Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The literature on Mandelbrot and Julia sets mentions the phase "critical point" quite a lot, but usually doesn't bother to define what it means.

As best as I can tell, a critical point is just any point where the function's derivative is zero.

This cannot be the whole story, however. For example, the same literature often states that the critical points of $f(z) = z^2 + c$ are "zero and infinity".

Clearly $f'(z) = 2z$. Obviously $f'(0)=0$. So zero is a critical point. But how on Earth do they figure that infinity is a critical point? $f'(\infty) = 2\infty = \infty \not= 0$. (??)

Edit: According to Wikipedia, a critical point is anywhere the function's derivative is zero or undefined.

Is this going to be one of those things involving the complex plane "with a point at infinity"? I'm thinking, perhaps as $z$ approaches $+\infty$ then $2z$ does to $+\infty$, but as $z$ approaches $-\infty$ then $2z$ approaches $-\infty$, which isn't the same limit? Or something like that?

share|cite|improve this question
Have you seen this? – J. M. May 4 '12 at 12:11
@J.M. I have now. :-) I don't really understand how the OP got to his conclusion, but I like Isaac's answer. Not 100% sure how what it implies for my question... – MathematicalOrchid May 4 '12 at 12:22
In a sense, $2z$ does have two roots: the obvious one and the infinite one. Why this is so is tackled in that question I linked to. – J. M. May 4 '12 at 12:25
Oh, I see what you're getting it. (I think.) Does this "sense" have a name, by any chance? – MathematicalOrchid May 4 '12 at 12:27

3 Answers 3

up vote 0 down vote accepted

I tend to think of critical points (in a calculus sense) as places where the derivative is zero or undefined/infinite (or possibly discontinuous).

share|cite|improve this answer
So does that mean that every polynomial function has a critical point at infinity then? – MathematicalOrchid May 4 '12 at 12:38
@MathematicalOrchid: I suppose so—I hadn't really thought much about it, but if $f(z)=z^2+c$ has a critical point at $\infty$, it would make sense that every polynomial would... or, at least, every non-constant polynomial... hmm, actually, maybe just for polynomials of degree greater than or equal to 2. – Isaac May 4 '12 at 12:40

From what I understand a critical number is a number $a$ where the derivative of the function is not defined or is equal to zero. So I would say (and I assume that most would agree with this definition) that when we are talking about real valued functions of one real variable, then $\infty$ is not a critical number simply because "infinity is not a number". I think that really is the end of it.

However, sometimes we are interested in what happens at infinity, but in those cases we start to talk about limits. But if you insisted on extending the definition of a critical number to: (a) any number where (a1) the function has no derivative or (a2) where the derivative is zero or (b) we call infinity ($\infty$) a critical number when $\lim_{x \to \infty} f'(x)$ does not exist (i.e. is not equal to a real number), then that might be OK (but that is really a question that depends on opinion).

At the bottom line, when people critical numbers in for example ca ollege calculus 1 class, they don't include $\infty$.

share|cite|improve this answer

Having a critical point "at infinity" makes sense when you're looking at the Riemann sphere, which you can think of as the union of $\mathbb C$ with a single point that we call $\infty$, and which is useful in complex dynamics (which includes the study of the Mandelbrot set and Julia sets). If you want to evaluate a rational function $\phi(z) = f(z)/g(z)$ whose numerator and denominator share no common factors, at a point $c\in\mathbb C$ where $g(c) = 0$, we define $\phi(c) = \infty$.

We say that a rational function $\phi(z)$ has a critical point at $\alpha\in\mathbb C\cup\{\infty\}$ if for all Möbius transformations $f(z)$ such that $\beta = f^{-1}(\alpha)\neq \infty$ and $\phi^f(\beta) = (f^{-1}\circ\phi\circ f)(\beta)\neq \infty$, $(\phi^f)'(\beta) = (f^{-1}\circ\phi\circ f)'(\beta) = 0$. One can show that this is a extension of the familiar definition of a critical point: if $\alpha\neq\infty$ and $\phi(\alpha)\neq\infty$, then $\alpha$ is a critical point of $\phi$ (according to the definition in the previous sentence) iff $\phi'(\alpha) = 0$. Note that we when refer to a derivative in this context, it is a formal derivative; for example, we are formally defining $$ \phi'(z) = \frac{g(z)f'(z) - f(z)g'(z)}{(g(z))^2}. $$ Now, the following lemma is quite handy:

Lemma. Let $\phi$ be a rational function and let $f$ be a Möbius transformation. Then $\forall \alpha\in\mathbb C\cup\{\infty\}$, $\alpha$ is a critical point of $\phi$ iff $\beta = f^{-1}(\alpha)$ is a critical point of $\phi^f$.

Using this lemma, we can verify Isaac's suggestion that every polynomial of degree at least two has infinity as a critical point. Consider a polynomial function $\phi(z) = a_n z^n + \cdots + a_1z + a_0$, where $n\geq 2$. Define the Möbius transformation $f(z) = 1/z$. Then $f^{-1}(\infty) = 0$, and \begin{align*} \phi^f(z) & = (f^{-1}\circ\phi\circ f)(z) \\ & = \frac{1}{\tfrac{a_n}{z^n} + \cdots + \tfrac{a_1}{z} + a_0} \\[3pt] & = \frac{z^n}{a_n + \cdots + a_1z^{n-1} + a_0z^n}. \end{align*} By writing out the formal derivative $(\phi^f)'(z)$, we can check that $(\phi^f)'(0) = 0$, so clearly $0$ is a critical point of $\phi^f$. It then follows from the lemma that $\infty$ is a critical point of $\phi$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.