我认为这实际上与浮点逼近误差没有任何关系，我认为还有另一个更重要的误差来源。
正如您自己所说，您的方法非常简单。您正在x=0对该函数进行泰勒级数逼近，然后在 处对其进行评估x=-100。
您实际上期望这种方法有多准确，为什么？
在高层次上，您应该只期望您的方法在 附近的狭窄区域内是准确的x=0。泰勒逼近定理告诉您，例如，如果您采用N围绕 的级数x=0，您的逼近将O(|x|)^(N+1)至少精确到 。因此，如果您采用 200 个术语，您应该准确到例如在10^(-60)range 内左右[-0.5, 0.5]。但是在x=100泰勒定理中只给了你一个非常可怕的界限。
从概念上讲，您知道随着趋向于负无穷大，它e^{-x}趋于零。x但是您的近似函数是一个固定次数的多项式，并且任何非常数多项式都趋向于渐近地正无穷或负无穷。因此，如果考虑 的可能值的整个范围，则相对误差必须是无限的x。
所以简而言之，我认为你应该重新考虑你的方法。您可能会考虑的一件事是，仅对x满足的值使用泰勒级数方法-0.5f <= x <= 0.5f。对于任何x大于0.5f，您尝试除以x二并递归调用函数，然后对结果进行平方。或者类似的东西。
为获得最佳结果，您可能应该使用已建立的方法。
编辑：
我决定写一些代码来看看我的想法到底有多好。它似乎与整个范围内的 C 库实现完美匹配x = -10000，x = 10000至少达到我所显示的精度。:)
x另请注意，即使对于大于 100的值，我的方法也是准确的，其中泰勒级数方法实际上在正端也失去了准确性。
#include <cmath>
#include <iostream>
long double taylor_series(long double x)
{
long double sum = 1.0, last = 1.0;
for(int i = 1; i < 200; i++) {
last *= x / i; //multiply the last term in the series from the previous term by x/n
sum += last; //add this new last term to the sum
}
return sum;
}
long double hybrid(long double x)
{
long double temp;
if (-0.5 <= x && x <= 0.5) {
return taylor_series(x);
} else {
temp = hybrid(x / 2);
return (temp * temp);
}
}
long double true_value(long double x) {
return expl(x);
}
void output_samples(long double x) {
std::cout << "x = " << x << std::endl;
std::cout << "\ttaylor series = " << taylor_series(x) << std::endl;
std::cout << "\thybrid method = " << hybrid(x) << std::endl;
std::cout << "\tlibrary = " << true_value(x) << std::endl;
}
int main() {
output_samples(-10000);
output_samples(-1000);
output_samples(-100);
output_samples(-10);
output_samples(-1);
output_samples(-0.1);
output_samples(0);
output_samples(0.1);
output_samples(1);
output_samples(10);
output_samples(100);
output_samples(1000);
output_samples(10000);
}
输出：
$ ./main
x = -10000
taylor series = -2.48647e+423
hybrid method = 1.13548e-4343
library = 1.13548e-4343
x = -1000
taylor series = -2.11476e+224
hybrid method = 5.07596e-435
library = 5.07596e-435
x = -100
taylor series = -8.49406e+24
hybrid method = 3.72008e-44
library = 3.72008e-44
x = -10
taylor series = 4.53999e-05
hybrid method = 4.53999e-05
library = 4.53999e-05
x = -1
taylor series = 0.367879
hybrid method = 0.367879
library = 0.367879
x = -0.1
taylor series = 0.904837
hybrid method = 0.904837
library = 0.904837
x = 0
taylor series = 1
hybrid method = 1
library = 1
x = 0.1
taylor series = 1.10517
hybrid method = 1.10517
library = 1.10517
x = 1
taylor series = 2.71828
hybrid method = 2.71828
library = 2.71828
x = 10
taylor series = 22026.5
hybrid method = 22026.5
library = 22026.5
x = 100
taylor series = 2.68812e+43
hybrid method = 2.68812e+43
library = 2.68812e+43
x = 1000
taylor series = 3.16501e+224
hybrid method = 1.97007e+434
library = 1.97007e+434
x = 10000
taylor series = 2.58744e+423
hybrid method = 8.80682e+4342
library = 8.80682e+4342
编辑：
对于谁感兴趣：
评论中提出了一些关于浮点错误在原始程序中的重要性的问题。我最初的假设是它们可以忽略不计——我做了一个测试，看看这是否正确。事实证明这是不正确的，并且存在严重的浮点错误，但即使没有浮点错误，也存在仅由泰勒级数引入的重大错误。泰勒级数在 200 项时的真实值x=-100似乎接近-10^{24}，而不是10^{-44}。我使用 进行了检查boost::multiprecision::cpp_rational，这是一种基于任意精度整数类型的任意精度有理类型。
输出：
x = -100
taylor series (double) = -8.49406e+24
(rational) = -18893676108550916857809762858135399218622904499152741157985438973568808515840901824148153378967545615159911801257288730703818783811465589393308637433853828075746484162303774416145637877964256819225743503057927703756503421797985867950089388433370741907279634245166982027749118060939789786116368342096247737/2232616279628214542925453719111453368125414939204152540389632950466163724817295723266374721466940218188641069650613086131881282494641669993119717482562506576264729344137595063634080983904636687834775755173984034571100264999493261311453647876869630211032375288916556801211263293563
= -8.46257e+24
library = 3.72008e-44
x = 100
taylor series (double) = 2.68812e+43
(rational) = 36451035284924577938246208798747009164319474757880246359883694555113407009453436064573518999387789077985197279221655719227002367495061633272603038249747260895707250896595889294145309676586627989388740458641362406969609459453916777341749316070359589697827702813520519796940239276744754778199440304584107317957027129587503199/1356006206645357299077422810994072904566969809700681604285727988319939931024001696953196916719184549697395496290863162742676361760549235149195411231740418104602504325580502523311497039304043141691060121240640609954226541318710631103275528465092597490136227936213123455950399178299
= 2.68812e+43
library = 2.68812e+43
代码：
#include <cmath>
#include <iostream>
#include <boost/multiprecision/cpp_int.hpp>
typedef unsigned int uint;
typedef boost::multiprecision::cpp_rational rational;
// Taylor series of exp
template <typename T>
T taylor_series(const T x) {
T sum = 1, last = 1;
for (uint i = 1; i < 200; i++) {
last = last * (x / i);
sum = sum + last;
}
return sum;
}
void sample(const int x) {
std::cout << "x = " << x << std::endl;
long double e1 = taylor_series(static_cast<long double>(x));
std::cout << "\ttaylor series (double) = " << e1 << std::endl;
rational e2 = taylor_series(static_cast<rational>(x));
std::cout << "\t (rational) = " << e2 << std::endl;
std::cout << "\t = " << static_cast<long double>(e2) << std::endl;
std::cout << "\tlibrary = " << expl(static_cast<long double>(x)) << std::endl;
}
int main() {
sample(-100);
sample(100);
}