Pi Formula
Name:
Anonymous
2018-10-08 21:55
Can you prove that
Pi = lim 2*(2^n / (n!/k!/(n/2)!))^2
[as n approaches infinity]
???
Name:
Anonymous
2018-10-08 21:56
! - factorial
^ - pow
lim means the bigger is n, the closer it gets to the value on the left.
Name:
Anonymous
2018-10-08 22:11
k = n/2
Name:
Anonymous
2018-10-09 0:11
The correct answer is:
no
Name:
Anonymous
2018-10-09 0:23
import math
def bin(n,k):
return math.factorial(n)/math.factorial(k)/math.factorial(n-k)
def pi(n):
v = math.pow(2.0, n)/bin(n,n/2)
return v*v*2/n
print("%f\n" % pi(256)) #prints 3.147735
Name:
Anonymous
2018-10-09 0:35
>>3so like (n! / k!^2)
n=2
2* (2^2 / (2! / 1!^2)) ^2
2* (4/ 2)^2 = 8
n=4
2* (4^2 / (4! / 2!^2)) ^2
2* (16/ (24/4))^2 =
2* (16/ 6)^2 = 14.222 (less than 16!?)
Name:
Anonymous
2018-10-09 0:53
for lim n
sqrt(pi / 2) = (2^n / (n!/k!/(n/2)!)
sqrt(pi / 2) * (n! / (1/2 n)! ^2) = 2^n
nth_rt( sqrt(pi / 2) * (n! / (1/2 n)! ^2) ) = 2
Name:
Anonymous
2018-10-09 1:33
So nth_rt( n! / k!^2 ) = 2 * 1/nth_rt(sqrt(pi/2))?
Name:
Anonymous
2018-10-09 1:49
lim n nth_rt(sqrt(pi/2))? ~ 1.000__1
lim n (n! / 1/2n!^2) = [1*5] * 2*6 * 3*7 * [4*8] / [1*1] * 2*2 * 3*3 * [4*4] ~2.000__1
Name:
Anonymous
2018-10-09 1:58
Is it false?
lim n nth_rt(2.000__1) != 2* 1/1.000__1
Name:
Anonymous
2018-10-09 4:04
Try again
lim n (n! / 1/2n!^2) = [1*5] * 2*6 * 3*7 * [4*8] / [1*1] * 2*2 * 3*3 * [4*4] ~2.000__1 ^n (-x?) ?
lim n nth_rt(2.000__1 ^n) ~= 2* 1/1.000__1
Name:
Anonymous
2018-10-09 4:09
Pi Formula: \(\frac{22}{7}\)
Name:
Anonymous
2018-10-09 6:50
>>12Might as well say
\(\frac{314}{100}\)
Name:
Anonymous
2018-10-09 7:10
You're all mental midgets. It was proven already in 1738 by Moivre, whose education consisted mainly of rote memorizing Bible:
https://en.wikipedia.org/wiki/Normal_distribution#Historyand you can't prove it using modern education.
Name:
Anonymous
2018-10-09 8:03
π = 4(½!)²
Name:
Anonymous
2018-10-09 8:12
>>14mental midgets
Admit your role
!
Name:
Anonymous
2018-10-09 8:50
>>15using gamma function is cheating.
Name:
Anonymous
2018-10-09 15:06
┌─────
│ ∞
π = │⎲
│⎳6/n²
⎷ n=1
Name:
Anonymous
2018-10-09 18:16
Name:
Anonymous
2018-10-09 19:20
[math] my anus
Name:
Anonymous
2018-10-09 19:41
Name:
Anonymous
2018-10-09 19:50
>>21Looks like a non
\(\TeX\) way of trying to restate the Basel problem.
Name:
Anonymous
2018-10-09 21:15
>>18Tried copy pasting that into text editor, and it crashed.
Math is hard.
Name:
Anonymous
2018-10-09 21:40
Name:
Anonymous
2018-10-10 1:04
define Pi = lim 2*(2^n / (n!/k!/(n/2)!))^2
Pi == Pi
Name:
Anonymous
2018-10-10 1:22
Is this a new one? Does it work?
define Pi =
nth_rt( 2 ^ floor(n/2) * 3 ^ floor(n/3) * 5 ^ floor(n/5) * ...)
Name:
Anonymous
2018-10-10 5:16
Name:
Anonymous
2018-10-11 10:57
>>8worth checking
x = 2^5
x^(1/5) = 2
(5x)^(1/5) = 2.759..
(5x)^(1/5) * (1/ 5^(1/5)) = 2
Name:
Anonymous
2018-10-11 11:30
>>26probably needs to be like 4nth_rt
define testFact =
2 ^ (floor(n/2) + sum([1:floor(log(2,n)] .- 1)) * 3 ...)
factx =
2 ^ (floor(n/2) + floor(n/2^2) + floor(n/2^3) + ..n/log(2,n)) * ...
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