1. ## 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.

2. 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!)

3. so far so good?

4. 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.

5. 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.

6. You can check that my answer (or any answer) is correct by plugging it into the recursion and seeing if both sides are equal.

7. Ill run a few variables through it. Thanks again!