Use results in the definition of my functions

I would like to use my X(i)s to define my function p, but it says that they are undefined variables or functions. How can I do?

A=[46/15, 5/2, 2/3; 5/2, 8/3, 1; 2/3, 1, 2];
B=[2*exp(1); 2*exp(1); 2*exp(1)-2];

X=linsolve(A,B)

integral(@(x) f(x),0,1)

function p=p(x)
    p=X(1)*x.^2+X(2)*x+X(3);
end

function f=f(x)
    f=(exp(x)-p(x)).^2;
end

Answer 1

[Star Strider]

Try this instead —

function pv=p(X,x)
    pv=X(1)*x.^2+X(2)*x+X(3);
end

Here, ‘X’ are the parameters and ‘x’ is the variable. Also, naming the variable inside the function to be the same as the function name confuses MATLAB (and everyone who reads the code).

Beyond that, what the problem and the desired result may be are both not possible to determine, and the parameters ‘X’ are nowhere either defined or calculated, so I’ll stop here.

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[Florian Spicher]

Thanks for the help. I'm not sure to understand something though. Do you mean that I can't compute what I would like to?

[Star Strider]

It is definitely possible to run that code and get results with some changes to the functions, those being to pass the parameter vector to both functions.

Try this —

A=[46/15, 5/2, 2/3; 5/2, 8/3, 1; 2/3, 1, 2];
B=[2*exp(1); 2*exp(1); 2*exp(1)-2];
X=linsolve(A,B)

X = 3×1

0.8451 0.8711 1.0011

integral(@(x) f(X,x),0,1)

ans = 8.3588e-05

function pv=p(X,x)
    pv=X(1)*x.^2+X(2)*x+X(3);
end

function f=f(X,x)
    f=(exp(x)-p(X,x)).^2;
end

Note — I added the parameter argument ‘X’ to both functions, because ‘f’ needs it in order to pass it to ‘p’. With those changes, it works!

.

[Florian Spicher]

Oh well yeah, you meant the code I published! Then definitely. I thought you meant "the actual purpose of the program makes no sense".

Thank you again for the help!

[Star Strider]

As always, my pleasure!

Since I don’t know the actual purpose of the code, I’m in no position to criticise its application. I’m just happy that I was able to help you get it to run! (I didn’t look carefully at the code the first time, so I didn’t notice that the ‘X’ parameters actually existed, and were calculated from the regression.)

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