How can I find $f(3)$ if $f(2x)=f^2(x)-2f(x)-\frac{1}{2}$?

Question

If $f(2x)=f^2(x)-2f(x)-\frac{1}{2}$ and $f(1) = 2$ then find $f(3)$.

Can you give me any hint that I can start with?

Answer 1

$f(2x)=f^2(x)-2f(x)-\frac{1}{2} $ and $f(1) = 2 $.

I can see how to get $f(2^n)$ and expressions for $f(2^{-n})$, but I don't see how to get $f(3)$.

I show what I've got so far. All fairly trivial.

$f(2x)+\frac32 =f^2(x)-2f(x)+1 =(f(x)-1)^2 $ so $f(x) =1\pm\sqrt{f(2x)+\frac32} $.

$x = 0 \implies f(0) = f^2(0)-2f(0)-\frac12 $ or $f^2(0)-3f(0) = \frac12 $ or $f^2(0)-3f(0)+9/4 =\frac12+9/4 =\frac{11}{4} $ or $(f(0)-\frac32)^2 =\frac{11}{4} $ or $f(0) =\frac32\pm\frac{\sqrt{11}}{2} $.

$x=1 \implies f(2) = -\frac12 $.

$x=\frac12 \implies f(\frac12) = 1\pm\sqrt{f(1)+\frac32} =1\pm\sqrt{\frac72} $.

And that's all.