# Is formula valid for relating $\pi$ with ALL of its OEIS A002485(n)/A002486(n) convergents?

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

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)

yields 618669248999119/196928538206400 - Pi

This question relates to my answer given in http://math.stackexchange.com/questions/1956/is-there-an-integral-that-proves-pi-333-106/127618#127618

UPDATE: Matt B. in his answer to my question on Mathematics Stack Exchange has provided analytical proof and improved my parametric formula by reducing the number of parameters from 5 to 4 (see http://math.stackexchange.com/questions/860499/seeking-proof-for-the-formula-relating-pi-with-its-convergents).

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

Below is the list of parameters in Matt B.'s formula for all cases, covered in Stephen Lucas' publications - the stuff to the right of the arrow sign is the actual Maple code, which one could copy (while in the "edit" mode) and then paste into (let say) Inverse Symbolic Calculator (it accepts Maple code) and run it there.

NB I replaced for brevity some parameter names used by Matt B: "alpha" by "a", "beta" 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)

Pi - 103993/33102 -> a:=124360;b:=77159;p:=6;c:=2;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)

355/113 - Pi -> a:=25;b:=816;p:=4;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)

Pi - 333/106 -> a:=197;b:=462;p:=3;c:=-1;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)

22/7 - Pi -> a:=1;b:=0;p:=2;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)

Obviously parameters in the formula somehow depend on "n". The most straight forward dependency on "n" is observed in what Matt B named as "m'" (and I call it "p") : {2,3,4,4,6,6} ... Note that when "n" -> infinity, then the integral should come to 0 ...

What computer algebra symbolic system (CAS - such as Axiom, Derive, ... etc) would be the best suited to take advantage of established parameterization - to calculate remaining (not covered by Stephen Lucas) A002485(n)/A002486(n) convergents ?

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Possibly somewhat related: mathoverflow.net/questions/67384/… –  j.c. Jul 10 '14 at 13:58
The integral $I_{l,m}=\int_0^1 \frac{x^l(1-x)^m}{1+x^2}\,dx$ is always a linear combination of $1$, $\pi$, and $\log 2$ with rational coefficients. So you can always find $k$ and $i+k$ so that $k I_{l,m}+(i+k)I_{l+2,m}$ is $\alpha+\beta\pi$. I'm not sure how you prove that $\beta/i$ is a power of two, though. –  Kirill Jul 10 '14 at 14:34
It does appear, though, that $x^l(1-x)^m\bmod 1+x^2$ has always zero or a power of two as its constant coefficient. –  Kirill Jul 10 '14 at 14:40
@Kirill, $x^{\ell}(1-x)^m=(1+x^2)p(x)+ax+b$; now evaluate at $x=i$, and note that $1-i=\sqrt2e^{-i\pi/4}$. –  Gerry Myerson Jul 13 '14 at 1:07
May I suggest, Alex, that you learn something about formatting mathematics on this site, as your latest edit is unreadable. Have a look at how the earlier parts of your question are formatted, and/or click on the "help" link on this page. –  Gerry Myerson Jul 24 '14 at 0:41