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



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pict

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

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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