/* D02BJ_P0W_F C++ Header Example Program.
*
* Copyright 2021 Numerical Algorithms Group.
*
* Mark 27.2, 2021.
*
*/
#include <iostream>
#include <math.h>
#include <nag.h>
#include <nagad.h>
#include <stdio.h>
using namespace std;
#ifdef __cplusplus
extern "C"
{
#endif
static void NAG_CALL fcn(void *& ad_handle,
const double &x,
const double y[],
double f[],
Integer iuser[],
double ruser[]);
static void NAG_CALL g(void *& ad_handle,
const double &x,
const double y[],
double & retval,
Integer iuser[],
double ruser[]);
#ifdef __cplusplus
}
#endif
int main(void)
{
int exit_status = 0;
const Integer n = 3;
cout << "D02BJ_P0W_F C++ Header Example Program Results\n\n";
Integer iw, iuser[1];
double pi, tol, x, xinit, xend, ruser[2];
double *y = 0, *w = 0;
iw = 20 * n;
w = new double[iw];
y = new double[n];
xinit = 0.0;
xend = 10.0;
pi = nag_math_pi;
cout << "\n\nCase: no intermediate output, root-finding\n";
tol = 1.0e-5;
cout << "\n Calculation with tol = " << tol << endl;
x = xinit;
y[0] = 0.5;
y[1] = 0.5;
y[2] = pi / 5.0;
const double alpha = -0.032;
const double beta = -0.02;
ruser[0] = alpha;
ruser[1] = beta;
Integer ifail = 0;
void * ad_handle = 0;
nag::ad::d02bj(ad_handle, x, xend, n, y, fcn, tol, "D", nullptr, g, w,
-1, iuser, -1, ruser, ifail);
cout.setf(ios::fixed);
cout.setf(ios::right);
cout.precision(3);
cout << "\n Root of Y(1) = 0.0 at ";
cout.width(5);
cout << x << endl;
cout << "\n Solution is ";
cout.precision(4);
for (int i = 0; i < 3; ++i)
{
cout.width(10);
cout << y[i];
}
cout << endl;
delete[] w;
delete[] y;
return exit_status;
}
static void NAG_CALL fcn(void *& ad_handle,
const double &x,
const double y[],
double f[],
Integer iuser[],
double ruser[])
{
double alpha, beta;
alpha = ruser[0];
beta = ruser[1];
f[0] = tan(y[2]);
f[1] = alpha * tan(y[2]) / y[1] + beta * y[1] / cos(y[2]);
f[2] = alpha / (y[1] * y[1]);
}
static void NAG_CALL g(void *& ad_handle,
const double &x,
const double y[],
double & retval,
Integer iuser[],
double ruser[])
{
retval = y[0];
}