1. ## Recurrence Relation

Hey guys, just hoping that I could get some help with these 2 questions.

Solve the recurrence relations for these two questions

2. Originally Posted by Storm20
Hey guys, just hoping that I could get some help with these 2 questions.

Solve the recurrence relations for these two questions

I just posted this

http://www.mathhelpforum.com/math-he...-sequence.html

- it might help.

Added note. In the case of repeated roots of the characteristic equation, solutions are of the form

$a_n = \left( c_1 n + c_2 \right) \rho^n$

3. Yeah I thought it might have had something to do with using the general form, but im not 100 percent sure where or how to use it.!

4. Originally Posted by Storm20
Yeah I thought it might have had something to do with using the general form, but im not 100 percent sure where or how to use it.!
Let's try the first one. Substitute $a_n = c \rho^n$ into

$a_n - 7 a_{n-1} + 10 a_{n-2} = 0$

gives

$c \rho^n - 7 c \rho^{n-1} + 10 c \rho^{n-2} = 0$

or

$(\rho^2 - 7 \rho + 10) c \rho^{n-2} = 0$

which gives
$(\rho^2 - 7 \rho + 10) = 0$ or $(\rho - 2) \rho - 5) = 0$ so $\rho = 2,\; 5$

So, the general solution to the difference equation is

$a_n = c_1 2 ^n + c_2 5^n$

Now the initial condition

$a_0 = c_1 + c_2 = -1$
$a_1 = c_1 2 + c_2 5 = 4$

Two equations for $c_1\; \text{and}\; c_2$. SOlving gives

$c_1 = -3,\;\;\;c_2 = 2$

$a_n = -3 \cdot 2 ^n + 2 \cdot 5^n$

5. Originally Posted by danny arrigo
Let's try the first one. Substitute $a_n = c \rho^n$ into

$a_n - 7 a_{n-1} + 10 a_{n-2} = 0$

gives

$c \rho^n - 7 c \rho^{n-1} + 10 c \rho^{n-2} = 0$

or

$(\rho^2 - 7 \rho + 10) c \rho^{n-2} = 0$

which gives
$(\rho^2 - 7 \rho + 10) = 0$ or $(\rho - 2) \rho - 5) = 0$ so $\rho = 2,\; 5$

So, the general solution to the difference equation is

$a_n = c_1 2 ^n + c_2 5^n$
hmm, I understand how you go to this part, but i'm a little bit confused after that. Your help is really appreciated .

Originally Posted by danny arrigo
Now the initial condition

$a_0 = c_1 + c_2 = -1$
$a_1 = c_1 2 + c_2 5 = 4$

Two equations for $c_1\; \text{and}\; c_2$. SOlving gives

$c_1 = -3,\;\;\;c_2 = 2$

$a_n = -3 \cdot 2 ^n + 2 \cdot 5^n$