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The Akra-Bazzi method

The Akra-Bazzi method gives asymptotics for recurrences of the form: $$T(x) = \sum_{1 \le i \le k} a_i T(b_i x + h_i(x)) + g(x) \quad \text{for x \ge x_0}$$ This covers the usual divide-and-conquer recurrences, but also cases in which the division is unequal. The "fudge terms" $$h_i(x)$$ can cater for divisions that don't come out exact, for example. The conditions for applicability are:

• There are enough base cases to get the recurrence going
• The $$a_i$$ and $$b_i$$ are all constants
• For all $$i$$, $$a_i > 0$$
• For all $$i$$, $$0 < b_i < 1$$
• $$\lvert g(x) \rvert = O(x^c)$$ for some constant $$c$$ as $$x \rightarrow \infty$$
• For all $$i$$, $$\lvert h_i(x) \rvert = O(x / (\log x)^2)$$
• $$x_0$$ is a constant

Note that $$\lfloor b_i x \rfloor = b_i x - \{b_i x\}$$, and as the sawtooth function $$\{ u \} = u - \lfloor u \rfloor$$ is always between 0 and 1, replacing $$\lfloor b_i x \rfloor$$ (or $$\lceil b_i x \rceil$$ as appropiate) satisfies the conditions on the $$h_i$$.

Find $$p$$ such that: $$\sum_{1 \le i \le k} a_i b_i^p = 1$$ Then the asymptotic behaviour of $$T(x)$$ as $$x \rightarrow \infty$$ is given by: $$T(x) = \Theta \left( x^p \left( 1 + \int _1^x \frac{g(u)}{u^{p + 1}} du \right) \right)$$

Example

As an example, take the recursion for $$n \ge 5$$, where $$T(0) = T(1) = T(2) = T(3) = T(4) = 17$$: $$T(n) = 9 T(\lfloor n / 5 \rfloor) + T(\lceil 4 n / 5 \rceil) + 3 n \log n$$ The conditions are satisfied, we need $$p$$: $$9 \left( \frac{1}{5} \right)^p + \left( \frac{4}{5} \right)^p = 1$$ As luck would have it, $$p = 2$$. Thus we have: $$T(n) = \Theta \left( n^2 \left(1 + \int_1^n \frac{3 u \log u}{u^3} du \right) \right) = \Theta(n^2)$$

(The help of maxima with the algebra is gratefully aknowledged)

 answered Mar 7 '13 at 18:22 vonbrand 9,07833 gold badges3131 silver badges4040 bronze badges