There is a bounded, continuous, even function of a variable t that satisfies the functional equations $$f(t)+f\bigg(t+\frac{1}{2}\bigg)=0$$ and $$2f\bigg(\frac{t}{4}\bigg)+f\bigg(t+\frac{1}{8}\bigg)=1$$
and $x=f(t)$ and $y=f(t-1/4)$ for $0 \leq t \leq 1$
$$f(t)=\frac{\Theta(t)}{2}-\frac{\Theta(t)\Theta(\tau_1(t))}{4}+\frac{\Theta(t)\Theta(\tau_1(t))\Theta(\tau_2(t))}{8}-...$$ $$\Theta(t)=-1$$ if $t\in [1/4,3/4]$ and $$\Theta(t)=1$$ if $t \in [0,1/4]$ or $t \in [3/4,1]$
$$\tau_1(t)=1/8+4t$$ if $t\in[0,1/4$ or $t \in [1/2,3/4]$ $$\tau_1(t)=1/8-4t$$ if $t \in [1/4, 1/2)$ or $ t \in [3/4,1)$
Note that $\tau_{k+1}(t)=\tau_k(\tau_1(t))$ for $k=1,2,3$
I am trying to use an example to show how to apply these equation.
In my example I let $t=0.123$
Because of my choice of $t$ I have $$\Theta(t)=1$$ and $$\tau_1(t)=1/8+4t$$ and so $$\tau_2(t)=\tau_1(\tau_1(t)) =(1/8+4t)^2$$
Hence from the equation for $f(t)$ above, $$\Theta(t)\Theta(\tau_1(t))\Theta(\tau_2(t))= 1 \times (1/8+4t)\times (1/8+4t)$$ and $$\Theta(t)\Theta(\tau_1(t))=1 \times 1 \times (1/8+4t)$$ hence
$$f(t)=\frac{1}{8}(1+t)+t(2t+1)$$
Is this correct?