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 Error of the inverse of Tau after curve fitting		  New Topic  Reply to Topic
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tony_lincoln

USA
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Posted - 10/04/2006 :  12:07:20 PM  Show Profile  Edit Topic  Reply with Quote  View user's IP address  Delete Topic
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

larry_lan

China
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Posted - 10/08/2006 :  02:42:04 AM  Show Profile  Edit Reply  Reply with Quote  View user's IP address  Delete Reply
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
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Mike Buess

USA
3037 Posts
Posted - 10/08/2006 :  07:14:17 AM  Show Profile  Edit Reply  Reply with Quote  View user's IP address  Delete Reply
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
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larry_lan

China
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Posted - 10/08/2006 :  07:49:13 AM  Show Profile  Edit Reply  Reply with Quote  View user's IP address  Delete Reply
Hi:

Sorry Tony, Mike is right, I made a mistake on the above post.

e(k) = error3 / Tau^2;

Larry
OriginLab Technical Services
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