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Everybody loves the good old quadratic Mandelbrot set. As you probably know, both it and the corresponding quadratic Julia sets are defined by the iteration $f(z) = z^2 + c$.

You might expect, however, that $f(z) = az^2 + bz + c$ would give you more possibilities. However, all the books on the subject assert that this is not the case. I'm trying to get to the bottom of why this is so.

For a start, you can see that if you divide the entire thing through $a$, then the specific values taken by $f$ would change, but their relationship would not. Hence, multiplying by $a$ is only scaling and/or rotating the system. It doesn't actually change its behaviour.

But what of the linear term, $bz$? Why is that redundant?

We can ask a similar question about the cubic Mandelbrot set. A lot of people define this as $g(z) = z^3 + c$, but the definition I like is $h(z) = z^3 - 3a^2z + b$. This has two critical points (whatever that means), which yields strange, shadowy images. More interestingly, with two complex-valued parameters, the corresponding Mandelbrot set is four-dimensional!

Again, we are told that $h(z)$ is the most general formulation. (In particular, you don't need a quadratic term.) The strange formulation $-3a^2z$ rather than just $az$ seems necessary to make the parameter plane plot correctly. (It also means that the critical points are $+a$ and $-a$ exactly.)

Does anybody know why this formulation is correct? What would the general 4th order case look like?

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You realize that if you complete the square in your general quadratic, your iteration ends up looking a lot like the simpler iteration by Mandelbrot, right? – J. M. May 4 '12 at 10:55
@J.M. If I knew what "completing the square" actually is, then perhaps the answer would be yes. :-} – MathematicalOrchid May 4 '12 at 10:56
Ah, you haven't seen the identity $$ax^2+bx+c=a\left(x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a}$$ before? – J. M. May 4 '12 at 10:58
"This has two critical points (whatever that means)" - it means that its derivative, which is a quadratic, has two roots. This corresponds to optima or inflexion points in the cubic. – J. M. May 4 '12 at 11:00
OK, ouch. It's going to take me at least a few minutes to congnitise that... – MathematicalOrchid May 4 '12 at 11:02
up vote 3 down vote accepted

For the first question, the key word is conjugation: there is an affine change of coordinates $\phi$ such that $f(z) = az^2 + bz + c$ becomes $g(z)=z^2+c$ in the sense that $f\circ\phi=\phi\circ g$. Since, $g=\phi^{-1} \circ f\circ\phi$, the iterates of $f$ are conjugated to the iterates of $g$ by the same change of coordinates.

For the second question, the formulation of cubic maps as $h(z) = z^3 - 3a^2z + b$ is probably just to highlight the critical point $a$, as you have noticed.

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OK, that's a pretty damned nice explanation. (Although I don't quite get what "affine" means; I'm presuming you just mean a simple geometric transformation that doesn't alter "structure" in some sense.) – MathematicalOrchid May 4 '12 at 11:58
@MathematicalOrchid, affine means $\phi(z)=u z + v$. – lhf May 4 '12 at 13:16

OK, I think I see what J.M. is getting at now:

If I write $g(z) = (z + \alpha)^2 + \beta$, then we have

$$(z + \alpha)^2 + \beta = z^2 + (2\alpha)z + (\alpha^2 + \beta)$$

In other words, taking $f(z) = z^2 + c$ and adding a linear term is like performing one addition before the squaring, and another addition after the squaring.

This means that $f$ and $g$ are different functions. However, when considering the iteration, you can coalesce the two additions into one step, such that both functions produce "the same" orbit, for a suitable definition of "same" (and suitable parameters, obviously).

This, then, is why $f(z)$ is all you need to care about. $g(z)$ truly adds no extra generality.

That seems to tie up the quadratic case. Perhaps there's an analogue for "completing the cube" which lets me see why the cubic case doesn't need a quadratic term...

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There is no such thing as "completing the cube"; there is, however, what is known as a depression substitution that can be done for polynomials of arbitrary order. Briefly, if you have a polynomial $ax^n+bx^{n-1}+\cdots$, and you let $u=x-\frac{b}{na}$, you obtain a polynomial in $u$ that has no $u^{n-1}$ term. Completing the square is the simplest special case. – J. M. May 4 '12 at 12:28

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