Generalized equation using multiple equation

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Raj Arora
Raj Arora 2023-5-1
编辑： Matt J 2023-5-2
x = [0.25 0.50 0.75 1 1.25 1.50 1.75 2 2.25 2..50 2.75 3]
y1 = [0.001524 0.00605 0.013592 0.024151 0.037728 0.054321 0.073921 0.096514 0.122102 0.150689 0.182278 0.21687]
y2 = [0.060598 0.14902 0.28793 0.42474 0.57648 0.78663 1.0389 1.6032 2.3667 3.1328 3.8971 4.6297]
y3 = [0.016373 0.0503 0.1896 0.60113 1.2101 1.8134 2.9318 4.0203 4.8728 6.1467 8.1357 10.277]
y4 = [0.11668 0.33853 0.66617 1.2037 1.6292 2.4379 3.6119 4.8274 6.0769 6.4846 8.064 9.6733]
y5 = [0.131518 0.418614 0.793038 1.33235 1.94051 2.54087 4.31947 5.25463 6.33347 7.82779 9.91558 12.4864]
Could someone provide guidance on how to derive a single equation that applies to all five curves, each of which includes a dimensionless parameter "z"? The goal is to have a generalized equation that can be used to obtain the corresponding values for all five cases by simply plugging in different values of "z" (e.g., 0, 0.5, 1, 2, and 2.5).
[Hint: When Z = 0 I will get black curve; When Z = 0.5 I will get blue curve; When Z = 1 I will get red curve; When Z = 2 I will get green curve; When Z = 2.5 I will get magenta curve] (I have also attached an image which also contains separate equations for all five curves in it)

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Matt J
Matt J 2023-5-1
编辑：Matt J 2023-5-1
Ran in:
You can use lsqcurvefit with model function,
xdata = [0.25 0.50 0.75 1 1.25 1.50 1.75 2 2.25 2.50 2.75 3];
y1 = [0.001524 0.00605 0.013592 0.024151 0.037728 0.054321 0.073921 0.096514 0.122102 0.150689 0.182278 0.21687];
y2 = [0.060598 0.14902 0.28793 0.42474 0.57648 0.78663 1.0389 1.6032 2.3667 3.1328 3.8971 4.6297];
y3 = [0.016373 0.0503 0.1896 0.60113 1.2101 1.8134 2.9318 4.0203 4.8728 6.1467 8.1357 10.277] ;
y4 = [0.11668 0.33853 0.66617 1.2037 1.6292 2.4379 3.6119 4.8274 6.0769 6.4846 8.064 9.6733] ;
y5 = [0.131518 0.418614 0.793038 1.33235 1.94051 2.54087 4.31947 5.25463 6.33347 7.82779 9.91558 12.4864];
ydata=[y1;y2;y3;y4;y5];
[x,fval]=lsqcurvefit(@F,ones(5,2),xdata,ydata,zeros(5,2))
Local minimum possible. lsqcurvefit stopped because the final change in the sum of squares relative to its initial value is less than the value of the function tolerance.
x = 5×2
0.0126 0.9669 0.1703 1.1217 0.3560 1.1323 0.6457 0.9198 0.6158 1.0116
fval = 8.4752
plot(xdata,ydata','x',xdata,F(x,xdata))
function out=F(x,xdata)
out=x(:,1).*exp( x(:,2).*xdata);
end
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Matt J
Matt J 2023-5-2
编辑：Matt J 2023-5-2
I dont have any relation of z with v and x.
If not, your problem is under-specified. There are infinitely many possible surfaces F(v,z) that coincide with your plots at the prescribed z.
Raj Arora
Raj Arora 2023-5-2
Okay Matt J thankyou for your valuable comments. I will try something then I know a way that is multiple non linear regression
I tried it like if T1, T2, T3, T4, T5 are 5 equation for all 5 cases then generalized equation will be
y = c1.T1.(z^1/4)+c2.T2.(z^1/5)+c3.T3.(z^1/6)+c4.T4.(z^1/7)+c5.T5.(z^1/8)
Through this I am getting approriate fit but not the exact result for that I have to put correct value of c1, c2 , c3, c4 and c5

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更多回答（1 个）

Matt J
Matt J 2023-5-2
编辑：Matt J 2023-5-2
Ran in:
[Hint: When Z = 0 I will get black curve; When Z = 0.5 I will get blue curve; When Z = 1 I will get red curve; When Z = 2 I will get green curve; When Z = 2.5 I will get magenta curve] (I have also attached an image which also contains separate equations for all five curves in it)
Here is one choice which fulfills this, but as I mentioned earlier, it is only one choice of infinitely many:
z=[0,0.5,1,2,2.5];
a=[0.051512, 0.32887, 0.67073, 1.1425, 1.1132];
b=[0.96062, 1.1142,1.1155,0.91651,0.99419];
pa=polyfit(z,a,4);
pb=polyfit(z,b,4);
f=@(z,v) polyval(pa,z).*exp(polyval(pb,z).*v); %joint function of z and v
%Visual check
for z0=[0,0.5,1,2,2.5]
fplot(@(v)f(z0,v)); hold on
end
hold off, xlim([0,3])

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