Given the recursive function:
$$f(0) = \frac{x^2}{2} + \frac{x}{2}, f(n) = \frac{f(n-1)}{2} + \frac{x}{2}$$
where $x$ = some integer
How would one rewrite this function to be strictly non-recursive? Eg. the non-recursive counterpart to the recursive function $f(0) = x$, $f(n) = \large \frac{f(n-1)}{2}$ is $f(x)$ = $\large\frac{1}{2^{x}}$.
Thank you.
$f(n)$ will be a linear function of $x^2$ and $x$. Looking at the first few terms $$f(1)=\frac{1}{4}x^2 + \frac{3}{4}x$$ $$f(2)=\frac{1}{8}x^2 + \frac{7}{8}x$$ etc., it is obvious $$f(n)=\frac{1}{2^{n+1}}x^2 + \left(1-\frac{1}{2^{n+1}}\right)x.$$ You can prove this by induction.
The terminology here is confused. Recursiveness is an intrinsic property of a function itself. Take for example the functions defined as follows:
$$f(0) = 1$$ $$f(n +1) = 2f(n)$$
and
$$g(n) = 2^n$$
It would be quite wrong to describe $g$ as a "non-recursive counterpart" to $f$. For $g$ is one and the very same recursive function as $f$. And $g$ is recursive precisely because it can be presented in the other way (which shows that $g$ belongs to the class of functions definable from the usual initial functions by composition, recursion and minimization),
We can talk about explicitly recursive vs closed form modes of presentation of a given function. But a function is recursive irrespective of how we present it. Thus take the function defined by
$$\mathit{fermat}(n) = 1 \mathrm{\ iff\ there\ are\ natural\ numbers\ } x, y, z \mathrm{\ such\ that\ } x^{n+3} +y^{n+3} = z^{n+3}$$ $$\mathit{fermat}(n) = 0 \mathrm{\ otherwise}$$
That's a bizarre presentation of what we now know, thanks to Andrew Wiles, is a recursive function -- the everywhere zero function!
