# recurrence relations...

• Jun 2nd 2009, 01:28 PM
ninobrn99
recurrence relations...

Find all solutions of the recurrence relation
a -subscript-n = n+2-2a-subscript-n-1, a-subscript 1 = 0

My instructor says it is similar to:
Let R be the relation R-{(a,b) | a divides b} on the set of positive integers. Find
A) r^-1
B)R^-

and

Which of the relations from Example 7 are symmetric and which are antisymmetric?

Do these indeed relate? I've read the entire chapter and haven't seen one problem setup like the one asked.
• Jun 2nd 2009, 02:22 PM
ninobrn99
great! I see that my version of the book is the special indian edition so chapters are mixed up. Now Im only behind a chapter ;) (being sarcastic!)
• Jun 4th 2009, 09:39 PM
ninobrn99
so far so good?
• Jun 4th 2009, 10:22 PM
Random Variable
Not quite.

$\displaystyle a_{n} = -2a_{n-1} + n + 2$

First find the solution to the homogeneous recursion

$\displaystyle a^{h}_{n} = -2a_{n-1}$

The characteristic equation of which is r+2=0, which has root r=-2 (multiplicity 1)

so $\displaystyle a^{h}_{n} = A (-2)^{n}$

Now we need a particular solution. For this problem we look for a solution of the form $\displaystyle a^{p}_{n} = Bn + C$

plugging it into the recursion we get $\displaystyle Bn + C = -2( B(n-1) + C)+ n + 2$

$\displaystyle Bn + C = -2Bn + 2B - 2C + n + 2$

equating like terms we get

B = -2B + 1
C = 2B -2C + 2

so B = $\displaystyle 1 \over 3$, and C = $\displaystyle 8 \over 9$

so the general solution is $\displaystyle a^{h}_{n} + a^{p}_{n} = A(-2^{n}) + \frac {n}{3} + \frac {8}{9}$

If the problem had an intial condition, we could solve for A.
• Jun 4th 2009, 10:27 PM
ninobrn99
Quote:

Originally Posted by Random Variable
Not quite.

$\displaystyle a_{n} = -2a_{n-1} + n + 2$

First find the solution to the homogeneous recursion

$\displaystyle a^{h}_{n} = -2a_{n-1}$

The characteristic equation of which is r+2=0, which has root r=-2 (multiplicity 1)

so $\displaystyle a^{h}_{n} = A (-2)^{n}$

Now we need a particular solution. For this problem we look for a solution of the form $\displaystyle a^{p}_{n} = Bn + C$

plugging it into the recursion we get $\displaystyle Bn + C = -2( B(n-1) + C)+ n + 2$

$\displaystyle Bn + C = -2Bn + 2B - 2C + n + 2$

equating like terms we get

B = -2B + 1
C = 2B -2C + 2

so B = $\displaystyle 1 \over 3$, and C = $\displaystyle 8 \over 9$

so the general solution is $\displaystyle a^{h}_{n} + a^{p}_{n} = A(-2^{n}) + \frac {n}{3} + \frac {8}{9}$

If the problem had an intial condition, we could solve for A.

Wow, I was way off! Guess I just can't learn some things from the book :(
Thanks for your assistance! Much appreciated.
• Jun 4th 2009, 10:32 PM
Random Variable
You can check that my answer (or any answer) is correct by plugging it into the recursion and seeing if both sides are equal.
• Jun 4th 2009, 10:54 PM
ninobrn99
Ill run a few variables through it. Thanks again!