c05avc attempts to locate an interval containing a simple zero of a continuous function using a binary search. It uses reverse communication for evaluating the function.
The function may be called by the names: c05avc, nag_roots_contfn_interval_rcomm or nag_interval_zero_cont_func.
3Description
You must supply an initial point x and a step h. c05avc attempts to locate a short interval $[{\mathbf{x}},{\mathbf{y}}]\subset [{\mathbf{boundl}},{\mathbf{boundu}}]$ containing a simple zero of $f\left(x\right)$.
(On exit we may have ${\mathbf{x}}>{\mathbf{y}}$; x is determined as the first point encountered in a binary search where the sign of $f\left(x\right)$ differs from the sign of $f\left(x\right)$ at the initial input point x.) The function attempts to locate a zero of $f\left(x\right)$ using h, $0.1\times {\mathbf{h}}$, $0.01\times {\mathbf{h}}$ and $0.001\times {\mathbf{h}}$ in turn as its basic step before quitting with an error exit if unsuccessful.
c05avc returns to the calling program for each evaluation of $f\left(x\right)$. On each return you should set ${\mathbf{fx}}=f\left({\mathbf{x}}\right)$ and call c05avc again.
4References
None.
5Arguments
Note: this function uses reverse communication. Its use involves an initial entry, intermediate exits and re-entries, and a final exit, as indicated by the argument ind. Between intermediate exits and re-entries, all arguments other thanfx must remain unchanged.
1: $\mathbf{x}$ – double *Input/Output
On initial entry: the best available approximation to the zero.
Constraint:
x must lie in the closed interval $[{\mathbf{boundl}},{\mathbf{boundu}}]$ (see below).
On intermediate exit:
contains the point at which $f$ must be evaluated before re-entry to the function.
On final exit: contains one end of an interval containing the zero, the other end being in y, unless an error has occurred. If ${\mathbf{fail}}\mathbf{.}\mathbf{code}=$NE_ZERO_NOT_FOUND, x and y are the end points of the largest interval searched. If a zero is located exactly, its value is returned in x (and in y).
2: $\mathbf{fx}$ – doubleInput
On initial entry: if ${\mathbf{ind}}=1$, fx need not be set.
If ${\mathbf{ind}}=\mathrm{-1}$, 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: $\mathbf{h}$ – double *Input/Output
On initial entry: a basic step size which is used in the binary search for an interval containing a zero. The basic step sizes ${\mathbf{h}},0.1\times {\mathbf{h}}$, $0.01\times {\mathbf{h}}$ and $0.001\times {\mathbf{h}}$ are used in turn when searching for the zero.
Constraint:
either ${\mathbf{x}}+{\mathbf{h}}$ or ${\mathbf{x}}-{\mathbf{h}}$ must lie inside the closed interval $[{\mathbf{boundl}},{\mathbf{boundu}}]$.
h must be sufficiently large that ${\mathbf{x}}+{\mathbf{h}}\ne {\mathbf{x}}$ on the computer.
On final exit: is undefined.
4: $\mathbf{boundl}$ – doubleInput
5: $\mathbf{boundu}$ – doubleInput
On initial entry: boundl and boundu must contain respectively lower and upper bounds for the interval of search for the zero.
On final exit: contains the closest point found to the final value of x, such that $f\left({\mathbf{x}}\right)\times f\left({\mathbf{y}}\right)\le 0.0$. If a value x is found such that $f\left({\mathbf{x}}\right)=0$, ${\mathbf{y}}={\mathbf{x}}$. On final exit with ${\mathbf{fail}}\mathbf{.}\mathbf{code}=$NE_ZERO_NOT_FOUND, x and y are the end points of the largest interval searched.
On final exit: if ${\mathbf{fail}}\mathbf{.}\mathbf{code}=$ NE_NOERROR or NE_ZERO_NOT_FOUND, ${\mathbf{c}}\left[0\right]$ contains $f\left({\mathbf{y}}\right)$.
8: $\mathbf{ind}$ – Integer *Input/Output
On initial entry: must be set to $1$ or $\mathrm{-1}$.
On intermediate exit:
contains $2$ or $3$. The calling program must evaluate $f$ at x, storing the result in fx, and re-enter c05avc with all other arguments unchanged.
On final exit: contains $0$.
Constraint:
on entry ${\mathbf{ind}}=\mathrm{-1}$, $1$, $2$ or $3$.
Note: any values you return to c05avc as part of the reverse communication procedure should not include floating-point NaN (Not a Number) or infinity values, since these are not handled by c05avc. If your code inadvertently does return any NaNs or infinities, c05avc is likely to produce unexpected results.
9: $\mathbf{fail}$ – NagError *Input/Output
The NAG error argument (see Section 7 in the Introduction to the NAG Library CL Interface).
6Error Indicators and Warnings
NE_ALLOC_FAIL
Dynamic memory allocation failed.
See Section 3.1.2 in the Introduction to the NAG Library CL Interface for further information.
NE_BAD_PARAM
On entry, argument $\u27e8\mathit{\text{value}}\u27e9$ had an illegal value.
NE_INT
On entry, ${\mathbf{ind}}=\u27e8\mathit{\text{value}}\u27e9$.
Constraint: ${\mathbf{ind}}=\mathrm{-1}$, $1$, $2$ or $3$.
NE_INTERNAL_ERROR
An internal error has occurred in this function. Check the function call and any array sizes. If the call is correct then please contact NAG for assistance.
See Section 7.5 in the Introduction to the NAG Library CL Interface for further information.
NE_NO_LICENCE
Your licence key may have expired or may not have been installed correctly.
See Section 8 in the Introduction to the NAG Library CL Interface for further information.
NE_REAL_2
On entry, ${\mathbf{boundl}}=\u27e8\mathit{\text{value}}\u27e9$ and ${\mathbf{boundu}}=\u27e8\mathit{\text{value}}\u27e9$.
Constraint: ${\mathbf{boundl}}<{\mathbf{boundu}}$.
On entry, ${\mathbf{x}}=\u27e8\mathit{\text{value}}\u27e9$ and ${\mathbf{h}}=\u27e8\mathit{\text{value}}\u27e9$.
Constraint: h must be sufficiently large that ${\mathbf{x}}+{\mathbf{h}}\ne {\mathbf{x}}$ on the computer.
NE_REAL_3
On entry, ${\mathbf{x}}=\u27e8\mathit{\text{value}}\u27e9$, ${\mathbf{boundl}}=\u27e8\mathit{\text{value}}\u27e9$ and ${\mathbf{boundu}}=\u27e8\mathit{\text{value}}\u27e9$.
Constraint: ${\mathbf{boundl}}\le {\mathbf{x}}\le {\mathbf{boundu}}$.
NE_REAL_4
On entry, ${\mathbf{x}}=\u27e8\mathit{\text{value}}\u27e9$, ${\mathbf{h}}=\u27e8\mathit{\text{value}}\u27e9$, ${\mathbf{boundl}}=\u27e8\mathit{\text{value}}\u27e9$ and ${\mathbf{boundu}}=\u27e8\mathit{\text{value}}\u27e9$.
Constraint: either ${\mathbf{x}}+{\mathbf{h}}$ or ${\mathbf{x}}-{\mathbf{h}}$ must lie inside the closed interval $[{\mathbf{boundl}},{\mathbf{boundu}}]$.
NE_ZERO_NOT_FOUND
An interval containing the zero could not be found. Try restarting with modified x and h.
7Accuracy
c05avc is not intended to be used to obtain accurate approximations to the zero of $f\left(x\right)$ but rather to locate an interval containing a zero. This interval can then be used as input to an accurate rootfinder such as c05aycorc05azc. The size of the interval determined depends somewhat unpredictably on the choice of x and h. The closer x is to the root and the smaller the initial value of h, then, in general, the smaller (more accurate) the interval determined; however, the accuracy of this statement depends to some extent on the behaviour of $f\left(x\right)$ near $x={\mathbf{x}}$ and on the size of h.
8Parallelism and Performance
Background information to multithreading can be found in the Multithreading documentation.
c05avc is not threaded in any implementation.
9Further Comments
For most problems, the time taken on each call to c05avc will be negligible compared with the time spent evaluating $f\left(x\right)$ between calls to c05avc. However, the initial value of x and h will clearly affect the timing. The closer x is to the root, and the larger the initial value of h then the less time taken. (However taking a large h can affect the accuracy and reliability of the function, see below.)
You are expected to choose boundl and boundu as physically (or mathematically) realistic limits on the interval of search. For example, it may be known, from physical arguments, that no zero of $f\left(x\right)$ of interest will lie outside $[{\mathbf{boundl}},{\mathbf{boundu}}]$. Alternatively, $f\left(x\right)$ may be more expensive to evaluate for some values of x than for others and such expensive evaluations can sometimes be avoided by careful choice of boundl and boundu.
The choice of boundl and boundu affects the search only in that these values provide physical limitations on the search values and that the search is terminated if it seems, from the available information about $f\left(x\right)$, that the zero lies outside $[{\mathbf{boundl}},{\mathbf{boundu}}]$. In this case (${\mathbf{fail}}\mathbf{.}\mathbf{code}=$NE_ZERO_NOT_FOUND on exit), only one of $f\left({\mathbf{boundl}}\right)$ and $f\left({\mathbf{boundu}}\right)$ may have been evaluated and a zero close to the other end of the interval could be missed. The actual interval searched is returned in the arguments x and y and you can call c05avc again to search the remainder of the original interval.
Though c05avc is intended primarily for determining an interval containing a zero of $f\left(x\right)$, it may be used to shorten a known interval. This could be useful if, for example, a large interval containing the zero is known and it is also known that the root lies close to one end of the interval; by setting x to this end of the interval and h small, a short interval will usually be determined. However, it is worth noting that once any interval containing a zero has been determined, a call to c05azc will usually be the most efficient way to calculate an interval of specified length containing the zero. To assist in this determination, the information in fx and in x, y and ${\mathbf{c}}\left[0\right]$ on successful exit from c05avc is in the correct form for a call to function c05azc with ${\mathbf{ind}}=\mathrm{-1}$.
If the calculation terminates because $f\left({\mathbf{x}}\right)=0.0$, then on return y is set to x. (In fact, ${\mathbf{y}}={\mathbf{x}}$ on return only in this case.) In this case, there is no guarantee that the value in x corresponds to a simple zero and you should check whether it does.
One way to check this is to compute the derivative of $f$ at the point x, preferably analytically, or, if this is not possible, numerically, perhaps by using a central difference estimate. If ${f}^{\prime}\left({\mathbf{x}}\right)=0.0$, then x must correspond to a multiple zero of $f$ rather than a simple zero.
10Example
This example finds a sub-interval of $[0.0,4.0]$ containing a simple zero of ${x}^{2}-3x+2$. The zero nearest to $3.0$ is required and so we set ${\mathbf{x}}=3.0$ initially.