On 05/02/2013 16:21, 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" ;-)

He is right though you have it for N=0 F(0), F(1) = 0.97*F(0)
and then the recurrence relation gives you F(N+1) and if paranoid
you can check it out formally by substituting your guess into it
explicitly

F(N) = k*(0.97^N)
F(N+1) = 0.97*k*(0.97^N) == k*(0.97^(N+1)) QED

which as Phil says is proof by induction.

This one is just a dull harmonic series.

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

...Jim Thompson

Not necessarily - the finite difference equation is always a discrete
approximation to the true full differential equation.

How well the finite difference equation reflects reality and what other
approximations you have made along the way affects whether or not you
can get back to the continuous version unambiguously.

Numerical approximations to derivatives are notoriously fickle.

--
Regards,
Martin Brown