Consider the Weierstrass function: $$\sum_{n=0}^{\infty}a^n\cos{b^n\pi x}$$ It is well-known as an example of a function that is everywhere continuous and nowhere differentiable. When reading about fractals for the first time, I quickly recalled the Weierstrass function, which indeed made understanding fractals more simple.

Recall that the Mandelbrot set is the set of values in the complex plane where $z_{n+1} = z_n^2 + c$ where $z_0 = 0$ and $c$ is a number so that the sequence remains bounded.

The Mandelbrot set is just one of many examples where self-similarity is a prominent feature. I chose it because its the only one I can define. Is there a purely mathematical connection/similarity between the Weierstrass function and the Mandelbrot set that is not a trivial consequence of self-similarity? (Note that I, as a first-year undergraduate, am very slack with the word "trivial." I would deem almost any example other than "they are similar in the sense that they both display self-similarity" non-trivial.) Can such connections be found between the Weierstrass function and (every other) fractal? Lastly, is there a connection between (the lack of) differentiability/slopes and self-similarity of functions in $\mathbb{R}^n$? In particular, are there examples of everywhere continuous, nowhere differentiable functions in all dimensions trivially similar to the Weierstrass function?

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    $\begingroup$ An interesting, more concrete question, might be to ask if there exist non-self-similar continuous non-differentiable functions. The only other one I'm aware of is the Blancmange curve, which is self-similar. $\endgroup$
    – Jack M
    Commented Apr 8, 2014 at 19:04
  • $\begingroup$ @Jack M: It seems to me that self similarity is a very special condition, one that you wouldn't expect in such functions unless specifically built in by some simple explicit construction process. As an analogy, consider sequences of real numbers that don't have a limit. We can obtain such sequences as simple as $a_1 = 1$ and $a_{n+1} = -a_{n},$ but we would expect that the typical sequence having no limit is not definable by even a finite term recursion, for any reasonable interpretation of typical. $\endgroup$ Commented Apr 8, 2014 at 19:36
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    $\begingroup$ The Mandelbrot set itself is not strictly self similar. But the Julia set for c=1/4 is self similar, in much the same way as the Weierstrass function. The boundary of this curve can be put into 1:1 correspondence with the unit circle with the Boetcher function. This unit circle boundary is self similar, continuous, and not differential. One can even express this as a Fourier series, by unwrapping this unit circle. I can post the equations later, if you like. There are also the Siegel disk Schroeder solutions, for irrationally indifferent cases, which have similar properties, can also post $\endgroup$
    – Sheldon L
    Commented Apr 8, 2014 at 19:49
  • $\begingroup$ I'd appreciate that, Sheldon. $\endgroup$ Commented Apr 8, 2014 at 21:23
  • $\begingroup$ I would love to see an example of a definable continuous nowhere-differential function that does not exhibit some form of self-similarity. I am curious whether or not they even could exist. $\endgroup$ Commented Apr 9, 2014 at 12:30

1 Answer 1


Since the op mentioned the Mandelbrot set, I decided to focus on the Cauliflower for the Julia from the c=1/4 Mandelbrot, and see what comparisons I could find with the Weiestrass function. First, lets define the Julia as the boundary of the set of points that escape when iterating $$z \mapsto z^2 + \frac{1}{4}$$

This mapping has a fixed point of z=0.5. On the real axis, anything bigger than 0.5 or smaller than -0.5 goes to infinity. At the imaginary axis, $z=\pm \frac{i}{2}\sqrt{3}$ is the boundary point, since $z^2+0.25=-0.5$. Then in the complex plane, we have this Cauliflower for the boundary of the Julia set for c=0.25. Next, we generate the Boettcher function for the boundary of the Cauliflower, which maps the Cauliflower to the unit circle, via a Taylor series, which has a fractal boundary at the unit circle radius of convergence, and we compare it to the Weiestrass function.

Caulifower Julia

I will use this form for the Weiestrass function $$\sum_{n=0}^{\infty} 2^{-n}\cos(2^nx) = \Re(\sum_{n=0}^{\infty} 2^{-n}\exp(i2^nx))$$ $$z=\exp(ix), \; \text{Weiestrass_circle}=\sum_{n=0}^{\infty}2^{-n}z^{2^n}=z+\frac{z^2}{2}+\frac{z^4}{4}+\frac{z^{8}}{8}+\frac{z^{16}}{16}+\frac{z^{32}}{32}+\frac{z^{64}}{64}...$$

The similarity I found involves wrapping the Weiestrass function around the unit circle as a Taylor series, where the Weiestrass function is the real part of the unit circle function. By comparison, the Cauliflower Julia set unit circle function is the reciprocal of the following series; see external ray on wikipedia for some background. Later I can add the derivation of the formal Boettcher series, if the op is interested.

$$\frac{1}{\Phi(z)}=z +\frac{1}{8}z^3 +\frac{11}{128}z^5 +\frac{29}{1024}z^7 +\frac{1619}{32768}z^9 +\frac{5039}{262144}z^{11}+\frac{75391}{4194304}z^{13}...$$ $$\Phi(\pm 1)=\pm \frac{1}{2}, \Phi(\pm i) =\pm \frac{i}{2}\sqrt{3}, \;\; \Phi(z^2)=\Phi(z)^2+\frac{1}{4}$$

Here is a graph of the Weiestrass "Taylor series" function, wrapped around the unit circle, followed by Julia set Cauliflower, reciprocal of the Taylor series, also wrapped around the unit circle. Both functions have a similar kind of fractal similarity. Both functions can be represented as a Taylor series which converges inside the unit circle, and on the boundary, but not outside the unit circle. Both functions are continuous on the unit circle boundary. Both functions are nowhere differentiable at the unit circle boundary, and cannot be extended outside the unit circle boundary.

Weiestrass Taylor series, where red is real (cosine part of original definition) and green is imag

Weiestrass Taylor series

Julia unit circle, reciprocal of the Boettcher Taylor series above, red is real, green is imag

Julia unit Circle

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    $\begingroup$ What is a quick way to compute the Boettcher function? I know that if $f(z)=z^d(1+b_1z+b_2z^2+\dots)$ then the Boettcher function for the superattracting point $z=0$ is the limit of $\phi_n(z)= (f^{\circ n}(z))^{1/d^n}$, but the computation is messy. $\endgroup$
    – Dimitris
    Commented Apr 23, 2015 at 2:03

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