Question

Discrete Math for COMP:

(12 points) For each sequence described below, first find a recurrence relation for that sequence and then solve your recurrence relation. (a) The sequence Sn where 80 = 0, si-l and, for n 〉 1, sn s the average of the previous two terms of the sequence. (b) The sequence bn whose nth term is the number of n-bit strings that don't have two zeros in a rovw (c) The sequence en whose nth term is the number of edges in Kn

Answer 1

a) $s_{n+1}=\frac{s_n+s_{n-1}}{2}$

Has the char equation $D^2=\frac{D}{2}+\frac{1}{2}$

So $D^2-\frac{D}{2}+\frac{1}{4}=\frac{3}{4}$

So that $\left(D-\frac{1}{2} \right )^2=\frac{3}{4}\Rightarrow D-\frac{1}{2}=\pm \frac{\sqrt{3}}{2}$

And so $D=\left(\frac{1\pm \sqrt{3}}{2} \right )$

General solution is $s_n=c_1\left(\frac{1+\sqrt{3}}{2} \right )^n+c_2\left(\frac{1-\sqrt{3}}{2} \right )^n$

Using initial conditions we get $s_0=c_1+c_2=0,\,s_1=c_1\left(\frac{1+\sqrt{3}}{2} \right )+c_2\left(\frac{1-\sqrt{3}}{2} \right )=1$

And so $c_1\left(\frac{1+\sqrt{3}}{2} \right )-c_1\left(\frac{1-\sqrt{3}}{2} \right )=1$

And so $c_1\left(\sqrt{3}\right )=1\Rightarrow c_1=\frac{1}{\sqrt{3}},\,c_2=-\frac{1}{\sqrt{3}}$

This means $s_n=\frac{1}{\sqrt{3}}\left(\frac{1+\sqrt{3}}{2} \right )^n-\frac{1}{\sqrt{3}}\left(\frac{1-\sqrt{3}}{2} \right )^n$

b) $b_{n+1}=b_n+b_{n-1}$ as the last character can be either a 0 or 1. If it is a 1, the previous n digit string can be made in $b_n$ ways. If it is 0 the its previous character must be 1 and the previous (n-1) digit string can be made in $b_{n-1}$ ways

Note that this is the Fibonacci recurrence with $b_0=1,b_1=2$ compared with $F_0=0,F_1=1,F_2=1,F_3=2$

So we must have $b_n=F_{n+2}$

c) $e_1=0,e_2=1$

Adding a new vertex to $K_n$ , we need to add n new edges connecting the previous n edges to the new vertex

Thus, $e_{n+1}=e_n+k$ so we have

$e_2=e_1+2=1+2,e_3=e_2+3=(1+2)+3,\,e_k=(1+2+3\cdots )+k$

And thus we have $e_n=\frac{n(n+1)}{2}$ using the sum of natural numbers formula