
Golden powers are nearly integers - adamnemecek
https://www.johndcook.com/blog/2017/03/22/golden-powers-are-nearly-integers/
======
Sniffnoy
Specifically, they're nearly Lucas numbers. For those unfamiliar, the Lucas
numbers are defined by the recursion L_0 = 2, L_1 = 1, and L_n = L_{n-1} +
L_n. (I.e., the Fibonacci recursion, but with different starting values.) They
can also be given explicitly by L_n = phi^n + (-1/phi)^n. Since (-1/phi)^n
goes rapidly to zero, phi^n will be very close to L_n. More specifically, for
n>=2, (1/phi)^n < 1/2, so L_n will be the nearest integer to phi^n.

The Fibonacci numbers, meanwhile, can be given by the formula F_n = (phi^n -
(-1/phi)^n) / sqrt(5). So a similar thing holds for them.

More generally, phi is a Pisot number:
[https://en.wikipedia.org/wiki/Pisot%E2%80%93Vijayaraghavan_n...](https://en.wikipedia.org/wiki/Pisot%E2%80%93Vijayaraghavan_number)
Any Pisot number will have its powers approach integers at an exponential
rate. (In that, the distance to the nearest integer decreases exponentially.)

~~~
jacobolus
Another way to get at this:

    
    
      φ^1 = φ
      φ^2 = φ + 1 [this is the definition of φ]
      φ^3 = φ(φ + 1) = 2φ + 1
      φ^4 = φ(2φ + 1) = 3φ + 2
      φ^5 = φ(3φ + 2) = 5φ + 3
      φ^6 = φ(5φ + 3) = 8φ + 5
      ...
      φ^n = F(n)φ + F(n-1)
    

Note that F(n+1)/F(n) → φ as n → ∞. Or flipped around, F(n)φ → F(n+1).

So φ^n = F(n)φ + F(n-1) → F(n+1) + F(n-1) = L(n) as n → ∞. This gets close to
being an integer because F(n+1) and F(n-1) are both integers.

~~~
adrianratnapala
I'm almost with you, but you need a bit more.

To show what we want, we must prove the stronger condition that F(n)φ - F(n+1)
→ 0. Merely knowing that F(n+1)/F(n) → φ only shows that the difference grows
more slowly than F(n).

~~~
winstonsmith
> _To show what we want, we must prove the stronger condition that F(n)φ -
> F(n+1) → 0._

Noting that {F(n+1)/F(n)} is the sequence of convergents of the continued
fraction expansion of φ, your stronger condition follows from Theorem 5 at
[https://en.wikipedia.org/wiki/Continued_fraction](https://en.wikipedia.org/wiki/Continued_fraction)
.

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drostie
If people are interested in this sort of thing, there's some nice linear
algebra underlying the idea.

The Fibonacci recurrence is a linear recurrence, meaning that if you have two
different sequences such that

    
    
        F[n] = F[n - 1] + F[n - 2]
        G[n] = G[n - 1] + G[n - 2]
    

then the scalar multiples (multiply every element of F by some value k) and
termwise sum of F and G will also solve this recurrence. One nice thing to do
then is to search for a base p such that one solution for the recurrence is
P[n] = p^n. This gives a formula for p, namely p² = p + 1, which also
characterizes the golden ratio φ = (1 + √5)/2 and its negative reciprocal -1/φ
= (1 − √5)/2.

Well, remember that the recurrence gives you everything if you also specify
the first two numbers F[0] and F[1]. We know that these two sequences that we
just computed are

    
    
        P1 = [1, (1 + √5)/2, ...]
        P2 = [1, (1 − √5)/2, ...]
    

and given F0, F1 you can use some simple linear algebra to solve for a and b
such that:

    
    
        a P1[0] + b P2[0] = F0,
        a P1[1] + b P2[1] = F1.
    

This gives a closed form solution for the Nth term in terms of these Ps. So
the sequences obeying the recurrence actually form a 2-dimensional vector
space and these polynomials just happen to be a nice basis for that vector
space.

Similarly for the case of the powers of phi, we can use this argument in
reverse to say "I know I want P1 plus something times P2 which causes all of
the terms to be integers," P2 will then approach zero (since it has modulus
less than 1).

For this, we can look at the recurrence and say "oh, we just need the first
two terms to be integers and then the recurrence makes everything else
integers," so choosing a = 1, b = 1 gives us F[0] = 2, F[1] = 1, and these
happen to be called the "Lucas numbers".

Similarly we could expect that the powers of any solution to p² = m p + n,
integer m, n, p > 1 will be close to the integers as long as the other
solution q lies somewhere in the unit interval -1 < q < 1\. Completing the
square gives (p − m/2)² = n + m²/4, the solutions are evenly spaced about m/2
with a distance from center to solution of √(m²/4 + n).

So for example the recurrence T[n] = 3 T[n-1] - T[n-2] is solved by the
numbers (3 ± √5)/2 and so (3 + √5)/2 must also have this property, whereas
T[n] = 3 T[n-1] + T[n-2] gives you (3 ± √13)/2 and has this property too.

------
xyzzyz
Suppose f(x) = x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0 is a monic polynomial
with integer coefficients. Let x_1, ..., x_n be all its (complex) roots. If it
happens that the absolute value of x_1 is larger than 1, but absolute values
of all x_2, x_3, ..., x_n are smaller than 1, the powers of x_1 will be close
to integers. The golden ratio (1+sqrt(5)/2 is a root of x^2 - x - 1, and the
other root (1-sqrt(5))/2 is smaller than 1 in absolute value, so the
observation follows.

One way to prove the assertion is following: note that for large n, x_1^n +
x_2^n + ... + x_3^n is very close in value to x_1^n, because all the other
summands are very small in absolute value, since they start out smaller than
1, and they decrease exponentially to 0. But it just so happens that x_1^n +
x_2^n + ... + x_3^n is an integer -- indeed, this is a symmetric polynomial
with integer coefficients in x_1, ..., x_n, and every such polynomial has a
unique expression as a polynomial in elementary symmetric polynomial with
integer coefficients -- this is the fundamental theorem of symmetric
polynomials[1]. But, if you plug the roots of a given polynomial into
elementary symmetric polynomials, you'll just get the coefficients of the
original polynomial times (-1)^k, that is, the original a_0, a_1, ..., a_{n-1}
-- for example, a_0 is clearly the product (-1)^n x_1 * x_2 * ... * x_n,
a_{n-1} is the sum (-1)^1 (x_1 + ... + x_n), a_{n-2} is a sum of two-products
(-1)^2 (x_1 x_2 + x_1 x_3 + ... + x_{n-1} x_n) and so on.

For concrete example, note that x_1^2 + ... + x_n^2 = (x_1 + ... + x_n)^2 -
2(x_1 x_2 + ... + x_{n-1} x_n) = a_{n-1} - 2 a_{n-2}, which is an integer. Try
expressing x_1^3 + ... + x_n^3 i n terms of a_{n-1}, a_{n-2} and a_{n-3}.

[1] -
[https://en.wikipedia.org/wiki/Elementary_symmetric_polynomia...](https://en.wikipedia.org/wiki/Elementary_symmetric_polynomial#The_fundamental_theorem_of_symmetric_polynomials)

------
got2surf
I don't have any advanced math experience, but this is one of the coolest
things I've read in a long time - kudos to the author for explaining a couple
interesting things about a complex topic in a simple way.

~~~
tedunangst
But he didn't explain why it happens at all?

~~~
amelius
At least he clearly explained the problem (or really phenomenon), so anyone
can understand it.

~~~
tedunangst
I mean, I'm not knocking the blog. It's fine. But I'm having a hard time
identifying what the complex topic was that was given a simple explanation.
"Here's a graph" isn't a complex topic, and saying "here's a graph" doesn't
really explain it either.

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ccvannorman
Another interesting relationship between powers and infinity is that on an
infinite N-dimensional lattice, taking an infinitely long random walk,
starting with dimension 3 you have a 34% chance to return to origin with each
successive dimension reducing your chance. The graph of this probability
follows a log curve. (dimensions 1 and 2 have a 100% probability of returning
to origin)

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Filligree
So what this is saying is, if you don't have "1" in your numerical system, you
can recover it by taking φ to the power of infinity and then dividing by
infinity?

~~~
pmalynin
Any "number system" (be it a field, ring, or multiplicative group) has a "1"

~~~
Smaug123
Depends on your definition of "ring"; for some people, a ring need not have a
multiplicative identity. But yes, any number system worth its salt has at
least a couple of integers in it.

~~~
drdeca
Wouldn't a ring without a multiplicative identity be a rng rather than a ring?

~~~
Filligree
Only if it started as a riiiing...

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Grue3
Yeah, it's really obvious too if you know basic algebra. The article doesn't
even attempt to explain it, so what's the point, except for blogspam?

~~~
recursive
I thought I knew basic algebra... I guess not. It definitely is not obvious to
me.

~~~
joelg
To be fair, "algebra" means _very_ different things to different people, so I
don't think Grue3 meant to be quite as brash as it may have sounded.

~~~
nandemo
What's the definition of basic algebra that makes that fact obvious? I took 3
semesters of algebra in college (Linear Algebra, Abstract Algebra I and II),
and don't see anything obvious about it. But then, neither does Terry Tao:

> _" the powers φ, φ2, φ3, … of the golden ratio lie_ unexpectedly _close to
> integers..._

~~~
Grue3
It becomes obvious when you know closed form expression for Fibonacci numbers
as a sum of powers of phi (which I'm sure Tao is aware of). I was taught that
in high school. The only reason this wouldn't be obvious to Tao is if he was
working on much harder problems for so long that he forgot the basic stuff.

