How does this implicit equation work? 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:


*

*How to works a right part of first equation?

*How to modify a first equation for getting sequence of radiuses: $\frac{1}{1}$,$\frac{1}{2}$,$\frac{1}{4}$,$\frac{1}{8}$,$\frac{1}{16}$,...  

*Can we call it a recursive or a fractal object?

 A: 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 $\sin\left(- \pi \log_2 \left(\sqrt{x^2+y^2}\right)\right)=0$

3.
I don't know what this kind of curve is called. It does seem to have self-similarity. 
