Just a few days ago I came up with a simple recursive process that defines an infinitely smooth, bounded curve, yet in some sense is self-similar:
Take a tent function with width 2 and heigth 1 and integrate over it. You'll end up with a piecewise quadratic curve that goes from 0 to 1 - just like the first half of the tent function does.
That made me think of a recursion, where you basically rescale the newly found function so it appears twice, integrate over it and renormalize the result to 1. Turns out, the renormalization factor always is 2, so this process can be described by a simple recursion. Since that worked with the definition, I later on added a simple box function into this definition, which became the first element of the recursion:
 =\begin{cases}<br />1/2 & |x|<1 \\<br />0 & |x|\geq1<br />\end{cases}<br />\\<br />f_n(x) = 2\int_{-1}^x(f_{n-1}(2t+1)-f_{n-1}(2t-1))\mathrm{d}t)
If you carefully think this through, you'll see that this definitions gives rise to smoother and smoother curves which each are
rescaled copies of the derivative of the next curve.
Here are two images, one showing the convergence and the other one showing this property of the derivatives:
This is an interesting version of self-similarity. Instead of while zooming in, you'll get more and more of the same structure, doubling the number of times it appears each time, just by taking the derivative of the original curve!
The limit of this sequence might in some sense be considered an infinitely smooth fractal.
I then decided to put this problem out there on Mathematics Stack Exchange. - wether or not there is a closed form to this curve.
http://math.stackexchange.com/q/240687/49989And relatively surprisingly, given the nature of this expression - namely that it's defined piecewise and each iteration gives two pieces - a closed form seems really near. One guy over there already came up with a sum with only one term not quite certain yet.
=\sum _{j=-2^{n-1}}^{2^{n-1}} \frac{2^{-\frac{1}{2} (n-1) n-1} c(j) \left(2 j+2^n x\right)^n \theta\left(2 j+2^n x\right)}{n!})
In here, Θ is the Heaviside-Θ-function, which is essentially a step-function.
c(j) is a set of constants depending on the summation index j. c takes values from {-2,-1,1,2} but it's not clear yet what pattern emerges for it, thus no closed form is rady.
I toyed around with that sum and it turns out, in simple cases for c(j), Mathematica can give a closed form for f
n(x).
Hopefully an expression for that term can be found so I can try the same for the actual values.
May 22, 2011, 09:54:19 AM
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