| T O P I C R E V I E W |
| tony_lincoln |
Posted - 10/04/2006 : 12:07:20 PM
Dear Colleagues,
I was using single exponential equation y= A0 + A1*(1-exp(tau1*x)) to do the curve fitting. I got the result like:
Tau = 0.02639 error3 = 3.41E-14
Error3 is the error for Tau. Now I need to get the inverse of the Tau: k = 1/Tau. How to get the error of k? I tried to use error3/Tau^2 as the error of k. Is this correct?
Thanks.
TOny
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| 3 L A T E S T R E P L I E S (Newest First) |
| larry_lan |
Posted - 10/08/2006 : 07:49:13 AM
Hi:
Sorry Tony, Mike is right, I made a mistake on the above post.
e(k) = error3 / Tau^2;
Larry
OriginLab Technical Services |
| Mike Buess |
Posted - 10/08/2006 : 07:14:17 AM
Hi Tony,
However, from a propagation of errors viewpoint... http://en.wikipedia.org/wiki/Propagated_error ...the error of k is indeed equal to error3/tau^2. Of course you can test that yourself merely by defining another version of your function with k rather than tau1. I find it true with the simpler ExpDec1 function.
Mike Buess
Origin WebRing Member
Edited by - Mike Buess on 10/08/2006 07:22:19 AM |
| larry_lan |
Posted - 10/08/2006 : 02:42:04 AM
Hi Tony:
The parameters' error of nonlinear regression is generated by matrix calculation, I posted a brief algorithm for calculating the error on this link:
http://www.originlab.com/forum/topic.asp?TOPIC_ID=5170
So in your case, error of k doesn't equals to the reciprocal of error of Tau.
Larry
OriginLab Technical Services |
