
Generating Functions and Fibonacci Numbers - MidsizeBlowfish
http://www.austinrochford.com/posts/2013-11-01-generating-functions-and-fibonacci-numbers.html
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sigil
Another way to derive a closed form for the Fibonacci sequence: put the
recurrence relation in vector form and solve by finding the eigenvalues.
(Which, beautifully, turn out to be the golden ratio and its rational
conjugate.)

[http://mathproofs.blogspot.com/2005/04/nth-term-of-
fibonacci...](http://mathproofs.blogspot.com/2005/04/nth-term-of-fibonacci-
sequence.html)

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MidsizeBlowfish
That's great. I had never seen/considered that.

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domdip
This is a fun example. The formula's easy to prove by induction, but
discovering it is less obvious and this is an interesting way.

It's the tip of the iceberg though - generatingfunctionology is worth
inspecting thoroughly.

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btilly
There are actually perfectly routine ways to discover formulas like this one
for this type of equation. See
[http://en.wikipedia.org/wiki/Recurrence_relation](http://en.wikipedia.org/wiki/Recurrence_relation)
for some of the theory behind how to do it.

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jordigh
I've always preferred to write Binet's formula like this:

(phi^n - psi^n)/(phi - psi)

The only difference is writing the sqrt(5) denominator as phi-psi. The reason
being, this exhibits that the Fibonacci numbers are a symmetric function of
the roots of a polynomial. Thus, even though phi and psi involve irrational
numbers, this symmetric function of them must be in the base rational field,
and since the polynomial x^2 - x - 1 is _monic_ , then these are plain
integers.

This way it looks less surprising that a formula involving irrationals is
always rational.

To show that the rational is actually an integer, you have to actually perform
the division to get

phi^n + phi^(n-1) psi + phi^(n-2) psi^2 + ... + psi^n

and use the fact that algebraic integers form a ring and the only rational
algebraic integers are plain integers.

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jkarni
This is pretty close to Knuth's discussion of Fibonacci Numbers in TAOCP. I'm
inclined that's where this came from, and that this is missing an attribution.

In either case, if you enjoyed this post add one to the "I-should-read-TAOCP"
(or "I-should-continue-reading-TAOCP") tally. I did...

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MidsizeBlowfish
OP here, I have actually not read TAOCP at all. It's just a derivation I
sometimes use to pass a few minutes while waiting/bored and I thought I'd blog
about it.

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jkarni
Yeah, sorry, I stand corrected.

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MidsizeBlowfish
No problem.

