Getting an indexing error when using functions

Question

Tyler Michael Paparella 2024-2-24

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2086403-getting-an-indexing-error-when-using-functions

回答： Walter Roberson 2024-2-24

I am trying to build a function uelem that takes the integral of another function ff. ff itself is the integral of a function f divided by another function A. I am getting the following error. It says I am trying to index something. I am guessing it has somthing to do with the way I am trying to define functions and use parameters. Can someone tell me what I am doing wrong?

Array indices must be positive integers or logical values.

Error in hw3_872>@(x)f(x) (line 133)

out = integral(@(x) f(x),0,1)/ @(x) A(x,x1,x2);

Error in integralCalc/iterateScalarValued (line 314)

fx = FUN(t);

Error in integralCalc/vadapt (line 132)

[q,errbnd] = iterateScalarValued(u,tinterval,pathlen);

Error in integralCalc (line 75)

[q,errbnd] = vadapt(@AtoBInvTransform,interval);

Error in integral (line 87)

Q = integralCalc(fun,a,b,opstruct);

Error in hw3_872>ff (line 133)

out = integral(@(x) f(x),0,1)/ @(x) A(x,x1,x2);

Error in hw3_872>@(x)ff(x,x1,x2,k,L) (line 137)

out = integral(@(x) ff(x,x1,x2,k,L),x1,x2);

Error in integralCalc/iterateScalarValued (line 314)

fx = FUN(t);

Error in integralCalc/vadapt (line 132)

[q,errbnd] = iterateScalarValued(u,tinterval,pathlen);

Error in integralCalc (line 75)

[q,errbnd] = vadapt(@AtoBInvTransform,interval);

Error in integral (line 87)

Q = integralCalc(fun,a,b,opstruct);

Error in hw3_872>uelem (line 137)

out = integral(@(x) ff(x,x1,x2,k,L),x1,x2);

Error in hw3_872 (line 32)

uexact = uexact + uelem(x1,x2,k,L);

N = 100;
L = 1;
uexact = 0;
x1 = 0;
x2 = L/N;
for e = 1:N
    uexact = uexact + uelem(x1,x2,k,L);
    x1 = x2;
    x2 = L/N*(e+1);
end
function out = A(xi,x1,x2)
    x = 0.5*(1 - xi)*x1 + 0.5*(1+xi)*x2;    
    if x < .1
        out = 2 + 0.*x;
    elseif x < .2
        out = 1.5 + 0.*x;
    elseif x < .3
        out = 1.75 + 0.*x;
    elseif x < .4
        out = 1.5 + 0.*x;
    elseif x < .5
        out = 3.75 + 0.*x;
    elseif x < .6
        out = .75 + 0.*x;
    elseif x < .7
        out = 1.25 + 0.*x;
    elseif x < .8
        out = .75 + 0.*x;
    elseif x < .9
        out = 2 + 0.*x;
    elseif x <= 1
        out = 1 + 0.*x;
    end
end
function out = ff(xi,x1,x2,k,L)
    f = k^2*cos(2*pi*k*xi/L);
    out = integral(@(x) f(x),0,1)/ @(x) A(x,x1,x2);
end
function out = uelem(x1,x2,k,L)
    out = integral(@(x) ff(x,x1,x2,k,L),x1,x2);
end

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Torsten 2024-2-24

编辑：Torsten 2024-2-24

在 MATLAB Online 中打开

It's not clear what x is in f and in A.

    f = k^2*cos(2*pi*k*xi/L);
    out = integral(@(x) f(x),0,1)/ @(x) A(x,x1,x2);   

And you cannot divide integral(@(x) f(x),0,1) by a function handle, only by the result of an evaluation of A. But what is x in A(x,x1,x2) ?

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Answer 1

Walter Roberson 2024-2-24

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2086403-getting-an-indexing-error-when-using-functions#answer_1415753

在 MATLAB Online 中打开

uexact = uexact + uelem(x1,x2,k,L);

Okay, you are invoking uelem() on a scalar x1

out = integral(@(x) ff(x,x1,x2,k,L),x1,x2);

uelem() takes the scalar x1 as the upper bound of the integral() and invokes ff() passing in a vector of values for the first parameter.

f = k^2*cos(2*pi*k*xi/L);

The vector of values is used to compute a vector of f.

out = integral(@(x) f(x),0,1)/ @(x) A(x,x1,x2);

The integral() call passes a vector of x values to the vector of scalars that is f. This fails because the vector of x values does not happen to be valid indices into the vector of f values.

... Even if the integral() just happened to work, you would not be able to divide the result of the integral by an anonymous function.