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authorDaniel Baumann <daniel.baumann@progress-linux.org>2024-04-27 18:24:20 +0000
committerDaniel Baumann <daniel.baumann@progress-linux.org>2024-04-27 18:24:20 +0000
commit483eb2f56657e8e7f419ab1a4fab8dce9ade8609 (patch)
treee5d88d25d870d5dedacb6bbdbe2a966086a0a5cf /src/boost/libs/math/example/numerical_derivative_example.cpp
parentInitial commit. (diff)
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+// Copyright Christopher Kormanyos 2013.
+// Distributed under 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).
+
+#ifdef _MSC_VER
+# pragma warning (disable : 4996) // assignment operator could not be generated.
+#endif
+
+# include <iostream>
+# include <iomanip>
+# include <limits>
+# include <cmath>
+
+#include <boost/static_assert.hpp>
+#include <boost/type_traits/is_floating_point.hpp>
+#include <boost/math/special_functions/next.hpp> // for float_distance
+
+//[numeric_derivative_example
+/*`The following example shows how multiprecision calculations can be used to
+obtain full precision in a numerical derivative calculation that suffers from precision loss.
+
+Consider some well-known central difference rules for numerically
+computing the 1st derivative of a function [f'(x)] with [/x] real.
+
+Need a reference here? Introduction to Partial Differential Equations, Peter J. Olver
+ December 16, 2012
+
+Here, the implementation uses a C++ template that can be instantiated with various
+floating-point types such as `float`, `double`, `long double`, or even
+a user-defined floating-point type like __multiprecision.
+
+We will now use the derivative template with the built-in type `double` in
+order to numerically compute the derivative of a function, and then repeat
+with a 5 decimal digit higher precision user-defined floating-point type.
+
+Consider the function shown below.
+!!
+(3)
+We will now take the derivative of this function with respect to x evaluated
+at x = 3= 2. In other words,
+
+(4)
+
+The expected result is
+
+ 0:74535 59924 99929 89880 . (5)
+The program below uses the derivative template in order to perform
+the numerical calculation of this derivative. The program also compares the
+numerically-obtained result with the expected result and reports the absolute
+relative error scaled to a deviation that can easily be related to the number of
+bits of lost precision.
+
+*/
+
+/*` [note Rquires the C++11 feature of
+[@http://en.wikipedia.org/wiki/Anonymous_function#C.2B.2B anonymous functions]
+for the derivative function calls like `[]( const double & x_) -> double`.
+*/
+
+
+
+template <typename value_type, typename function_type>
+value_type derivative (const value_type x, const value_type dx, function_type function)
+{
+ /*! \brief Compute the derivative of function using a 3-point central difference rule of O(dx^6).
+ \tparam value_type, floating-point type, for example: `double` or `cpp_dec_float_50`
+ \tparam function_type
+
+ \param x Value at which to evaluate derivative.
+ \param dx Incremental step-size.
+ \param function Function whose derivative is to computed.
+
+ \return derivative at x.
+ */
+
+ BOOST_STATIC_ASSERT_MSG(false == std::numeric_limits<value_type>::is_integer, "value_type must be a floating-point type!");
+
+ const value_type dx2(dx * 2U);
+ const value_type dx3(dx * 3U);
+ // Difference terms.
+ const value_type m1 ((function (x + dx) - function(x - dx)) / 2U);
+ const value_type m2 ((function (x + dx2) - function(x - dx2)) / 4U);
+ const value_type m3 ((function (x + dx3) - function(x - dx3)) / 6U);
+ const value_type fifteen_m1 (m1 * 15U);
+ const value_type six_m2 (m2 * 6U);
+ const value_type ten_dx (dx * 10U);
+ return ((fifteen_m1 - six_m2) + m3) / ten_dx; // Derivative.
+} //
+
+#include <boost/multiprecision/cpp_dec_float.hpp>
+ using boost::multiprecision::number;
+ using boost::multiprecision::cpp_dec_float;
+
+// Re-compute using 5 extra decimal digits precision (22) than double (17).
+#define MP_DIGITS10 unsigned (std::numeric_limits<double>::max_digits10 + 5)
+
+typedef cpp_dec_float<MP_DIGITS10> mp_backend;
+typedef number<mp_backend> mp_type;
+
+
+int main()
+{
+ {
+ const double d =
+ derivative
+ ( 1.5, // x = 3.2
+ std::ldexp (1., -9), // step size 2^-9 = see below for choice.
+ [](const double & x)->double // Function f(x).
+ {
+ return std::sqrt((x * x) - 1.) - std::acos(1. / x);
+ }
+ );
+
+ // The 'exactly right' result is [sqrt]5 / 3 = 0.74535599249992989880.
+ const double rel_error = (d - 0.74535599249992989880) / 0.74535599249992989880;
+ const double bit_error = std::abs(rel_error) / std::numeric_limits<double>::epsilon();
+ std::cout.precision (std::numeric_limits<double>::digits10); // Show all guaranteed decimal digits.
+ std::cout << std::showpoint ; // Ensure that any trailing zeros are shown too.
+
+ std::cout << " derivative : " << d << std::endl;
+ std::cout << " expected : " << 0.74535599249992989880 << std::endl;
+ // Can compute an 'exact' value using multiprecision type.
+ std::cout << " expected : " << sqrt(static_cast<mp_type>(5))/3U << std::endl;
+ std::cout << " bit_error : " << static_cast<unsigned long>(bit_error) << std::endl;
+
+ std::cout.precision(6);
+ std::cout << "float_distance = " << boost::math::float_distance(0.74535599249992989880, d) << std::endl;
+
+ }
+
+ { // Compute using multiprecision type with an extra 5 decimal digits of precision.
+ const mp_type mp =
+ derivative(mp_type(mp_type(3) / 2U), // x = 3/2
+ mp_type(mp_type(1) / 10000000U), // Step size 10^7.
+ [](const mp_type & x)->mp_type
+ {
+ return sqrt((x * x) - 1.) - acos (1. / x); // Function
+ }
+ );
+
+ const double d = mp.convert_to<double>(); // Convert to closest double.
+ const double rel_error = (d - 0.74535599249992989880) / 0.74535599249992989880;
+ const double bit_error = std::abs (rel_error) / std::numeric_limits<double>::epsilon();
+ std::cout.precision (std::numeric_limits <double>::digits10); // All guaranteed decimal digits.
+ std::cout << std::showpoint ; // Ensure that any trailing zeros are shown too.
+ std::cout << " derivative : " << d << std::endl;
+ // Can compute an 'exact' value using multiprecision type.
+ std::cout << " expected : " << sqrt(static_cast<mp_type>(5))/3U << std::endl;
+ std::cout << " expected : " << 0.74535599249992989880
+ << std::endl;
+ std::cout << " bit_error : " << static_cast<unsigned long>(bit_error) << std::endl;
+
+ std::cout.precision(6);
+ std::cout << "float_distance = " << boost::math::float_distance(0.74535599249992989880, d) << std::endl;
+
+
+ }
+
+
+} // int main()
+
+/*`
+The result of this program on a system with an eight-byte, 64-bit IEEE-754
+conforming floating-point representation for `double` is:
+
+ derivative : 0.745355992499951
+
+ derivative : 0.745355992499943
+ expected : 0.74535599249993
+ bit_error : 78
+
+ derivative : 0.745355992499930
+ expected : 0.745355992499930
+ bit_error : 0
+
+The resulting bit error is 0. This means that the result of the derivative
+calculation is bit-identical with the double representation of the expected result,
+and this is the best result possible for the built-in type.
+
+The derivative in this example has a known closed form. There are, however,
+countless situations in numerical analysis (and not only for numerical deriva-
+tives) for which the calculation at hand does not have a known closed-form
+solution or for which the closed-form solution is highly inconvenient to use. In
+such cases, this technique may be useful.
+
+This example has shown how multiprecision can be used to add extra digits
+to an ill-conditioned calculation that suffers from precision loss. When the result
+of the multiprecision calculation is converted to a built-in type such as double,
+the entire precision of the result in double is preserved.
+
+ */
+
+/*
+
+ Description: Autorun "J:\Cpp\big_number\Debug\numerical_derivative_example.exe"
+ derivative : 0.745355992499943
+ expected : 0.745355992499930
+ expected : 0.745355992499930
+ bit_error : 78
+ float_distance = 117.000
+ derivative : 0.745355992499930
+ expected : 0.745355992499930
+ expected : 0.745355992499930
+ bit_error : 0
+ float_distance = 0.000000
+
+ */
+