Does ArrayValued Affect the Efficiency of Integral()

Suppose I want to numerically integrate a function of a scalar that returns a vector

f = @(x) [1 ; sin(1000*x)];

To integrate that function I use ArrayValued = true

y = integral(f,0,2,'ArrayValued',true)

y = 2×1

2.0000 0.0014

Presumably, the integration of the second element of f requires a much smaller step size than does the first. So does integral() choose the smallest step size based on evaluations of all components of f and then apply that same step to integrate all of f, even though the first element of f doesn't need such a small step? Or does it magically integrate each compoent of f separately with different step sizes (though it's not clear to me how it could do so)?

Answer 1

[Matt J]

Based on the tests below, it appears that integral() does not develop a separate mesh for each element of f. Otherwise, I would expect the final test to have nearly double the time of the first. Also, I would expect it to be an option to specify waypoints for every element of f individually, but the documentation does not indicate such an option.

f = @(x) [1 ; sin(1000*x)];

tic;
y = integral(f,0,2,'ArrayValued',true);
toc

Elapsed time is 0.099493 seconds.

f = @(x) [1 ; 1];

tic;
y = integral(f,0,2,'ArrayValued',true);
toc

Elapsed time is 0.013101 seconds.

f = @(x) [sin(1000*x) ; sin(1000*x)];

tic;
y = integral(f,0,2,'ArrayValued',true);
toc

Elapsed time is 0.101991 seconds.

[Paul]

Thanks Matt J. The timing results are persuasive, even if I find the waypoints point less so.

Do the results below suggest it's always best to break up the vector function into indvidual scalar functions?

f1 = @(x) ones(1,numel(x));
f2 = @(x) sin(1000*x);
f  = @(x) [f1(x);f2(x)];

t1 = timeit(@() integral(f,0,2,'ArrayValued',true))

t1 = 0.0176

t2a = timeit(@() integral(f1,0,2));
t2b = timeit(@() integral(f2,0,2));
t2 = t2a + t2b

t2 = 0.0011

[Matt J]

Do the results below suggest it's always best to break up the vector function into indvidual scalar functions?

No, because you only have two functions. If you had 100 or 1000, I speculate that it wouldn't be so advantageous to split them up.

e=ones(100,1)*1000;

f1 = @(x) sin(1000*x);

f  = @(x) sin(e*x(:).');

t1 = numel(e) * timeit(@() integral(f1,0,2))

t1 = 0.1479

t2 = timeit(@() integral(f,0,2,'ArrayValued',true))

t2 = 0.0376