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It's been a while since I had to solve a recurrence and I wanted to make sure I understood the iterative method of solving these problems. Given:

$$T(n) = 3T(n-2)$$

My first step was to iteratively substitute terms to arrive at a general form:

$$T(n-2) = 3T(n-2 -2) = 3T(n-4)$$ $$T(n) = 3 *3T(n-4)$$

leading to the general form:

$$ T(n) = 3^k T(n-2k) $$

Now I solve $n-2k = 1$ for $k$, which is the point where the recurrence stops (where $T(1)$) and insert that value ($n/2 - 1/2 = k$) into the general form:

$$T(n) = 3^{n/2-1/2}$$ $$T(n) = O(3^n)$$

I'm not sure about that last step:

I would just "argue" that as $n \to \infty$ one can ignore $-1/2$ and $n/2 \to n$ ? Is that assumption correct?

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Note that $3^{n/2-1/2} = \frac{1}{\sqrt{3}} 3^{n/2} = \frac 1{\sqrt{3}} \sqrt{3^n}$. So the $-\frac 12$ indeed becomes a constant factor that is absorbed by the $O()$, but $\frac n2$ in the exponent changes the base and changing the base changes the $O$-class.

The correct answer thus is $T(n) = O(3^{n/2}) = O(\sqrt{3^n})$.

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  • $\begingroup$ Thank you both @FrankW and @Jared! I marked this answer as correct because it contains an explanation (changing the base changes O-class). $\endgroup$ – wpp May 1 '14 at 11:36
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Everything was correct up until the last step...you found:

$$ T(n) \propto 3^{\frac{n}{2}- \frac{1}{2}} = \left(3^{n - 1}\right)^{\frac{1}{2}} = \frac{1}{\sqrt{3}}\left(\sqrt{3}\right)^n $$

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