std.numeric.find_root - multiple declarations

Given a function f and a range [a..b] such that f(a) and f(b) have opposite signs or at least one of them equals ±0, returns the value of x in the range which is closest to a root of f(x). If f(x) has more than one root in the range, one will be chosen arbitrarily. If f(x) returns NaN, NaN will be returned; otherwise, this algorithm is guaranteed to succeed.

Uses an algorithm based on TOMS748, which uses inverse cubic interpolation whenever possible, otherwise reverting to parabolic or secant interpolation. Compared to TOMS748, this implementation improves worst-case performance by a factor of more than 100, and typical performance by a factor of 2. For 80-bit reals, most problems require 8 to 15 calls to f(x) to achieve full machine precision. The worst-case performance (pathological cases) is approximately twice the number of bits.

Prototypes

T findRoot(T, DF, DT)(
  DF f,
  T a,
  T b,
  DT tolerance
)
if (isFloatingPoint!T && is(typeof(tolerance(T.init, T.init)) : bool) && is(typeof(f(T.init)) == R, R) && isFloatingPoint!R);

T findRoot(T, DF)(
  DF f,
  T a,
  T b
);

References

"On Enclosing Simple Roots of Nonlinear Equations", G. Alefeld, F.A. Potra, Yixun Shi, Mathematics of Computation 61, pp733-744 (1993). Fortran code available from www.netlib.org as algorithm TOMS478.

Prototypes

Tuple!(T,T,R,R) findRoot(T, R, DF, DT)(
  DF f,
  T ax,
  T bx,
  R fax,
  R fbx,
  DT tolerance
)
if (isFloatingPoint!T && is(typeof(tolerance(T.init, T.init)) : bool) && is(typeof(f(T.init)) == R) && isFloatingPoint!R);

Tuple!(T,T,R,R) findRoot(T, R, DF)(
  DF f,
  T ax,
  T bx,
  R fax,
  R fbx
);

T findRoot(T, R)(
  R delegate(T) f,
  T a,
  T b,
  bool delegate(T lo, T hi) tolerance = 
);

Parameters

Returns

A tuple consisting of two ranges. The first two elements are the range (in x) of the root, while the second pair of elements are the corresponding function values at those points. If an exact root was found, both of the first two elements will contain the root, and the second pair of elements will be 0.