recurrence relation and initial condition

**papa_chango123** · December 4th 2006, 04:50 PM

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.

**ThePerfectHacker** · December 4th 2006, 04:56 PM

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?

**papa_chango123** · December 4th 2006, 05:06 PM

Both if you don't mind. So is problem one right? Is the answer zero?

**ThePerfectHacker** · December 4th 2006, 07:12 PM

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

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