MATLAB: Does trapz yield infinity

trapz

Hi there…

Why does the following code yield infinity

x      = 0:0.001:1;
ytDiff = 1./(2*(x).^(1/2));
ArcEq  = sqrt(1+ytDiff.^2);
s      = trapz(x,ArcEq)

when the actual value should be s = 1.4789!?

Thanks!

Because that function approaches infinity at x=0, and you have included that endpoint of the interval explicitly. Try this instead:

x      = 0.001:0.001:1;
ytDiff = 1./(2*(x).^(1/2));
ArcEq  = sqrt(1+ytDiff.^2);
s      = trapz(x,ArcEq)

A very good question. You need to use QUADV here instead of QUAD.

quadv(@(x) exp(x.^2).*quad(@(z) exp(z.^2),-2,x),-2,2)

QUAD expects vector inputs and outputs, and so when it goes to evaluate the inner QUAD, it sends in a vector which is why you get that error.

Notice that QUAD will error on a problem like this one here below for the same reason (but QUADV will work fine):

quad(@(x) det([x 1; 1 1]),0,1)

QUADV does it without trying to send in vectors.

Just as a sanity check, compare what you get from QUADV with what you get if you did it using TRAPZ

f = @(x) exp(x.^2).*quad(@(z) exp(z.^2),-2,x);
trapz(arrayfun(f,-2:.001:2)) * 0.001
quadv(f,-2,2)

Both answers are very similar, about 541.38.

syms x(t) y(t)  
eq1 = x == y+2;  
deq = diff(eq1)
deq(t) = 
isolate(deq, diff(y))
ans = 
subs(ans, diff(x),5)
ans =