This question is related to Master Theorem on oscillating function.

Consider a recurrence of the form

$T(n) = a T(n/b) + f(n)$

Master Theorem's regularity condition excludes some cases (for example, when $f(n)$ is oscillating).

Suppose, however, that $f(n)=\Theta(g(n))$ for a function $g(n)$ that does not violate the regularity condition, so that the Master Theorem is applicable if $g(n)$ is used instead of $f(n)$. Consider then the following recurrence:

$T'(n) = a T'(n/b) + g(n)$

and assume that the master theorem gives the solution $T'(n)=\Theta(g(n))$.

Can I then safely conclude that $T(n)=\Theta(g(n))$? Or there are some reasonable conditions on $g(n)$ we can add (I suppose that if $g(n)$ is polinomially bounded then maybe Akra-Bazzi method will apply, even if one have to swich integration and $\Theta$, and I'm not sure this is sound)?

Notice that from $f(n)=\Theta(g(n))$ we cannot deduce that $f(n)$ is always less than or equal to $g(n)$ or to a $d\cdot g(n)$ for some fixed $d$ (indeed, $g(n)$ can be equal to 0 for some value of $n$, so I'm not totally convinced by the answer to Regularity condition in the master Theorem in the presence of Landau notation for f as it is not possible to prove the base case of the induction, unless one changes the initial condition). So I cannot prove by induction that $T(n)<T'(n)$ for all $n$ and, anyway, I only want to prove $T(n)=\Theta(T'(n))$.

To give some context, this is the case I want to apply the result to: consider the recurrence

$T(n)=2\cdot T(n/2)+f(n)$

where I only know that $f(n)=\Theta(n\sqrt{n})$. Then I can unfold the recurrence obtaining \begin{equation*} \begin{split} T&(n)=2T\left(\frac{n}{2}\right)+\Theta(n\sqrt{n})= 2\left(2T\left(\frac{n}{4}\right)+\Theta\left(\frac{n}{2}\sqrt{\frac{n}{2}}\right)\right)+\Theta(n\sqrt{n}) =\\ &=\ldots=2\left(2\left(2\ldots \left(2T\left(\frac{n}{2^h}\right)+\Theta\left(\frac{n}{2^{h-1}}\sqrt{\frac{n}{2^{h-1}}}\right)\right)\ldots\right)\right)+\Theta(n\sqrt{n}) \end{split} \end{equation*}

until $2^h=n$, i.e. $h=\log(n)$. Then $T(n)=\sum_{i=0}^{\log(n)-1} 2^i\cdot\Theta\left(\frac{n}{2^i}\sqrt{\frac{n}{2^i}}\right)$.

Performing the calculations I get \begin{equation*} \begin{split} T(n)&=\sum_{i=0}^{\log(n)-1} 2^i\cdot\Theta\left(\frac{n}{2^i}\sqrt{\frac{n}{2^i}}\right)=\Theta\left(\sum_{i=0}^{\log(n)-1} 2^i\cdot\frac{n}{2^i}\sqrt{\frac{n}{2^i}}\right)=\\ &=\Theta\left(n\sqrt{n}\cdot\sum_{i=0}^{\log(n)-1} \frac{1}{\sqrt{2^i}}\right) \end{split} \end{equation*} The series $\sum_{i=0}^{+\infty} \frac{1}{\sqrt{2^i}}$ is convergent, so I can conclude (I hope correctly) that $T(n)=\Theta(n\sqrt{n})$.

On the other hand, I cannot directly apply Master Theorem, as from $f(n)=\Theta(n\sqrt{n})$ I cannot conclude $2f\left(\frac{n}{2}\right)<cf(n)$ for some $c<1$. Indeed, if definitively we have $c_1 n\sqrt{n}\leq f(n)\leq c_2 n\sqrt{n}$, then

$f(n)\geq c_1 n\sqrt{n}= 2\sqrt{2}c_1 \frac{n}{2}\sqrt{\frac{n}{2}}\geq\frac{2\sqrt{2}c_1}{c_2} f\left(\frac{n}{2}\right)=\frac{\sqrt{2}c_1}{c_2} 2f\left(\frac{n}{2}\right)$

and so

$2f\left(\frac{n}{2}\right)\leq \frac{c_2}{\sqrt{2}c_1} f(n)$

but $\frac{c_2}{\sqrt{2}c_1}$ is not necessarily less than 1.


1 Answer 1


Yes. Here is the idea.

If $f(n) = \Theta(g(n))$ then there exist constants $C_1,C_2>0$ such that for large $n$, $$ C_1g(n) \leq f(n) \leq C_2g(n). $$ Now consider the two recurrences $$ T_1(n) = aT_1(n/b) + C_1g(n) \\ T_2(n) = aT_2(n/b) + C_2g(n) $$ Choosing suitable initial conditions, you can prove inductively that $$ T_1(n) \leq T(n) \leq T_2(n). $$ You consider a regime in which applying the master theorem gives you that $T_1(n) = \Theta(g(n))$ and $T_2(n) = \Theta(g(n))$. It follows that $T(n) = \Theta(g(n))$.


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