Syntax error which I cannot find ( find the minimum of a function using Quadratic Approximation Method )

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Janidu
Janidu 2022-11-9
评论： Janidu 2022-11-10
Ran in:
Hey there!
Im trying to find the minimum of a function using quadratic method and I'm getting some errors which I cannot find.
And maybe there are mistakes in my code also. Hope your help to find out my mistakes.
thanks.
clc
clear
f = @(x) exp((x.^4 + x.^2 -x + sqrt(5))./5) + sinh((x.^3 + 21.*x + 9)./(21.*x + 6)) -3 ;
%QUADRATIC APPROXIMATION METHOD
a = 0;
b = 1;
e = 0.01;
x1=a; x2=0.5; x3=b;
line=x3 - x1;
figure; X = [a,b]; Y = [0,0];
plot(X,Y,'black','linewidth',4);
hold on;
grid on;
n = 0;
while (line>e)
n=n+1;
f1 = f(x1);
f2 = f(x2);
f3 = f(x3);
p = ( f1*(x2.^2-x3.^2) + f2*(x3.^2-x1.^2) + f3*(x1.^2x2.^2) ) / (2*(f1*(x2-x3)+f2*(x3-x1)+f3*(x1-x2)));
Invalid expression. Check for missing multiplication operator, missing or unbalanced delimiters, or other syntax error. To construct matrices, use brackets instead of parentheses.

Error in connector.internal.fevalMatlab

Error in connector.internal.fevalJSON
if p >= x2
x1 = x2;
x2 = p;
else
x3 = x2;
x2 = p;
end
fprintf('------%.10f\n',x1,x2,x3);
line = x3-x1;
fprintf('line=%.10f\n',line);
X = [x1,x3]; Y = Y +n;
plot(X,Y,'black','linewidth',4);
end
fprintf('The number of iterations is %d\n',n);
fprintf('Number of calculated values is %d\n',n+3);
fprintf('The point of minimum is %.4f\n',(x1+x3)/2);
fprintf('Minimum is %.4f\n',f((x1+x3)/2));

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采纳的回答

Walter Roberson
Walter Roberson 2022-11-9
p = ( f1*(x2.^2-x3.^2) + f2*(x3.^2-x1.^2) + f3*(x1.^2x2.^2) ) / (2*(f1*(x2-x3)+f2*(x3-x1)+f3*(x1-x2)));
^^^^^^^^^^
Taking something ^2x2 is not valid syntax. Looking at the other clauses you probably forgot a - and wanted x1.^2-x2.^2

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