Finding Function for a recurrance relation

**hockey777** · March 17th 2008, 12:18 PM

I need help find a function for the following recurrence relation

h $_{n}$ =2*h $_{n-1}$ +2*h $_{n-2}$

It's getting to big and messy for me to recognize any pattern.

**galactus** · March 17th 2008, 01:39 PM

I am assuming you have some initial condtions?. Something like

$h_{n}=0, \;\ h_{1}=1$ or something like that.

You can write it as $h^{2}-2h-2=0$

Solving, we see $h=\sqrt{3}+1, \;\ h=1-\sqrt{3}$

Then, $h_{n}=a(\sqrt{3}+1)^{n}+b(1-\sqrt{3})^{n}$

Now apply your initial condtions to find a and b.

**hockey777** · March 17th 2008, 03:14 PM

h $_{1}$ =3
h $_{2}$ =8

**galactus** · March 17th 2008, 03:25 PM

In that event, you have two equations to solve for a and b. Just use your conditions.

$a(\sqrt{3}+1)+b(1-\sqrt{3})=3$

$a(\sqrt{3}+1)^{2}+b(1-\sqrt{3})^{2}=8$

**CaptainBlack** · March 17th 2008, 03:46 PM

Originally Posted by hockey777

I need help find a function for the following recurrence relation

h $_{n}$ =2*h $_{n-1}$ +2*h $_{n-2}$

It's getting to big and messy for me to recognize any pattern.

This is a linear constant coefficient homogeneous difference equation. You treat it like a linear constant coefficient homogeneous ODE.

Take trial solution $h_n=\mu^n$ , substitute this into the recurrence to get a quadratic equation for $\mu$ . Solve the quadratic to get two basic solutions, form a general solution as a linear combination of these two solutions.

RonL

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