Question

There are many "strange" functions to choose from and the deeper you get involved with math the more you encounter. I consciously don't mention any for reasons of bias. I am just curious what you consider strange and especially like.

Please also give a reason why you find this function strange and why you like it. Perhaps you could also give some kind of reference where to find further information.

As usually: Please only mention one function per post - and let the votes decide :-)

Answer 1

Characteristic p commutative algebra leads naturally to the construction of various continuous functions on [0,1]^m that have beautiful self-similarity properties; for explication and some pictures see:

Pedro Teixeira, Syzygy gap fractals--I, arXiv 1008.0583

Answer 2

How about the function given by the Banach-Tarski paradox? This maps a ball into two copies of the same size ball, and is composed of isometries on subsets of $\mathbb{R}^3$.

Answer 3

Nonconstant continuous locally recurrent functions are quite unintuitive. A real-valued function is locally recurrent on $\mathbb R$ if for every $x_0\in\mathbb R$ and every deleted neighborhood $N(x_0)$ of $x_0$, there exists $x\in N(x_0)$ for which $f(x)=f(x_0)$. Thus in some sense a nonconstant continuous locally recurrent function looks everywhere like $x\sin(1/x)$ at $x=0$. See papers in the American Math. Monthly of Bush (1962), Marcus (1963), and Mauldon (1965).

Answer 4

The formula for the nth term in the Fibonacci sequence

$F_{n} = \cfrac{1}{\sqrt{5}}\cdot\left(\cfrac{1+\sqrt{5}}{2}\right)^n-\cfrac{1}{\sqrt{5}}\cdot\left(\cfrac{1-\sqrt{5}}{2}\right)^n$

This is interesting because it is a non-recursive expression for the Fibonacci sequence and also because it involves the golden ratio.

Answer 5

Just a simple construction to illustrate Nate Eldredge's answer about functions with dense graphs. Pick any $\mathbb{R}$-vector space E with a norm. On E, choose a non-continuous linear form $L: E \to \mathbb{R}$; now this can only be done if $\dim(E)=\infty$, of course.

Then, pick y such that $L(y)=1$, and let $T: E \to E$ be defined by $Tx=x-L(x)y$. Then obviously T maps E onto the kernel of L; it is not difficult to prove that $\ker (L)$ must be dense in E for any non-continuous L (the two conditions are even equivalent), and thus the graph of T must be dense in $E \times E$.

Answer 6

How about a function f: f(f(x)) = exp(x).

Answer 7

I like the beauty and mysticism of Euler's identity:

$$f(\theta) = e^{i\theta} = \cos\theta + i \sin\theta$$