On Tue, 05 Feb 2013 13:26:23 -0600, Tim Wescott
wrote:

On Tue, 05 Feb 2013 12:18:20 -0700, Jim Thompson wrote:

On Tue, 05 Feb 2013 13:16:39 -0600, Tim Wescott
wrote:

On Tue, 05 Feb 2013 09:21:34 -0700, Jim Thompson wrote:

On Tue, 05 Feb 2013 11:13:45 -0500, Phil Hobbs
wrote:

On 2/5/2013 11:05 AM, Jim Thompson wrote:
It's been too long ago, I can't remember how to reduce difference
equations.

Suppose I have...

f(N+1) = 0.97*f(N)

Now I know, off the seat of my pants (I think :-), that...

f(N) = Const*(0.97^N)

But I can't remember how to rigorously get to that conclusion.

...Jim Thompson


Mathematical induction. It's true for N=0, by construction, and if
it's true for N, it's true for N+1 by the use of your recurrence
relation.

Cheers

Phil Hobbs

"Truth" isn't "solution" ;-)

Isn't there some rigorous way to derive the "continuous" equation from
the difference equation?

I'm not sure what you mean by "continuous" equation -- a difference
equation lives in discrete-time, and has no direct relationship to a
differential equation in continuous time. (You can _make_ direct
relationships, but to do so you have to specify how sampling and
reconstruction are carried out).

I was taught in differential equations class that when you dig right
down to the bottom of things, you solve a differential equation by
guessing an answer, then proving that you were right. Finding symbolic
solutions to difference equations is the same general procedure. So I
think that's as much rigor as you're going to find in this.

Yes, there are recipes for these solutions, but all of of them
(including the use of the z transform) are just the results of the
guess-then-prove technique being carried out for a whole class of
difference equations, rather than any one specific one.

In the case of a linear, shift-invariant difference equation, the recipe
is to find the auxiliary polynomial of the difference equation, and
"posit" that the solution is the sum of A_k * d_k^N, where d_k is the
k'th root of the auxiliary polynomial and A_k is a constant that goes
with it.

In your case (assuming that f(N) is the N'th element in f, which is a
possibly infinitely long vector of values), then your auxiliary
polynomial is z - 0.97, your "posited" values of d are just d = 0.97,
and your "posited" solution is

f(N) = A * (0.97)^N

Note, too, that just like differential equations, linear difference
equations can be homogeneous or non-homogeneous (yours is homogeneous).
You can find the non-homogeneous solutions to difference equations the
same way as you do for differential equations.

And, finally, if you do this a lot with linear, shift-invariant
difference equations, it pays to learn how to use the z-transform. It
simplifies things almost as much as the Laplace transform does for
linear time-invariant differential equations, and makes all of this
folderol much easier to remember.

Yep, It's all coming back to me... guess a solution and prove it fits
:-(

The guy who taught my second term of diff eqs clearly wanted us to
remember this for all time, because he made this a mantra. Nearly every
class meeting he would put some new form of differential equation up on
the board, and he'd say "Now, how do we solve this differential
equation?" then (because we didn't all shout it out in unison) he'd
answer himself: "We guess, and prove that we're right!"

That was 30 years ago. It's stuck with me, so I guess he met his goal in
my case.

That's probably what surprised me most... start the class with, "There
are these few basic equations we can solve exactly..." :-(

...Jim Thompson
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