Is there a mistake in my code?

Hello everyone,

I need to plot a graph and after many trials and errors I came up with this code:

syms A B
D1= 12*A + 19.7891*B == 78.8820;
D2= 19.7891*A + 449*B == 44700;
sol=solve([D1,D2], [A,B]);
A=sol.A % A= -169.9537
B=sol.B % B= 107.0451

a=exp(A)
x=2.5:0.1:10
y=a*(x.^B);
plot(x,y)

However the graph I end up with is nothing like the graph I need. This is the graph I get:

And this is the graph I need:

I double-checked the values of all the variables and they are all correct, so I'm thinking I must have made a mistake with the syntax or the commands themselves, but I can't figure it out. I would appreciate any help you can give me.

Answer 1

[Star Strider]

I might be better to estimate the parameters directly —

x = [2.5 6 9];                                                  % Excerpt From Plot
y = [1500 600 400];                                             % Excerpt From Plot

pwrfcn = @(b,x) b(1) .* x.^b(2);                                % Objective Function

[B,fv] = fminsearch(@(b)norm(y - pwrfcn(b,x)), rand(2,1))

B = 2×1

1.0e+03 * 3.8860 -0.0010

fv = 5.3873

xv = 2.5:0.1:10;

figure
plot(x, y, 'pk')
hold on
plot(xv, pwrfcn(B, xv), '-b')
hold off
grid

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[Arsel Tanriverdi]

I'm afraid I can't understand some parts of your solution, would you care to explain what this line means?

[B,fv] = fminsearch(@(b)norm(y - pwrfcn(b,x)), rand(2,1)

[Hiro Yoshino]

[B,fv] = fminsearch(@(b)norm(y - pwrfcn(b,x)), rand(2,1))

fminsearch: rand(2,1) corresponds to the initial point.

Is there any reasons for using syms to solve the equation? It seems to be too much for what you are trying to do...

[Arsel Tanriverdi]

I think I understand fminsearch now, thank you.

As for why I used syms, it's the only way I know for solving linear equations. I would love to know if there is a better option.

[Star Strider]

A more appropriate option would be to use mldivide,\ for example —

D1= 12*A + 19.7891*B == 78.8820;

D2= 19.7891*A + 449*B == 44700;

becomes —

AB = [12  19.7891; 19.7891  449] \ [78.8820; 44700]

AB = 2×1

-169.9536 107.0451

fprintf('A = %10.4f\nB = %10.4f', AB)

A =  -169.9536
B =   107.0451

@Hiro — Thank you!

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[Arsel Tanriverdi]

That looks much easier, thank you both for your time and effort! You've been of great help!

[Star Strider]

As always, (our) pleasure!

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