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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 $\mathrm{A002485}(n)/\mathrm{A002486}(n)$ ?

$$ (-1)^n\cdot\left( \pi - \frac{\mathrm{A002485}(n)}{\mathrm{A002486}(n)} \right) = \frac{1}{|i|\cdot2^j} \int_0^1 \frac{x^l(1-x)^{2(j+2)}(k+(i+k)x^2)}{1+x^2} \,\mathrm{d}x \tag{1} $$

and in Maple notations:

$$(-1)^n*(Pi−A002485(n)/A002486(n))=(abs(i)*2^{j})^{}(-1)Int((x^{l}(1-x)^{(2*(j+2))}*(k+(i+k)*x^{2}))/(1+x^{2}),x=0...1)$$

where integer $n = 0,1,2,3,\cdots$ serves as the index for terms in OEIS $\mathrm{A002485}(n)$ and $\mathrm{A002486}(n)$, and $\{i, j, k, l\}$ are some integer parameters (which are some implicit functions of $n$ and so far to be found experimentally for each value of $n$), .

The "interesting" (I think) part of my generalization conjecture is that both "$i$" and "$j$" are present in both: denominator of the coefficient in front of the integral and in the body of the integral itself.

It is shown in examples below that the formula under question is applicable for some first few convergents (of the $\mathrm{A002485}(n)/\mathrm{A002486}(n)$ 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 $n=3, i=-1, j=0, k=1, l=4$ - with regards to my above suggested generalization.

In Maple notation,

i:=-1; j:=0; k:=1; l:=4;
Int(x^l*(1-x)^(2*(j+2))*(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 $n=4, i=265, j=1, k=197, l=5$ - with regards to my above suggested generalization.

In Maple notation

i:=265; j:=1; k:=197; l:=5;
Int(x^l*(1-x)^(2*(j+2))*(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 $n=5, i=791, j=2, k=25, l=8$ - with regards to my above suggested generalization.

In Maple notation

i:=791; j:=2; k:=25; l:=8;
Int(x^(2*(j+2))*(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 $n=6, i= -47201, j=4, k=124360, l=14$ -with regards to my above suggested generalization.

In Maple notation

i:=-47201; j:=4; k:=124360; l:=14;
Int(x^l*(1-x)^(2*(j+2))*(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 $n=7, i= -2409, j=4, k=1349, l=12$ - with regards to my above suggested generalization.

In Maple notation

i:=-2409; j:=4; k:=1349; l:=12;
Int(x^l*(1-x)^(2*(j+2))*(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 $\frac{618669248999119}{196928538206400}$

which, by the way, is not part of A002485/A002486 OEIS sequences:

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

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

In Maple notation

i:=47201; j:=4; k:=77159; l:=14;
Int(x^l*(1-x)^(2*(j+2))*(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 $\pi > 333/106$?

Update#1:

Recently Thomas Baruchel (see his answer below) has conducted extensive calculations and found that four parameters formula yields infinite number of solutions for each $n$.

Thomas shared with me his calculations results and supplied me with quite a few of valid combinations of $i, j, k, l$ values - so now I have a lot of experimentally found five-tuples $\{ n,i, j, k, l\}$, which satisfy above parametrization, where $n$ varies in the range from 2 to 26.

Based on this data, of course, it would be nice to find how (if at all) $i, j, k, l$ are inter-related between each other and with "$n$" - but such inter-relation (if exists) is not obvious and difficult to derive just by observation ... (though it is clearly seen that an absolute value of "$i$" is strongly increasing as "$n$" is growing from 2 to 26).

If I did not make a mistake, RHS could be reduced (after performing integration) to:

(abs(i)*2^j)^(-1)*Gamma(2*j+5)*((k+i)*Gamma(l+3)*HypergeometricPFQ(1,l/2+3/2,l/2+2;j+l/2+4,j+l/2+9/2;-1)/Gamma(2*j+l+8)+k*Gamma(l+1)*HypergeometricPFQ(1,l/2+1/2,l/2+1;j+l/2+3, j+l/2+7/2;-1)/Gamma(2*j+l+6))

May be from discussed parametric identity one could derive irrationality measure for pi, if to assume that RHS in this identity holds true, when the rational fraction on the LHS is equal to 0, then we have:

Pi = (abs(i)*2^j)^(-1)*Gamma(2*j+5)*((k+i)*Gamma(l+3)*HypergeometricPFQ(1,l/2+3/2,l/2+2;j+l/2+4,j+l/2+9/2;-1)/Gamma(2*j+l+8)+k*Gamma(l+1)*HypergeometricPFQ(1,l/2+1/2,l/2+1;j+l/2+3, j+l/2+7/2;-1)/Gamma(2*j+l+6))

Perhaps someone could programmatically check if there are any {i,j,k,l}, which would satisfy above?

Update #2:

Thanks to Jaume Oliver Lafont, at least one case, answering affirmatively to the last question, is identified: i=-1, j=-2, k=1, l=0

$$\pi = \int_{0}^{1}\frac{4}{1+x^2}\,\mathrm{d}x$$

Should there be infinite number of such cases?

share|cite|improve this question
4  
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
3  
@Winther - done – Alex Jul 9 '14 at 5:21
    
Also posted at mathoverflow.net/q/175762/12357 – Joel Reyes Noche Jul 14 '14 at 0:43
    
I fixed some formatting for you; hope you like it. – Sangchul Lee Jan 25 at 19:44
    
Thanks, Sangchul. – Alex Jan 27 at 3:42
up vote 6 down vote accepted
+50

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 http://www.math.binghamton.edu/dikran/478/Ch7.pdf (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

I computed some more values and got convinced this formula isn't relevant yet unless you remove at least the three variables $j$, $l$ and $m$. It is actually very easy to trivially find arbitrarily many tuple $(i,j,k,l,m)$. First of all, if I understand well, your $j$ value seems to be $(m/2-2)$, or am I wrong?

Even after having discarded $j$, it looks like you can find $(i,k)$ solutions for almost any arbitrary tuple $(l,m)$; thus I suggest you find some interesting rule on $(l,m)$ in order to minimize $(i,k)$ if you want your formula to be strong enough. Of course using only one $i$ variable instead of $(i,j,k,l,m)$ would be a significant improvement.

The code below can help you; it is Maxima code yielding thousands and thousand $(i,j,k,l,m)$:

display2d:false$

A002485: [ 3,22,333,355,103993,104348,208341,312689,833719,1146408,4272943,
           5419351,80143857,165707065,245850922,411557987,1068966896,2549491779,
           6167950454,14885392687,21053343141,1783366216531,3587785776203,
           5371151992734,8958937768937 ] $

A002486: [ 1,7,106,113,33102,33215,66317,99532,265381,364913,1360120,
           1725033,25510582,52746197,78256779,131002976,340262731,811528438,
           1963319607,4738167652,6701487259,567663097408,1142027682075,
           1709690779483,2851718461558 ] $

for n:2 thru 26 do (
  for l:1 thru 36 do (
    for m:1 thru 36 do (
      e: expand(integrate(x^l *(1-x)^m*(k+(i+k)*x^2)/(1+x^2), x, 0, 1)),
      myi: coeff(e, i, 1),
      p: coeff(myi, %pi, 1),
      myi2: myi - p*%pi,
      if coeff(myi2, log(2), 1) = 0 then (
        myk: coeff(coeff(e, i, 0),k, 1),
        if myk # 0 then (
          j: round(float(log(abs(p))/log(2))),
          kipos: (-1)^n * (-A002485[n-1]/A002486[n-1])*(2^j) - myi2,
          if kipos # 0 then (
            k1: denom(myk/kipos), i1: num(myk/kipos),
            check1: expand( (k1*myk+p*i1*%pi+myi2*i1)/( abs(i1)*(2^j))*(-1)^(n) - %pi + A002485[n-1]/A002486[n-1] ),
            if check1 = 0 then
              print ("n =",n,"==> (",i1,j,k1,l,m,")" )),
          kineg: -myk/((-1)^n * (-A002485[n-1]/A002486[n-1])*(2^j) + myi2),
          if kineg # 0 then (
            k2: denom(kineg), i2: num(kineg),
            check2: expand( (k2*myk+p*i2*%pi+myi2*i2)/( abs(i2)*(2^j))*(-1)^(n) - %pi + A002485[n-1]/A002486[n-1] ),
            if check2 = 0 then 
              print ("n =",n,"==> (",i2,j,k2,l,m,")" )))))))$

You asked me by mail if I could find some $(i,j,k,l,m)$ solutions for $n=8$, but there is an infinite number of such solutions; for instance:

n = 8 ==> ( -66317 -1 9977 1 2 ) 
n = 8 ==> ( -6963285 3 212651 1 10 ) 
n = 8 ==> ( -66383317 7 833127 1 18 ) 
n = 8 ==> ( -103605391175 11 720252257 1 26 ) 
n = 8 ==> ( -55884747999795 15 517817918873 1 34 ) 
n = 8 ==> ( -66317 0 8805 2 4 ) 
n = 8 ==> ( -2188461 4 89050 2 12 ) 
n = 8 ==> ( -16380299 8 317214 2 20 ) 
n = 8 ==> ( -1305427928805 12 23297755114 2 28 ) 
n = 8 ==> ( 896546759355 14 27371124886 2 32 ) 
n = 8 ==> ( -23756674250925 16 5393931750178 2 36 ) 
n = 8 ==> ( -66317 1 8246 3 6 )
etc.
share|cite|improve this answer
    
Yes, indeed, for all cases covered (included your results for n=8) it appears that j = m/2 -2 and that is why Matt B reduced number of parameters from 5 to 4. – Alex Jan 9 at 17:47
    
And, of course, it would be nice to find some interesting rule on (l,m) in order to minimize (i,k) and, even better, it would be real cool to reduce number of parameters just to one. I was hoping to try to achieve empirically/experimentally - by collecting more data. – Alex Jan 9 at 18:10
    
I appreciate your input, Thomas. – Alex Jan 9 at 21:49

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