The Synergistic Dynamics of Art, Mathematics, & Science
An Interdisciplinary Project at Stowe Middle School
[©Frederick David Abraham, project contributors, and sources]

Difference Equations for Archimedean Spirals

That name sounds kind of scary, but it is really easy. In making the Archimedean sprials, we added an equal amount of angle and an equal amount of distance at each step of the way to the values of angle and distance that we had at the previous step. This is a procedure of repeating (mathematicians say iterating) the same thing (mathematicians say operation) over and over in several successive steps. When we summarize such an iterative procedure in an algebraic way, that is with symbols, we call these difference equations. Difference means that something is made different by a certain operation or amount; that amount is the difference. (Review the figure whenever you wish while working on this section.)

Let's number the step of the iteration, 0,1,2,3, . . . n for steps 0 to n. In exercise 8 we went from 0 to 360 degrees in 36 steps, so n = 36. Actually at step 0, we do not perform the operation, we just establish the initial values of the angle and distance that we start with, 0° and 0 distance for these spirals.

So, in English:

The angle at any step n equals the angle at the previous step, step n-1, plus a constant; the constant equalled 10° for our one-turn Archimedean spirals of Exercises 8-10. Below we will use n and n-1 as subscripts to act as an index (numerical reminder) for which step of the iteration we are doing.

To say this algebraically, mathematicians like to use language something like this:

If we let the anlgle, a, at step n be called a_n, and
if we let the constant amount of angle to add at each step be called D(a), then

a_n = a_n-1 + D(a)

We call this equation a difference equation because it uses a difference D. In algebra, we could change this equation around to read:

D(a) = (a_n) - (a_n-1)

For those who have not yet learned algebra (until now!), you can figure that these two equations are equivalent by seeing that if you subtract (a_n-1) from both sides of the first equation (you can substract the same thing from both sides of an equation and it is still true), and you can switch the right and left sides (you can do that too in algebra), then you get the second equation. The first form of the equation is more like we actually built the spiral, adding to the length of the radius as we went around the circle. The second form of the equation simply reminds us what we already knew: the difference between the one at any step is 10° more than the previous one. Similarly, the distance at step n equalled the distance at step n-1 (the previous step) plus a constant distance. In Exercise 8, this constant we added was .2777... units of distance. So after 9 steps (not counting step 0, and starting with 0° and 0 distance at step 0), we were at 90 degrees and a distance of 2.5 (9 x .27778 = 2.5, one fourth of the distance of 10, the 9 because it was 9 steps after step 0 at 0,0). So the difference equation for the distance is, turning this English into algebra:

If we let the distance at step n be r_n, and
if we let the constant distance to be added at each step to be D(r), then

r_n = r_n-1 + D(r)

Again the other form of the equation would be:

D(r) = r_n - r_n-1

Taking the first form of each of these equations, we now have two simultaneous difference equations to specify the vector in polar coordinates at each step. Thats fancy mathematical talk for the very simple procedure we just did. There is one equation for for each variable, the angle variable, a, and the distance variable, r. Let's see how they work out if used with Exercise 8.

We let:

D(a) = 10°
D(r) = .27778
a₀ = 0 (we make it so, or pretend that a_-1 = -10°)
r₀ = 0 (we make it so, or pretend that r_-1 = -.27778)

Then using the pair of simultaneous equations:

a_n = a_n-1 + D(a)
r_n = r_n-1 + D(r),

We find that:

a₁ = 0 + 10 = 10 degrees
r₁ = 0 + .2778 = .2778 units of distance

a₂ = 10 + 10 = 20
r₂ = .2778 + .2778 =.5556

a₃ = 20 + 10 = 30
r₃ = .5556 + .2778 =.5556

Exercise 12. What are they at step 4? Step 9, step 18, step 28, and step 36? [Hint: you can always add to get the answers, but you can also use multiplication as a shortcut to addition. You can enter the r values in the memory of a hand caclulator, or you can write computer programs to do the work for you. The computers can do the geometry as well as the algebra, and can also make graphs and tables of the results. If you know any computer programs, you might want to try this for the rest of us.