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I experimented with mathematics and Mathematica and got a very interesting curve, which defined as: $$\sqrt{x^{2}+y^{2}}=\cos\left(\log_{2}\left(\sqrt{x^{2}+y^{2}}\right)\right)$$ After plotting this equation, we get infinite concentric circles(whose radius is less than or equal to one). It looks like a circle equation: $\sqrt{x^{2}+y^{2}}=R$

Questions:

  1. How to works a right part of first equation?
  2. How to modify a first equation for getting sequence of radiuses: $\frac{1}{1}$,$\frac{1}{2}$,$\frac{1}{4}$,$\frac{1}{8}$,$\frac{1}{16}$,...
  3. Can we call it a recursive or a fractal object?
PavelDev
  • 397

1 Answers1

3

1.

You're right about the links with the circle.

Let $r=\sqrt{(x^2+y^2)}$

Then your equation is $r-\cos (\log_2(r))=0$

This has an infinite number of roots that are packed in closer and closer to zero. Let's call them $r_1, r_2, r_3, ...$

Each of these roots corresponds to one of your circles. $r_1^2=x^2+y^2$

The largest value is when $r_1=1$.

That's because $\log 1=0$ and $\cos 0=1$

$1-\cos(\log(1))=0$ as required.

2.

To achieve circles with radius $\frac 11, \frac 12, \frac 14, ...$ you need a function $f(r)$ that has has zeroes at those values.

We want roots $r_0=\frac 11, r_1=\frac 12, r_2=\frac 14, ...$

In general, $r_p=\frac 1{2^p}$, which we can rearrange to give $p=\log_2\left(\frac 1r\right)=-\log_2 r$

So now we need a function $g(p)$ that has roots for $p=0,1,2,3,...$ which is much easier.

Let $g(p)=\sin \pi p$

Combine all these to get $\sin\left(- \pi \log_2 \left(\sqrt{x^2+y^2}\right)\right)=0$

enter image description here

3.

I don't know what this kind of curve is called. It does seem to have self-similarity.

tomi
  • 9,594