[Edu-sig] Fibonacci Numbers and Phi (again)

[Litvin]

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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