Adding upstream version 16.2.11+ds.upstream/16.2.11+dsupstream

Signed-off-by: Daniel Baumann <daniel.baumann@progress-linux.org>

1 files changed, 485 insertions, 0 deletions

diff --git a/src/boost/libs/math/example/root_finding_fifth.cpp b/src/boost/libs/math/example/root_finding_fifth.cpp
new file mode 100644
index 000000000..6c9811475
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+++ b/src/boost/libs/math/example/root_finding_fifth.cpp
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+// root_finding_fith.cpp
+
+// Copyright Paul A. Bristow 2014.
+
+// Use, modification and distribution are subject to the
+// Boost Software License, Version 1.0.
+// (See accompanying file LICENSE_1_0.txt
+// or copy at http://www.boost.org/LICENSE_1_0.txt)
+
+// Example of finding fifth root using Newton-Raphson, Halley, Schroder, TOMS748 .
+
+// Note that this file contains Quickbook mark-up as well as code
+// and comments, don't change any of the special comment mark-ups!
+
+// To get (copious!) diagnostic output, add make this define here or elsewhere.
+//#define BOOST_MATH_INSTRUMENT
+
+
+//[root_fifth_headers
+/*
+This example demonstrates how to use the Boost.Math tools for root finding,
+taking the fifth root function (fifth_root) as an example.
+It shows how use of derivatives can improve the speed.
+
+First some includes that will be needed.
+Using statements are provided to list what functions are being used in this example:
+you can of course qualify the names in other ways.
+*/
+
+#include <boost/math/tools/roots.hpp>
+using boost::math::policies::policy;
+using boost::math::tools::newton_raphson_iterate;
+using boost::math::tools::halley_iterate;
+using boost::math::tools::eps_tolerance; // Binary functor for specified number of bits.
+using boost::math::tools::bracket_and_solve_root;
+using boost::math::tools::toms748_solve;
+
+#include <boost/math/special_functions/next.hpp>
+
+#include <tuple>
+#include <utility> // pair, make_pair
+
+//] [/root_finding_headers]
+
+#include <iostream>
+using std::cout; using std::endl;
+#include <iomanip>
+using std::setw; using std::setprecision;
+#include <limits>
+using std::numeric_limits;
+
+/*
+//[root_finding_fifth_1
+Let's suppose we want to find the fifth root of a number.
+
+The equation we want to solve is:
+
+__spaces ['f](x) = x[fifth]
+
+We will first solve this without using any information
+about the slope or curvature of the fifth function.
+
+If your differentiation is a little rusty
+(or you are faced with an equation whose complexity is daunting,
+then you can get help, for example from the invaluable
+
+http://www.wolframalpha.com/ site
+
+entering the command
+
+  differentiate x^5
+
+or the Wolfram Language command
+
+  D[x^5, x]
+
+gives the output
+
+  d/dx(x^5) = 5 x^4
+
+and to get the second differential, enter
+
+  second differentiate x^5
+
+or the Wolfram Language
+
+  D[x^5, {x, 2}]
+
+to get the output
+
+  d^2/dx^2(x^5) = 20 x^3
+
+or
+
+  20 x^3
+
+To get a reference value we can enter
+
+  fifth root 3126
+
+or
+
+  N[3126^(1/5), 50]
+
+to get a result with a precision of  50 decimal digits
+
+  5.0003199590478625588206333405631053401128722314376
+
+(We could also get a reference value using Boost.Multiprecision).
+
+We then show how adding what we can know, for this function, about the slope,
+the 1st derivation /f'(x)/, will speed homing in on the solution,
+and then finally how adding the curvature /f''(x)/ as well will improve even more.
+
+The 1st and 2nd derivatives of x[fifth] are:
+
+__spaces ['f]\'(x) = 2x[sup2]
+
+__spaces ['f]\'\'(x) = 6x
+
+*/
+
+//] [/root_finding_fifth_1]
+
+//[root_finding_fifth_functor_noderiv
+
+template <class T>
+struct fifth_functor_noderiv
+{ // fifth root of x using only function - no derivatives.
+  fifth_functor_noderiv(T const& to_find_root_of) : value(to_find_root_of)
+  { // Constructor stores value to find root of.
+    // For example: calling fifth_functor<T>(x) to get fifth root of x.
+  }
+  T operator()(T const& x)
+  { //! \returns f(x) - value.
+    T fx = x*x*x*x*x - value; // Difference (estimate x^5 - value).
+    return fx;
+  }
+private:
+  T value; // to be 'fifth_rooted'.
+};
+
+//] [/root_finding_fifth_functor_noderiv]
+
+//cout  << ", std::numeric_limits<" << typeid(T).name()  << ">::digits = " << digits
+//   << ", accuracy " << get_digits << " bits."<< endl;
+
+
+/*`Implementing the fifth root function itself is fairly trivial now:
+the hardest part is finding a good approximation to begin with.
+In this case we'll just divide the exponent by five.
+(There are better but more complex guess algorithms used in 'real-life'.)
+
+fifth root function is 'Really Well Behaved' in that it is monotonic
+and has only one root
+(we leave negative values 'as an exercise for the student').
+*/
+
+//[root_finding_fifth_noderiv
+
+template <class T>
+T fifth_noderiv(T x)
+{ //! \returns fifth root of x using bracket_and_solve (no derivatives).
+  using namespace std;  // Help ADL of std functions.
+  using namespace boost::math::tools; // For bracket_and_solve_root.
+
+  int exponent;
+  frexp(x, &exponent); // Get exponent of z (ignore mantissa).
+  T guess = ldexp(1., exponent / 5); // Rough guess is to divide the exponent by five.
+  T factor = 2; // To multiply and divide guess to bracket.
+  // digits used to control how accurate to try to make the result.
+  // int digits =  3 * std::numeric_limits<T>::digits / 4; // 3/4 maximum possible binary digits accuracy for type T.
+  int digits = std::numeric_limits<T>::digits; // Maximum possible binary digits accuracy for type T.
+
+  //boost::uintmax_t maxit = (std::numeric_limits<boost::uintmax_t>::max)();
+  // (std::numeric_limits<boost::uintmax_t>::max)() = 18446744073709551615
+  // which is more than anyone might wish to wait for!!!
+  // so better to choose some reasonable estimate of how many iterations may be needed.
+
+  const boost::uintmax_t maxit = 50; // Chosen max iterations,
+  // but updated on exit with actual iteration count.
+
+  // We could also have used a maximum iterations provided by any policy:
+  // boost::uintmax_t max_it = policies::get_max_root_iterations<Policy>();
+
+  boost::uintmax_t it = maxit; // Initially our chosen max iterations,
+
+  bool is_rising = true; // So if result if guess^5 is too low, try increasing guess.
+  eps_tolerance<double> tol(digits);
+  std::pair<T, T> r =
+    bracket_and_solve_root(fifth_functor_noderiv<T>(x), guess, factor, is_rising, tol, it);
+  // because the iteration count is updating,
+  // you can't call with a literal maximum iterations value thus:
+  //bracket_and_solve_root(fifth_functor_noderiv<T>(x), guess, factor, is_rising, tol, 20);
+
+  // Can show how many iterations (this information is lost outside fifth_noderiv).
+  cout << "Iterations " << it << endl;
+  if (it >= maxit)
+  { // Failed to converge (or is jumping between bracket values).
+    cout << "Unable to locate solution in chosen iterations:"
+      " Current best guess is between " << r.first << " and " << r.second << endl;
+  }
+  T distance = float_distance(r.first, r.second);
+  if (distance > 0)
+  { //
+    std::cout << distance << " bits separate the bracketing values." << std::endl;
+    for (int i = 0; i < distance; i++)
+    { // Show all the values within the bracketing values.
+      std::cout << float_advance(r.first, i) << std::endl;
+    }
+  }
+  else
+  { // distance == 0 and  r.second == r.first
+    std::cout << "Converged to a single value " << r.first << std::endl;
+  }
+
+  return r.first + (r.second - r.first) / 2;  // return midway between bracketed interval.
+} // T fifth_noderiv(T x)
+
+//] [/root_finding_fifth_noderiv]
+
+
+
+// maxit = 10
+// Unable to locate solution in chosen iterations: Current best guess is between 3.0365889718756613 and 3.0365889718756627
+
+
+/*`
+We now solve the same problem, but using more information about the function,
+to show how this can speed up finding the best estimate of the root.
+
+For this function, the 1st differential (the slope of the tangent to a curve at any point) is known.
+
+[@http://en.wikipedia.org/wiki/Derivative#Derivatives_of_elementary_functions Derivatives]
+gives some reminders.
+
+Using the rule that the derivative of x^n for positive n (actually all nonzero n) is nx^n-1,
+allows use to get the 1st differential as 3x^2.
+
+To see how this extra information is used to find the root, view this demo:
+[@http://en.wikipedia.org/wiki/Newton%27s_methodNewton Newton-Raphson iterations].
+
+We need to define a different functor that returns
+both the evaluation of the function to solve, along with its first derivative:
+
+To \'return\' two values, we use a pair of floating-point values:
+*/
+
+//[root_finding_fifth_functor_1stderiv
+
+template <class T>
+struct fifth_functor_1stderiv
+{ // Functor returning function and 1st derivative.
+
+  fifth_functor_1stderiv(T const& target) : value(target)
+  { // Constructor stores the value to be 'fifth_rooted'.
+  }
+
+  std::pair<T, T> operator()(T const& z) // z is best estimate so far.
+  { // Return both f(x) and first derivative f'(x).
+    T fx = z*z*z*z*z - value; // Difference estimate fx = x^5 - value.
+    T d1x = 5 * z*z*z*z; // 1st derivative d1x = 5x^4.
+    return std::make_pair(fx, d1x); // 'return' both fx and d1x.
+  }
+private:
+  T value; // to be 'fifth_rooted'.
+}; // fifth_functor_1stderiv
+
+//] [/root_finding_fifth_functor_1stderiv]
+
+
+/*`Our fifth root function using fifth_functor_1stderiv is now:*/
+
+//[root_finding_fifth_1deriv
+
+template <class T>
+T fifth_1deriv(T x)
+{ //! \return fifth root of x using 1st derivative and Newton_Raphson.
+  using namespace std; // For frexp, ldexp, numeric_limits.
+  using namespace boost::math::tools; // For newton_raphson_iterate.
+
+  int exponent;
+  frexp(x, &exponent); // Get exponent of x (ignore mantissa).
+  T guess = ldexp(1., exponent / 5); // Rough guess is to divide the exponent by three.
+  // Set an initial bracket interval.
+  T min = ldexp(0.5, exponent / 5); // Minimum possible value is half our guess.
+  T max = ldexp(2., exponent / 5);// Maximum possible value is twice our guess.
+
+  // digits used to control how accurate to try to make the result.
+  int digits = std::numeric_limits<T>::digits; // Maximum possible binary digits accuracy for type T.
+
+  const boost::uintmax_t maxit = 20; // Optionally limit the number of iterations.
+  boost::uintmax_t it = maxit; // limit the number of iterations.
+  //cout << "Max Iterations " << maxit << endl; //
+  T result = newton_raphson_iterate(fifth_functor_1stderiv<T>(x), guess, min, max, digits, it);
+  // Can check and show how many iterations (updated by newton_raphson_iterate).
+  cout << it << " iterations (from max of " << maxit << ")" << endl;
+  return result;
+} // fifth_1deriv
+
+//] [/root_finding_fifth_1deriv]
+
+//  int get_digits = (digits * 2) /3; // Two thirds of maximum possible accuracy.
+
+//boost::uintmax_t maxit = (std::numeric_limits<boost::uintmax_t>::max)();
+// the default (std::numeric_limits<boost::uintmax_t>::max)() = 18446744073709551615
+// which is more than we might wish to wait for!!!  so we can reduce it
+
+/*`
+Finally need to define yet another functor that returns
+both the evaluation of the function to solve,
+along with its first and second derivatives:
+
+f''(x) = 3 * 3x
+
+To \'return\' three values, we use a tuple of three floating-point values:
+*/
+
+//[root_finding_fifth_functor_2deriv
+
+template <class T>
+struct fifth_functor_2deriv
+{ // Functor returning both 1st and 2nd derivatives.
+  fifth_functor_2deriv(T const& to_find_root_of) : value(to_find_root_of)
+  { // Constructor stores value to find root of, for example:
+  }
+
+  // using boost::math::tuple; // to return three values.
+  std::tuple<T, T, T> operator()(T const& x)
+  { // Return both f(x) and f'(x) and f''(x).
+    T fx = x*x*x*x*x - value; // Difference (estimate x^3 - value).
+    T dx = 5 * x*x*x*x; // 1st derivative = 5x^4.
+    T d2x = 20 * x*x*x; // 2nd derivative = 20 x^3
+    return std::make_tuple(fx, dx, d2x); // 'return' fx, dx and d2x.
+  }
+private:
+  T value; // to be 'fifth_rooted'.
+}; // struct fifth_functor_2deriv
+
+//] [/root_finding_fifth_functor_2deriv]
+
+
+/*`Our fifth function is now:*/
+
+//[root_finding_fifth_2deriv
+
+template <class T>
+T fifth_2deriv(T x)
+{ // return fifth root of x using 1st and 2nd derivatives and Halley.
+  using namespace std;  // Help ADL of std functions.
+  using namespace boost::math; // halley_iterate
+
+  int exponent;
+  frexp(x, &exponent); // Get exponent of z (ignore mantissa).
+  T guess = ldexp(1., exponent / 5); // Rough guess is to divide the exponent by three.
+  T min = ldexp(0.5, exponent / 5); // Minimum possible value is half our guess.
+  T max = ldexp(2., exponent / 5); // Maximum possible value is twice our guess.
+
+  int digits = std::numeric_limits<T>::digits / 2; // Half maximum possible binary digits accuracy for type T.
+  const boost::uintmax_t maxit = 50;
+  boost::uintmax_t it = maxit;
+  T result = halley_iterate(fifth_functor_2deriv<T>(x), guess, min, max, digits, it);
+  // Can show how many iterations (updated by halley_iterate).
+  cout << it << " iterations (from max of " << maxit << ")" << endl;
+
+  return result;
+} // fifth_2deriv(x)
+
+//] [/root_finding_fifth_2deriv]
+
+int main()
+{
+
+  //[root_finding_example_1
+  cout << "fifth Root finding (fifth) Example." << endl;
+  // Show all possibly significant decimal digits.
+  cout.precision(std::numeric_limits<double>::max_digits10);
+  // or use   cout.precision(max_digits10 = 2 + std::numeric_limits<double>::digits * 3010/10000);
+  try
+  { // Always use try'n'catch blocks with Boost.Math to get any error messages.
+
+    double v27 = 3125; // Example of a value that has an exact integer fifth root.
+    // exact value of fifth root is exactly 5.
+
+    std::cout << "Fifth root  of " << v27 << " is " << 5 << std::endl;
+
+    double v28 = v27+1; // Example of a value whose fifth root is *not* exactly representable.
+    // Value of fifth root is 5.0003199590478625588206333405631053401128722314376 (50 decimal digits precision)
+    // and to std::numeric_limits<double>::max_digits10 double precision (usually 17) is
+
+    double root5v2 = static_cast<double>(5.0003199590478625588206333405631053401128722314376);
+
+    std::cout << "Fifth root  of " << v28 << " is "  << root5v2 << std::endl;
+
+    // Using bracketing:
+    double r = fifth_noderiv(v27);
+    cout << "fifth_noderiv(" << v27 << ") = " << r << endl;
+
+    r = fifth_noderiv(v28);
+    cout << "fifth_noderiv(" << v28 << ") = " << r << endl;
+
+    // Using 1st differential Newton-Raphson:
+    r = fifth_1deriv(v27);
+    cout << "fifth_1deriv(" << v27 << ") = " << r << endl;
+    r = fifth_1deriv(v28);
+    cout << "fifth_1deriv(" << v28 << ") = " << r << endl;
+
+    // Using Halley with 1st and 2nd differentials.
+    r = fifth_2deriv(v27);
+    cout << "fifth_2deriv(" << v27 << ") = " << r << endl;
+    r = fifth_2deriv(v28);
+    cout << "fifth_2deriv(" << v28 << ") = " << r << endl;
+  }
+  catch (const std::exception& e)
+  { // Always useful to include try & catch blocks because default policies
+    // are to throw exceptions on arguments that cause errors like underflow, overflow.
+    // Lacking try & catch blocks, the program will abort without a message below,
+    // which may give some helpful clues as to the cause of the exception.
+    std::cout <<
+      "\n""Message from thrown exception was:\n   " << e.what() << std::endl;
+  }
+  //] [/root_finding_example_1
+  return 0;
+}  // int main()
+
+//[root_finding_example_output
+/*`
+Normal output is:
+
+[pre
+1>  Description: Autorun "J:\Cpp\MathToolkit\test\Math_test\Release\root_finding_fifth.exe"
+1>  fifth Root finding (fifth) Example.
+1>  Fifth root  of 3125 is 5
+1>  Fifth root  of 3126 is 5.0003199590478626
+1>  Iterations 10
+1>  Converged to a single value 5
+1>  fifth_noderiv(3125) = 5
+1>  Iterations 11
+1>  2 bits separate the bracketing values.
+1>  5.0003199590478609
+1>  5.0003199590478618
+1>  fifth_noderiv(3126) = 5.0003199590478618
+1>  6 iterations (from max of 20)
+1>  fifth_1deriv(3125) = 5
+1>  7 iterations (from max of 20)
+1>  fifth_1deriv(3126) = 5.0003199590478626
+1>  4 iterations (from max of 50)
+1>  fifth_2deriv(3125) = 5
+1>  4 iterations (from max of 50)
+1>  fifth_2deriv(3126) = 5.0003199590478626
+[/pre]
+
+to get some (much!) diagnostic output we can add
+
+#define BOOST_MATH_INSTRUMENT
+
+[pre
+1>  fifth Root finding (fifth) Example.
+1>  I:\modular-boost\boost/math/tools/toms748_solve.hpp:537 a = 4 b = 8 fa = -2101 fb = 29643 count = 18
+1>  I:\modular-boost\boost/math/tools/toms748_solve.hpp:340  a = 4.264742943548387 b = 8
+1>  I:\modular-boost\boost/math/tools/toms748_solve.hpp:352  a = 4.264742943548387 b = 5.1409225585147951
+1>  I:\modular-boost\boost/math/tools/toms748_solve.hpp:259  a = 4.264742943548387 b = 5.1409225585147951 d = 8 e = 4 fa = -1714.2037505671719 fb = 465.91652114644285 fd = 29643 fe = -2101
+1>  I:\modular-boost\boost/math/tools/toms748_solve.hpp:267 q11 = -3.735257056451613 q21 = -0.045655399937094755 q31 = 0.68893005658139972 d21 = -2.9047328414222999 d31 = -0.18724955838500826
+1>  I:\modular-boost\boost/math/tools/toms748_solve.hpp:275 q22 = -0.15074699539567221 q32 = 0.007740525571111408 d32 = -0.13385363287680208 q33 = 0.074868009790687237 c = 5.0362815354915851
+1>  I:\modular-boost\boost/math/tools/toms748_solve.hpp:388  a = 4.264742943548387 b = 5.0362815354915851
+1>  I:\modular-boost\boost/math/tools/toms748_solve.hpp:259  a = 4.264742943548387 b = 5.0362815354915851 d = 5.1409225585147951 e = 8 fa = -1714.2037505671719 fb = 115.03721886368339 fd = 465.91652114644285 fe = 29643
+1>  I:\modular-boost\boost/math/tools/toms748_solve.hpp:267 q11 = -0.045655399937094755 q21 = -0.034306988726112195 q31 = 0.7230181097615842 d21 = -0.1389480117493222 d31 = -0.048520482181613811
+1>  I:\modular-boost\boost/math/tools/toms748_solve.hpp:275 q22 = -0.00036345624935100459 q32 = 0.011175908093791367 d32 = -0.0030375853617102483 q33 = 0.00014618657296010219 c = 4.999083147976723
+1>  I:\modular-boost\boost/math/tools/toms748_solve.hpp:408  a = 4.999083147976723 b = 5.0362815354915851
+1>  I:\modular-boost\boost/math/tools/toms748_solve.hpp:433  a = 4.999083147976723 b = 5.0008904277935091
+1>  I:\modular-boost\boost/math/tools/toms748_solve.hpp:434  tol = -0.00036152225583956088
+1>  I:\modular-boost\boost/math/tools/toms748_solve.hpp:259  a = 4.999083147976723 b = 5.0008904277935091 d = 5.0362815354915851 e = 4.264742943548387 fa = -2.8641119933622576 fb = 2.7835781082976609 fd = 115.03721886368339 fe = -1714.2037505671719
+1>  I:\modular-boost\boost/math/tools/toms748_solve.hpp:267 q11 = -0.048520482181613811 q21 = -0.00087760104664616457 q31 = 0.00091652546535745522 d21 = -0.036268708744722128 d31 = -0.00089075435142862297
+1>  I:\modular-boost\boost/math/tools/toms748_solve.hpp:275 q22 = -1.9862562616034592e-005 q32 = 3.1952597740788757e-007 d32 = -1.2833778805050512e-005 q33 = 1.1763429980834706e-008 c = 5.0000000047314881
+1>  I:\modular-boost\boost/math/tools/toms748_solve.hpp:388  a = 4.999083147976723 b = 5.0000000047314881
+1>  I:\modular-boost\boost/math/tools/toms748_solve.hpp:259  a = 4.999083147976723 b = 5.0000000047314881 d = 5.0008904277935091 e = 5.0362815354915851 fa = -2.8641119933622576 fb = 1.4785900475544622e-005 fd = 2.7835781082976609 fe = 115.03721886368339
+1>  I:\modular-boost\boost/math/tools/toms748_solve.hpp:267 q11 = -0.00087760104664616457 q21 = -4.7298032238887272e-009 q31 = 0.00091685202154135855 d21 = -0.00089042779182425238 d31 = -4.7332236912279757e-009
+1>  I:\modular-boost\boost/math/tools/toms748_solve.hpp:275 q22 = -1.6486403607318402e-012 q32 = 1.7346209428817704e-012 d32 = -1.6858463963666777e-012 q33 = 9.0382569995250912e-016 c = 5
+1>  I:\modular-boost\boost/math/tools/toms748_solve.hpp:592 max_iter = 10 count = 7
+1>  Iterations 20
+1>  0 bits separate brackets.
+1>  fifth_noderiv(3125) = 5
+]
+*/
+//] [/root_finding_example_output]