Pi and Log(2) in a mirror

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The world of Pi - V2.57
modif. 13/04/2013

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Benoit Cloitre

$p$ and $Log(2)$ in a mirror

Benoit Cloitre, still very creative, is pursuing his chronicles about resemblance between famous constants, after those between $e$ et $p$ .

1 For $p$
2 For $Log(2)$
Proof for Pi
Proof for $Log(2)$
Checking under Pari-GP

1 For $p$

Let $f1 = x /= 2$ and $fn = n(nfn--11)+ 1$ then $( ) limn --> oo ff2n2n---2n2n--12-= xx--12 p2$ . Thus, if we write down

---n(n---1)--- fn = 1+ 1 + -(n-(1n-) 2(n)-(n2-)3) 1+...+1+...2.1- x

then we obtain an inverse Brounker-like continued fraction.

2 For $Log(2)$

Let $u1 = x /= 2$ and $2 un = (nu-n1-)1 + 1$ then $limn -->o o 2n-1u2n = Log(2) + x1-1$ . The equivalent continued fraction is

2 un = 1 +---(n---1)2--- 1+ 1+(n(-n2-)3.)2..- ...+1+1.x1

Proof for Pi

It is easy to see by induction that $f2n-2n-1 ( 1 )(x-1)p rod n 4k2 f2n-2n-2 = 1 + 2n x-2 k=14k2--1$ and we recognise the Wallis product.

Proof for $Log(2)$

Let $---1-- -1- Q(n) = 2n-u2n- x-1$ . We easily see that for $n > 1$ , $sum k+1 Q(n + 1)- Q(n) = 2n(21n+1) = 21n - 21n+1 ==> Q(n +1) = 2nk+=11 (-1)k-----> Log(2)$ .

Checking under the software Pari-GP

For Pi

: x=171679;u=x;for(n=2,100,u=n*(n-1)/u+1;if(n%2==0,print1((u-n-1)/(u-n)-prod(k=1,n/2,4*k^2/(4*k^2-1))*(1+1/n)*(x-1)/(x-2),",")))
,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,

For log(2)

: f(n)=if(n<2,171679,(n-1)^2/f(n-1)+1);Q(n)=1/(2*n-f(2*n))-1/(171679-1);for(n=1,10,print1(Q(n+1)-sum(k=1,2*n+1,(-1)^(k+1)/k),","))
0,0,0,0,0,0,0,0,0,0,

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and in a mirror

1 For

2 For