Math Is Fun Forum

wingardio

Guest

difference equation help.

hello,

I don't know how to solve difference equations it turns out.

Example 1:

Consider the sequence 1,2,4,8,16,32,64...

The recursive relation is F[sub]n+1[/sub] = 2F[sub]n[/sub]

If we let F[sub]n[/sub] = k[sup]n[/sup]

Then k[sup]n+1[/sup] = 2k[sup]n[/sup]

Therefore k = 2.

With difference equations you then say that F[sub]n[/sub] = A(2)[sup]n[/sub]... or something... and I dunno where to go from there. Please help.

Example 2:

Consider the sequence 1,2,2,4,8,32,256...

F[sub]n+2[/sub] = F[sub]n+1[/sub]F[sub]n[/sub]

So k[sup]n+2[/sup] = k[sup]n+1[/sup]k[sup]n[/sup]

So k[sup]2[/sup] = k

So k[sup]2[/sup] - k = 0

Roots of equation = 1, -1, and 0, right?

So then we take the roots and raise them to the power of n...times some constants A, B and C:

F[sub]n[/sub] = A(1)[sup]n[/sup] + B(-1)[sup]n[/sup] + C(0)[sup]n[/sup]

and then I don't know where to go from there...help please.

Also, for a difference equation with homogeneous and inhomogeneous parts, for example, F[sub]n+1[/sub] = F[sub]n[/sub] + 2n - 3, what do I do there?

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: difference equation help.

Hi wingardio;

Fn+1 = 2Fn

The solution to this is done using the characteristic polynomial, the solution is.

F(n) = c1*2^(n-1) where c1 is determined from initial conditions.

The second one should be linearized before you try to solve it by the method of the charcteristic equation.

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