Here's something mathematical from the Poly list [1], adapting
something by David Koski.
The code shows how one might use a program to express a
cool relationship, without offering a proof.
The unit-test shows use raising PHI from the -6 to the 29th
power and finding an equivalent expression built solely from
three consecutive Fibonacci numbers and the sqrt(5).
Something like:
LOOP:
PHI ** E = (FIB[N] + FIB[N-2] + FIB[N-1]*RT5) / 2
E += 1
N += 1
Start with:
E = -6
FIB[N-2] = 13
FIB[N-1] = -8
FIB[N] = 5
in . . . 13, -8, 5, -3, 2, -1,1, 0, 1, 2, 3, 5, 8,13 . . .
===
"""
Encapsulating a discovery -- not a proof.
By David Koski (Python by K. Urner)
Failure at around 37th power is due to floating point limitations.
"""
import unittest
def fibonacci(a=0, b=1):
while True:
yield a
a, b = b, a + b
series1 = fibonacci(5, -3) # 5, -3, 2, -1, 1, 0 , 1, 2, 3, 5, 8
series2 = fibonacci(-8, 5) # -8, 5, -3, 2, -1, 1, 0 , 1, 2, 3, 5, 8
series3 = fibonacci(13,-8) # 13, -8, 5, -3, 2, -1, 1, 0 , 1, 2, 3, 5, 8
RT5 = pow(5, .5) # "square root" of five
PHI = (1 + RT5)/2 # golden proportion
def cool_formula():
"""
Two fib numbers, two apart, plus the one in the middle * sqrt(5)
all over 2, gives PHI to a power. Advancing all three sequences
gives successive powers.
"""
while True:
yield (next(series1) + next(series3) + next(series2) * RT5)/2.0
class TestPhi(unittest.TestCase):
def test_loop(self):
gem = cool_formula()
for e in range(-6, 30): # adjust range (fails about 37th power)
answer = next( gem )
self.assertAlmostEqual( PHI ** e, answer)
if __name__ == "__main__":
unittest.main()
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