I recently stumbled on a paper [1] that looks at a cubic equation that comes out of a problem in orbital mechanics: σx³ = (1 + x)² Much of the paper is about the derivation of the equation, but here I’d like to focus on a small part of the paper where the author looks at two ways to go about solving this equation by looking for a fixed point.

If you wanted to isolate x on the left side, you could divide by σ and get x = ((x + 1)² / σ)1/3.

If you work in the opposite direction, you could start by taking the square root of both sides and get x = √(σx3) – 1.

Both suggest starting with some guess at x and iterating.

There is a unique solution for any σ > 4 and so for our example we’ll fix σ = 5.

We define two functions to iterate, one for each approach above.

sigma = 5 x0 = 0.

1 def f1(x): return sigma**(-1/3)*(x+1)**(2/3) def f2(x): return (sigma*x**3)*0.

5 – 1 Here’s what we get when we use the cobweb plot code from another post.

cobweb(f1, x0, 10, “ccube1.

png”, 0, 1.

2) This shows that iterations converge quickly to the solution x = 0.

89577.

Now let’s try the same thing for f2.

When we run cobweb(f2, x0, 10, “ccube2.

png”, 0, 1.

2) we get an error message OverflowError: (34, Result too large) Let’s print out a few values to see what’s going on.

x = 0.

1 for _ in range(10): x = f2(x) print(x) This produces -0.

9975 -3.

4812968359375005 -106.

4783129145318 -3018030.

585561691 -6.

87243939752166e+19 -8.

114705541507359e+59 -1.

3358518746543001e+180 before aborting with an overflow error.

Well, that escalated quickly.

The first iteration converges to the solution for any initial starting point in (0, 1).

But the solution is a point of repulsion for the second iteration.

More on fixed points Kepler’s equation Contractions and weak contractions Fixed point in logistic regression [1] C.

W.

Groetsch.

A Celestial Cubic.

Mathematics Magazine, Vol.

74, No.

2 (Apr.

, 2001), pp.

145–152.

.