numerical integration with nonarray function
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Hello, guys,
I have a trouble in using numerical integration command. The integrand in my case is det(x*A),x is the variable, and A is a n*n matrix. I noticed that nearly all the numerical integration command has requirement of array function, which means, they have to use .*, ./ when needed? do we have numerical integration command without such requirement?
Thank you very much!
Clair
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Réponse acceptée
Mike Hosea
le 12 Avr 2013
It is, indeed, unnecessary to perform numerical integration on this integrand. However, to answer the question in general, The INTEGRAL function has an option called 'ArrayValued' that allows you to integrate array-valued functions. When this option is set to true, the integrator will only call the integrand function with scalar inputs, and it doesn't matter if the "array value" only has one element. So, this does it
integral(@(x)det(A*x),a,b,'ArrayValued',true)
For example
>> rng(0)
>> A = round(100*rand(3))
A =
81 91 28
91 63 55
13 10 96
>> integral(@(x)det(A*x),0,1,'ArrayValued',true)
ans =
-7.050624999999999e+04
>> det(A)/4
ans =
-7.050624999999999e+04
2 commentaires
Mike Hosea
le 15 Avr 2013
Modifié(e) : Mike Hosea
le 15 Avr 2013
ARRAYFUN is usually a faster way for scalar-valued problems, and since INTEGRAL2 and INTEGRAL3 do not support an 'ArrayValued' option, you will have to do something like that (or write a wrapper function with a loop). Here's how to use ARRAYFUN. If f(x,y) is a bivariate integrand but that only works with scalar inputs, integrate
g = @(x,y)arrayfun(f,x,y)
Try this technique with the univariate function as well if speed is an issue, i.e. integrate
g = @(x)arrayfun(f,x);
Here I am assuming in both cases that f is defined as an anonymous function. If f is defined in a MATLAB program file, f.m, then of course you need @f instead of just f as the first argument to ARRAYFUN.
Plus de réponses (2)
Yao Li
le 12 Avr 2013
Assuming
x=[x0,x1,x2];
B=[det(x0*A) det(x1*A) det(x2*A)];
trapz(x,B)
0 commentaires
Roger Stafford
le 12 Avr 2013
With x used as a vector, your integrand can be written as:
(x.^n)*det(A)
with the x factored out, which should integrate very nicely. However, why bother to do numerical integration when the indefinite integral is already known from elementary calculus, namely
x^(n+!)/(n+1)*det(A)
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