# Recurrence Relation

• Jan 14th 2009, 05:16 AM
Storm20
Recurrence Relation
Hey guys, just hoping that I could get some help with these 2 questions.

Solve the recurrence relations for these two questions

http://imagehost.platinum.net.au/pic...1bb125fd2d.jpg
• Jan 14th 2009, 05:20 AM
Jester
Quote:

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

http://imagehost.platinum.net.au/pic...1bb125fd2d.jpg

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

$a_n = \left( c_1 n + c_2 \right) \rho^n$
• Jan 14th 2009, 12:10 PM
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.!
• Jan 14th 2009, 12:23 PM
Jester
Quote:

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

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

$a_n = -3 \cdot 2 ^n + 2 \cdot 5^n$
• Jan 14th 2009, 06:15 PM
Storm20
Quote:

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 :).

Quote:

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

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

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