How can i solve the recurrence $T(n)=2T(n-2)+n\log n$

Question

What would be the solution of the following recurrence? $$T(n) = 2T(n-2) + n\log n.$$

I've managed to get:

$$T(n) = 2^{n/2}T(0)+\sum_{i=0}^{n/2} 2^i (n-2i) \log(n-2i). $$

The sum cannot be easily bounded for estimating the recurrence, because part of it increases while the other part decreases as the index grows.

Look at the following: https://cs.stackexchange.com/questions/1957/master-theorem-not-applicable — FunnyBuzer, Apr 25 '18 at 10:46
This is a different case, the Master theorem is not applicable here. — Binyamin Pinnick, Apr 25 '18 at 11:09
What prevents you from simplifying that sum? Some rearranging and known closed forms should do the trick. — Raphael, Apr 25 '18 at 11:31
Possible duplicate of Solving or approximating recurrence relations for sequences of numbers — David Richerby, Apr 25 '18 at 12:58
@Raphael My main difficulty with the sum is that 2^i increases while the 2nd part of the sum decreases as i grows — Binyamin Pinnick, Apr 25 '18 at 15:45
@BinyaminPinnick Common trick: revert the sum so you have $2^{\dots} \cdot 2i \cdot \log (2i)$. Transform further to get $2^i i \log i$. Then apply know formulae. — Raphael, Apr 25 '18 at 19:39
That is a first order linear recurrence (just that $n$ advances by 2), it can be solved explicitly. — vonbrand, Apr 27 '18 at 14:09

score 2 · Answer 1 · answered Apr 25 '18 at 14:50

2

Hint:

Let $n=2m$.

$$T(2m)=2T(2m-2)+2m\log(2m)$$

or

$$T'(m)=2T'(m-1)+2m\log(2m)$$

which has the classical form.

answered Apr 25 '18 at 14:50

I still reach a similar sum like the one I posted which I can't seem to figure out – Binyamin Pinnick Apr 25 '18 at 15:42

score 1 · Answer 2 · answered Apr 26 '18 at 10:26

Let $S(m) = T(2m)/2^m$. Then $$ S(m) = \frac{T(2m)}{2^m} = \frac{2T(2m-2) + (2m)\log(2m)}{2^m} = S(m-1) + \frac{m\log(2m)}{2^{m-1}}. $$ It follows that $$ S(m) = S(0) + \sum_{t=1}^m \frac{t\log(2t)}{2^{t-1}}. $$ If we extend the sum on the right to infinity, then we get a convergent series. In other words, $S(m) = \Theta(1)$, and so $T(n) = \Theta(2^{n/2})$.

How can i solve the recurrence $T(n)=2T(n-2)+n\log n$

2 Answers2