# The silver ratio

Most people have heard of the golden ratio, but have you ever heard of the silver ratio? I only heard of it this week. The golden ratio can be expressed by a continued fraction in which all coefficients equal 1.

The silver ratio is the analogous continued fraction with all coefficients equal to 2.

You might think for a moment that the silver ratio should be just twice the golden ratio, but the coefficients contribute to the series in a non-linear way. The silver ratio actually equals 1 + √2. The golden ratio has a simple geometric interpretation. I don’t know of a geometric interpretation of the silver ratio. (Update: See Maxwell’s Demon for geometric applications of the silver ratio.)

A previous post mentioned that the golden ratio and related numbers are the worst case for Hurwitz’s theorem. The silver ratio and its cousins are the second worst case for the theorem.

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# Breastfeeding, the golden ratio, and rational approximation

Gil Kalai’s blog featured a guest post the other day about breastfeeding twins.

# Golden ratio and special angles

The golden ratio comes up in unexpected places. This post will show how the sines and cosines of some special angles can be expressed in terms of the golden ratio and its complement.

Recall the golden ratio is

φ = (1 + √ 5)/2

and the complementary golden ratio is

φ’ = (1 – √ 5)/2.

The derivation begins by solving the trigonometric equation

sin 2θ = cos 3θ

in two different ways. To make the solution unique, we look for the smallest positive solution.

First, note that the sine of an angle is the cosine of its complement, i.e.

sin(x) = cos(π/2 – x).

So our equation can be written as

cos(π/2 – 2θ) = cos 3θ.

The smallest positive solution satisfies π/2 – 2θ = 3θ, and so θ = π/10 or 18°.

Now let’s solve the same equation another way. First, we use the double and triple angle identities.

sin 2θ = 2 sin θ cos θ
cos 3θ = 4 cos3 θ – 3 cos θ

Set the two equations above equal to each other and divide by cos θ. Then we have

4 cos2 θ – 3 = 2 sin θ.

Substitute 1 – sin2 θ for cos2 θ and the result is a quadratic equation in sin θ:

4 sin2 θ + 2 sin θ – 1 = 0.

From the quadratic equation, the solutions are sin θ = (-1 ± √ 5)/2. The positive solution is

sin θ = (-1 + √ 5)/2 = -φ’/2.

Setting the solutions obtained from both methods equal to each other,

sin π/10 = sin 18°= -φ’/2.

We can now use common trig identities and the above result to express the sines and cosines of other angles in terms of  φ. Switching to degrees will make the following a little easier to read.

We know sin 18° = -φ’/2, and so cos 72° = -φ’/2. We can use the sum angle identities to express the sine and cosine of every multiple of 18° in terms of φ. Also, we could apply the half angle identities to express the sine of cosine of 9° in terms of φ, and then again by addition formula we could extend this to all multiples of 9°.

This post was an expanded form of a derivation given in The Divine Proportion.

# Connecting Fibonacci and geometric sequences

Here’s a quick demonstration of a connection between the Fibonacci sequence and geometric sequences.

The famous Fibonacci sequence starts out 1, 1, 2, 3, 5, 8, 13 … The first two terms are both 1, then each subsequent terms is the sum of the two preceding terms.

A generalized Fibonacci sequence can start with any two numbers and then apply the rule that subsequent terms are defined as the sum of their two predecessors. For example, if we start with 3 and 4, we get the sequence 3, 4, 7, 11, 18, 29, …

Let φ be the golden ratio, the positive solution to the equation 1 + x = x2. Let φ’ be the conjugate golden ratio, the negative solution to the same quadratic equation. Then φ = (1 + √ 5)/2 and φ’ = (1 – √ 5)/2.

Now let’s look at a generalized Fibonacci sequence starting with 1 and φ. Then our terms are 1, φ, 1 + φ, 1 + 2φ, 2 + 3φ, 3 + 5φ, … Let’s see whether we can simplify this sequence.

Now 1 + φ = φ2 because of the quadratic equation φ satisfies. That tells us the third term equals φ2. So our series starts out 1, φ, φ2. This is looking like a geometric sequence. Could the fourth term be φ3? In fact, it is. Since the fourth term is the sum of the second and third terms, it equals φ +φ2 = φ(1 + φ) = φ(φ2) = φ3. You can continue this line of reasoning to prove that the generalized Fibonacci sequence starting with 1 and φ is in fact the geometric sequence 1, φ, φ2, φ3, …

Now start a generalized Fibonacci sequence with φ’. Because φ’ is also a solution to 1 + x = x2, it follows that the sequence 1, φ’, 1 + φ’, 1 + 2φ’, 2 + 3φ’, … equals the geometric sequence 1, φ’, (φ’)2, (φ’)3, …

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# Fibonacci numbers at work

Today I needed to use Fibonacci numbers to solve a problem at work. Fibonacci numbers are great fun, but I don’t recall needing them in an applied problem before.

I needed to compute a series of integrals of the form

f(x, y) = xa (1-x)b yc (1-y)d p(x, y)

over the unit square for a statistical application. The function p(x, y) is a little complicated but its specific form is not important here. If the constants a, b, c, and d are all positive, as they usually are in my application, the integrand can be extended to a continuous periodic function in the plane. Lattice rules are efficient for such integration problems, and the optimal lattice rules for two-variable integration are given by Fibonacci lattice rules.

If Fk is the kth Fibonacci number and {x} is the remainder when subtracting off the greatest integer less than x, the kth Fibonacci lattice rule approximates the integral of f by

This rule worked well. I compared the results to those using the two-variable trapezoid rule and the lattice rule always gave more accuracy for the same number of integration points. (Normally the trapezoid rule is not very efficient. However, it can be surprisingly efficient for periodic integrands. See point #2 here. But the Fibonacci lattice rule was even more efficient.)

To implement this integration rule, I first had to write a function to compute Fibonacci numbers. This is such a common academic exercise that it is almost a cliché. I was surprised to have a practical need for such a function. My implementation was

```    int Fibonacci(int k)
{
double phi = 0.5*(1.0 + sqrt(5.0));
return int( pow(phi, k)/sqrt(5.0) + 0.5 );
}```

The reason this works is that

where

is the golden ratio. The second term in the formula for Fk is smaller than 1, and Fk is an integer, so by rounding the first term to the nearest integer we get the exact result. Adding 0.5 before casting the result to an integer causes the result to be rounded to the nearest integer.

If you’re interested in technical details of the numerical integration method, see Lattice Methods for Multiple Integration by I. H. Sloan and S. Joe.