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I've been working on a problem set for a bit now and I seem to have gotten the master method down for recurrence examples. However, I find myself having difficulties with other methods (recurrence trees, substitution). here is the question I am stuck on: $$T(n) = T(n-2) + n^2$$ Is there a pattern as follows? $$n^2 + T(n-2) + T(n-4) +...$$ where it goes until there is no more n left. so around n/2 times and would that mean that $$n^2 + (n-2)^2 + (n-i) ^2$$ so the asymptotic bound would be $\theta(n^2)$?

I am honestly taking a shot in the dark here, so I was hoping someone could help guide me in how to approach these questions.

Thank you,

Tyler

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perhaps an indirect answer would even do, something to show how to solve questions of form t(n-i) + f(n) –  Tyler Feb 5 '13 at 9:31
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2 Answers

$$T(n) = T(n-2) + n^2 = T(n-4) + (n-2)^2 + n^2 = T(n-2k) + \sum\limits_{i = 0}^{k - 1}(n - 2i)^2$$

This goes down till $n - 2k \ge 0$. Assuming even $n$ (for asymptotic complexity, it does not really matter, and you can do similar calculations for odd $n$ also, with the same asymptotic results), we have $k = \frac{n}{2}$ at the end.

$$T(n) = T(0) + \sum\limits_{i = 0}^{\frac{n}{2} - 1}(n - 2i)^2 = \sum\limits_{i = 0}^{\frac{n}{2} - 1}(n^2 - 4ni + 4i^2) + C$$ $$T(n) = n^2\cdot\left(\frac{n}{2}-1\right) - 4n\cdot\frac{1}{2}\cdot\frac{n}{2}\cdot\left(\frac{n}{2} - 1\right) + 4\cdot\frac{1}{6}\cdot\left(\frac{n}{2} - 1\right)\cdot\frac{n}{2}\cdot n + C$$ $$\therefore \ T(n) = \Theta(n^3)$$

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Note that if $n=2k$ is even, then $$ T(n)+T(n-1) = n^2+(n-1)^2+ \cdots+4^2+3^2 + T(2)+T(1) =\frac{n(n+1)(2n+1)}{6} + C. $$ Here $C=T(2)+T(1) -2^2-1^2$ and we used the formula $\sum_{i=0}^n i^2 = \frac{n(n+1)(2n+1)}{6}$. We also note that $T(n) \sim T(n-1)$, so we may conclude that $$ T(n) \sim n^3/12. $$

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