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A sequence of semiconvergent approximations to $\pi$ is given by fractions

$$3, 4, \frac{7}{2}, \frac{10}{3}, \frac{13}{4}, \frac{16}{5}, \frac{19}{6}, \frac{22}{7}, \frac{25}{8}, \frac{47}{15}, \frac{69}{22}, \frac{91}{29}, \frac{113}{36}, \frac{135}{43}, \frac{157}{50}$$

Some related Dalzell-type integrals are

$$\pi=3+2\int_0^1\frac{x(1-x)^2}{1+x^2}dx$$

$$\pi=4-4\int_0^1 \frac{x^2}{1+x^2}dx$$

$$\pi=\frac{10}{3}-\int_0^1 \frac{(1-x)^4}{1+x^2}dx$$

$$\pi=\frac{16}{5}-\int_0^1 \frac{x^2(1-x)^2(1+2x+x^2)}{1+x^2}dx$$

$$\pi=\frac{19}{6}-2\int_0^1 \frac{x^3(1-x)^2}{1+x^2}dx$$

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

$$\pi=\frac{22}{7}-\frac{1}{28}\int_0^1\frac{x(1-x)^8(2+7x+2x^2)}{1+x^2}dx$$

$$\pi=\frac{25}{8}+\frac{1}{4}\int_0^1\frac{x(1-x)^4(1+4x+x^2)}{1+x^2}dx$$

$$\pi=\frac{47}{15}+\int_0^1 \frac{x^2(1-x)^4}{1+x^2}dx$$

$$\pi=\frac{47}{15}+2\int_0^1 \frac{x^5(1-x)^2}{1+x^2}dx$$

$$\pi=\frac{157}{50}+\frac{1}{5}\int_0^1 \frac{x^3(1-x)^4(4-5x+4x^2)}{1+x^2} dx$$

The latter one was obtained as an answer to a recent question by Paramanand Singh.

Which are similar integrals for the other fractions listed?

Jaume Oliver Lafont
  • 5,212

1 Answers1

1

What about

$\displaystyle \pi=\frac{16}{5}+\frac{1}{10}\int_0^1 \frac{(1-x)^5(2x^2-5x-3)}{1+x^2}\,dx$

$\displaystyle \pi=\frac{7}{2}-\int_0^1 \frac{(1-x)(3x^2-2x+1)}{1+x^2}\,dx$

?

ADDENDUM 1:

What about

$\displaystyle \pi=\frac{7}{2}-\frac{1}{2}\int_0^1 \frac{(1-x)^2(3x^2-4x+3)}{1+x^2}\,dx$

?

ADDENDUM 2:

$\displaystyle \pi=\frac{16}{5}-\int_0^1 \frac{x(1-x)^4(2x^2-x+2)}{1+x^2}\,dx$

ADDENDUM 3:

$\displaystyle \pi=\frac{13}{4}-\frac{1}{10}\int_0^1 \frac{(1-x)^6(7x^2+5x+7)}{1+x^2}\,dx$

$\displaystyle \pi=\frac{13}{4}-\frac{1}{2}\int_0^1 \frac{x^2(1-x)^2(5x^2+4x+5)}{1+x^2}\,dx$

ADDENDUM 4:

$\displaystyle \pi=\frac{69}{22}+\frac{2}{11}\int_0^1 \frac{x^2(1-x)^2(8x^3+x^2-3x+1)}{1+x^2}\,dx$

ADDENDUM 5:

I have obtained all the results above using a PARI GP script.

\p 30;WX(m,n)={intnum(x=0,1,x^m*(1-x)^n/(1+x^2))};WY(m,n,a,b,c)={intnum(x=0,1,x^m*(1-x)^n*(a*x^2+b*x+c)/(1+x^2))};pisearch(mm,nn,aa,bb,cc,p,q)={T=matrix(mm+1,nn+1,m,n,WX(m-1,n-1));print("Initialisation completed");for(m=0,mm,for(n=0,nn,for(a=-aa,aa,for(b=-bb,bb,for(c=-cc,cc,if(a!=0 || b!=0 || c!=0,if(length(polrootsreal(a*x^2+b*x+c,[0,1]))==0,U=lindep([Pi,1,a*T[m+3,n+1]+b*T[m+2,n+1]+c*T[m+1,n+1]]);if(U[1]!=0 && abs(U[1])<700 && abs(U[2])<700 && abs(U[3])<700 && (q*U[2]-p*U[1]==0 || q*U[2]+p*U[1]==0),print(m," ",n," ",a," ",b," ",c," ",-U[2]/U[1])))))))));}

parameters: $mm,nn$ are the range of powers in integrals:

$\displaystyle WY(m,n,a,b,c):=\int_0^1 \frac{x^m(1-x)^n(ax^2+bx+c)}{1+x^2}\,dx$

aa,bb are the range respectively for $a,b$. ($-aa\leq a\leq aa$...)

$p,q$ is for the fraction $\frac{p}{q}$.

Suppose you want to search for a rational dependence for $\pi,\frac{13}{4}$ and integral WY(m,n,a,b,c).

launch,

pisearch(8,8,20,20,20,13,4)

one of the results you obtain is:

3 3 14 1 15 13/4

To verify this result:

increase precision,

\p 100

lindep([Pi,1,WY(3,3,14,1,15)])

you obtain:

[-4,13,-4]

Therefore (probably),

$-4\pi+13-4\times WY(3,3,14,1,15)=0$

Therefore (probably),

$\displaystyle \pi=\frac{13}{4}-\int_0^1 \frac{x^3(1-x)^3(14x^2+x+15)}{1+x^2}\,dx$

NB1: All polynomial $ax^2+bx+c$ considered don't have any roots on $[0;1]$.

NB2: all results are only conjectures.

ADDENDUM 6:

$\displaystyle \pi=\frac{2}{7}\int_0^1 \frac{11x^2+25}{1+x^2}\,dx-\frac{22}{7}$

(sorry not useful)

This one is more useful,

$\displaystyle \pi=\frac{22}{7}-\frac{1}{28}\int_0^1 \frac{x(1-x)^8(2x^2+7x+2)}{1+x^2}\,dx$

ADDENDUM 7:

$\displaystyle \pi=\frac{22}{7}-\frac{1}{10}\int_0^1 \frac{x^2(1-x)^5(8x^2-5x+3)}{1+x^2}\,dx$

FDP
  • 13,647
  • All three integrals are proofs, yes. So $a=c$ is not needed. – Jaume Oliver Lafont Oct 25 '17 at 19:57
  • Nice addenda. Are there two integrals with $a=c$ per fraction? – Jaume Oliver Lafont Oct 26 '17 at 08:30
  • i haven't understood the question. There are several others "a=c" integrals for 13/4 and 16/5. – FDP Oct 26 '17 at 16:17
  • Are there? I am not aware of them... Your integral for $\frac{69}{22}$ resembles this one for $\frac{333}{106}$ – Jaume Oliver Lafont Oct 28 '17 at 01:27
  • The last one is already in the question... These integrals can be proven as in https://en.wikipedia.org/wiki/Proof_that_22/7_exceeds_%CF%80 – Jaume Oliver Lafont Oct 31 '17 at 23:43
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    Sorry i didn't notice. There are several other ones for 22/7. I have added one that is not in your list. – FDP Nov 01 '17 at 06:56
  • Nice one. Probably decomposing the polynomial into (3x^2+3) and (5x^2-5x) makes some sense. The corresponding series is

    $$\frac{22}{7}-\pi=\sum_{k=0}^\infty \frac{1152 (48 k^3 + 200 k^2 + 337 k + 255)}{(4 k + 3) (4 k + 4) (4 k + 5) (4 k + 6) (4 k + 7) (4 k + 8) (4 k + 9) (4 k + 10) (4 k + 11) (4 k + 12)}$$

    The numerator is not constant. Next version of the question: are there more than two integrals with constant numerator in the corresponding series?

     – Jaume Oliver Lafont Nov 01 '17 at 08:43
  • Maybe you have to consider : $\displaystyle S_{r,s}:=\sum_{k=0}^\infty \frac{\sum_{n=0}^r a_nk^n}{(4 k + r) (4 k + r+1) ... (4 k + s)}$ and search for rational dependence for numbers $1,\pi,S_{r,s}$. – FDP Nov 01 '17 at 09:49
  • i mean $\displaystyle S_{r,s,t}:=\sum_{k=0}^\infty \frac{k^t}{(4 k + r) (4 k + r+1) ... (4 k + s)}$ – FDP Nov 01 '17 at 09:53