Kirby,
I am sorry to spoil all the fun, but this is a not a gem but a
mathematical trick, and I think the way to deal with mathematical
tricks is to explain them, not to verify with code.
There are two Fibonacci sequences that are also geometric sequences
1, x, x^2, ..., x^n,...: when x = phi or x = -1/phi, because phi and
-1/phi are the roots of the quadratic equation x^2 = x+1. Any
Fibonacci sequence is a linear combination of these two: F[n] =
a*phi^n + b*(-1/phi)^n. a and b can be determined from the first two
terms of the sequence:
a + b = F[0]
a*phi - b/phi = F[1]
If you choose F[0] and F[1] (not necessarily integers) in a tricky
way, so that b = 0, then F[n] = a*phi^n. In addition, if a = phi^e,
then F[n] = phi^(e+n). Working in reverse, you can find F[0] and
F[1] that satisfy these conditions and get the trick going. Here
F[0] = (13 + 5 - 8*sqrt(5))/2 = 9 - 4*sqrt(5)
F[1] = (-8 - 3 + 5*sqrt(5))/2 = (-11+5sqrt(5))/2
Easy to verify that F0*phi = F1, so, indeed, b = 0 and F[n] is a
geometric sequence. Also, phi^6 = 9 + 4*sqrt(5), so F[0]*phi^6 = 1
and a = F[0] = phi^(-6). You get F[n] = phi^(n-6).
Gary Litvin
www.skylit.com
At 10:04 PM 11/22/2013, you wrote:
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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At 10:04 PM 11/22/2013, kirby urner wrote:
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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