new formula for pi

[Martin Doina]

We have :

import numpy as np

import matplotlib.pyplot as plt

# Define the number of iterations (nested loops) to compute

num_iterations = 20

# Initialize the wave function values

psi_values = np.zeros(num_iterations)

psi_values[0] = 1 # Initial condition

# Compute the evolution of the recursive wave equation

for i in range(1, num_iterations):

psi_values[i] = np.sin(psi_values[i-1]) + np.exp(-psi_values[i-1])

print("Computed Ψ(n) values:")

print(psi_values)

Computed Ψ(n) values:

[1. 1.20935043 1.23377754 1.23493518 1.23498046 1.23498221

1.23498228 1.23498228 1.23498228 1.23498228 1.23498228 1.23498228

1.23498228 1.23498228 1.23498228 1.23498228 1.23498228 1.23498228

1.23498228 1.23498228]

we have :

import numpy as np

import matplotlib.pyplot as plt

# Define the number of iterations (nested loops) to compute

num_iterations = 20

# Initialize the wave function values

psi_values = np.zeros(num_iterations)

psi_values[0] = 0.8934691018292812244027 # Initial condition

# Compute the evolution of the recursive wave equation

for i in range(1, num_iterations):

psi_values[i] = np.sin(psi_values[i-1]) + np.exp(-psi_values[i-1])

print("Computed Ψ(n) values:")

print(psi_values)

Computed Ψ(n) values:

[1. 1.20935043 1.23377754 1.23493518 1.23498046 1.23498221

1.23498228 1.23498228 1.23498228 1.23498228 1.23498228 1.23498228

1.23498228 1.23498228 1.23498228 1.23498228 1.23498228 1.23498228

1.23498228 1.23498228]

Eventually with any start number the formula will converge to same value I called Loop Zero LZ=1.23498228

And the start value I called lz0 the next lz1, lz2 ...till LZ.

We observe:

PI EMEERGE AS QUANTIZATION FROM LZ

LZ = 1.23498228

φ = 1.618033988749895 sqrt(φ) = 1.272019649514069 2 * LZ * sqrt(φ) = 2 * 1.23498228 * 1.272019649514069 = 3.141592653589793

It matches π to 15 decimal places.

If we start with

lzo =

0.8934691018292812244027957267340518204164769216500536082639661202175013678

652728144116855653516467769449418678645614476686312366345874100712097550257

565621279831814210603530359668474308122648409382707760279298797639244570277

317763777871619616097501217

lz3 =

1.234883696486107689368331045920205838137290185188156247179344662717329133134006518

14491636715009386015694979913335663300729179674965466765580954136273249418734151988

47687368635527576621019716226571355

Golden ratio φ =

1.618033988749894848204586834365638117720309179805762862135448622705260462818902449

70720720418939113748475408807538689175212663386222353693179318006076672635443338908

65959395829056383226613199282902679

sqrt(φ) =

1.272019649514068964252422461737491491715608041840096248616640382539297575536068011

83038421498846025853851414763672802650571033811881483526492194484574461860433489454

90171119065937154727384550376304265

2 * lz3* sqrt(φ) =

3.141592653589793238462643383279502884197169399375105820974944592307816406286208998

62803482534211706798214808651328230664709384460955058223172535940812848111745028410

27019385211055596446229489549303819

Actual π =

3.141592653589793238462643383279502884197169399375105820974944592307816406286208998

62803482534211706798214808651328230664709384460955058223172535940812848111745028410

2701938521105559644622948954930382

Difference =

6.205495838993736767068679487061693641541207689100011563625855380210530798881524805

30869882429367098306744684140936521781559976510831065397628163802482560518846377146

38370751311799417689080356571908466e-200

With more decimals:

Calculated π =

3.14159265358979323846264338327950288419716939937510582097494459230781640628620899862803482534211706798214808651328230664709384460955058223172535940812848111745028410

27019385211055596446229489549303819

Actual π =

3.141592653589793238462643383279502884197169399375105820974944592307816406286208998

62803482534211706798214808651328230664709384460955058223172535940812848111745028410

2701938521105559644622948954930382

Your take on this please.

[Gareth McCaughan]

On 18/11/2025 14:01, Martin Doina wrote:

We have :

import numpy as np

import matplotlib.pyplot as plt

# Define the number of iterations (nested loops) to compute

num_iterations = 20

# Initialize the wave function values

psi_values = np.zeros(num_iterations)

psi_values[0] = 1 # Initial condition

# Compute the evolution of the recursive wave equation

for i in range(1, num_iterations):

psi_values[i] = np.sin(psi_values[i-1]) + np.exp(-psi_values[i-1])

So you're setting a0 = 1 and a_{n+1} = sin a_n + exp -a_n.

This iteration has a stable fixed point at a value y such that sin y + exp -y = y.

If a_n is between 0 and pi then a_{n+1} is between 0 and 1+1/e, and if a_n is between 0 and 1+1/e then a_{n+1} is between 1 and 1+1/e (and in particular is between 0 and 1+1/e). Further, the derivative of the function taking a_n to a_{n+1} is cos x - exp -x which on that interval has |derivative| < 0.3, so the function is a contraction mapping, and therefore an iteration starting anywhere will converge to y.

(However, you quote two runs, one starting from 1 and one starting from 0.893..., and it looks as if for the latter you have copied-and-pasted the same values as one gets starting from 1.)

PI EMEERGE AS QUANTIZATION FROM LZ

LZ = 1.23498228

φ = 1.618033988749895 sqrt(φ) = 1.272019649514069 2 * LZ * sqrt(φ) = 2 * 1.23498228 * 1.272019649514069 = 3.141592653589793

It matches π to 15 decimal places.

How are you doing that calculation? I get 3.14184545... which doesn't match pi to even 4 decimal places.

Your take on this please.

Apologies for being rude, but my take on it is that if you're going to post about numerical coincidences then you should take the trouble to get the arithmetic right.

(If it turns out that I'm the one who's got the arithmetic wrong and your thing really does match pi to 15 places then of course I apologize profusely and indeed your finding is provocative.)

You also give some results starting from the mysterious value 0.893469..., taking _just three_ iterations, and getting something very close to pi.

If that mysterious value was just chosen to make the result after three iterations very close to pi, this seems entirely uninformative; one could pick pretty much any convergent iteration, any computation that happens to map the limit to something close to an interesting number, and then find a suitable starting point to hit it on the nose. Observe:

Let's take the iteration x -> cos(x), which for all starting values converges rather slowly to about 0.739085; call this value z. 4z is a (very poor) approximation to pi. (It's actually a bit less than 3.)

Can we find a starting value for which 4 cos(cos(cos(a0))) is exactly pi? Yes, we can. I'm too lazy to find the value accurately, but if you start at 0.5734992, do three iterations, and multiply by 4 then you'll get a result beginning 3.1415926... .

This doesn't tell us anything very exciting about this particular iteration or the particular value pi. It just happens that if you start at 0 and iterate three times you get something a bit over pi/4, if you start at 1 and iterate three times you get something a bit under pi/4, and so there must be a value somewhere in between that hits pi/4 on the nose.

Similarly, for your iteration, if you start at 0 and iterate three times you get something a bit smaller than pi/2sqrt(phi), if you start at 1 and iterate three times you get something a bit bigger than pi/2sqrt(phi), so there must be a value somewhere in between that hits pi/2sqrt(phi) on the nose.

--
g

[Martin Doina]

do not matter where you start the final values is 1.23498228... yes I may paste the wrong output from the bash, and is not the formula  

psi_values[i] = np.sin(psi_values[i-1]) + np.exp(-psi_values[i-1]) this formula gives the LZ final that is constant but the intermediate LZ3 give pi by  π = 2 * LZ * sqrt(φ), complete here https://zenodo.org/records/17302392

[Stephen Lucas]

Martin, I had written a reply, then saw that Gareth has been me to asking exactly the same questions.
With your converged values of LZ, I also only get pi to 3 decimal places. So I looked at your paper online, and saw that you have done exactly what Gareth guessed. Looks like you have defined LZ3 as pi/(2*sqrt(phi)), then backtracked to the given starting value. So you only get pi because you set it up so you would get pi exactly! This is not particularly useful number theory: you can always do this sort of thing with iterations for any value, and if we replaced sqrt(phi) by something else (5/4 is close) we would have a slightly different LZ0 that also goes to pi.

Now if you can get pi as an iteration that is the fixed point that isn’t artificially constructed, then that would be interesting!

Steve