How to find the value from equations?
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Chuanjung Lin
le 5 Août 2022
Commenté : Walter Roberson
le 8 Août 2022
Hi all,
I have a dataset (Ids) and equestion as the attachment.
Ids = [6.35E-06
6.18E-06
6.01E-06
5.84E-06
5.68E-06
5.52E-06
5.36E-06
5.20E-06
5.05E-06
4.89E-06
4.75E-06
4.60E-06
4.45E-06
4.31E-06
4.17E-06
4.03E-06
3.90E-06
3.76E-06
3.63E-06
3.50E-06
3.38E-06
3.25E-06
3.13E-06
3.01E-06
2.90E-06
2.78E-06
2.67E-06
2.56E-06
2.45E-06
2.35E-06
2.25E-06
2.15E-06
2.05E-06
1.95E-06
1.86E-06
1.77E-06
1.68E-06
1.60E-06
1.51E-06
1.43E-06
1.35E-06
1.28E-06
1.20E-06
1.13E-06
1.06E-06
9.95E-07
9.30E-07
8.68E-07
8.08E-07
7.50E-07
6.95E-07
6.41E-07
5.90E-07
5.41E-07
4.95E-07
4.51E-07
4.09E-07
3.69E-07
3.31E-07
2.95E-07
2.62E-07
2.31E-07
2.02E-07
1.75E-07
1.50E-07
1.27E-07
1.06E-07
8.77E-08
7.09E-08
5.60E-08
4.31E-08
3.20E-08
2.28E-08
1.52E-08
9.39E-09
5.11E-09
2.28E-09
7.14E-10
1.12E-10
6.28E-12
2.10E-13
6.54E-15
3.27E-16
-2.03E-16
-2.06E-16
4.78E-16
3.29E-16
-3.03E-16
4.10E-16];
where, the Wg = 6; Lg = 2.3;
may I know how to obtain the Ids'?
Here, I set Ids' = a , and code as following:
Wg = 6;
Lg = 2.3;
syms a
eqn = (1*10^(-5)).*(Wg./Lg)* log10( 1 + exp(a./5)) ./ ( 1 + exp(a./5))==Ids
S = solve(eqn);
S
when I click the run, it seems no response. Does this approach is correct?
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Réponse acceptée
Walter Roberson
le 5 Août 2022
eqn = (1*10^(-5)).*(Wg./Lg)* log10( 1 + exp(a./5)) ./ ( 1 + exp(a./5))==Ids
IDS is a vector, so eqn is a vector. But syms a is a scalar symbolic variable.
S = solve(eqn);
You are asking to solve the vector of equations for the single scalar value for a that solves all of the equations simultaneously. That is not going to exist.
Are you looking for "best fit" or are you looking to solve each of the equations independently ?
3 commentaires
Walter Roberson
le 8 Août 2022
Are you looking for "best fit" or are you looking to solve each of the equations independently ? The code I posted shows the best fit in a least-squares sense. If you want to solve each individually then loop or use arrayfun.
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