A similar question was asked here: Solving recurrences using substitution method, but I am still somewhat hazy as to how this process works.

Say, for $T(n) = T(\lceil n/5 \rceil + 36) + n \log n$

Statement: we wish to prove that $T(n) \le 5 \ n \log n$ for $n \ge 0$

Base case, $n = 0$ and $T(0) \le 0$

$T(0) = 0$

$T(0) \le 0 \log 0$

Holds, since $0 \log 0 = 0$

Inductive step, for $n > 0$

(Inductive Hypothesis is: $T(n) = 5 \ n \log n$)

$T(n) = T(\lceil n/5 \rceil + 36) + n \log n$

$\le 5 \, (\lceil n/5 \rceil + 36) \log (\lceil n/5 \rceil + 36) + n \log n$ by Ind. Hyp.

$\le 5 \, (n/5 + 36) \log (n/5 + 36) + n \log n$ by property of ceiling

$= (n + 180) \log ((n + 180)/5) + n \log n$ by algebra

This is the point where it gets hazy. According to the link, what I should do is say that for all $n \ge 180$, then

$\le (n + n) \log ((n + n)/5) + n \log n$

$= 2n \log (2n/5) + n \log n$

$= 2n \, [\log n + \log (2/5)] + n \log n$

$= 2n \log n + 2n \log (2/5) + n \log n$

$= 3n \log n + kn$ where $k = \log (2/5) * 2$

$\le 5n \log n$

Does this complete the proof? Or am I in error? I still don't understand why I can just set $n \ge 180$ and therefore get rid of the irritating arithmetic within the log.


1 Answer 1


The link doesn't say what you say it does. I don't see the number 180 appearing anywhere on that link. And that link doesn't say the purpose is to prove $T(n) \le 5 n \log n$ for all $n \ge 0$; it says to prove that $T(n) = O(n \log n)$. Check the definition of big-O notation; it doesn't require proving $T(n) \le c n \log n$ for all $n \ge 0$; it suffices to prove it for all $n \ge 180$ (or for all $n \ge n_0$ for any other constant $n_0$ of your choice).


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