Simplifying a recursive function

[randommacuser]

I am trying to find a formula for f(x) = a*f(x-1) + b*f(x-2) for x≥1, assuming f(-1) and f(0) are known. In other words, I need some expression for f(x) just in terms of a, b, f(-1), and f(0).

I tried plugging in stuff and factoring on paper for the first few iterations, but it quickly got out of control. Then I let Mathematica do the grunt work and tried to see some patterns as x increases. But so far I am only seeing the obvious first couple of terms: a^x * f(0) - (x-1) * a^(x-1) * b * f(0)

Any clues? Should I even be doing the problem this way, or is there some simpler method?

 

[moe_3_moe]

just a question...the function is x and the variable is n?

 

[randommacuser]

Well, that was the way the problem was stated. I've edited the OP to make it more standard. Now f is a function of the variable x, as usual.

 

[moe_3_moe]

hey man did u try to get f(1) and then replace it in f(2) and so on u will find a relation between f(2) with just f(0) and f(-1) in it with some factorize and like that ... try it

 

[Data]

Try substituting $f(x) = m^x$ for some m, and see where that gets you (this is a standard method for solving homogeneous constant coefficient linear recurrences. It's analogous to solving homogeneous constant coefficient linear ODEs using an exponential substitution).

 

[randommacuser]

moe_3_moe: Yes, I did this by hand to about f(5) and on Mathematica to f(10). But the goal is one expression of f(x), for all x, in terms of the four given parameters.

 

[moe_3_moe]

ok get it ... let's see what we can do ... at least now we know how mathematica works hard lol ...

 

[moe_3_moe]

data...can we have more details about ur suggestion

 

[Data]

Using that substitution you can find the values for m that give you solutions. There will be two of them. The general solution is a linear combination of the two; use the given values for f(0) and f(-1) to find the particular solution.

 

[moe_3_moe]

right thanks ..it did work ...

 

[randommacuser]

Perfect. Thanks for your help, Data.