On 2024-01-05, Kaz Kylheku <433-929-6894@kylheku.com> wrote:
> Puzzle:
>
> Could we use this information somehow to neatly build an amplifier whose
> closed-loop gain is the Golden Ratio, using only resistors that have
> integer-valued resistances in Ohms?
>
> I might post this to the Electronics Stack Exchange.

Someone beat me to it, seven years ago:

"How can i make a circuit such that output voltage is reduced by golden
ratio, using feedback"

https://electronics.stackexchange.com/questions/261616/how-can-i-make-a-circuit-such-that-output-voltage-is-reduced-by-golden-ratio-us

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Kaz Kylheku wrote:

> On 2024-01-05, mhx <mhx@iae.nl> wrote:
>> Kaz Kylheku wrote:
>>
>>> Sometimes, the Fibonacci recurrence is expressed as a matrix
>>> multiplication on a state vector [a b]
>>
>>> [1 1][a] => [a + b]
>>> [1 0][b] [ a ]
>>
>>> a receives the sum of a + b, while b receives the prior value of a.
>>
>>> Raising this matrix to an arbitrary integer power is how we arrive
>>> at the closed form solution.
>>
>> Really nice! I'll keep this general idea in a safe place.

> You don't have to; it's found in textbooks.

It strongly connects to the control engineering approaches I use for
analyzing recurrences in switched electronic systems.

> Using linear algebra techniques found in introductory texts, you can do
> this this:

> n
> [1 1]
> [1 0]

> The details escape me, much like a departing youthful appearance, but
> basically we can factor it into a form

> n
> U X V

> where U, X, V are all 2x2 matrices. The middle one X is special
> in that it is diagonal (consisting of a diagonal trace of eigenvalues).
> A diagonal matrix is easy to exponentiate.

Tricky one to write from scratch :--) I implemented the Pade approximation
(the stablest one) for it when my appearance was still youthful.

> Something like that.

> That gives us that closed form formula for Fibonacci with those
> square roots of five in it.

> Closely relating to the Golden Ratio! Which is (/ (+ 1 (sqrt 5)) 2).
> The ratio of successive values of the Fibonacci numbers approaches
> the golden ratio.

Almost like a party trick. I love to find elegant techniques in
unrelated fields of research that solve (my) practical problems in
unexpected ways.

> 1> (/ (fib 50) (fib 49))
> 1.61803398874989

-marcel

mhx@iae.nl (mhx) writes:
>Kaz Kylheku wrote:
>> Closely relating to the Golden Ratio! Which is (/ (+ 1 (sqrt 5)) 2).
>> The ratio of successive values of the Fibonacci numbers approaches
>> the golden ratio.
>
>Almost like a party trick. I love to find elegant techniques in
>unrelated fields of research that solve (my) practical problems in
>unexpected ways.

Not sure if you wrote this about the golden ratio, but given the way
the golden ratio phi is defined, I find it unsurprising that

phi = lim(n->inf) fib(n+1)/fib(n)

The definition of the golden ratio is

(a+b)/a = a/b = phi (with a>b>0)

or, as described in <https://en.wikipedia.org/wiki/Golden_ratio>:

| a+b is to a as a is to b

Given that's exactly how Fibonacci numbers are formed, if the sequence
of ratios fib(n+1)/fib(n) converges, the result must be the golden
ratio.

It's equally unsurprising that for the Lucas numbers, which are formed
in the same way, but with different base values (2 and 1 instead of 0
and 1), the limit of L(n+1)/L(n) is also the golden ratio.

Conversely, you can compute the nth Fibonacci number using a closed
formula that involves phi.

- anton
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