NAG Toolbox: nag_roots_contfn_cntin_rcomm (c05ax)

Purpose

Syntax

Description

References

Parameters

Compulsory Input Parameters

1:     $\mathrm{x}$ – double scalar

On initial entry: an initial approximation to the zero.

2:     $\mathrm{fx}$ – double scalar

On initial entry: if $\mathbf{ind}=1$, fx need not be set.

If $\mathbf{ind}=-1$, fx must contain $f\left(\mathbf{x}\right)$ for the initial value of x.

On intermediate re-entry: must contain $f\left(\mathbf{x}\right)$ for the current value of x.

3:     $\mathrm{tol}$ – double scalar

On initial entry: a value that controls the accuracy to which the zero is determined. tol is used in determining the convergence of the secant iteration used at each stage of the continuation process. It is used directly when solving the last problem (${\theta }_m=0$ in Description), and $\sqrt{\mathbf{tol}}$ is used for the problem defined by ${\theta }_r$, $r<m$. Convergence to the accuracy specified by tol is not guaranteed, and so you are recommended to find the zero using at least two values for tol to check the accuracy obtained.

Constraint: $\mathbf{tol}>0.0$.

4:     $\mathrm{ir}$ – int64int32nag_int scalar

On initial entry: indicates the type of error test required, as follows. Solving the problem defined by ${\theta }_r$, $1\le r\le m$, involves computing a sequence of secant iterates $x_r^0,x_r^1,\dots$. This sequence will be considered to have converged only if:

for $\mathbf{ir}=0$,

$$\left|x_r^{\left(i+1\right)}-x_r^{\left(i\right)}\right|\le \mathit{eps}\times \max \left(1.0,\left|x_r^{\left(i\right)}\right|\right)\text{,}$$

for $\mathbf{ir}=1$,

$$\left|x_r^{\left(i+1\right)}-x_r^{\left(i\right)}\right|\le \mathit{eps}\text{,}$$

for $\mathbf{ir}=2$,

$$\left|x_r^{\left(i+1\right)}-x_r^{\left(i\right)}\right|\le \mathit{eps}\times \left|x_r^{\left(i\right)}\right|\text{,}$$

for some $i>1$; here $\mathit{eps}$ is either tol or $\sqrt{\mathbf{tol}}$ as discussed above. Note that there are other subsidiary conditions (not given here) which must also be satisfied before the secant iteration is considered to have converged.

Constraint: $\mathbf{ir}=0$, $1$ or $2$.

5:     $\mathrm{c}\left(26\right)$ – double array

($\mathbf{c}\left(5\right)$ contains the current ${\theta }_r$, this value may be useful in the event of an error exit.)

6:     $\mathrm{ind}$ – int64int32nag_int scalar

On initial entry: must be set to $1$ or $-1$.

$\mathbf{ind}=1$

fx need not be set.

$\mathbf{ind}=-1$

fx must contain $f\left(\mathbf{x}\right)$.

Constraint: on entry $\mathbf{ind}=-1$, $1$, $2$, $3$ or $4$.

Optional Input Parameters

1:     $\mathrm{scal}$ – double scalar

Suggested value: $\sqrt{\epsilon }$, where $\epsilon$ is the machine precision returned by nag_machine_precision (x02aj).

Default: $\sqrt{\mathit{machine\; precision}}$

On initial entry: a factor for use in determining a significant approximation to the derivative of $f\left(x\right)$ at $x=x_0$, the initial value. A number of difference approximations to $f^{\prime }\left(x_0\right)$ are calculated using

$$f^{\prime }\left(x_0\right)\sim \left(f\left(x_0+h\right)-f\left(x_0\right)\right)/h$$

where $\left|h\right|<\left|\mathbf{scal}\right|$ and $h$ has the same sign as scal. A significance (cancellation) check is made on each difference approximation and the approximation is rejected if insignificant.

Constraint: $\mathbf{scal}$ must be sufficiently large that $\mathbf{x}+\mathbf{scal}\ne \mathbf{x}$ on the computer.

Output Parameters

1:     $\mathrm{x}$ – double scalar

On intermediate exit: the point at which $f$ must be evaluated before re-entry to the function.

On final exit: the final approximation to the zero.

2:     $\mathrm{c}\left(26\right)$ – double array

3:     $\mathrm{ind}$ – int64int32nag_int scalar

On intermediate exit: contains $2$, $3$ or $4$. The calling program must evaluate $f$ at x, storing the result in fx, and re-enter nag_roots_contfn_cntin_rcomm (c05ax) with all other arguments unchanged.

On final exit: contains $0$.

4:     $\mathrm{ifail}$ – int64int32nag_int scalar

On final exit: $\mathbf{ifail}=\mathbf{0}$ unless the function detects an error (see Error Indicators and Warnings).

Error Indicators and Warnings

   $\mathbf{ifail}=1$

On entry,	$\mathbf{tol}\le 0.0$,
or	$\mathbf{ir}\ne 0$, $1$ or $2$.

   $\mathbf{ifail}=2$

The argument ind is incorrectly set on initial or intermediate entry.

   $\mathbf{ifail}=3$

scal is too small, or significant derivatives of $f$ cannot be computed (this can happen when $f$ is almost constant and nonzero, for any value of scal).

   $\mathbf{ifail}=4$

The current problem in the continuation sequence cannot be solved, see $\mathbf{c}\left(5\right)$ for the value of ${\theta }_r$. The most likely explanation is that the current problem has no solution, either because the original problem had no solution or because the continuation path passes through a set of insoluble problems. This latter reason for failure should occur rarely, and not at all if the initial approximation to the zero is sufficiently close. Other possible explanations are that tol is too small and hence the accuracy requirement is too stringent, or that tol is too large and the initial approximation too poor, leading to successively worse intermediate solutions.

   $\mathbf{ifail}=5$

Continuation away from the initial point is not possible. This error exit will usually occur if the problem has not been properly posed or the error requirement is extremely stringent.

   $\mathbf{ifail}=6$

The final problem (with ${\theta }_m=0$) cannot be solved. It is likely that too much accuracy has been requested, or that the zero is at $\alpha =0$ and $\mathbf{ir}=2$.

   $\mathbf{ifail}=-99$

An unexpected error has been triggered by this routine. Please contact NAG.

   $\mathbf{ifail}=-399$

Your licence key may have expired or may not have been installed correctly.

   $\mathbf{ifail}=-999$

Dynamic memory allocation failed.

Accuracy

Further Comments

Example

Open in the MATLAB editor: c05ax_example

function c05ax_example


fprintf('c05ax example results\n\n');

fx = 0;
c = zeros(26, 1);
for k=3:4
  x = 1;
  tol = 10^-k;
  ir = int64(0);
  ind = int64(1);

  while (ind ~= 0)
    [x, c, ind, ifail] = c05ax(x, fx, tol, ir, c, ind);
    fx = x - exp(-x);
  end

  if ifail == 4 || ifail ==6
    fprintf('FTol = %11.4e, final value = %11.4e, theta = %10.2e\n', tol, ...
             x, c(5));
  elseif ifail == 0
    fprintf('Tol is %11.4e, Root is %11.4e\n', tol, x);
  end
end

c05ax example results

Tol is  1.0000e-03, Root is  5.6715e-01
Tol is  1.0000e-04, Root is  5.6715e-01