Anonymous functions and integration
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I want to integrate an anonymous function but be able to manipulate it first. For example
f = @(x) [x -x sin(x)];
r = integral (f'*f, 0, 1, 'ArrayValued', true);
This isn't possible. I would have to define a new function but this isn't flexible. Any alternatives to directly manipulate f?
1 个评论
Either
r = integral (@(x)f(x)'*f(x), 0, 1, 'ArrayValued', true);
or
g = @(x)f(x)'*f(x) ;
r = integral (g, 0, 1, 'ArrayValued', true);
but that's what you call defining a new function I guess.
采纳的回答
Kye Taylor
2013-4-22
Try
r = integral (@(x)f(x)'*f(x), 0, 1, 'ArrayValued', true);
2 个评论
Jim
2013-4-22
Kye Taylor
2013-4-22
编辑：Kye Taylor
2013-4-22
Well, the integral function actually evaluates the function handle many times!
But, to answer your question, each time the function handle g = @(x)f(x)'*f(x) is evaluated by the integral function, the function handle f will be evaluated twice, though this will hardly affect performance.
If you're unconvinced, you could instead define the entire outer product:
g = @(x)[x^2 -x^2 x*sin(x);-x^2 x^2 -x*sin(x);x*sin(x) -x*sin(x) sin(x)^2]
then integrate
integral (g, 0, 1, 'ArrayValued', true);
更多回答（1 个）
Mike Hosea
2013-4-22
MATLAB files can be flexible when they are combined with the use of anonymous functions. Anonymous functions can also be supplied as parameters to anonymous functions. The example given can be handled with simple nesting:
f1 = @(x)[x -x sin(x)]
f2 = @(x)x'*x;
g = @(x)f2(f1(x));
r = integral (g, 0, 1, 'ArrayValued', true);
More cleverness may be required in some cases, I guess. You can extend f2 to accept multiple inputs based on x or nest deeper, constructing what amounts to an evaluation tree to minimize redundant computations. -- Mike
2 个评论
Jim
2013-4-22
Mike Hosea
2013-4-22
编辑：Mike Hosea
2013-4-22
I expect no significant difference on a simple function like this. It should be more valuable if the real f1 is rather expensive to evaluate or if the real f2 involves many uses of the input rather than just a pair. I didn't intend my response to be so much a competing answer to Kye's as a response to your concern with his solution. I probably should have made it a comment under his answer.
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