Recurrence Relation

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January 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
January 14th 2009, 05:20 AM
Danny

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

$a_n = \left( c_1 n + c_2 \right) \rho^n$
January 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.!
January 14th 2009, 12:23 PM
Danny

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

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

leading to the solution

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

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

leading to the solution

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