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Think in terms of the definition: $f(m)=O(g(m))$ means that there exists some constants $c>0, M\ge 0$ such that $f(m) \le c\cdot g(m)$ for all $m\ge M$ so when you write $$ \sum_{i=1}^nO(1) $$ you mea …

Here's a hint: we'll show that Definition 1 (the usual definition) implies Definition 2 (Papadimitriou's). Suppose that $f\in\mathcal{O}(g)$, so for some $c, k>0$ we have $f(n)\le c\cdot g(x)$ for al …

Let $f(n) = 1-1/n$ and $g(n) = 1/n$, so we've violated the condition that $g$ is non-decreasing. However, it's not hard to show that $f(n) \in O(g(n)+2)$ but we also have $f(n)\notin O(g(n)$, invalida …

You're right about the second one, but your reasoning for both is sketchy. The definition of $f(n)=O(g(n))$ is that there exist constants $c,N$ both greater than zero such that $f(n)\le c\cdot g(n)$ f …

You're almost there. You correctly have $$\begin{align} T(n) &= T(n-1)+2n\\ &= [T(n-2)+2(n-1)] + 2n = T(n-2)+2(n-1)+2n\\ &= T(n-3)+2(n-2)+2(n-1)+2n\\ &= T(n-4)+2(n-3)+2(n-2)+2(n-1)+2n \ …

[Note: this probably should be a comment, but it's too long and I don't like comments that need to be split over multiple entries.] You need to be a bit careful with these arguments, since it's not a …

Hint: for $a,b>0$ we have $\log_an=(\log_ab)(\log_bn)$.

We'll show two results. The first is that for any positive constant $k$, we'll have $$ k\cdot f(n) = O(f(n)) $$ To see this, recall what it means for $g(n)=O(f(n))$: this is true when we can find two …

We want to find a $c$ such that $4+4n\le cn$ for all $n$ greater than or equal to some bound $N$. So we need a $c$ such that $(4+4n)/n \le c$, or $(4/n)+4\le c$. Now while it's true that $$ \lim_{n\ri …