Find a function mapping from the real number line to itself such that $f(2x+1) = 3f(x)+5 \space\space\space\space\forall x \epsilon \mathbb{R}$ provided that $f(0) = 0$.
I ran through a solution of this problem with a recursion. Let:
$a_{n+1} = 2a_n + 1$
$b_{n+1} = 3b_n + 5$
Note that $f(a_{n}) = b_n$
$f(0) = 0$ demands that $a_0 = b_0 = 0$
$a_1 = 1$
$a_2 = 1 + 2$
$a_3 = 1 + 2 +4$
from which we can see:
$a_n = 2^n - 1$
A similar argument can be made for $b_n$ to get:
$b_n = \frac{5}{2}(3^n-1) $
Noting that $x = a_n$, we get that:
$n = \frac{ln(x+1)}{ln(2)}$, so that
$f(x) = \frac{5}{2}(3^{\frac{ln(x+1)}{ln(2)}} -1) $
to make it work for $x \le -1$, we just take the absolute value of the argument.
$f(x) = \frac{5}{2}(3^{\frac{ln(|x+1|)}{ln(2)}} -1) $
This function can be easily checked to satisfy the the original equation, so it is a solution. I understand the methodology of this argument quite well, but am somewhat flustered over the idea that we can write a discrete recursion for a continuous variable and still obtain a solution for all $x$. In some ways, it seems appropriate, as we do have an infinite number of values for which the solution is valid (i.e. all values $2^n - 1$ for an integer $n$), but it seems a little handwavy to me. I see that we essentially convert $n$ to a continuous variable by letting $a_n=x$, but why exactly is this valid?