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# Lill’s Method and the Golden Ratio

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Lill’s method is a remarkable visual method of finding the real roots of polynomials of any degree. This method was developed by the Austrian engineer Eduard Lill in 19th century. In this hub I will show how to construct the golden ratio using Lill’s method. More specifically I will use Lill’s method to find the roots of the polynomial P(z)=z2 -z -1. P(z)=0 when z= ϕ (golden ratio)≈1.61803 and when z=-Φ (negative of golden ratio conjugate)≈-0.61803. I will also introduce Lill’s circle, which will help us solve the polynomial geometrically. At the end of the hub I will add useful sources where you can learn more about Lill’s method and Lill’s circle. This hub can be treated as a introduction to Lill’s method using a specific example.

## Lill’s Method Setup

Let’s have the general polynomial of order n, P(z)=anzn+an-1zn-1+…+a1z+a0 , where the coefficients {an, an-1, … , a1, a0} are real numbers. In Lill’s method these n coefficients are represented by n “reflective” segments that are connected in a link, and each segment will have its length equal to the absolute value of one of the n coefficients. Each of these n “reflective” segments must also be extended (at infinity) in both directions by “refractive” lines. Also each of these segments must have a direction, and I will use the following formula for determining the length and the direction of a segment : akei(n-k)π/2, where ak is the coefficient corresponding to the segment, i is the imaginary number and 0≤k≤n. To get an idea, we have e0=1 (points right), eiπ/2=0+1i (points up), ei2π/2= e=-1 (points left) , ei3π/2=-i (points down), ei4π/2 =1 (points right) etc. We see that after 4 consecutive steps, the formula gives the same direction.

To connect the segments, first we start with a point O. Then we construct the segment corresponding to an by using the formula for length and direction anei(n-n)π/2= an e0=an. Thus from O we will construct a segment of length abs(an) that will point in the right direction if an is positive and it will point in the left direction if an is negative. The terminus point of the segment corresponding to an is going to be the starting point of the segment corresponding to an-1. This process is repeated until we construct the segment corresponding to a0.

Now let’s make the initial setup for the polynomial P(z)=z2-z-1. The order of the polynomial is 2, thus n=2. We will have the following segment lengths and directions: ei(2-2)π/2=2 e0=1 (points right), (-1) ei(2-1)π/2=-eiπ/2=-1i (points down), (-1) ei(2-0)π/2=-e=1 (points right). Thus starting from point O, the segment corresponding to a2=1 will go one unit to the right. The terminus of the first segment can be labeled as point A. From point A we will go one unit down, and mark the new terminus as point B. Finally from B we go one unit to the right and mark the final terminus point as C. Image 1 shows the initial setup. All 3 segments were extended by dotted lines, and these dotted lines are “refractive” lines.

## Finding a Solution

A polynomial is solved using Lill’s method if you can find a continuous “ray” path made of n segments that starts at point O and ends at the terminus point of the segment corresponding to the coefficient a0. In our case we must find a 2 segment ray path that starts at O and ends at C. Thus starting at O we will draw a ray that makes an angle θ with an, and we want this ray to intersect the segment corresponding to an-1 or the dotted line extension of this segment. In our case we will mark the intersection as D. From point D we will make another ray that makes the same angle θ, and similarly we want this to be in the direction such that it will intersect the segment corresponding to an-2 or the dotted line extension of this segment. One important rule is that if point D is on the segment AB, the second ray must be a “reflection”. If D is on the dotted line, the second ray segment will be a refraction. Another important rule is that if our ray segment goes from segment ak to ak-1 we can ignore the other segments or the dotted lines corresponding to the other segments if they block the path of our ray. Taking in consideration these rules from D we make a second ray that intersects an-2=a0 or its extension at point E. We have a solution if point E falls over point C (they are the same point). In Image 2 and 3 I show two paths that don’t give a solution, but the important rules are illustrated. In image 3 I pointed specifically the fact that 2 consecutive ray segments are always perpendicular. Thus in our case, segment OD will be always perpendicular to DE. In Image 2 I specified the fact that angle ODE=90.

If the points E and C are the same, than we have a solution. In numerical terms the root of the polynomial will be z=-tan(θ), where θ is positive if the first ray OD goes counterclockwise with respect to OA and negative otherwise (see image 4). If we use a transparent graph paper with small squares we may be able to approximate very accurately the angles that give the real roots. But it is preferable to have a method that gives the exact angles. By drawing a Lill circle we will obtain the desired solution.

## Lill’s Circle

Lill’s circle is very similar to Carlyle circle, but Lill’s circle can be applied to any second degree polynomial that has real roots. Our second degree polynomial P(z)=z2-z-1 has real roots, thus we are able to use this method. The first step in this method is to connect by a line the origin point O to the terminus point of a0, which is C in our case. Thus we obtain a segment OC. The next step is find the midpoint M of segment OC. Now we will draw a circle centered at M and with radius equal to r= OM=MC. This circle will intersect segment AB or its dotted line extension at points D1 and D2. Finally we have AD1=abs(a2)abs(first root)=1(ϕ)=ϕ (golden ratio) , and AD2=abs(a2)abs(second root)=1abs(-Φ)=Φ (conjugate of golden ratio). Image 5 shows Lill’s circle for our polynomial and the points of intersection.

Thus angles AOD11 and AOD22 should be the angles that give the 2 roots. In image 6 I show that the 2 angles are indeed solutions. I included also the angle values so that you can check that –tan(θ1)=ϕ and –tan(θ2)=-Φ. Remember that according to the rules θ1 should be negative and θ2 should be positive (in image 6 you can see the absolute values).

## Lill’s Circle And Thales’ Theorem

In one of the previous sections I mentioned that 2 consecutive segments of a ray path are always perpendicular. In our example we had to find a ray path made of 2 perpendicular segments that started at O and ended at C. Why Lill’s circle (and Carlyle circle) works can be explained by Thales’ theorem. To remind you, Thales’ theorem states the following: : if A,B and C are points on a circle where the segment AC is a diameter of the circle, then the angle ABC is a right angle. In image 7 you can see the theorem illustrated.

In image 8 we have again the circle with center at M and with the radius r= OM=MC. Our points O,D1 and C are on the circle and OC is a diameter of the circle. Thus by Thales’ theorem the angle OD1C should be a right angle, and this gives as a ray paths that solves our polynomial. Similarly by Thales’ theorem, the angle OD2C is a right angle.