In our analysis book it says that the following function proposed by Weierstrass is continuous everywhere, but nowhere differentiable.

$$f(x)=\sum _{n=1}^{∞}\frac{1}{2^n}\cos \left(3^nx\right)$$

The series must converge because it is the product of a convergent series and an alternating series, but the exponent inside cosine makes it tricky to evaluate.

When I graph it, it looks very strange indeed.

graph of function (n=1 to 100)

Would this be considered fractal? Is it differentiable analytically?

  • 2
    $\begingroup$ Related, the Weierstrass function $\endgroup$ Mar 8 '17 at 13:23
  • $\begingroup$ I've always thought of it as a fractal myself, but I'm not sure if that squares up with the official definition of the term. $\endgroup$
    – The Count
    Mar 8 '17 at 13:25
  • $\begingroup$ A fractal is a mathematical set that exhibits a repeating pattern displayed at every scale. This function definitely has that. $\endgroup$ Mar 8 '17 at 13:31
  • $\begingroup$ It's actually not quite alternating; the signs oscillate but not in a regular manner except for particular values of $x$. But since $|\cos| \leq 1$ it's easy to see that it is absolutely convergent. Also, a finite truncation like you plotted is actually still differentiable, but the derivative is very oscillatory. The infinite sum will not be differentiable (because $\frac{3}{2} \geq 1$). $\endgroup$
    – Ian
    Mar 8 '17 at 13:34
  • $\begingroup$ It's not an alternating series (the cosine term will be close to one for large n). It converges uniformly, because the cosine is bounded and the other factor decreases rapidly. $\endgroup$
    – Dirk
    Mar 8 '17 at 13:41

Your function is a Weierstrass function, which are of the form

$$W(x)=\sum_{k=0}^\infty a^k\cos(b^n\pi x)$$

Your function is of this form with $a=\tfrac12$ and $b=3$, since then $W(\frac x\pi)=f(x)$. Weierstrass functions are nowhere differentiable yet continuous, and so is your $f$. A quote from wikipedia:

Like fractals, the function exhibits self-similarity: every zoom is similar to the global plot.

So yes, it would be considered a fractal.

Read more about Weierstrass functions here.


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