FindFit with conditionals
- To: mathgroup at smc.vnet.net
- Subject: [mg57307] FindFit with conditionals
- From: Edward Peschko <esp5 at pge.com>
- Date: Tue, 24 May 2005 05:12:44 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
hey all,
I'd like to do a FindInstance on a function with conditionals, something like
FindInstance
[
NIntegrate[m * (e^(-lambda) * lambda^x) / x! + n, {x,1,2}] == 1 &&
m * (e^-1 * 1^1)/1! + n == 0 &&
m * (e^-2 * 2^2)/2! + n == 0,
{ lambda, m, n}
]
What I was *hoping* that mathematica would do would be to substitute
lambda, m, and n inside the expression passed to NIntegrate, which would then
calculate the integral based on individual values, and then find an instance
which fits all three criteria. However, it doesn't seem to do this, instead
it complains that 'Integrand ... is not numerical'.
A side note - can you do interval arithmetic in FindInstance? ie: could you say:
FindInstance
[
NIntegrate[m * (e^(-lambda) * lambda^x) / x! + n, {x,1,2}] ==
Interval[.999,1.001] &&
m * (e^-1 * 1^1)/1! + n == Interval[-.001, .001] &&
m * (e^-2 * 2^2)/2! + n == Interval[-.001, .001] &&
{ lambda, m, n}
]
to get around machine precision issues?
Anyways, this would be much more helpul behavior (evaluating lambda, m, and n
in subexpressions) than what it does now. Any ideas on how to implement this?
Ed
(
ps - any ideas on how to solve something like this in other ways, aside
from FindInstance?
)
- Follow-Ups:
- Re: FindFit with conditionals
- From: Andrzej Kozlowski <andrzej@akikoz.net>
- Re: FindFit with conditionals