# Asymptotic approximation of a recurrence relation (Akra-Bazzi doesn't seem to apply)

Suppose an algorithm has a runtime recurrence relation:

$T(n) = \left\{ \begin{array}{lr} g(n)+T(n-1) + T(\lfloor\delta n\rfloor ) & : n \ge n_0\\ f(n) & : n < n_0 \end{array} \right.$

for some constant $0 < \delta < 1$. Assume that $g$ is polynomial in $n$, perhaps quadratic. Most likely, $f$ will be exponential in $n$.

How would one go about analyzing the runtime ($\Theta$ would be excellent)? The master theorem and the more general Akra-Bazzi method do not seem to apply.

• Finding good lower bound is easy but finding good upper bound is hard, but roughly speaking seems to be close to $T(n) = a\cdot T(n/a) + g(n)$. – user742 Sep 16 '13 at 21:02
• If you are at still looking for an answer you should check Graham, Knuth, and Patashnik, "Concrete Mathematics". – Kaveh Nov 13 '13 at 6:41
• Assuming that $n_0$ is constant, we don't need any assumptions on $f$, or do we? – Raphael Nov 13 '13 at 21:55
• The parameter $n_0$ may be instance-specific. It would be nice to see how the runtime depends on $n_0$. – Austin Buchanan Nov 13 '13 at 21:59
• I asked a related question that, so far, has not brought forth any general theorem for recurrences of this kind. – Raphael Dec 9 '14 at 18:03

One possible approach might be by analogy to differential equations. Let $T'(n) = T(n)-T(n-1)$. Here $T'(n)$ is a discrete analog of the first derivative of $T(n)$. We get the following relationship: $$T'(n) = T(\lfloor \delta n \rfloor) + g(n).$$ The continuous analog of this is the differential equation $$t'(x) = t(\delta x) + g(x),$$ or, if you prefer to see it written differently: $${d \over dx} t(x) = t(\delta x) + g(x).$$ That's a differential equation.
Now you could try to solve the differential equation for the continuous function $t(x)$, then hypothesize that a similar function will be the solution to your original recurrence relation, and try to prove your hypothesis. At least, this is one general approach you could follow.
• Without looking further. It is not easy to solve that differential equation. I am not too convinced that it has a closed form solution after trying a few basic form for $t(x)$. – InformedA Dec 10 '14 at 18:26