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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 :-)

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m.reddit.com/r/math/comments/9txzv/… – Regenbogen Apr 22 2010 at 14:21
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Counterexamples in analysis has some nice ones: books.google.nl/… – skupers Aug 21 2010 at 18:40
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closed as no longer relevant by Felipe Voloch, Mark Meckes, Henry Cohn, unknown (google), S. Sra Jul 17 at 16:21

37 Answers

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A Brownian motion sample path.

These are about the most bizarrely behaved continuous functions on $\mathbb{R}^+$ that you can think of. They are nowhere differentiable, have unbounded variation, attain local maxima and minima in every interval... Many, many papers and books have been written about their strange properties.

Edit: As commented, I should clarify the term "sample path". Brownian motion is a stochastic process $B_t$. We say a sample path of Brownian motion has some property if the function $t \mapsto B_t$ has that property almost surely. So, run a Brownian motion, and with probability 1 you will get a function with all these weird properties.

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Pedantic, but maybe worth mentioning that these functions have these properties almost surely rather than certainly. – Tom Smith Apr 22 2010 at 20:48
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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.

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

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

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The Conway base 13 function has to be the weirdest function I know. This function is continuous nowhere, yet it satisfies the intermediate value theorem. Only John Conway...

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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).

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Thomae's function, also called the "popcorn function". It's continuous at all irrationals and discontinuous at all rationals. Here a picture:

Thomae's function

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The function defined by the power series $f(x)=x-x^2+x^4-x^8+x^{16}-\dots$ What is its limit as $x$ approaches $1$ from below? EDIT (This answer is a trick question.)

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SPOILER (albeit 1.5+ years late): math.harvard.edu/~elkies/Misc/sol8.html (and the PDF graph at math.harvard.edu/~elkies/Misc/gamma_pic0.pdf) – Noam D. Elkies Jun 8 at 23:06
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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 $\mathbb{R}^3$.

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

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I can't believe no one has mentioned the Dirichlet function:
(I guess it's up to me to bring it up...)

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If you plant a post of unit height at every point in $\{(x, y) : x, y \in \mathbb{N}^+\}$ and stand at the origin, looking in the direction of $(1,1)$, you will see this picture. – Max Nov 13 2010 at 13:09
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The Fabius function

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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 $E \times E$.

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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
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Any of the isomorphisms $\mathbb{C}'\to S^{1}$, where $S^{1}$ is the unit circle and $\mathbb{C}'$ is the non-zero complex numbers, with the group operation for both being multiplication.

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Fix a probability $p < 1/2$ of winning an unfair coin toss. For $x \in [0,1]$ rational, let $f(x)$ be the probability that, if you started with $x$ dollars, you could make it to 1 dollar through optimal betting* on the outcome of these coin flips. This function $f(x)$ is obviously weakly increasing on $[0,1]$ (in fact strictly). Less obvious is that it extends to a continuous function on $[0,1]$, whose derivative exists almost everywhere, but that derivative is $0$.

http://www.maa.org/joma/Volume8/Siegrist/RedBlack.pdf

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The Banach limit assigns to every bounded sequence of real numbers a real number "limit" in a way that is linear, shift invariant, and agrees with the usual limit whenever it exists. Banach limits are among the mysterious examples of continuous linear functionals on $\ell^\infty$ that aren't represented by elements of $\ell^1$. Unfortunately, the Hahn-Banach theorem is used in the construction of the Banach limit, and the values aren't canonical. There's a precise definition at http://en.wikipedia.org/wiki/Banach_limit .

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These functions like the Cantor function and the continuous-but-not-differentiable function are all well and good, but contrived - the only place you ever see them is as counterexamples. Here is a function that has many uses in Number Theory, and still manages to have a strange property or two. Let $x=h/k$ with $h$ and $k$ integers, $k>0$. Define $$s(x)=\sum_{c=1}^{k-1}((c/k))((ch/k))$$ where $((y))=0$ if $y$ is an integer, $((y))=\lbrace y\rbrace-1/2$ otherwise. It is easily proved that the sum depends only on the ratio of $h$ and $k$, not on their individual values, so $s$ is a well-defined function from the rationals to the rationals. It is known as the Dedekind sum; it came up originally in Dedekind's study of the transformation formula of the Dedekind $\eta$-function.

Now for the strange properties.

Hickerson, Continued fractions and density results for Dedekind sums, J Reine Angew Math 290 (1977) 113-116, MR 55 #12611, proved that the graph of $s$ is dense in the plane.

With Nick Phillips, I proved (Lines full of Dedekind sums, Bull London Math Soc 36 (2004) 547-552, MR 2005m:11075) that, with the exception of the line $y=x/12$, every line through the origin with rational slope passes through infinitely many points on the graph of $s$. We suspect that the points are dense on those lines, though we could only prove it for the line $y=x$.

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A canonical example from elementary real analysis - the Blancmange function. Consider $f$ defined piecewise by

$f(x) = x - [x], \quad \text{if} \quad 0 \leq x- [x] \leq \frac{1}{2}$,

and

$f(x) = 1 - (x - [x]), \quad \text{if} \quad \frac{1}{2} < x - [x] < 1$,

(where $[x]$ is the integer part of $x$). Then define the Blancmange function, $B$

$B(x) = \sum_{n=0}^{\infty}\dfrac{1}{2^n} f(2^{n}x)$.

The series converges by the Comparison Test, since $|f(2^{n}x)| \leq \frac{1}{2}$, for all $x \in \mathbb{R}$, and it can be shown that $B$ is uniformly continuous but nowhere differentiable. Here a picture of the function:

Blancmange function

A tasty counterexample to the converse of "differentiability $\implies$ continuity".

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Closely related is the tavuk göğsü function, replete with shredded chicken. – Tom LaGatta Jul 1 2010 at 23:34
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f(x) = sin (1/x): (x not 0); f(x) = 0 (x equals 0)

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The functions one learns about early in studying mathematics are chosen to illustrate various "issues:' continuity, having a derivative, being periodic, etc. One of the functions one learns about in this way is y = sin(x). So while there are many functions that are "strange," the transition from y = sin (x) to y = sin (1/x) offers I feel lots of nice lessons about functions and their behavior. There are many web sites that use graphics to help one understand what is going on here. One such site is: math.washington.edu/~conroy/general/sin1overx – Joseph Malkevitch Apr 23 2010 at 15:32
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Interpreting your questions a bit liberally, I suggest the Goodstein sequence:

http://en.wikipedia.org/wiki/Goodstein%27s_theorem

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Since Mariano took my favorite already, I'll go with the stopping time function for the 3x+1 problem: http://www.ieeta.pt/~tos/3x+1.html

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The Osgood curve ("A Jordan Curve of Positive Area") is an injective map from [0,1] into $\mathbb{R}^2$ which traces out an image of positive area. (This differs from standard space-filling curves, which are not injective.)

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I'm still quite impressed about $f(x)=\mathrm e^x$ …

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Very true, me too! Could you please elaborate on why you consider this function "strange" - thank you. – vonjd Apr 23 2010 at 9:29
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Because of its stubborn nature: whether you differentiate it, or integrate it, it remains unmoved ;-) – S. Sra Nov 13 2010 at 17:39
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There's a part of me which has never quite gotten over the Euler identity $e^{ix} = \cos(x) + i\sin(x)$, which was perhaps the biggest intellectual thrill of my early teenage years... – Todd Trimble May 29 2011 at 23:33
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It is pretty obvious after you've seen it, but I like the crinkled curve from Halmos's Hilbert Space Problem book:

Let $f:\mathbb{R}\rightarrow(0,\infty)$ be an $L^2$ function, and define $t\mapsto g_t:\mathbb{R}\rightarrow L^2(\mathbb{R})$ by $$g_t(x)=\chi_{(-\infty,t)}(x) \times f(x).$$

Then $g_t$ has the property that for all $t_1 < t_2 < t_3$ the secants $g_{t_2}-g_{t_1}$ and $g_{t_3}-g_{t_2}$ are mutually orthogonal. (The curve turns a corner at every point.)

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I suppose the strangest function in mainstream mathematics is Riemann zeta function http://en.wikipedia.org/wiki/Riemann_zeta_function

$ \zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s} = \frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \cdots \;\;\;\;\;\;\;{Re}(s) >1. $

It is part of one of the most important hypothesis and is very influential in many branch of moder mathematics. It is actively used in many areas and is researched in many ways, it is not curiosity, or exotic example, but important mathematical being!

And is mysterious and strange! Take a look: alt text

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See also en.wikipedia.org/wiki/Zeta_function_universality . – Qiaochu Yuan Apr 23 2010 at 6:17
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@Qiaochu Yuan: THAT IS SO COOL! – Vectornaut Apr 23 2010 at 18:27
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The Ackermann function $A(n,m)$ is defined on the natural numbers by a very simple recursion, but the values grow enormously, almost beyond conception. This function completely transcends any simple-minded system of rates-of-growth based on polynomial, exponential, double-exponential and so on.

The first few values of the diagonal function $A(n) = A(n,n)$ are:

  • $A(0) = 1$
  • $A(1) = 3$
  • $A(2) = 7$
  • $A(3) = 61$
  • $A(4) = 2^{2^{2^{65536}}}-3$
  • $A(5)$ is vast, and can be described in terms of exponential stacks of $2$s, whose height is a stack of $2$s, etc. 5 times.
  • $A(6)$ is so vast, it is best described using the Ackermann function itself.

The levels of the Ackerman function $A_n(m)=A(n,m)$ stratify the primitive recursive functions, in the sense that they are each primitive recursive, but every primitive recursive function is bounded by such a level of the Ackermann function. Thus, the Ackermann function itself is not primitive recursive, although it is computable in the sense of computability theory.

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I never really thought of the Ackermann function as being strange, only big. But maybe that's just me. – Ketil Tveiten Apr 23 2010 at 12:13
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Ketil, yes, perhaps I agree. But what is strangely wonderful about it is that the recursive definition A(n+1,m+1)=A(n,A(n+1,m)) is so simple, and yet leads immediately to such incomprehensible growth. – Joel David Hamkins Apr 23 2010 at 12:21
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I can't remember where (probably tvtropes), but when reading something about the Ackermann numbers (1 ^ 1, 2 ^^ 2, 3 ^^^ 3, etc), which are related to the Ackermann function, the joke was "it's always weird when looking at a sequence of numbers that goes: 1, 4, too big to count." – Gabriel Benamy Jul 4 2010 at 15:39
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The empty function $\emptyset:\emptyset\to\emptyset$ is quite strange when you first meet it.

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For me, it makes a good argument that the "correct" value for $0^0$ is 1: it's the number of functions from a set with 0 elements to a set with 0 elements. – Nate Eldredge Apr 22 2010 at 19:13
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And, of course, it gives combinatorial meaning to the fact that 0! = 1, since the empty function is a bijection! – Qiaochu Yuan Apr 23 2010 at 6:36
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@Tom: …but it can always be very easily computed! – Peter LeFanu Lumsdaine Nov 13 2010 at 7:53
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Elsewhere I have raised the question of whether this function should be considered a constant function. On the one hand, $f(x_1)=f(x_2)$ for every $x_1$ and $x_2$ in the domain; on the other hand, there is no $y$ in the codomain such that for every $x$ we have $f(x)=y$. I consider it non-constant. – Tom Goodwillie Nov 14 2010 at 3:29
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A morphism in a category with terminal object $t$ may be called constant if it factors through $t$. According to this definition, $\emptyset \to S$ is constant iff $S$ is nonempty. – Martin Brandenburg Jul 16 at 16:50
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One can construct a natural 'metric' for the Riemann sphere which is equivalent to the spherical metric but which is singular on a dense set of points of the Riemann sphere though remains $L^1$ integrable.

These are built from degree 2 rational maps (first constructed by Mary Rees) which have the whole Riemann sphere as their Julia sets, and have the orbits of their critical points also dense. The Carlesson-Jones-Yoccoz construction of a expanding metric for critically-finite rational maps actually extends to this case, and we get a metric in which this Julia set actually looks as if it was hyperbolic!

[The details are worked out in my PhD thesis, never published as I decided that computer algebra suited me better than complex dynamics].

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Please excuse me if I include two related functions in one answer. Any space filling curve is rather strange, at least for me. Let $\gamma\colon[0,1]\to[0,1]^2$ be such a curve, that is, $\gamma$ is continuous and surjective. Let $\gamma(t)=(x(t),y(t))$; then $x(t)$ (or $y(t)$) is my other candidate for strangest function: given any $z\in[0,1]$, $x^{-1}(z)$ has the cardinality of the continuum.

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I like the Cantor function. A continuous, increasing function $f:[0,1]\rightarrow[0,1]$ with derivative $0$ almost everywhere. See wiki article here.

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The same is true of $f(x)=\sum_{0<p/q<x} \frac{1}{2^q-1}$, where the sum is over all irreducible fractions $p/q$. But this function is also strictly increasing! – Kevin O'Bryant Apr 23 2010 at 12:36
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