Anonymous functions and integration

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?

[Cedric]

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.

Answer 1

[Kye Taylor]

Try

 r = integral (@(x)f(x)'*f(x), 0, 1, 'ArrayValued', true);

[Jim]

This works, but does it evaluate f(x) twice?

[Kye Taylor]

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);

Answer 2

[Mike Hosea]

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

[Jim]

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]

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.