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$3^n$ grows faster than any exponential function with a base of $2$. True. This implies that $3^n = O(2^n)$ cannot be true. But what you have here is $2^{O(n)}$. Recall that $O(f(n))$ is really …

Yes, it's possible to have an infinite chain. I'm sure you're already familiar with some examples: $$ O(x) \subseteq O(x^2) \subseteq \ldots \subseteq O(x^{42}) \subseteq \ldots $$ You have an infin …

The two definitions are not equivalent. However, the case where they aren't equivalent is one that we aren't interested in for algorithm analysis. The standard definition implies the Papadimitriou de …

Let's start with the simple case $g = 1$, and $f$ having positive values only (that's all we care about with functions that represent complexity). $f \in O(1)$ means that $f$ is bounded: there exist …

First, as other answers have already explained, $O(3n) = O(n)$, or to put it in words, a function is $O(3n)$ if and only if it is $O(n)$. $f = O(3n)$ means that there exists a point $N$ and a factor $ …

Simplifying the left-hand side is a good strategy. You aren't going to find a simpler expression that's equal to $\log(n+1)$, though. However, since you're only interested in proving that $\log(n+1)$ …

You've omitted a few steps. It looks like you're attempting to prove by induction that $T(n) = O(n)$, and your proof goes: Suppose $T(k) = O(k)$ for $k<n$. This means $T(k) \le c \, k$ for some $c …

Big O: upper bound “Big O” ($O$) is by far the most common one. When you analyse the complexity of an algorithm, most of the time, what matters is to have some upper bound on how fast the run time¹ g …