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-   -   NEW YOU FORMULATE THAT THEY CALCULATE PI (https://www.mersenneforum.org/showthread.php?t=7806)

ewmayer 2007-04-09 16:30

NEW YOU FORMULATE THAT THEY CALCULATE PI
 
I received the following in an e-mail over the weekend (during which time I was offline) - anybody care to see if there is anything new or at all interesting (e.g. in the sense of "yes, this is the well-known Fu-Bar series expansion...") here? I'm at work, no time to examine it in detail at this time. The integral formulae at least seem likely to be fairly trivial, possibly Fourier series (which often yield Pi-containing results by way of series expansions) in disguise. Anyway, might be some fun math exercises here.

-Ernst

p.s.: please try not to let laughter at the Borat-style mangled english subject line distract you from your work here. ;)

p.p.s.: If someone would be kind enough to format the attached in nice LaTeX-style, that would be appreciated. Off to work...

[code]
==========================================================================
NEW YOU FORMULATE THAT THEY CALCULATE PI
==========================================================================

FIRST FORMULA

infinity n! PI
1 - SUM (--------------------------------) = ------
n=0 n 4
2 * PRODUCT ( 2 * p + 3 )
p=0


==========================================================================

SECOND FORMULA

H[0] = 1


H[n]
H[n + 1] = ---------------------------

1 + ( 1 + H[n]^2 )^(1/2)


2^(n + 2) * H[n] > PI


==========================================================================

THIRD FORMULA

1
INTEGRAL ( ----------------------- ) between -1 and 0
X^2
1 + X + ------
2

that he is equal to PI/2

developing in series the integral it is obtained formulates it

1 1 infinity
1 + ----- + ------- - SUM [ (-1)^(n+1) *
2 2*3 n=1

1 1 1
( ------------------- + ------------------ + -------------------- ) ]
2^(2*n) * (4*n+1) 2^(2*n) * (4*n+2) 2^(2*n+1) * (4*n+3)



that he is equal to PI/2

==========================================================================

FOURTH FORMULA

1
INTEGRAL (----------------------) between -1 and 0
x x^2
1 + ----- + ------
2 4

that he is equal to (2 * SQRT(3) * PI ) / 9



developing in series the integral it is obtained formulates it



1 infinity
1 + ------ + SUM [ (-1)^n *
2*2 n=1

1 1
( ------------------- + -------------------- ) ]
2^(3*n) * (3*n+1) 2^(3*n+1) * (3*n+2)



that he is equal to ( 2 * SQRT(3) * PI ) / 9


==========================================================================

FIFTH FORMULA

A[0] = 4 B[0] = SQRT( 1/2 )


2 * A[n] * B[n]
A[n + 1] = -------------------
1 + B[n]

1 + B[n]
B[n + 1] = SQRT ( ------------ )
2

A[n] > A[n + 1] > PI

==========================================================================
[/code]

T.Rex 2007-04-09 21:54

First formula works
 
PARI/gp:

[CODE]G(N)=
{x=0;
for(n=0,N,y=1;for(p=0,n,y=y*(2*p+3));x=x+(factorial(n)/(2*y)));print((1-x)*4/Pi)}[/CODE]
In TeX, the formula 1 is: [TEX]1-\sum_{n=0}^{\infty}\frac{n!}{\ \ \ \ 2 \prod_{p=0}^{n} (2 p+3)}=\frac{\pi}{4}[/TEX]

Now time to go to bed ...

Tony


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