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Let $(r_i)_{i=1}^m$ be a sequence of positive reals such that $\sum_i r_i < 1$ and let $t$ and $u$ be positive reals. Consider the sequence $T(n)$ defined by $T(0) = t$, $T(n) = un + \sum_i T(\lfloor r_i n \rfloor) $ for $n \ge 1$.

  1. Show that $T(n) = \Theta(n)$, that is $T(n) = O(n)$ and $T(n) = \Omega(n)$.

  2. Show that there exists a real number $L$ such that $\lim_{n \to \infty} \dfrac{T(n)}{n} = L $.

  3. Specify $L$ in terms of $t$, $u$, and the $r_i$.

Note: This is a generalization of https://math.stackexchange.com/questions/504495/recurrence-relation-using-induction-method/504536#504536.

I have a proof of part 1 which is a fairly straightforwand generalization of my solution to that question. I do not yet have a solution to part 2. I have a conjecture as to the solution to part 3 which should be pretty evident from my solution to part 1, which I will submit if no one submits a solution within two days.

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marty cohen
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1 Answers1

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Here is my proof of part 1, as promised.

Let $r = \sum_i r_i$ and $s = 1-r$ so that $0 < s < 1$. Let $r_0 = \min_i r_i$.

First, the upper bound.

Suppose that $T(k) \le bk$ for $r_0n \le k < n$. We want to be able to deduce that $T(n) \le bn$.

$\begin{align} T(n) &=un + \sum_i T(\lfloor r_i n \rfloor)\\ &\le un + \sum_i b r_i n\\ &= n(u + \sum_i b r_i )\\ &= n(u + b r)\\ \end{align} $

and this is $\le bn$ if $n(u + b r) \le bn$ or $u+br \le b$ or $u \le b(1-r) = bs$ or $b \ge \dfrac{u}{s}$.

For a particular $n$, if we choose $b \ge \max\left(\dfrac{u}{s}, \max_i \left( \dfrac{T(\lfloor r_i n \rfloor)}{\lfloor r_i n \rfloor}\right)\right) $, all subsequent $T(n)$ will be $\le bn$.

The only problem might be if some $r_i n < 1$, so that $\lfloor r_i n \rfloor = 0$. To avoid this, start with $n$ such that $r_0 n \ge 1$, or $n \ge \dfrac1{r_0}$.

Then, the lower bound.

Suppose that $T(k) \ge ak+c$ for $r_0n \le k < n$. We want to be able to deduce that $T(n) \ge an+c$. The constant term $c$ is needed because of the floor functions in the recurrence for $T(n)$.

$\begin{align} T(n) &=un + \sum_i T(\lfloor r_i n \rfloor)\\ &\ge un + \sum_i (a(\lfloor r_i n \rfloor+c)\\ &> un + \sum_i (a (r_i n-1)+c)\\ &= n(u + \sum_i (a r_i ) - \sum_i (c-a) \\ &= n(u + a r) + m(a-c)\\ \end{align} $

and this is $\ge an+c$ if $n(u + a r)+m(a-c) \ge an$ or $n( a (1-r)-u)+m(a-c) \le 0$ or $n( as-u)\le m(c-a)$.

For this to be true, we must have $as-u \le 0$ or $a \le \dfrac{u}{s}=v$. Otherwise, the left side would get arbitrarily large for large enough $n$.

If, in addition, $a < \dfrac{u}{s}$, then this will be true for all large enough $n$ (i.e., all $n$ if $c \ge a$ and $n > \dfrac{a-c}{as-u}$ if $c < a$.

We now choose $a$ small enough so that $a < \dfrac{u}{s}=v$ (as before, remembering that $vs = u$). To do this, let $a = v-d$ where $v > d > 0$ so $a > 0$.

Since we want $n( as-u)\le m(c-a)$, or $n( as-u)+am\le mc$, this will be true if $n( (v-d)s-u)+am\le mc$, or $n( -ds)+am\le mc$.

Since $-dsn < 0$, this will be true if $c \ge a$, and this, in turn, will be true if $c \ge v$.

We also need, for the induction hypothesis, that $T(k) \ge ak+c$ for $r_0n \le k < n$.

Since $c = v$ works for the recurrence to be true, we want $a \le \dfrac{T(k)-c}{k}$ for $r_0 n \le k < n$. To ensure that $a > 0$, we need $T(k) > c$. Since $T(k) \ge uk$, this will be true if $uk > c$ or $k > \dfrac{c}{u}$ and this will be true if $k > \dfrac{v}{u}$. Since we want to look at $r_0 n \le k$, we need $r_0 n > \dfrac{v}{u}$ or $n > \dfrac{v}{u r_0}$.

Following all these prescriptions will allow us to find a value of $n$ that will allow us to compute a value of $a$ such that $T(k) \ge ak+c$ for $r_0 n \le k < n$ will imply that $T(n) \ge an+c$ for all larger $n$.

For problem 2, I do not know how to refine this solution so show how the values for $a$ and $b$ can be made arbitrarily close.

For problem 3, my obvious conjecture is that $\lim_{n \to \infty} \dfrac{T(n)}{n} = v = \dfrac{u}{s} = \dfrac{u}{1-\sum_i r_i} $.

marty cohen
  • 107,799