# recurrence relation and initial condition

• Dec 4th 2006, 04:50 PM
papa_chango123
recurrence relation and initial condition
Could someone make sure my answers are correct. Not sure about the first one but after the S0 s1 = 0? Here are the questions:

6.1 In the following sequences determine s5 if s0, s1, ... sn, ... is a sequence satisfying the given recurrence relation and initial condition.
a. sn= -sn-1 - n2 for n >= 1, s0 = 3

-3(3) - 33 = -9 + 9 - 0

S1 = 0
The rest of the answers up to S5 would be zero correct?

b. sn = 5sn-1 - 3sn-2 for n >= 2, s0 = -1, s1 = -2

S2 = 5(-2) - 3(-1) = -10 + 3 = -7
S3 = 5(-7) - 3(-2) = -35 + 6 = -29
S4 = 5(-29) - 3(-7) = -145 + 21 = 124
S5 = 5(-124) - 3(29) = -620 + 87 = -533

Any help would be much appreciated.
• Dec 4th 2006, 04:56 PM
ThePerfectHacker
Quote:

Originally Posted by papa_chango123
The rest of the answers up to S5 would be zero correct?

It seems thus.
In the second problem do you want me to solve the recurrence relation or just compute it?
• Dec 4th 2006, 05:06 PM
papa_chango123
Both if you don't mind. So is problem one right? Is the answer zero?
• Dec 4th 2006, 07:12 PM
ThePerfectHacker
Quote:

Originally Posted by papa_chango123

b. sn = 5sn-1 - 3sn-2 for n >= 2, s0 = -1, s1 = -2

As I already said the first one is correct.

The charachteristic equation is,
$\displaystyle k^2=5k-3$
Thus,
$\displaystyle k^2-5k+3=0$
Thus,
$\displaystyle k=\frac{5\pm \sqrt{13}}{2}$
Thus the general solution to the recurrence relation is,
$\displaystyle C_1\left( \frac{5+\sqrt{13}}{2} \right)^n+C_2 \left( \frac{5-\sqrt{13}}{2} \right)^n$
When,
$\displaystyle n=0$
We have,
$\displaystyle s_0=C_1+C_2=-1$
When,
$\displaystyle n=1$
We have,
$\displaystyle s_1=C_1 \left( \frac{5+ \sqrt{13}}{2} \right)+C_2\left( \frac{5-\sqrt{13}}{2} \right)=-2$
Multiply by 2,
$\displaystyle C_1(5+\sqrt{13})+C_2(5-\sqrt{13})=-4$
$\displaystyle 5(C_1+C_2)+\sqrt{13}(C_1-C_2)=-4$
From first equation,
$\displaystyle -5+\sqrt{13}(C_1-C_2)=-4$
$\displaystyle \sqrt{13}(C_1-C_2)=1$
Thus,
$\displaystyle \left\{ \begin{array}{c}C_1+C_2=-1\\ C_1-C_2=\frac{\sqrt{13}}{13} \end{array} \right\}$
Add then subtract, respectively,
$\displaystyle 2C_1=\frac{-13+\sqrt{13}}{13}$
$\displaystyle 2C_2=\frac{-13-\sqrt{13}}{13}$
Thus,
$\displaystyle (C_1,C_2)=\left( \frac{-13+\sqrt{13}}{26},\frac{-13-\sqrt{13}}{26} \right)$