Bounding dimension of IFS

Question

Given the IFS $\{\frac x {2+x},\frac 2 {2+x}\}$ ($0\le x \le 1$) with attractor K prove that $0.53<\dim_HK<0.8$

I thought using the results from my last question by saying $$\bigg|\frac{2}{2+y}-\frac{2}{2+x}\bigg|=\bigg|\frac{y}{2+y}-\frac{x}{2+x}\bigg|=2\frac{|x-y|}{|(2+x)(2+y)|}$$ but when I try to bound it I get $0.46<\dim_HK<1$ (the right for substituting $x=1$ and the left for $x=0$) which are not tight enough. How can I get better bounds?

EDIT: maybe taking IFS which contains this and another one which is contained may be helpful but how do I pick them?

Answer 1

Hint: If $K$ is the invariant set of the IFS $\{f_1,f_2\}$, then it is also the invariant set of the IFS $$\{f_1\circ f_1, f_1\circ f_2, f_2\circ f_1, f_2\circ f_2\}.$$

---

In fact, the following Mathematica code computes upper and lower bounds for the dimension using $2^{10}$ functions obtained by iterating the IFS 10 times. Setting depth=2 yields the results you asked for.

depth = 10;
funcs = Simplify[
  Through[(Composition @@@ Tuples[{f1, f2}, depth])[x]]
];
ratiosBig = First /@
  (Maximize[{#, 0 <= x <= 1}, x] & /@ Abs[D[funcs, x]]);
ratiosSmall = First /@
  (Minimize[{#, 0 <= x <= 1}, x] & /@ Abs[D[funcs, x]]);
FindRoot[Total[ratiosBig^s] == 1, {s, 1}]
FindRoot[Total[ratiosSmall^s] == 1, {s, 1}]

(* Out: 
  {s -> 0.664199}
  {s -> 0.608226}
*)