| Author |
Comment/Response |
sk597653
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 10/23/12 10:40am
In Response To 'Re: Re: Re: Re: Re: Plotting non-linear values ...'
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Here we are, still not solved but a bit of progress made with the help of Mathematica.
The equation was not correctly cited. The paper I pulled it from placed the square root over the incorrect values. The actual equation should be:
y2 = (b - (Sqrt[b^2 - 2 k^2 c x/s]))/(2 k )
With that correction, now we are able to solve for k and s!!!
For the bad news- it does not calculate k and s to what they are known to be with a set of data that has been previously used to determine k as 2900000 and s= 0.84 with the use of SigmaPlot.
In Mathematica, the same data:
data = {{0.0, 0.0}, {0.0000079333, 0.4176}, {0.0000119,
0.5385}, {0.000015867, 0.6374}, {0.000019833, 0.7473}, {0.0000238,
0.8571}, {0.000027767, 0.9231}, {0.000031733, 0.9560}, {0.0000357,
0.9780}, {0.000039667, 0.9780}, {0.000043633, 0.9890}, {0.0000476,
1.0220}, {0.000051567, 0.9890}, {0.000055533,
1.0110}, {0.000063467, 1.0000}, {0.000075367,
1.0000}, {0.000087267, 1.0110}, {0.00010313, 1.0000}}
Using the FindFit function:
fit = FindFit[data, y2, {k, s}, x]
{k -> 1.57811, s -> 0.421896}
Why?
How does Sigmaplot get 2900000 and 0.84?
This was also done:
fit1 = y2 /. fit
2
k x -0.0000264 k x k x 2
1 + 0.0000132 k + --- - Sqrt[--------------- + (1 + 0.0000132 k + ---) ]
2 s s 2 s
------------------------------------------------------------------------ /. fit
2 k
Plot[fit1, {x, 0, 0.000098}]
which produced a graph with the straight line and the y-axis scaling from 10^-10 -> 10^-9 while the x-axis scaled 0.00001 -> 0.0001.
I did try a simple fit, such as the one you described and ws able to solve for it. My advisor suggested the same thing!
Thank you again, I look forward to hearing what you make of this.
Attachment: FindFitSOLVED shannons.nb, URL: , |
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