Finding a suitable function in a specified domain

Question

Find a fuction $f$ $:$ $[0, \infty) \rightarrow \mathcal{R}$ so that :-

$f(2x+1)=3f(x)$ $\forall x \in [0, \infty) $ .

Here ,$\mathcal{R}$ denotes the set of all real numbers .

Answer 1

I'd try an experimental approach here. Let's insert a linear function $f(x)=mx+b$ to see if that will work: $$f(2x+1)=m(2x+1)+b=3(mx+b)$$ $$2mx+m+b=3mx+b$$ which yields $m=0$,$b=0$, so no non-trivial solution there. Let's try quadratic: $f(x)=ax^2+bx+c$. Then $$a(2x+1)^2+b(2x+1)+c=3(ax^2+bx+c)$$ $$2ax^2+(4a+2b)x+a+b+c=3ax^2+3bx+3c$$ which again leads to no non-trivial solutions. The pattern is clear here: the leading order always leads to a zero coefficients, so there is no non-trivial polynomial.

Let's try exponential next: $f(x)=ae^{bx}$. Then: $$ae^{b(2x+1)}=3ae^{bx}$$ or $$ae^{2b}e^{2bx}=3ae^{bx}$$ which again leads to no-nontrivial solutions.

Nice continuous functions are not cooperating. So let's try something else. The pattern suggests perhaps a piecewise definition. In particular, from $f(0)$ we know $f(1)$, from which we know $f(3)$, from which we know $f(7)$, etc. So we can use these as the breaks of the piecewise definition. You can let $f(x)$ equal anything on $0 \leq x <1$, and f(x) can be defined recursively on the other intervals. So for example let $f(x)=1$ on $0 \leq x <1$, $f(x)=3$ on $1 \leq x <3$, etc. In general, you can write $$f(x)=3^{a-1}$$, $$2^{a-1}-1 \leq x \leq 2^a-1, a \in \mathbb{N}$$ will give you a simple, but non-trivial solution.