How can I construct a triangle function, when I tested one using 'integral(fun.-pi,pi)', error occurred.

function [a] = triangle(n)

a=zeros(size(n));

a(n>=0 & n<=pi)=pi-n;

end

--------------------

>> fun = @(x)triangle(x)

fun =

@(x)triangle(x)

>> integral(fun,-pi, pi)

In an assignment A(I) = B, the number of elements in B and I must be the same.

Error in triangle (line 3)

a(n>=0)=pi-n;

Error in @(x)triangle(x)

Error in integralCalc/iterateScalarValued (line 314)

fx = FUN(t);

Error in integralCalc/vadapt (line 132)

[q,errbnd] = iterateScalarValued(u,tinterval,pathlen);

Error in integralCalc (line 75)

[q,errbnd] = vadapt(@AtoBInvTransform,interval);

Error in integral (line 88)

Q = integralCalc(fun,a,b,opstruct);

I used to define a step function like this and it works. Why did it not work this time?

Answer 1

[VBBV]

integral(@triangle,-pi,pi,'ArrayValued',1)  % call function using function handle

ans = 0

function [a] = triangle(n)
  a=zeros(size(n));
  if a(n>=0 & n<=pi)  % use an if else statement to execuete 
      a = pi-n;
  else
      a = 0; %n; % 0 otherwise
  end 
end

[Thamrongsin Siripongsakul]

Thank you very much. I removed 'a' from the condition. Everthing works.

if (n>=0 & n<=pi) % use an if else statement to execuete

>> integral(fun,-pi,pi,'ArrayValued',1)

ans =

4.9348

[Thamrongsin Siripongsakul]

May I ask one more question?

Later I try to calculate a vector y = triangle(x). It doesn't work. The argument x is a vector.

x = -pi:0.01:pi;

The result is

y = 0

---------------------------------------------------------------------------------------------

function [a] = triangle(n)

a=zeros(size(n));

if (n>=0 & n<=pi) % use an if else statement to execuete

a = pi-n;

else

a = 0; %n; % 0 otherwise

end

end