| T O P I C R E V I E W |
| D_Mike |
Posted - 12/10/2007 : 7:03:28 PM
Origin Version (Select Help-->About Origin): 7.5
Operating System: windows Xp
Hi chaps,
I have some data that I want to fit in origin. I have already successfully done this excel but for two reasons I need to use something else (I am a bit untrusting of solver's ability to find the minimum and also there are statistical parameters outputted by it).
so, the problem is:
I have a chemical reaction where A = B --> C (= denotes in equilibrium with).
I have monitored the species of all three reactants A, B and C (these are my dependant variables) over time. I have equations derived to calculate A, B and C at any time, but there is a separate equation for each. However each equation contains three parameters, k1, k2 and k3 which are shared between the equations and must be the same for each.
So what I want to do is fit three sets of data, with three different functions but with global parameters. Is this possible in origin, and if so, how? I know it is possible to fit multiple sets of data with shared parameters and one function, but I haven't read anything about being able to use multiple functions.
Thanks |
| 5 L A T E S T R E P L I E S (Newest First) |
| D_Mike |
Posted - 12/11/2007 : 06:44:02 AM
Hi Mike,
I did.
so it actually turned out to be a very trivial problem. Thanks for your help... I am somewhat of an origin novice and feel a bit like I've jumped in at the deep end.
Cheers,
Mike |
| Mike Buess |
Posted - 12/11/2007 : 06:28:04 AM
I'm not even going to try to check the expressions but did you leave out the semicolons like you did above?
Mike Buess
Origin WebRing Member |
| D_Mike |
Posted - 12/11/2007 : 04:58:27 AM
Ok, 2nd problem... my functions all compile ok individually but when i but them all together they won't... I can't see a problem with any of these... they have a strange form because i put them together in maple and then copied and pasted from there...
A = 1.000000000*A0*((-.5*k1+.5*k2+.5*k3+.5*((k1+k2+k3)^2-4*k1*k3)^.5)*exp(-(.5*k1+.5*k2+.5*k3-.5*((k1+k2+k3)^2-4*k1*k3)^.5)*t)-(-.5*k1+.5*k2+.5*k3-.5*((k1+k2+k3)^2-4*k1*k3)^.5)*exp(-(.5*k1+.5*k2+.5*k3+.5*((k1+k2+k3)^2-4*k1*k3)^.5)*t))/((k1+k2+k3)^2-4*k1*k3)^.5
B = -1.000000000*A0*k1*(exp(-(.5*k1+.5*k2+.5*k3-.5*((k1+k2+k3)^2-4*k1*k3)^.5)*t)-exp(-(.5*k1+.5*k2+.5*k3+.5*((k1+k2+k3)^2-4*k1*k3)^.5)*t))/((k1+k2+k3)^2-4*k1*k3)^.5
C = A0*(1-1.000000000*(.5*k1+.5*k2+.5*k3+.5*((k1+k2+k3)^2-4*k1*k3)^.5)*exp(-(.5*k1+.5*k2+.5*k3-.5*((k1+k2+k3)^2-4*k1*k3)^.5)*t)/((k1+k2+k3)^2-4*k1*k3)^.5+1.000000000*(.5*k1+.5*k2+.5*k3-.5*((k1+k2+k3)^2-4*k1*k3)^.5)*exp(-(.5*k1+.5*k2+.5*k3+.5*((k1+k2+k3)^2-4*k1*k3)^.5)*t)/((k1+k2+k3)^2-4*k1*k3)^.5)
admittedly they are hard to check... can anyone see anything glaringly wrong wit them?
Edited by - D_Mike on 12/11/2007 05:00:53 AM |
| D_Mike |
Posted - 12/11/2007 : 03:46:25 AM
Thanks, Mike! This occured to me overnight... It was rather obvious. thanks for the help though. if only my equations were that simple! |
| Mike Buess |
Posted - 12/10/2007 : 11:17:19 PM
A fitting function can have multiple dependent variables so I'd try something like this...
Mike Buess
Origin WebRing Member |
