# Thread: solving recurrences - I keep gettin' alpha = some fraction???

1. ## solving recurrences - I keep gettin' alpha = some fraction???

ok, so I have this recurrence:

a(n) = 2(a(n-1)) + 3(a(n-2))

so the degree is k, & k=2, & I used the formula r^2 - 2r - 3, to get the roots r=1 & r= -3.

then I used the general solution technique

a(n) = alpha1(r1^n) + alpha2(r2^n)

& subbed in my "r" values to get:

a(n) = alpha1[(1)^n] + alpha2[(-3)^n]

then I used the initial conditions a(0) = 0 & a(1) = 2 (subbed in n=0 & n=1) to the equation:

a(0) = alpha1[(1)^0] + alpha2[(-3)^0] = 0

a(1) = alpha1[(1)^1] + alpha2[(-3)^1] = 2

to try & solve for alpha1 & alpha2, but I keep gettin' fractions for both alphas...(I got alpha1 = 1/2, & alpha2 = -(1/2))

is this right?!??! b/c most of the examples in my textbook have alpha equaling an integer....

I'm not sure if I messed up the calculations.

2. Hello amyu2005
Originally Posted by amyu2005
ok, so I have this recurrence:

a(n) = 2(a(n-1)) + 3(a(n-2))

so the degree is k, & k=2, & I used the formula r^2 - 2r - 3, to get the roots r=1 & r= -3.

then I used the general solution technique

a(n) = alpha1(r1^n) + alpha2(r2^n)

& subbed in my "r" values to get:

a(n) = alpha1[(1)^n] + alpha2[(-3)^n]

then I used the initial conditions a(0) = 0 & a(1) = 2 (subbed in n=0 & n=1) to the equation:

a(0) = alpha1[(1)^0] + alpha2[(-3)^0] = 0

a(1) = alpha1[(1)^1] + alpha2[(-3)^1] = 2

to try & solve for alpha1 & alpha2, but I keep gettin' fractions for both alphas...(I got alpha1 = 1/2, & alpha2 = -(1/2))

is this right?!??! b/c most of the examples in my textbook have alpha equaling an integer....

I'm not sure if I messed up the calculations.
Thanks for showing us all your working. You just made a simple slip in solving the quadratic. The roots are $-1, 3$. The solution is then

$a_n = (-1)^n - 3^n$

Hello amyu2005Thanks for showing us all your working. You just made a simple slip in solving the quadratic. The roots are $-1, 3$. The solution is then

$a_n = (-1)^n - 3^n$

thanks~~~

4. I used the correct roots, but I'm still gettin' alpha as 0.5 & -0.5.....

is this possible?

5. Hello amyu2005

Sorry, I got a sign wrong in my working as well! I think it should be:

$r = -1, 3$ (That bit was correct!)

$n= 0 : a_0 = 0 = \alpha_1 + \alpha_2$

$n=1: a_1 = 2 = -\alpha_1 +3\alpha_2$ (I originally had $2 = -\alpha_1 -3\alpha_2$ here, which gave me my first answer.)

$\Rightarrow 4\alpha_2 = 2$

$\Rightarrow \alpha_2 = \tfrac12$

$\Rightarrow \alpha_1 = -\tfrac12$

$\Rightarrow a_n = \tfrac12\Big(3^n - (-1)^n\Big)$

Check this out to see that the original equation $a_n = 2a_{n-1} + 3a_{n-2}$ is satisfied:

$a_0 = 0,\, a_1 = 2,\, a_2 = 4 = 2a_1 + 3a_0,\, a_3 = 14 = 2a_2 +3a_1,\, a_4 = 40 = 2a_3+3a_2,\,...$