# Solving a simple recurrence

I'm having a real hard time solving recurrences using the substitution method.

Show that: $T(n) = T(n/2) + 1$ is $O(\lg n)$

I thought this to be relatively easy:

We have to show that $T(n) \leq c \lg n$

Substitution gives me:

\qquad \begin{align} T(n) &\leq c \lg(n/2) + 1 \\ &= c \lg n - c \lg 2 + 1 \\ &= c \lg n - c + 1 \\ &\leq c \lg n \end{align}

for every c.

I was under the impression this was it, but when I was looking for an answer, I came around a much more elaborate answer on the web, given involving subtracting a constant. I don't get why that's needed, I thought I had shown what was needed.

Any help would be greatly appreciate, starting Monday I'm enrolled in an algorithms class and I don't want to get behind!

We are using the CLRS book (surprise) and though I appreciate the amount of information in it, I'd rather have some more resources. I've really enjoyed a datastructures class and I really think I can enjoy this as well, but more resources would be very much appreciated.

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We have a reference question with ample material about solving recurrences, in particular this answer. – Raphael Jan 30 at 20:54
Your substitution proves nothing. You use the claim to derive the claim -- that's not very helpful. – Raphael Jan 30 at 20:56
@Raphael This is proof by induction. – Yuval Filmus Jan 31 at 0:27
Your solution looks fine, though you'd better write it as an induction, i.e $T(n) = T(n/2) + 1 \leq c\lg(n/2) + 1$ and so on. You also need to take care of the base case, and notice that you get a condition on $c$ (not all $c$ work). – Yuval Filmus Jan 31 at 0:28
@YuvalFilmus No, it's not. What is written there could be used as the inductive step, true. Given the level of the question, I would not assume that Oxymoron knows what happens there; the question even says "substitution method". – Raphael Jan 31 at 9:34