Suppose $f(x)$ is a function that satisfies $f(x) = O(1)$ (big oh of $1$)? Is it possible that $e^{f(x)} = O(1)$?
I think I've figured out the first part of the answer.
We know $(f(x)/g(x)) \leq c$ for all $n \geq k$,
where $f(x) = f(x)$ and $g(x) = 1$
$(f(x)/g(x)) = (f(x)/1) \leq c$ (some constant $c$) for all $n \geq k$ (Assume $k = 1$)
$(f(x)/g(x)) \leq c*1$ for all $n \geq 1$
So $f(x)$ is indeed part of $o(1)$
Assume the top is correct
Is it possible that $e^{f(x)} = O(1)$ (big oh of $1$)
$(f(x)/g(x)) \leq c$ Assume $f(x) = 1$ $(e^1)/1 \leq c$ for all $n \geq k$
$e$ (Euler constant) $\leq c$ for all $n \geq k$
Assume $k = 1$
$e$ (Euler constant) $\leq c$ for all $n \geq 1$
$e \leq c*1$ for all $n \geq 1$
Does this prove that $e^{f(x)} = O(1)$? Is my approach correct?
$f(x) = O(1)$ means that there is a constant $a$ such that $|f(x)| < a$.
$e^{f(x)} = O(1)$ means that there is a constant $b$ such that $|e^{f(x)}| < b$.
Since $|e^{f(x)}| < e^a$, $e^{f(x)} = O(1)$ with constant $b = e^a$.