1 \frac{x^l(1-x)^m(\alpha + \beta x^2) }{\gamma(1+x^2)}dx$.

As previously discussed, you want to be able to represent this as follows: $\int_0^1 \frac{P(x)(1+x^2) + C}{\gamma(1+x^2)}dx$ with $P(x)$ a polynomial and $C$ a constant.

With some algebra on the polynomials you know that $P(x)= Q(x)(1+x^2) + Ax+B$. So you want to find $A=0$ and if possible get some ideas on $B$.

You need to evaluate the above polynomial on $x=i=e^{i\frac{\pi}{2}}$ and $-i$ to identify the coefficient.

$ A=0 \Leftrightarrow P(i)=P(-i)=0 \Leftrightarrow A = \Im (i^l(1-i)^m(\alpha - \beta) ) = (\alpha - \beta) 2 ^\frac{m}{2} \sin(\frac{l \pi}{2} + \frac{-m\pi}{4} ) = (\alpha - \beta) 2 ^\frac{m}{2} \sin(\frac{\pi}{4} (2l-m) )$

So we have a condition on $2l-m$ to zero $A$ which is $\frac{\pi}{4} (2l-m) = K\pi \Leftrightarrow 2l-m \equiv 0 [4]$.

In particular, $m=2m'$ which we will use going forward and $l-m' \equiv 0[2]$. We can set $l-m' = 2 \epsilon$

Second part is to look at $B$

$B= \Re (i^l(1-i)^m(\alpha - \beta) ) = (\alpha - \beta) 2 ^\frac{m}{2} \cos(\frac{\pi}{4} (2l-m) ) = (\alpha - \beta) 2 ^{m'} \cos(\frac{\pi}{2} (l-m') ) = (\alpha - \beta) 2 ^{m'} (-1)^\epsilon$

So, to sum up, we have $I_n = \int_0^1 Q(x) + \frac{B}{\gamma(1+x^2)}$. As you're trying to approximate $\pi$, you need to take $\gamma = B/4$.

So, provided that $\int_0^1 Q(x) dx$ can be used to approximate fractions of $\pi$, which is likely given the numbers of degress of freedom, we've proved that $I_n$ would be of the following form, which is slightly better as you can drop $j$ from your variables and it shows some relationships better (provided that $\alpha - \beta \not = 0)$:

$(-1)^n (\pi- \frac{p_n}{q_n}) = \int_0^1 \frac{x^{\epsilon+2m'}(1-x)^{2m'}(\alpha + \beta x^2) }{(\alpha - \beta) 2 ^{m'-2} (-1)^{\epsilon}(1+x^2)}dx$.

In your last example, that yields: $n=6,m'=6,\epsilon = -8,\alpha = 77159, \beta = 124360$

I leave it to you to verify it works on the rest of them.

share|cite|improve this answer
@Alex : I've amended the end of my answer so you can see it better. – Matt B. Jul 16 '14 at 13:07
I am not sure you understood my proof. I started with a more general integral than yours, but as you have already computed the values for particular cases, we know that this integral yields the approximation you want as there is a direct relationship between the 2. Then I have shown that for my general integral to work (and by extension yours which a particular case), I had to have several conditions on the different coefficient. This proves that your coefficients will translate into mines for sure. (apart from the case $\alpha = \beta$ that I have left alone. – Matt B. Jul 16 '14 at 17:04
@(Matt B.) In the next two comments I list parameters in your formula for all cases (I replaced alpha by a, bets by b, epsilon by c, m' by p) 104348/33215 - Pi -> a:=1349;b:=-1060;p:=6;c:=0;int((x^(c+2*p)*(1-x)^(2*p)*(a+b*x^2))/((a-b)*2^(p-2)*‌​((-1)^(c)*(1+x^2))),x=0...1) – Alex Jul 18 '14 at 0:59
Pi - 103993/33102 -> a:=124360;b:=77159;p:=6;c:=2;int((x^(c+2*p)*(1-x)^(2*p)*(a+bx^2))/((a-b)*2^(p-2‌​‌​)*((-1)^(c)*(1+x^2))),x=0...1) 355/113 - Pi -> a:=25;b:=816;p:=4;c:=0;int((x^(c+2*p)*(1-x)^(2*p)*(a+bx^2))/((a-b)*2^(p-2)*((-1‌​‌​)^(c)*(1+x^2))),x=0...1) Pi - 333/106 -> a:=197;b:=462;p:=3;c:=-1;int((x^(c+2*p)*(1-x)^(2*p)(a+bx^2))/((a-b)*2^(p-2)*((-‌​‌​1)^(c)*(1+x^2))),x=0...1) 22/7 - Pi -> a:=1;b:=0;p:=2;c:=0;int((x^(c+2*p)*(1-x)^(2*p)*(a+bx^2))/((a-b)*2^(p-2)*((-1)‌​‌​^(c)*(1+x^2))),x=0...1) – Alex Jul 22 '14 at 18:28
Obviously parameters in the formula somehow depend on "n". The most straight forward dependency on "n" is observed in what you call "m'" (and I call "p") ;{2,3,4,4,6,6} ... It appears that when "n" -> infinity - then the integral should come to 0 ... – Alex Jul 22 '14 at 18:29

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.