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Integral of matrix determinant

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I found that it seems the
integral
fucntion cannot integrate determinant. For example
f=@(x)det([1,x;0,2]);
integral(f,0,1)
does not return a proper result. Of course we can mannually take
f=@(x)(2*1-0*x);
but this can be hardly applied to larger matrices. Is there a solution to this problem?

Accepted Answer

Torsten
Torsten on 6 Mar 2023
f=@(x)det([1,x;0,2]);
integral(f,0,1,'ArrayValued',true)
ans = 2.0000
  2 Comments
Zhenghao Yang
Zhenghao Yang on 6 Mar 2023
I see. What about 2 variables? Seems not directly applicable to integral2.
Torsten
Torsten on 6 Mar 2023
Edited: Torsten on 6 Mar 2023
1d:
f = @(x) det([1,x;0,2]);
value = integral(@(X)arrayfun(@(x)f(x),X),0,1)
value = 2.0000
2d:
f = @(x,y)det([x 2*y^2;0 3*sin(x*y)]);
value = integral2(@(X,Y)arrayfun(@(x,y)f(x,y),X,Y),0,1,0,1)
value = 0.4756
3d:
f = @(x,y,z) det([x 2*y^2 z;0 3*cos(x*z) log(z+1);4*exp(y+z) cos(z) 1/(x+1)]);
value = integral3(@(X,Y,Z)arrayfun(@(x,y,z)f(x,y,z),X,Y,Z),0,1,0,1,0,1)
value = -13.6865

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More Answers (2)

Askic V
Askic V on 6 Mar 2023
I would do it symbolically:
syms x
f = det([1,x;0,2]);
Q = int(f,0,1)
Q = 
2
  2 Comments
Zhenghao Yang
Zhenghao Yang on 6 Mar 2023
Will this method run quick enough when the matrix is considerably large?
Askic V
Askic V on 6 Mar 2023
For sure this is slower approach, but you need to test it and see if this would be sufficient for your particular application.

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John D'Errico
John D'Errico on 6 Mar 2023
For this specific problem, I might just suggest that the determinant of an upper triangular matrix is just the product of the diagonal elements.
syms x
A = [1 x;0 2]
A = 
As such, the use of det or integral is wild overkill here, since the determinant is independent of the value of x.
det(A)
ans = 
2
Why does integral fail? Because it tries to pass in a list of points to evaluate the function at. And det is not vectorized. So this is not a problem of integral in reality, but of the interaction between integral and det. Integral insists on the function being vectorized. Det insists is it NOT vectorized. Failure!
The solution is to evaluate the determinant as a polynomial in symbolic form, and then integrate, or do as @Torsten has shown, to use the 'arrayvalued' option in integral.
help integral
INTEGRAL Numerically evaluate integral. Q = INTEGRAL(FUN,A,B) approximates the integral of function FUN from A to B using global adaptive quadrature and default error tolerances. FUN must be a function handle. A and B can be -Inf or Inf. If both are finite, they can be complex. If at least one is complex, INTEGRAL approximates the path integral from A to B over a straight line path. For scalar-valued problems the function Y = FUN(X) must accept a vector argument X and return a vector result Y, the integrand function evaluated at each element of X. For array-valued problems (see the 'ArrayValued' option below) FUN must accept a scalar and return an array of values. Q = INTEGRAL(FUN,A,B,PARAM1,VAL1,PARAM2,VAL2,...) performs the integration with specified values of optional parameters. The available parameters are 'AbsTol', absolute error tolerance 'RelTol', relative error tolerance INTEGRAL attempts to satisfy |Q - I| <= max(AbsTol,RelTol*|Q|), where I denotes the exact value of the integral. Usually RelTol determines the accuracy of the integration. However, if |Q| is sufficiently small, AbsTol determines the accuracy of the integration, instead. The default value of AbsTol is 1.e-10, and the default value of RelTol is 1.e-6. Single precision integrations may require larger tolerances. 'ArrayValued', FUN is an array-valued function when the input is scalar When 'ArrayValued' is true, FUN is only called with scalar X, and if FUN returns an array, INTEGRAL computes a corresponding array of outputs Q. The default value is false. 'Waypoints', vector of integration waypoints If FUN(X) has discontinuities in the interval of integration, the locations should be supplied as a 'Waypoints' vector. Waypoints should not be used for singularities in FUN(X). Instead, split the interval and add the results from separate integrations with singularities at the endpoints. If A, B, or any entry of the waypoints vector is complex, the integration is performed over a sequence of straight line paths in the complex plane, from A to the first waypoint, from the first waypoint to the second, and so forth, and finally from the last waypoint to B. Examples: % Integrate f(x) = exp(-x^2)*log(x)^2 from 0 to infinity: f = @(x) exp(-x.^2).*log(x).^2 Q = integral(f,0,Inf) % To use a parameter in the integrand: f = @(x,c) 1./(x.^3-2*x-c) Q = integral(@(x)f(x,5),0,2) % Specify tolerances: Q = integral(@(x)log(x),0,1,'AbsTol',1e-6,'RelTol',1e-3) % Integrate f(z) = 1/(2z-1) in the complex plane over the % triangular path from 0 to 1+1i to 1-1i to 0: Q = integral(@(z)1./(2*z-1),0,0,'Waypoints',[1+1i,1-1i]) % Integrate the vector-valued function sin((1:5)*x) from 0 to 1: Q = integral(@(x)sin((1:5)*x),0,1,'ArrayValued',true) Class support for inputs A, B, and the output of FUN: float: double, single See also INTEGRAL2, INTEGRAL3, FUNCTION_HANDLE Documentation for integral doc integral Other uses of integral gpuArray/integral
And, to be honest, the help does not really state that this is how you can avoid the need for your objective function to be vectorized.
f=@(x) det([1 x;0 2]);
integral(f,0,1,'ArrayValued',true)
ans = 2.0000

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