# If f(n) = Θ(g(n)) then what all can one say about f(h(n)) and Θ(g(h(n))

I was going across some problems which translated the fact that $\log (n!) = \Theta(n\log n)$ to give $$\log(\lceil\log n\rceil!) = \Theta(\lceil\log n\rceil \log \lceil\log n\rceil)$$ and therefore $\log(\lceil \log n\rceil!) = \omega(\lg n)$.

Now I just have this intuition that this sort of conversions shouldn't and wouldn't be always right.

Would it be right to think that.

Any justification as to why am I wrong in thinking that and guidance to what I should have a look at, what texts I should go through to understand it better for myself are very much appreciated.

Thank you.

• The title you have chosen is not well suited to representing your question. Please take some time to improve it; we have collected some advice here. Thank you! – Raphael Sep 15 '16 at 17:07
• See bullet two here. – Raphael Sep 15 '16 at 17:08

In general, you can prove nothing. You know $f(n) = \Theta(g(n))$. Now assume f (n) = g (n) = n for all n except g (n) = 0 if n ≤ 1000. And h (n) = 1000 for all n.

So we now have f (h (n)) = 1000 and g (h (n)) = 0.

The problem is that $O (n)$ and $\Theta (n)$ only work for large n. You are fine if h (n) -> inf, or if you look specifically at f (n) and h (n).

If the limit of h (n) is infinity, then you overcome the problem that $\Theta(g(n))$ only gives you information for n ≥ N for some possibly large N, because h (n) will eventually be greater than that N.

• I get your point. Just couldn't understand the last part ! h(n) -> inf ?. – ps_ Sep 16 '16 at 2:42

Why would there be a problem? You know that $\log (t!) = \Theta(t\log t)$. Why would it matter whether you call the thing $t$ or $\lceil\log n\rceil$? (As long as you can make $\lceil\log n\rceil$ large enough for the asymptotic relationship to kick in; and you can.)

• I'm not aware of formal proofs for claims stronger than bullet two here. – Raphael Sep 15 '16 at 17:11

Below, I'm assuming that the domain of all function is the non-negative reals.

By definition, $f(n) = \Theta(g(n))$ if there exist $N \geq 0$ and $A,B > 0$ such that $$n \geq N \Longrightarrow A g(n) \leq f(n) \leq B g(n).$$ If there exists a constant $c > 0$ such that $f(n),g(n) \geq c$ for all $n$ and $f,g$ are continuous, then we can in fact take $N = 0$ (perhaps modifying $A,B$ in the process). This means that for all $n$, it holds that $$A g(n) \leq f(n) \leq B g(n).$$ Assuming that $h(n) \geq 0$ for all $n$, this implies that $$A g(h(n)) \leq f(h(n)) \leq B g(h(n)),$$ and so $f(h(n)) = \Theta(g(h(n))$.

Even without assuming that $N = 0$, we can make the same conclusion if $\lim_{n\to\infty} h(n) = \infty$. In that case, there exists $M$ such that $n \geq M$ implies $h(n) \geq N$, and so $$n \geq M \Longrightarrow A g(h(n)) \leq f(h(n)) \leq B g(h(n)),$$ again implying $f(h(n)) = \Theta(g(h(n))$.

• I haven't gone through the details, but I think there are some subtleties involved. Compare with bullet two here, a statement we found a rigorous proof for (but not for stronger claims). – Raphael Sep 15 '16 at 17:10