A tale of two iterations

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

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)

cobweb plot for iterations of f1

This shows that iterations converge quickly to the solution x = 0.89578.

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)

This produces


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.

If we started exactly on the solution, the unstable iteration wouldn’t move. But if we start as close to the solution as a computer can represent, the iterations still diverge quickly. When I changed the starting point to 0.895781791526322, the correct root to full floating point precision, the script crashed with an overflow error after 9 iterations.

More on fixed points

[1] C. W. Groetsch. A Celestial Cubic. Mathematics Magazine, Vol. 74, No. 2 (Apr., 2001), pp. 145–152.

2 thoughts on “A tale of two iterations

  1. I hoped you would throw in :
    if r is a root of x = f(x) then
    f(x) – r = f(x) – f(r) = f'(c)(x-r) for c between x and r
    so estimating |f’| 1 tells you converge or diverge

  2. Additionally, since (f^-1)’ = (f’)^-1, you should always expect that one will converge and the other will not, barring some pathology with |f’| = 1.

    This is in contrast to the situation with a function of two or more variables, where you can have something like f(x, y) = (x/2, 2y) such that neither f nor f^-1 converges to the fixed point in general.

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