On 2/5/2013 1:49 PM, Jim Thompson wrote:
On Tue, 05 Feb 2013 13:41:05 -0500, bitrex
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
You've received good answers so far, but another way to do it is through
generating functions. The differential equation for the "exponential
generating function" of this recurrence can be written by inspection;
it's y' = 0.97*y.
Assuming F(0) = 1, the solution to this differential equation is y =
e^(0.97*x). The solution to your recurrence is then the coefficient of
x^k/k! in the Taylor series expansion of e^(0.97*x), or 0.97^k.
Aha! Thanks, bitrex, you rang the right chime :-}
I'm slow this morning, that should have been obvious to me :-(
...Jim Thompson
You can also "solve" it just by plugging in a trial solution r^n into
the equation - any linear recurrence with constant coefficients can be
solved this way. See:
http://en.wikipedia.org/wiki/Recurre...eneral_methods