It’s cool that
x = 1/(ϕ-(1/(ϕ-(1/(ϕ-(1/(ϕ-(1/(ϕ-x)))))))))
where ϕ is the golden ratio,
x = 1/(√2-(1/(√2-(1/(√2-(1/(√2-x)))))))
and
x = 1/(1-(1/(1-1/(1-x)))).
But why?
Consider the matrix:
cos θ -sin θ
sin θ cos θ
which rotates by θ, so call this R.
Every matrix is a zero of its own characteristic polynomial, so:
(I cos θ - R)^2 + (I sin θ)^2 = 0
R^2 = (2 cos θ) R - I
Now, pick any vector v, and define:
w = -R v
Then:
R v = -w
R w = -R^2 v
= v - (2 cos θ) R v
= v + (2 cos θ) w
This tells us that if we change to the basis {v,w}, the matrix for R becomes:
0 1
-1 2 cos θ
which we will call S.
R and S are conjugate matrices; they do the same thing, but they just describe it in different bases.
Suppose we set θ to π/n. Then:
R^n = -I
so:
S^n = -I
Now, any invertible 2×2 matrix:
a b
c d
can be used to define a rational function:
f(x) = (a x + b)/(c x + d)
If we *compose* the functions we get from two matrices, the result is the same as *multiplying* the matrices.
The function for -I is just:
f(x) = (-x+0)/(0x-1) = x
So the equation satisfied by S:
S^n = -I
means that if we take the function corresponding to S:
0 1
-1 2 cos θ
f(x) = 1/(2 cos θ - x)
and compose it with itself n times … we just end up with x.
The formulas at the top of this post are saying this for n=5, n=4 and n=3.
Greg Egan /
npub1qq…j5nhv
2023-09-27 08:05:12
Author Public Key
npub1qqm7nu2qf25xd3mw6y6cyp4v2wr7ktf43ymp5wqz4u8jvz76ymtsjj5nhvPublished at
2023-09-27 08:05:12Event JSON
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"content": "It’s cool that\n\nx = 1/(ϕ-(1/(ϕ-(1/(ϕ-(1/(ϕ-(1/(ϕ-x)))))))))\n\nwhere ϕ is the golden ratio,\n\nx = 1/(√2-(1/(√2-(1/(√2-(1/(√2-x)))))))\n\nand\n\nx = 1/(1-(1/(1-1/(1-x)))).\n\nBut why?\n\nConsider the matrix:\n\ncos θ -sin θ\nsin θ cos θ\n\nwhich rotates by θ, so call this R.\n\nEvery matrix is a zero of its own characteristic polynomial, so:\n\n(I cos θ - R)^2 + (I sin θ)^2 = 0\n\nR^2 = (2 cos θ) R - I\n\nNow, pick any vector v, and define:\n\nw = -R v\n\nThen:\n\nR v = -w\n\nR w = -R^2 v \n= v - (2 cos θ) R v\n= v + (2 cos θ) w \n\nThis tells us that if we change to the basis {v,w}, the matrix for R becomes:\n\n0 1\n-1 2 cos θ\n\nwhich we will call S.\n\nR and S are conjugate matrices; they do the same thing, but they just describe it in different bases.\n\nSuppose we set θ to π/n. Then:\n\nR^n = -I\n\nso:\n\nS^n = -I\n\nNow, any invertible 2×2 matrix:\n\na b\nc d\n\ncan be used to define a rational function:\n\nf(x) = (a x + b)/(c x + d)\n\nIf we *compose* the functions we get from two matrices, the result is the same as *multiplying* the matrices.\n\nThe function for -I is just:\n\nf(x) = (-x+0)/(0x-1) = x\n\nSo the equation satisfied by S:\n\nS^n = -I\n\nmeans that if we take the function corresponding to S:\n\n0 1\n-1 2 cos θ\n\nf(x) = 1/(2 cos θ - x)\n\nand compose it with itself n times … we just end up with x.\n\nThe formulas at the top of this post are saying this for n=5, n=4 and n=3.",
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