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NAG Toolbox: nag_roots_contfn_cntin_rcomm (c05ax)
Purpose
nag_roots_contfn_cntin_rcomm (c05ax) attempts to locate a zero of a continuous function using a continuation method based on a secant iteration. It uses reverse communication for evaluating the function.
Syntax
[
x,
c,
ind,
ifail] = c05ax(
x,
fx,
tol,
ir,
c,
ind, 'scal',
scal)
[
x,
c,
ind,
ifail] = nag_roots_contfn_cntin_rcomm(
x,
fx,
tol,
ir,
c,
ind, 'scal',
scal)
Description
nag_roots_contfn_cntin_rcomm (c05ax) uses a modified version of an algorithm given in
Swift and Lindfield (1978) to compute a zero
$\alpha $ of a continuous function
$f\left(x\right)$. The algorithm used is based on a continuation method in which a sequence of problems
are solved, where
$1={\theta}_{0}>{\theta}_{1}>\cdots >{\theta}_{m}=0$ (the value of
$m$ is determined as the algorithm proceeds) and where
${x}_{0}$ is your initial estimate for the zero of
$f\left(x\right)$. For each
${\theta}_{r}$ the current problem is solved by a robust secant iteration using the solution from earlier problems to compute an initial estimate.
You must supply an error tolerance
tol.
tol is used directly to control the accuracy of solution of the final problem (
${\theta}_{m}=0$)
in the continuation method, and
$\sqrt{{\mathbf{tol}}}$ is used to control the accuracy in the intermediate problems
(
${\theta}_{1},{\theta}_{2},\dots ,{\theta}_{m1}$).
References
Swift A and Lindfield G R (1978) Comparison of a continuation method for the numerical solution of a single nonlinear equation Comput. J. 21 359–362
Parameters
Note: this function uses
reverse communication. Its use involves an initial entry, intermediate exits and reentries, and a final exit, as indicated by the argument
ind. Between intermediate exits and reentries,
all arguments other than fx must remain unchanged.
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 reentry: 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
${x}_{r}^{0},{x}_{r}^{1},\dots \text{}$. This sequence will be considered to have converged only if:
for
${\mathbf{ir}}=0$,
for
${\mathbf{ir}}=1$,
for
${\mathbf{ir}}=2$,
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
where
$\lefth\right<\left{\mathbf{scal}}\right$ 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 reentry 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 reenter
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
Errors or warnings detected by the function:
 ${\mathbf{ifail}}=1$

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
The accuracy of the approximation to the zero depends on
tol and
ir. In general decreasing
tol will give more accurate results. Care must be exercised when using the relative error criterion (
${\mathbf{ir}}=2$).
If the zero is at
${\mathbf{x}}=0$, or if the initial value of
x and the zero bracket the point
${\mathbf{x}}=0$, it is likely that an error exit with
${\mathbf{ifail}}={\mathbf{4}}$,
${\mathbf{5}}$ or
${\mathbf{6}}$ will occur.
It is possible to request too much or too little accuracy. Since it is not possible to achieve more than machine accuracy, a value of
${\mathbf{tol}}\ll \mathit{machineprecision}$ should not be input and may lead to an error exit with
${\mathbf{ifail}}={\mathbf{4}}$,
${\mathbf{5}}$ or
${\mathbf{6}}$. For the reasons discussed under
${\mathbf{ifail}}={\mathbf{4}}$ in
Error Indicators and Warnings,
tol should not be taken too large, say no larger than
${\mathbf{tol}}=\text{1.0e\u22123}$.
Further Comments
For most problems, the time taken on each call to
nag_roots_contfn_cntin_rcomm (c05ax) will be negligible compared with the time spent evaluating
$f\left(x\right)$ between calls to
nag_roots_contfn_cntin_rcomm (c05ax). However, the initial value of
x and the choice of
tol will clearly affect the timing. The closer that
x is to the root, the less evaluations of
$f$ required. The effect of the choice of
tol will not be large, in general, unless
tol is very small, in which case the timing will increase.
Example
This example calculates a zero of $x{e}^{x}$ with initial approximation ${x}_{0}=1.0$, and ${\mathbf{tol}}=\text{1.0e\u22123}$ and $\text{1.0e\u22124}$.
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 1.0000e03, Root is 5.6715e01
Tol is 1.0000e04, Root is 5.6715e01
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, 64bit version, 64bit version)
© The Numerical Algorithms Group Ltd, Oxford, UK. 2009–2015