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Anonymous functions and integration

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Jim
Jim on 22 Apr 2013
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 Comment

Cedric Wannaz
Cedric Wannaz on 22 Apr 2013
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.

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Accepted Answer

Kye Taylor
Kye Taylor on 22 Apr 2013
Try
r = integral (@(x)f(x)'*f(x), 0, 1, 'ArrayValued', true);

  2 Comments

Jim
Jim on 22 Apr 2013
This works, but does it evaluate f(x) twice?
Kye Taylor
Kye Taylor on 22 Apr 2013
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);

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More Answers (1)

Mike Hosea
Mike Hosea on 22 Apr 2013
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 Comments

Jim
Jim on 22 Apr 2013
That's an elegant solution. I wonder if there is any difference in performance from Kye's answer. I did a basic timing test of the two approaches and there isn't any meaningful difference in computation times.
Mike Hosea
Mike Hosea on 22 Apr 2013
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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