# Second Order Recurrence relation

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• Feb 12th 2008, 11:42 AM
frostking2
Second Order Recurrence relation
I can usually solve second order ones fine but this one has me stumped!

a(n) = 6a(n-1) - 9a(n-2) a(0) = 1 a(1) = 1

Using t^2 = 6t^(n-1) - 9t^(n-2)
I get t = 3 and only 3

When I try to plug in S(n) = 3^n and T(n) = 3^n and solve for

U(n) = bS(n) + dT^n with b3^n + d3^n and use my initial values of 1 for U(0) and for U(1) I can not find a solution for the values of b and d that meet both of these!!!!!! PLease give me a clue or two or three.....
• Feb 12th 2008, 12:25 PM
galactus
Running it through Maple, I get:

$\displaystyle a_{n}=(1-\frac{2n}{3})\cdot{3^{n}}$

That's something to shoot for.
• Feb 12th 2008, 12:59 PM
Soroban
Hello, frostking2!

I think we speak the same language ... We just use different symbols.

Quote:

$\displaystyle a(n) \:= \:6a(n-1) - 9a(n-2),\quad a(0) = 1,\;\;a(1) = 1$
Here's the way I was taught to handle these . . .

We conjecture that: .$\displaystyle a(n) \:=\:X^n$ . . . that the function is exponential.

The equation becomes: .$\displaystyle X^n \:=\:6X^{n-1} - 9X^{n-2}\quad\Rightarrow\quad X^n - 6X^{n-1} + 9X^{n-2}\;=\;0$

Divide by $\displaystyle X^{n-2}\!:\;\;X^2 - 6X + 9 \:=\:0\quad\Rightarrow\quad (X-3)^2\:=\:0\quad\Rightarrow\quad X \:=\:3,\,3$

With repeated roots, the function is: .$\displaystyle a(n) \;=\;A\!\cdot\!3^n + B\!\cdot\!n\!\cdot\!3^n$

Plug in the first two values of the sequence . . .

. . $\displaystyle a(0) = 1:\;A\!\cdot\!3^0 + B(0)(3^0) \:=\:1 \quad\Rightarrow\quad A \:=\:1$

. . $\displaystyle a(1) = 1:\;A\!\cdot\!3^! + B(1)(3^1) \:=\:1\quad\Rightarrow\quad B \:=\:-\frac{2}{3}$

Hence: .$\displaystyle a(n) \;=\;3^n - \frac{2}{3}n\cdot3^n \;=\;3^n\left(1 - \frac{2}{3}n\right) \;=\;3^n\left(\frac{3-2n}{3}\right)$

Therefore: .$\displaystyle \boxed{a(n) \;=\;3^{n-1}(3 - 2n)}$

• Feb 12th 2008, 01:59 PM
frostking2
Second order problem solved!
Thank you soooooo much. Stupid me I NEVER considered 0 for one of the values at the last portion of the problem, second roots and values for a and b. Of course it does work and so the regular process yields an answer. Thanks so much for your time and helpful attitude!!!!!