All about Golden Ratio

Continued fraction

Continued fraction Think about expressing some decimals as a fraction. For example, 0.351351351… The reciprocal is 1/0.351351… = 2.8461… So that can be approximated by 1/2.

Continued fraction What if you want to be more precise that fractions? 0.351351… = 1/2.8461… = 1/(2+0.8461…) The reciprocal of a decimal fraction 1/0.8461… = 1.18181818... so, 0.351351… = 1/(2+1/1) Keep doing this, and end up with… 1/(2+1/(1+(1/(5+(1/(1+1/1)))))) This form called “Continued fraction( 連分数 )”.

Continued fraction Convert this fraction to the form A/B. 1/(2+1/(1+(1/(5+(1/(1+1/1)))))) = 13/37 We get 0.351351… = 13/37 For readability, use the following symbols. [0;2,1,5,1,1]

Continued fraction Is it just useful for rational number guessing quizzes? No! It can also be used with irrational numbers. For example, 1.41421356237… [1; 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2…] x = 1 + 1/(1+x) (x - 1)(x + 1) = 1 x² = 2 x = √2 We realize 1.41421356237… = √2

Continued fraction(連分数) In general, a solutions to the quadratic equation is a circulating continued fraction. √2 : [1; 2, 2, 2, 2, 2, 2, …] √7 : [2; 1, 1, 1, 4, 1, 1, 1, 4, ...] (7+√13)/3: [3; 1, 1, 6, 1, 1, 1, 1, 6, 1, 1, 1, 1, 6, …] But non-quadratic irrational numbers do not circulating. ∛ 2 : [1; 3, 1, 5, 1, 1, 4, 1, 1, …] e : [2; 1, 2, 1, 1, 4, 1, 1, 6, …] And what is terminated becomes an approximation. e ≒ [2; 1, 2, 1] = 2.75

Continued fraction approximation

Continued fraction approximation, 1.23 != 123/100 Think about the value “1.23”. Assuming rounding of decimal representation, the range of this is [1.225, 1.235). The error is “0.005”

Continued fraction approximation, 1.23 != 123/100 On the other hand, it is expressed as a continued fraction. [1; 4, 2, 1, 7] = 123/100 = 1.23 change it to a slightly different value [1; 4, 2, 1, 7, 2] = 1.2300469... The error is “0.0000469...” This is much smaller than decimal representation one(0.005).

Continued fraction approximation, 1.23 != 123/100 In general, the error bound in the… decimal representation of m/n is 1/(2n) continued fraction approximation of m/n is 1/n². The error “bound” is 1/n², The actual error will be even smaller.

Continued fraction approximation, 1.23 != 123/100 Think about ∛ 19 = 2.66840164872... = [2; 1, 2, 63, 1, 2, 2, 2...] [2; 0] = 2.0000 Error: 0.6684... [2; 1] = 3.0000 Error: 0.3315... [2; 1, 2] = 2.6666… Error: 0.001734... [2; 1, 2, 63] = 2.6684… Error: 0.0000024... The “63” means "Oh, it's almost approximated!" The higher the next number, the smaller the error. The error is actually 1/(n² * next number)

Continued fraction approximation, 1.23 != 123/100 Summarizing the discussion so far… 1. Continued fraction is making a good approximation!(In fact, the best!) 2. Huge fraction means “Almost approximated”. Since the opposite of huge is 1, continued fraction that are all composed of 1 could be a very difficult number to approximate??? 🤔🤔🤔 Let's simulate it!

Phyllotaxy simulation

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Phyllotaxy simulation Let's simulate the way the leaves grow! ➢ Let a leaf be a circle. ➢ The i-th leaf grows at (√i, i* α *2π) in the polar coordinate system. ➢ Find an alpha that places the leaves most evenly. What would be a alpha? * "√i" is the radius of a circle equal to the area of i circles of radius 1. * the alpha called phyllotaxy( 葉序 )

Phyllotaxy simulation You can see 12 “arms”. 12 comes from 1/12 What happens if (big) irreducible fraction?

Phyllotaxy simulation Still see “arms”. It consists of two parts. 3 and 22 arms. 64/201 = [0; 3, 7, 9] 64/201 ≒ 1/3 ≒ 7/22.

Phyllotaxy simulation The rational number n/m will produce m arms, and p arms of its approximate fraction q/p. If it's not good because it's a rational number, would it be better if it were an irrational number? For instance, π?

Phyllotaxy simulation π has a good fractional approximation with a very small denominator! π = [3; 7, 15, 1, 292, 1,...] π ≒ 22/7 = 3.1428… That's why you can see the seven arms.

Phyllotaxy simulation e also has a good fractional approximation. e = [2; 1, 2, 1, 1, 4, 1...] e ≒ 19/7 = 2.7142…

Phyllotaxy simulation It seems that it is not enough to just simply be irrational. Let's intentionally create numbers. Remember the nice features of continued fractions! 1. The accurate approximation. 2. Huge fraction means “Almost approximated”. In other words, [1; 1, 1, 1, 1…] would be the hardest to approximate.

Golden Ratio

Golden Ratio A continued fraction consisting only of 1 φ = [1; 1, 1, 1, 1, 1,...] Circulating! As mentioned earlier, cyclic continued fraction is a solution of a quadratic equation. φ = 1 + 1/ φ φ ² - φ - 1 = 0 φ = (1 + √5)/2 = 1.618033… *Negative solution (1 - √5)/2 = -0.61803 are not often referred to as the golden ratio.

Golden Ratio The limit of the Fibonacci sequence also leads to Golden Ratio. 1, 1, 2, 3, 5, 8, 13, 21, 34, 55... An ₊₂ = An + An ₊₁ Suppose as a geometric sequence... An ₊₁ = φ An An ₊₂ = φ ² An φ ² An = An + φ An φ = (1 + √5) / 2

Golden Ratio α = φ , It looks perfect! If you look carefully, you can see the 13, 21, 34, 55 arms. 2π φ = 582.49°, 720° - 582.49° = 137.5° 137.5° called golden angle. * Definition is 2π(1-1/ φ ), but same value.

Golden Ratio BTW, sqrt of little integer possibly be a candidate? √2 = [1; 2, 2, 2, 2, …] √3 = [1; 1, 2, 1, 2, …] √5 = [1; 4, 4, 4, 4, …] √6 = [1; 2, 4, 2, 4, …]

Golden Ratio

Golden Ratio The idea of simply using the square root of an integer doesn't sound good. But we know that the golden ratio seems to be the “best” number. Is there any way to create the same kind of numbers???

Diophantine approximation

Diophantine approximation The approximation of some number by a fraction is called Diophantine approximation. Let's calculate the performance of that approximation with the following equation. Abs(Error) * denominator² For example, e = [2; 1, 2, 1, 1, 4, …] [2;1]: Abs(e - 3/1)*1² = 0.2817... [2;1,2]: Abs(e - 8/3)*3² = 0.4794... [2;1,2,1]: Abs(e - 19/7)*7² = 0.1958... [2;1,1,1,4]: Abs(e - 87/32)*32² = 0.4794...

The continued fraction approximation of the transcendental numbers( 超越数 ). The score varies from small to large. Smaller scores indicate better approximation. The score is not above 1.

The continued fraction approximation of the solutions to quadratic functions. The score is very stable.

Diophantine approximation Dirichlet's approximation theorem has shown that there are an infinite number of diophantine approximation with this score below 1. It has been proven that this score is always below 1 for continued fraction. Thue–Siegel–Roth theorem proved that algebraic numbers are more difficult to approximate than transcendental numbers.

Diophantine approximation Anyway, it has been proven that algebraic numbers are difficult to approximate. It is also shown that the score of a quadratic function is 1/√(b²-4ac). In case of √2, √2 is a solution of x² = 2. a = 1, b = 0, c=-2 so 1/√(b²-4ac) = 1/√8 = 0.35355...

Diophantine approximation Let's find the pair of a, b, and c that minimize (b²-4ac)! Golden Ratio comes from “x² - x - 1 = 0” so (-1)² - 4*1*(-1) = 5 5 is the baseline. (This upper bound was given by Hurwitz's theorem.) x² - 7x - 11 = 0, x = (7 + √5)/2 = 4.6180… x² - 5x + 5 = 0, x = (5 + √5)/2 = 3.6180… x² - 3x + 1 = 0, x = (3 + √5)/2 = 1.6180... But same decimal...it doesn't feel useful...

Diophantine approximation Serrat’s theorem claims that with the constraint of ad-bc=±1, equation y=(ax+b)/(cx+d) becomes equivalence class!! I don't know why...but let's make (a,b,c,d) combinations!! (0 φ +1)/(1 φ +5) = [0; 6, 1, 1, 1, 1,...] = 0.15110... (3 φ +1)/(2 φ +1) = [1; 2, 1, 1, 1, 1,...] = 1.3819... (3 φ -2)/(2 φ -1) = [1; 3, 1, 1, 1, 1,...] = 1.2763... (1 φ +2)/(1 φ +3) = [0; 1, 3, 1, 1, 1,…] = 0.7834... (1 φ +4)/(1 φ +5) = [0; 1, 5, 1, 1, 1,…] = 0.8488… (3 φ +2)/(7 φ +5) = [0; 2, 2, 1, 1, 1,…] = 0.4198...

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There are combinations that fill in two dimensions. In some cases, artifacts are visible, so there may be some correlation?

Conclusion

Conclusion ➢ Approximating a number with continued fractions reveals that properties. ➢ Golden ratio is the most difficult number to express in fractions. ➢ We can create an infinite number of numbers with the same strength as the φ .

Things I couldn't talk about. ➢ Silver ratio ➢ Plastic Number

References

Helpful Sites ➢ A Continued Fraction Calculator. ➢ Going beyond the Golden Ratio.