MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 :-)

share|cite|improve this question

closed as no longer relevant by Felipe Voloch, Mark Meckes, Henry Cohn, unknown (google), Suvrit Jul 17 '12 at 16:21

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

2… – Regenbogen Apr 22 '10 at 14:23
thank you... would like to see this community's answers and votes! – vonjd Apr 22 '10 at 14:24
What restrictions, if any, are you placing on the functions? Do they have to be defined (or valued) on the reals or complex numbers, or are functions between arbitrary topological spaces, algebraic objects, unstructured sets etc. acceptable? What about functors between categories? – Robin Saunders Apr 22 '10 at 14:39
Counterexamples in analysis has some nice ones:… – skupers Aug 21 '10 at 18:40

37 Answers 37

up vote 28 down vote accepted

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.

share|cite|improve this answer
Pedantic, but maybe worth mentioning that these functions have these properties almost surely rather than certainly. – Tom Smith Apr 22 '10 at 20:48

The empty function $\emptyset:\emptyset\to\emptyset$ is quite strange when you first meet it.

share|cite|improve this answer
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 '10 at 19:13
And, of course, it gives combinatorial meaning to the fact that 0! = 1, since the empty function is a bijection! – Qiaochu Yuan Apr 23 '10 at 6:36
@Tom: …but it can always be very easily computed! – Peter LeFanu Lumsdaine Nov 13 '10 at 7:53
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 '10 at 3:29
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 '12 at 16:50

The Busy Beaver function

Let Σ be a finite alphabet, for instance {0, 1}; let M be the set of Turing machines with alphabet Σ, and let HM be the set of Turing machines that halt when given the empty string ε as input.

For each MH, Let s(M) be the number of steps performed by M before halting (when given ε as input).

Finally, let S : ℕ → ℕ be the function defined by

S(n) = max {s(M) : MH and M has n states}

Notice that S is well-defined, since only finitely many Turing machines with n states exist.

In other words, S(n) is the maximum number of steps performed on ε among all halting Turing machines with n states. S is called the Busy Beaver function.

It turns out that S is uncomputable because it grows faster than any computable function, that is, for all recursive functions f : ℕ → ℕ we have S(n) > f(n) for large enough n, and in particular f is o(S).

share|cite|improve this answer
In particular, it grows faster than the Ackermann function. – Joel David Hamkins Apr 22 '10 at 19:17
@JoelDavidHamkins It's faster than the Ackermann function iterated Graham's number of times. – PyRulez Sep 8 '15 at 0:24

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

share|cite|improve this answer
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 '10 at 12:36
@KevinO'Bryant I realize it's been over four years, but 18 upvotes notwithstanding, for sure this function is discontinuous at every positive rational number? But I grant you that it is not only continuous, but differentiable at irrational points. – Harald Hanche-Olsen Aug 17 '14 at 11:55
It is definitely discontinuous at rational points. At $p/q$ the difference between the limit from the left and from the right is $1/(2^q-1)$, which is definitely not 0. It is differentiable at irrational $x$, and moreover its derivative is 0 whenever it has one! – Kevin O'Bryant Aug 28 '14 at 2:47

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.

share|cite|improve this answer
I never really thought of the Ackermann function as being strange, only big. But maybe that's just me. – Ketil Tveiten Apr 23 '10 at 12:13
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 '10 at 12:21
@Joel I HATE the Ackermann function-I lost a whole letter grade on my set theory final because I couldn't construct the damn thing. – The Mathemagician Jul 1 '10 at 23:20
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 '10 at 15:39

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

share|cite|improve this answer
Very true, me too! Could you please elaborate on why you consider this function "strange" - thank you. – vonjd Apr 23 '10 at 9:29
Because of its stubborn nature: whether you differentiate it, or integrate it, it remains unmoved ;-) – Suvrit Nov 13 '10 at 17:39
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 '11 at 23:33

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...

share|cite|improve this answer
The function is pretty funny too. It adds +, -, and . as digits in base-13, and then his function interprets it as a base-10 number. – PyRulez Sep 8 '15 at 0:29

I suppose the strangest function in mainstream mathematics is 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

share|cite|improve this answer
@Qiaochu Yuan: THAT IS SO COOL! – Vectornaut Apr 23 '10 at 18:27

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

share|cite|improve this answer
I wonder who Thomae was? – Koundinya Vajjha Mar 10 '11 at 10:04

The Weierstrass function is particularly intriguing, as it's a function that's everywhere continuous, but nowhere differentiable.
$f(x)= \sum_{n=0} ^\infty a^n \cos(b^n \pi x)$
where 0<a<1, and b is a positive odd integer such that $ab > 1 + \frac{3\pi}{2}$.
It challenges the notion that, just because a function is continuous, it must also be differentiable in most places, which I think is pretty cool.

share|cite|improve this answer
So is the Brownian sample path, also mentioned in this thread. – Michael Hardy Nov 13 '10 at 19:16

An unbounded operator with dense graph.

In functional analysis, one deals with unbounded operators on Hilbert space, but usually ones that are closed, or are at least closable. At the opposite end of the spectrum, one can construct linear operators whose graph is dense: for any pair $(x,y)$, there is a sequence $x_n$ such that $x_n \to x$ and $A x_n \to y$ ! It's not so easy even on $\mathbb{R}$ to come up with a function whose graph is dense, and the examples I think of aren't measurable. But in infinite dimensions, you can find one that is linear! It's just an illustration that facts that are trivial in finite dimensions can be horribly, horribly false in infinite dimensions.

A family of examples of this is constructed in

MR0782615 (86i:47052) Lindsay, J. M. A family of operators with everywhere dense graphs. Exposition. Math. 2 (1984), no. 4, 375--378.

Interestingly, as an appplication, Lindsay uses such operators to prove that a Brownian motion sample path is nowhere differentiable --- which is my other favorite strange function!

share|cite|improve this answer
Thank you - perhaps you could create a new post for Brownian Motion then, too?!? – vonjd Apr 22 '10 at 14:48
The exotic solutions $f : \mathbb{R}\to\mathbb{R}$ to Cauchy's functional equation $f(x+y)=f(x)+f(y)$ have dense graphs. They are obviously linear, but not continuous. (Not sure about measurability). – Greg Graviton Nov 13 '10 at 14:47
(I mean $\mathbb{Q}$-linear.) – Greg Graviton Nov 13 '10 at 14:48
There's one acting in $\mathcal{C}([0,1])\cup\mathcal{C}([2,3])$. I forgot what it acts like, though. :/ Can you help me? (I'm pretty sure I saw it on wiki.) – Alexander Frei Jul 2 '15 at 0:19

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.

share|cite|improve this answer

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.)

share|cite|improve this answer
SPOILER (albeit 1.5+ years late): (and the PDF graph at – Noam D. Elkies Jun 8 '12 at 23:06

Minkowski's question mark function if only for the strange $?(\cdot)$ notation.

share|cite|improve this answer
Could you give some more details - thank you – vonjd Apr 22 '10 at 17:59
1 - it is closely related with Cantor function mentioned elsewhere. – kakaz Apr 22 '10 at 20:58
This article explains it well: – Progo Jan 11 '15 at 13:46

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.)

share|cite|improve this answer
Heh, that's cute. – Todd Trimble May 29 '11 at 23:39

Dirac delta function seems strange to me since the first time I saw it.

share|cite|improve this answer
Well, technically the Dirac delta function is a distribution, and not a function. – J.C. Ottem Apr 22 '10 at 14:57
Well, it is a function, just not on the space you might think... – Nate Eldredge Apr 22 '10 at 15:42
Haha, well everything is a function not on the space you might think. – Tom Ellis Apr 22 '10 at 22:08
I heard that after Schwartz got the Fields medal someone quipped "Now they're giving the Fields medal for integration by parts." – Jon Apr 23 '10 at 6:18
We used to say that Dirac's delta is the characteristic function of physicists: if you're a physicist, it's a function, otherwise, it isn't. – Federico Poloni Aug 21 '10 at 20:02

Interpreting your questions a bit liberally, I suggest the Goodstein sequence:

share|cite|improve this answer

f(x) = sin (1/x): (x not 0); f(x) = 0 (x equals 0)

share|cite|improve this answer
Could you please elaborate a little on that... thanks! – vonjd Apr 23 '10 at 14:20
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: – Joseph Malkevitch Apr 23 '10 at 15:32
As an added bonus, you get that the function is connected but not path connected... – Andrew Parker Nov 13 '10 at 0:57

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 $y=x$.

share|cite|improve this answer

I like the Theta functions which are given by Fourier-type series. They show up in many areas in mathematics. For example:

i)They are very important in the study of abelian varieties in algebraic geometry (for example, in the case of elliptic curves they are used in the proof of Abel's theorem and are related to Weierstrass $\mathcal{P}-$function).

ii) They satisfy a number of interesting indentities. For example, in the one-dimensional case, they satisfy Jacobi's triple product identity which can be used to show Jacobi's four square theorem

iii) They can be used to solve algebraic equations degree equations explicitly (see this link)

iv) In the one-dimensional case, they solve the heat equation.

share|cite|improve this answer
But are they strange? – Gerry Myerson Apr 23 '10 at 4:41
Well, they are definitely unusual in the sense that they are not taught in school. Also, the fact that they appear in such a variety of mathematical disciplines is also rather surprising. But I see your point. – J.C. Ottem Apr 23 '10 at 7:31
I was going to write modular forms, but this is good enough. – David Corwin Jul 15 '10 at 15:21

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 .

share|cite|improve this answer

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

share|cite|improve this answer
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 '10 at 13:09
@Max It's been 3 years since your comment, but I don't think that's precisely the case. Standing at the origin you will see a pole at x-coordinate $\arctan \frac{m}{n}$ with height $\frac{1}{n}$. Here, the pole of height $\frac{1}{n}$ is at x-coordinate $\frac{m}{n}$ instead. – 6005 Feb 19 '14 at 12:06

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.)

share|cite|improve this answer

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

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


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

(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".

share|cite|improve this answer
Huh? The explicit you description you gave is just a sawtooth function. – Harry Altman Jul 1 '10 at 23:17
My bad - I compiled my answer using LaTeX, however I didn't copy the complete post across properly. I will edit now. – Colin Pratt Jul 1 '10 at 23:25
Closely related is the tavuk göğsü function, replete with shredded chicken. – Tom LaGatta Jul 1 '10 at 23:34
one can describe $f(x)$ as the distance function from $\mathbb{Z}$. – Pietro Majer Nov 2 '14 at 19:44

Among the "special" functions encountered in analysis in my view Ingrid Daubechies' waveletes with compact support are the strangest.

share|cite|improve this answer
Could you please give some more details? Thanks – vonjd Apr 22 '10 at 18:00

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].

share|cite|improve this answer

Since Mariano took my favorite already, I'll go with the stopping time function for the 3x+1 problem:

share|cite|improve this answer

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 $0$.

share|cite|improve this answer

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).

share|cite|improve this answer

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