continuation of integer-valued solution I have a recurrence relation that is given by
$$f(n) = 2af(n-1)-a,\quad f(1) = a.$$
The solution given in Wolfram alpha is
$$f(n) = \frac{(a-1)2^na^n+a}{2a-1}.$$
(This is not a homework problem).  My question is, supposing that I want to determine the continuous solution $f(x)$, can I find it by knowing $f(n)$ - it's values at the positive integers?  If so, any suggestions on how to do this?
 A: Well you could just set $f(x):= \frac{(a-1)2^xa^{x}+ a}{2a-1}$. This is continuous.... (assuming $a\neq 0.5$ and $a>0$.)
A: Your question is about a function $\,f(x)\,$ that satisfies
$$f(n) = 2af(n-1)-a\quad \forall n\in\mathbb{Z},\quad f(1) = a. \tag{1} $$
You asked about continuous solutions to equation $(1)$. Let
$\,F(x)\,$ be any continuous function on the interval $\,[0,1]\,$
with $\,F(0)=1,\; F(1)=a.\,$ Use the recurrence relation
$$ F(x) = 2a\,F(x-1)-a\quad \forall x\in\mathbb{R} \tag{2} $$ to extend the function
uniquely and continuously to all reals. Verify using induction that
$$ f(n) = F(n) \quad \forall n\in\mathbb{Z}. \tag{3} $$
Note that this method does not give an explicit formula for any of the
continuous solutions, because allowing $\,F(x)\,$ to be any
continuous function on the interval $\,[0,1]\,$ with
$\,F(0)=1,\,$ $\,F(1)=a.\,$makes this impossible. Given
a particular solution to equation $(1)$ such as
$$ f(n) := \frac{(a-1)(2a)^n+a}{2a-1} \tag{4} $$
it is easy to extend this to all real numbers if $\,a>0\,$ because
$\,(2a)^n\,$ simply becomes $\,(2a)^x.$ In general, it is not possible
to get a unique solution. For a simple example, $\,f(n)=f(n-1)\,$ has
period $1$ functions as solutions and there are very many such functions.
