Asymptotic. If f(n) = theta(g(n)) and g(n) = theta(h(n)), then why h(n) = theta(f(n))

Question

it is f(n)=theta(h(n)) as theta is transitive. But Can any one explain why h(n)=theta(f(n)).

Answer 1

Expanding the Big-O notation by its definition usually makes things easy.

Answer 2

If k1.h(n) <= f(n) <= k2.h(n) for large n, then (1/k2)f(n) <= h(n) <= (1/k1)f(n).

Answer 3

That's basically simply because f(n) € theta(h(n)) is equivalent to h(n) € theta(f(n)) because of the following:

f(n) € O(h(n)) => h(n) € Omega(f(n))
f(n) € Omega(h(n)) => f(n) € O(h(n))