The term Gibbs Phenomenon refers to the peculiar way Fourier Series behave at sharp changes in a function's value.

However, this problem becomes particularly annoying to deal with when trying to estimate the discrete length functional of a function $f(x)$, given by,

$$(1) \quad L^b_a[f(x)]=\frac{b-a}{\omega} \cdot \sum_{n=1}^{\omega} \sqrt{1+\omega^2 \cdot \left(f \left(\frac{(b-a)\cdot n}{\omega} \right)-f\left(\frac{(b-a)\cdot (n-1)}{\omega}\right) \right)^2}$$


$$(2) \quad f(x)=\sum_{k=1}^{\infty} k^{-p} \cdot \sin(k^p \cdot x)$$

I want to know the asymptotic behavior of $(1)$ as $\omega \rightarrow \infty$. Specifically, I want the techniques used to be clearly explained and possibly reproducible for other functions.

Because of Gibb's Phenomenon, I can't fully trust my numerical results that find $L \sim \omega^{2^{-p}}, \ p \gt v$

However, I do have formal methods pointing towards a power law with the exponent $\beta$ being bounded by,

$$(3) \quad 0\lt \beta \lt 1/p$$


The motivation behind the problem is to find the box dimension of the graph of $f(x)$. So a good criteria for a good answer is whether or not it is accuarate enough to yield this number. As mentioned in comments, I don't currently know of a way to approximate the sum except in special cases. However,

$$\sqrt{1+x^2} \sim |x|$$

Is useful since, the derivatives involved are much larger than unity.

(I intend to make this a bounty and add more information)

  • $\begingroup$ do u have a good approximation for the inner sum? $\endgroup$ – tired Jan 5 '16 at 22:56
  • $\begingroup$ I only have that the sum, $S$ is bounded by, $0 \lt | S | \lt (2 \cdot c_l)^{1/p} \cdot \omega^{1/p-1}$ if $\omega$ is large. If you apply the sum rule for $\sin$ to the inner sum, $c_l$ becomes the constant for how large the argument of $\sin$ can get. So not really, unless you want the upper bound or expansion about $x=0$; still looking though... $\endgroup$ – Zach466920 Jan 6 '16 at 0:09
  • $\begingroup$ @tired added a bounty, and wrote this to ping you. (Forgot to do so earlier, sorry) $\endgroup$ – Zach466920 Jan 7 '16 at 21:11
  • $\begingroup$ i will think about this problem a little more tomorrow. But it seems to be really though $\endgroup$ – tired Jan 8 '16 at 10:17

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