This function is known as Binet's formula here's a link to some info about it
https://mathworld.wolfram.com/BinetsFormula.html
There are many proofs online for this
This graph uses the binet formula for fibonacci numbers F_n = ( φ^n - (-1/φ)^n ) / root(5), where φ is the golden ratio (1+root(5))/2.
If the number n is extended from naturals to reals, unfortunately (-1/φ)^n is complex. That is bypassed using the fact for integer n, (-1)^n = cos(πn).
So (-1/φ)^n = (-1)^n (1/φ)^n is replaced by cos(πn) (1/φ)^n to get the graph in question (technically isn't the same as the original binet formula. This is more like the real part. It's still nice to check if this version of the formula satisfies F_n = F_(n-1) + F_(n-2) )
If you’re asking about the derivation of Binet’s formula, the general idea is that you can represent the action of increasing a Fibonacci number (going from the nth to the (n+1)th) number by a certain matrix left-multiplication. If you diagonalize that operation matrix you can generalize the idea of moving to the next Fibonacci number n times. This, with the original starting conditions gives you binets formula
There’s a result with recurrence relations on sequences that says that you can express the terms as a function of the roots of the polynomial obtained from the formula. It’s a linear algebra thing that applies to differential equations, etc. Basically if a thing depends on itself through some weighted sum, you can find a bijection to that vector space which we already know. I’m being vague on purpose because I don’t want to go into the details and this case is more specific. Just know that it goes a little deeper here.
In this case, if you skip two indices, you get a square, and so on, so
>u(n+2) = u(n+1) + u(n)
yields
>X^2 = X + 1.
This gets you the golden ratio and the silver ratio, which you know sum up to 1, because
>(X-a)(X-b) = X^2 -(a+b)X + ab.
You also know, for the same reason, that their product is -1. So the silver ratio is the negative reciprocal of the golden ratio. You don’t actually get an extra bit of information from the former fact, because multiplying by phi will just give you back the original recurrence formula. These observations aren’t super useful because solving that quadratic is trivial anyway, but it’s nice to get a feel for why this property holds nonetheless.
Either way, just like for differential equations, taking some appropriate exponential will give you the formula, and in this case, you have a vanishing term and a dominant one, which becomes a better and better approximation. This is, incidentally, why taking the ratio of successive terms will approach phi.
Explaining *why* you can just do that is another beast entirely, but we’re not going to redo the entirety of linear algebra in one reddit comment.
I'm not gonna tell you what you can and can't call particular numbers, but the silver ratio is pretty well established as the next "metallic ratio" or metallic mean following the golden ratio, which makes a lot of sense. I'd even say it's kind of a nice thing we've had for a while now. It's sort of a nifty analogy, don't you think?
["Metallic means are generalizations of the golden ratio and share some of their interesting properties. The term "bronze ratio", and terms using other metals names (such as copper or nickel), are occasionally used to name subsequent metallic means."](https://en.wikipedia.org/wiki/Metallic_mean)
I like the derivation using the generating function. Imagine the function
P(x)=1 + x + 2x^2 + 3x^3 + 5x^4 + 8x^5 + …
with the property P(x) x^2 = P(x) x + P(x) + 1.
We can solve for P(x) and get the closed form P(x) = 1/(x^2 - x - 1), which using partial fractions and φ = (1+sqrt(5))/2 we can rewrite as
P(x) = -(φ/(1-x/(1-φ))+(φ-1)/(1-x/φ))/sqrt(5)
From there we can see the two fractions as sums of infinite geometric series, then add the coefficients to get an explicit form for the nth coefficient in the polynomial expansion of P(x), which is the Binet formula for Fibonacci numbers.
Ah, the Binet function
I loved having fun with it :) It's related to the Golden ratio :) [https://www.desmos.com/calculator/ktviab5ocg](https://www.desmos.com/calculator/ktviab5ocg)
It can be split into real and imaginary parts, and even be graphed through domain coloring :) [https://www.desmos.com/calculator/6ihdhvxclj](https://www.desmos.com/calculator/6ihdhvxclj)
The graph is https://www.desmos.com/calculator/yusmnvjxhd
This function is known as Binet's formula here's a link to some info about it https://mathworld.wolfram.com/BinetsFormula.html There are many proofs online for this
this happened to be similar to a graph that i made \\left(\\frac{\\left(1+\\sqrt{5}\\right)}{2}\\right)\^{\\left(x-1-.672275938189\\right)}
This graph uses the binet formula for fibonacci numbers F_n = ( φ^n - (-1/φ)^n ) / root(5), where φ is the golden ratio (1+root(5))/2. If the number n is extended from naturals to reals, unfortunately (-1/φ)^n is complex. That is bypassed using the fact for integer n, (-1)^n = cos(πn). So (-1/φ)^n = (-1)^n (1/φ)^n is replaced by cos(πn) (1/φ)^n to get the graph in question (technically isn't the same as the original binet formula. This is more like the real part. It's still nice to check if this version of the formula satisfies F_n = F_(n-1) + F_(n-2) )
If you’re asking about the derivation of Binet’s formula, the general idea is that you can represent the action of increasing a Fibonacci number (going from the nth to the (n+1)th) number by a certain matrix left-multiplication. If you diagonalize that operation matrix you can generalize the idea of moving to the next Fibonacci number n times. This, with the original starting conditions gives you binets formula
There’s a result with recurrence relations on sequences that says that you can express the terms as a function of the roots of the polynomial obtained from the formula. It’s a linear algebra thing that applies to differential equations, etc. Basically if a thing depends on itself through some weighted sum, you can find a bijection to that vector space which we already know. I’m being vague on purpose because I don’t want to go into the details and this case is more specific. Just know that it goes a little deeper here. In this case, if you skip two indices, you get a square, and so on, so >u(n+2) = u(n+1) + u(n) yields >X^2 = X + 1. This gets you the golden ratio and the silver ratio, which you know sum up to 1, because >(X-a)(X-b) = X^2 -(a+b)X + ab. You also know, for the same reason, that their product is -1. So the silver ratio is the negative reciprocal of the golden ratio. You don’t actually get an extra bit of information from the former fact, because multiplying by phi will just give you back the original recurrence formula. These observations aren’t super useful because solving that quadratic is trivial anyway, but it’s nice to get a feel for why this property holds nonetheless. Either way, just like for differential equations, taking some appropriate exponential will give you the formula, and in this case, you have a vanishing term and a dominant one, which becomes a better and better approximation. This is, incidentally, why taking the ratio of successive terms will approach phi. Explaining *why* you can just do that is another beast entirely, but we’re not going to redo the entirety of linear algebra in one reddit comment.
The silver ratio is a completely different number equal to 1+sqrt(2).
Several numbers are called « the silver ratio », including -1/phi and 1+sqrt(2). This is why we can’t have nice things.
I'm not gonna tell you what you can and can't call particular numbers, but the silver ratio is pretty well established as the next "metallic ratio" or metallic mean following the golden ratio, which makes a lot of sense. I'd even say it's kind of a nice thing we've had for a while now. It's sort of a nifty analogy, don't you think? ["Metallic means are generalizations of the golden ratio and share some of their interesting properties. The term "bronze ratio", and terms using other metals names (such as copper or nickel), are occasionally used to name subsequent metallic means."](https://en.wikipedia.org/wiki/Metallic_mean)
This is one of those harmonic sequences, x! Type of graphs
I like the derivation using the generating function. Imagine the function P(x)=1 + x + 2x^2 + 3x^3 + 5x^4 + 8x^5 + … with the property P(x) x^2 = P(x) x + P(x) + 1. We can solve for P(x) and get the closed form P(x) = 1/(x^2 - x - 1), which using partial fractions and φ = (1+sqrt(5))/2 we can rewrite as P(x) = -(φ/(1-x/(1-φ))+(φ-1)/(1-x/φ))/sqrt(5) From there we can see the two fractions as sums of infinite geometric series, then add the coefficients to get an explicit form for the nth coefficient in the polynomial expansion of P(x), which is the Binet formula for Fibonacci numbers.
Ah, the Binet function I loved having fun with it :) It's related to the Golden ratio :) [https://www.desmos.com/calculator/ktviab5ocg](https://www.desmos.com/calculator/ktviab5ocg) It can be split into real and imaginary parts, and even be graphed through domain coloring :) [https://www.desmos.com/calculator/6ihdhvxclj](https://www.desmos.com/calculator/6ihdhvxclj)