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 A problem of a sigmoidal shape
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segovi01

Mexico
Posts

Posted - 10/06/2006 :  2:34:00 PM  Show Profile  Edit Topic  Reply with Quote  View user's IP address  Delete Topic
Origin Version (Select Help-->About Origin): 7.5pro
Operating System: windows xp

Hi All

im trying to fit data which generates a S shape graphic to a equation
of Y=(A0+B0)/(1+(A0/B0)exp(-k(A0+B0)X))
with just one parameter to be fitted, k, A0 and B0 are both constants and their value dont need to be fitted
since im not familiar with origin c i do not know how to program this in origin, could any body help me

thank you

Sincerely

Jose


Mike Buess

USA
3037 Posts

Posted - 10/07/2006 :  08:03:30 AM  Show Profile  Edit Reply  Reply with Quote  View user's IP address  Delete Reply
Hi Jose,

Learn about the fitting procedure by starting with the Advanced Fitting Tool topic here...
Help > Tutorials > Origin Tutorials > Tutorial 11

You can use Origin's built-in SLogistic3 function in the Growth/Sigmoidal category which is defined as y=a/(1+b*exp(-k*x)). Start by fixing a to A0+B0 and b to A0/B0 and divide the final value of k by A0+B0.

Mike Buess
Origin WebRing Member
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segovi01

Mexico
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Posted - 10/23/2006 :  06:41:35 AM  Show Profile  Edit Reply  Reply with Quote  View user's IP address  Delete Reply
thank you for your response, but my equation is Y=(A0+B0)/(1+(A0/B0)exp(-k(A0+B0)X))
and the
SLogistic3 function in the Growth/Sigmoidal is y=a/(1+b*exp(-k*x))
and it would be true just when A0+B0=1, i can fit using this because and need to modify in order to obtain y=a/(1+b*exp(-k*a*x))
that would be what i need, how can i do that??

thank you


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Mike Buess

USA
3037 Posts

Posted - 10/23/2006 :  07:37:02 AM  Show Profile  Edit Reply  Reply with Quote  View user's IP address  Delete Reply
quote:
...it would be true just when A0+B0=1
As I said above, just divide the value of k obtained with SLogistic3 by A0+B0 to get your k.

Mike Buess
Origin WebRing Member
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segovi01

Mexico
Posts

Posted - 10/23/2006 :  12:14:48 PM  Show Profile  Edit Reply  Reply with Quote  View user's IP address  Delete Reply
dear sir:

yes, you are right,thank you for your help and pacience

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