Kenneth Rosen Edition 7 Exercise 8.2 Question 3 (Page No. 524)

Question

Solve these recurrence relations together with the initial conditions given.

• A. $a_{n} = 2a_{n−1}\:\text{for}\: n \geq 1, a_{0} = 3$ 
• B. $a_{n} = a_{n−1} \:\text{for}\: n \geq 1, a_{0} = 2$ 
• C. $a_{n} = 5a_{n−1} − 6a_{n−2} \:\text{for}\: n \geq 2, a_{0} = 1, a_{1} = 0$ 
• D. $a_{n} = 4a_{n−1} − 4a_{n−2} \:\text{for}\: n \geq 2, a_{0} = 6, a_{1} = 8$ 
• E. $a_{n} = −4a_{n−1} − 4a_{n−2} \:\text{for}\: n \geq 2, a_{0} = 0, a_{1} = 1$
• F. $a_{n} = 4a_{n−2} \:\text{for}\: n \geq 2, a_{0} = 0, a_{1} = 4$ 
• G. $a_{n} = a_{n−2} /4 \:\text{for}\: n \geq 2, a_{0} = 1, a_{1} = 0$

Answer 1

3. a) The characteristic equation is$ r - 2 = 0$, so the only root is $ r = 2.$ Therefore the general solution to
the recurrence relation is

$a_{n}= a2^{n},a0 = 3$

$3 = a2^{0} ,  a= 3 $

Therefore the solution is $a_{n} = 3 · 2^{n}$

 

b).The characteristic equation is
$r - 1 = 0$ , so the only root is $r = 1$

$a_{n}= a1^{n},a0 = 2$

$2 = a1^{0} ,  a= 2 $

Therefore the solution is $a_{n} = 2 1^{n}$

 

c)

The characteristic equation is$ r ^{2} - 5r + 6 = 0$, which factors as $(r - 2)(r - 3) = 0$ , so the roots are $r = 2$
and $r = 3$

general solution to the recurrence relation is $a_{n} = a2^{n} + b3^{n}$

after put value of $n=0,n=1$.

$1=a+b,0=2a+3b,a=3,b=-2$

Therefore the solution is$ a_{n} = 3 · 2^{n} - 2 · 3^{n}$.

 

d).The characteristic equation is $r ^{2} - 4r +4 = 0$ , which factors as $(r - 2)^{ 2} = 0$, so there is only one root, $r = 2$,
which occurs with multiplicity 2.

general solution to the recurrence relation is$a_{n} = a2^{n} + bn2^{n} $

$6=a+b,n=0$

$8=2a+4b,n=1$

$a=8,b=-2$

$a_{n} = 82^{n} -2n2^{n} $

 

e).This time the characteristic equation is $r^{2} + 4r + 4 = 0$, which factors as$ (r + 2)^{2} = 0$, so again  there
is only one root, $r = -2$, which occurs with multiplicity 2.

recurrence relation is$ a_{n} = a (-2)^{n} + bn( -2)^{n}$

after put value of $n=0,n=1$

so value of$ a=0,b=-1/2$

$ a_{n} = 0 (-2)^{n} -1/2n( -2)^{n}$

$ a_{n} =  -1/2n( -2)^{n}$
 

f) The characteristic equation is$ r ^{2} - 4 = 0$, so the roots are $r = 2 ,r = -2$.

Therefore the solution is $a_{n} = a(2)^{n} + b(-2)^{n}$

after put value of $n=0,n=1$

so value of$ a=1,b=-2$

$ a_{n} = 1 (2)^{n} -2( -2)^{n}$

 

g).The characteristic equation is $r ^{2} - 1/ 4 = 0$, so the roots are $r = 1/2 and r = -1/2. $

Therefore the solution is$ a_{n} = a(1/2)^{n} + b(-l/2){n} $

after put value of $n=0,n=1$

so value of$ a=1/2,b=1/2$

$ a_{n} = 1/2 (1/2)^{n} +1/2( -1/2)^{n}$