y = f (x)
where y and x are scalars.
The second and third arguments are limits of integration. Either or both may be infinite.
The optional argument tol is a vector that specifies the desired accuracy of the result. The first element of the vector is the desired absolute tolerance, and the second element is the desired relative tolerance. To choose a relative test only, set the absolute tolerance to zero. To choose an absolute test only, set the relative tolerance to zero.
The optional argument sing is a vector of values at which the integrand is known to be singular.
The result of the integration is returned in v and ier contains an integer error code (0 indicates a successful integration). The value of nfun indicates how many function evaluations were required, and err contains an estimate of the error in the solution.
quad
. Given one argument,
quad_options
returns the value of the corresponding option. If
no arguments are supplied, the names of all the available options and
their current values are displayed.
Here is an example of using quad
to integrate the function
This is a fairly difficult integration (plot the function over the range of integration to see why).
The first step is to define the function:
function y = f (x) y = x .* sin (1 ./ x) .* sqrt (abs (1 - x)); endfunction
Note the use of the `dot' forms of the operators. This is not necessary
for the call to quad
, but it makes it much easier to generate a
set of points for plotting (because it makes it possible to call the
function with a vector argument to produce a vector result).
Then we simply call quad:
[v, ier, nfun, err] = quad ("f", 0, 3) => 1.9819 => 1 => 5061 => 1.1522e-07
Although quad
returns a nonzero value for ier, the result
is reasonably accurate (to see why, examine what happens to the result
if you move the lower bound to 0.1, then 0.01, then 0.001, etc.).
Here is an example of using colloc
to generate weight matrices
for solving the second order differential equation
First, we can generate the weight matrices for n points (including the endpoints of the interval), and incorporate the boundary conditions in the right hand side (for a specific value of
n = 7; alpha = 0.1; [r, a, b] = colloc (n-2, "left", "right"); at = a(2:n-1,2:n-1); bt = b(2:n-1,2:n-1); rhs = alpha * b(2:n-1,n) - a(2:n-1,n);
Then the solution at the roots r is
u = [ 0; (at - alpha * bt) \ rhs; 1] => [ 0.00; 0.004; 0.01 0.00; 0.12; 0.62; 1.00 ]
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