Discover the Cute and Simple Formula for Pi Derived from the Dirchlet Function

[Goongyae]

Not sure if this interests anyone (maybe it's too basic?), but

pi = (5^1.25/2) * (2/sqrt(5)) * (3/sqrt(10)) * (5/sqrt(25)) * (7/sqrt(50)) * (11/sqrt(120)) * ...

Each radicand is the square of the corresponding prime number numerator, but rounded to the nearest mutliple of 5.

I derived this by evaluating the Dirchlet function with character (1,-1,-1,1,0), basically 1 - 2^-2 - 3^-2 + 4^-2 + 6^-2 -7^-2 -8^-2 +9^-2 +...

 

[Raphie]

pi = (5^1.25/2) * (2/sqrt(5)) * (3/sqrt(10)) * (5/sqrt(25)) * (7/sqrt(50)) * (11/sqrt(120)) * ...

A very cool formula.

Far less sophisticated, but still perhaps "cute," is the following exact formula for 2*pi by substitution:

2*pi
= 10 arccos (Golden Ratio /2)
= (sqrt (4*|T_13 - 1| + 1) + 1)/2))*arccos (sqrt (4*|T_0 - 1| + 1) + 1)/2))/(sqrt (4*|T_2 - 1| + 1) + 1)/2))

Interesting to me only insofar as A) this construction led to the following: Conjecture: Sophie Germain Triangles & x | 2y^2 + 2y - 3 = z^2 https://www.physicsforums.com/showthread.php?t=462793 Note: the sqrt (4*T_x + 1) is in N form relates to Triangular Numbers such that (T_x -1)/2 is Triangular (Sophie Germain Triangular numbers) and B) 0 and 13 are the only two integer solutions to:

(p'_x*p'_2x)*(M_x - |T_x - 1|)/(T_(M_x) - T_|T_x - 1|) = z = 1

NOTE: If we allow the solution to be any n in N, then the solutions are x = 0, 1, 2, 3, 13 and z = 1,6,7,10,1; and the Ramanujan-Nagell Triangular numbers 1, 0, 3, 15, 4095 = T_|T_x - 1|

p'_x = n| 0< d(n)<3 {1 Union Primes}
M_x = Mersenne #
T_x = Triangular #

And, coincidentally...
p'_(2*13) = partition_13 = 101
p'_(2*0) = partition_0 = 1

When you sum all the associated numerators and denominators, you get -1 + -1 + 33546241 + 33546241 = 67092480

67092480 == M_13^d(13) - M_0^d(0) = M_13^2 - M_0^0 = 8191^2 - 0^0
67092480 == 1*(2T_(T_13 - T_1))^2 + 2*(2T_(T_13 - T_1))^1 = 1*8190^2 + 2*8190^1; (also, 8190 == K_24/24 = 196560/24)
67092480 == 2^13 - 2*2^13 = 8192^2 - 2*8192

Or, alternatively...

67092480 == 1 * 2 * 4 * 8 * 15 * 26 * 42 * 64, this series of numbers being the Cake Numbers (Division of 3-Space) from Cake_0 through Cake_7 [totient (7) + 7 = 13 & 13 - pi (13) = 7 & 7 * 13 = T_(13 + 0) = (7+0)*F_(7+0) for F_n a Fibonacci Number]

 

[Goongyae]

Cool, I don't know much about Sophie Germaine numbers but I'll look into it!

Here's a forumla made using stuff I learned on Domingo Gomez's page http://domingogomez.web.officelive.com/gcf.aspx . This formula amounts to treating cos(x) as an infinite polynomial in x^2 and then extracting the minimum root from it

It's hard to write though (please ignore the underscores):


pi^2 = ___________________________ 4
________________---------------------------------------------------------
___________________________________(-1/8!)+...
___________________________ (1/6!)+ ----------------
___________________________________ (1/2!)+...
______________________(-1/4!)+ -------------------------
___________________________________ (-1/4!)+...
___________________________ (1/2!) + --------------
___________________________________ (1/2!)+...
_______________ (1/2!)+ ------------------------------------
___________________________________ (1/6!)+...
___________________________ (-1/4!)+ ----------------
___________________________________ (1/2!)+...
______________________ (1/2!)+ ---------------
___________________________________ (-1/4!)+...
___________________________ (1/2!) + --------------
___________________________________ (1/2!)+...


Using only the terms shown leads to the approximation pi = sqrt(13664/1385)=3.1409...

 

[Raphie]

Cool, I don't know much about Sophie Germaine numbers but I'll look into it!

The real question for me is this: Is there a way to relate Sophie Germain Triangles and Sophie Germain primes, associated with safe primes (2p + 1). The first 3 Sophie Germain primes are 2, 3 & 5, corresponding with the safe primes 5, 7, 11, which might aptly be termed "Ono Primes" of Hausdorff Dimension 0 (Geometric Correspondence --> Point).

And what role, if so, might pi or 2*pi play in that?

We already know that Sophie Germain Triangles are related to the square root of 2, Triangular numbers of the form 3*T_n + 1 with the square root of 3, and I'm pretty sure that Triangular numbers of the form 5*T_n + 1 can be associated with the square root of 5.

Worth mentioning since you are already playing around with pi in relation to the primes.RF

2*3*5, by the way, == totient 31, while 13 == sqrt (pi (31) + 31 + p_31) = sqrt (11 + 31 + 127), 13 & 31 being the first and last Ono primes associated with Hausdorff Dimension 1 (Geometric Correspondence --> Line). Oh, and as an interesting curio... (31 + 127 + 8191) = 8349 == M_5 + M_7 + M_13 == partition_32 == partition (sigma 31), where 31 is the 11th prime number.

 

[Goongyae]

Hmmm... I understand the definition of Sophie Germain or safe primes but having trouble relating them to anything else in a significant way :(
13,17, and 19 are neither Sophie Germain nor safe and would probably be excluded from any pattern you are suggesting.

BTW, a regularized value for the product 2x3x5x7x11x... is 4pi^2. While for 1x2x3x4x... it is sqrt(2pi). The former is based on expressing the prime zeta function as a Moebius inversion of log(zeta(s)) and taking the derivative. The latter is just exp(-zeta'(0)).

 

[Goongyae]

>pi = (5^1.25/2) * (2/sqrt(5)) * (3/sqrt(10)) * (5/sqrt(25)) * (7/sqrt(50)) * (11/sqrt(120)) * ...
>Each radicand is the square of the corresponding prime number numerator, but rounded to the nearest mutliple of 5.

FYI I was able to calculate the following.

With prime numbers, as above, it was:

(2/sqrt(5)) * (3/sqrt(10)) * (5/sqrt(25)) * (7/sqrt(50)) * (11/sqrt(120)) * ...
= 2pi * 5^-1.25
= 0.840363...

But if we admit all integers >= 2, it is:

(2/sqrt(5)) * (3/sqrt(10)) * (4/sqrt(15)) * (5/sqrt(25)) * ...
=sqrt{ [5^1.5/pi * sqrt((5-sqrt(5))/2)] / [exp(2pi/5)+exp(-2pi/5)+ ((sqrt(5)+1)/2)]}
= sqrt(5^(7/4) / (pi*sqrt(G)*(2cosh(pi/5)+G))) where G=golden ratio=1.618...
=0.878877...

I think this can be done for any Dirichlet function and positive integral parameter.
For example, for the character 1,0,-1,0:
pi/4 = 2/2 * 3/4 * 5/4 * 7/8 * 11/12 * ...
where each denominator is the corresponding prime rounded to the nearest multiple of 4 (and in the case of 2, which is equidistant from 0 and 4, we use the average value 2).

and if we use all integers >= 2,
2/2*3/4*4/4*5/4*6/6*7/8*8/8*9/8*...= 0.75/1*1.25/1*1.75/2*2.25/2*...=gamma(1)*gamma(1)/gamma(0.75)/gamma(1.25)
=sqrt(8)/pi=0.9003...

Another way to see the above is to use divergent series (care must be exercised) and note that
2*3*4*5*... = sqrt(2pi)
2*4*4*4*6*8*8*8... = 2*4*6*8*... * (4*8*12*...)^2
=2^(1+1+1+...) * sqrt(2pi) * (4^(1+1+1+...)*sqrt(2pi))^2
=sqrt(pi) * pi/2
Hence the quotient (2*3*4*5*...)/(2*4*4*4*6*8*8*8...) = sqrt(8)/pi

 

[Raphie]

Here's a "fun" formula for pi, based on the Central Binomial Coefficients, that might be of interest to someone or the other...

lim n --> infinity 2^((8k + 1)/2)/(sqrt (2*k^2)*C(2k, k)^2) = pi

This derivation takes its own sweet time converging to pi. For instance, if one looks at "where one is" for n = 38...

((sqrt(2)^((8 * 38) + 1)) / (sqrt (2*38^2)*6892620648693261354600^2))
= 3.1623285635

... one could easily be fooled into thinking one was looking at a formula convergent to the square root of 10 (= 3.1622776601683...). The numbers have to get pretty big before you start to see where you're headed.

((sqrt(2)^((8 * 200) + 1)) / (sqrt (2*200^2)*102952500135414432972975880320401986757210925381077648234849059575923332372651958598336595518976492951564048597506774120^2))
= 3.14552209570

One kind of "cool" (to me, anyway...) coincidence related to this specific derivation is that if you take the floor of the first differences of a modified version of the formula convergent to sqrt (2*n^2) * pi...

e.g.

floor [2^((8(k+1) + 1)/2)/(C(2(k+1), (k+1))^2)]
-
floor [2^((8k + 1)/2)/(C(2k, k)^2)]

Then the first 15 values you get are as follows:
4, 5, 4, 4, 5, 4, 5, 4, 5, 4, 4, 5, 4, 5, 4

The musically inclined may well recognize the distribution of those 4's and 5's because it's the Piano Keyboard Sequence starting from D going either forwards or backwards (for 4, then a White Key. For 5, then a black key). It breaks down after that, but if one were to set k = to n (modulo 12), you'd not only have a nice little programmable musical loop that you could use to play the chromatic scale in ascending or descending order, but you'd also have a number sequence (less 4) that shows up in the expansions of both pi and pi^2 in the Golden base.