C++ exp（-x）和exp（+；x）之间的泰勒级数差
我正试图编写一个程序来计算exp（-x）和exp（x）的泰勒级数，对于大x，最多迭代200次。（exp（x）=1+x+x^2/2+…） 我的程序真的很简单，看起来应该可以完美地工作。然而，对于exp（-x）它会发散，但是对于exp（+x）它会收敛得很好。以下是我目前的代码：C++ exp（-x）和exp（+；x）之间的泰勒级数差,c++,iteration,rounding,series,function-approximation,C++,Iteration,Rounding,Series,Function Approximation,我正试图编写一个程序来计算exp（-x）和exp（x）的泰勒级数，对于大x，最多迭代200次。（exp（x）=1+x+x^2/2+…） 我的程序真的很简单，看起来应该可以完美地工作。然而，对于exp（-x）它会发散，但是对于exp（+x）它会收敛得很好。以下是我目前的代码： long double x = 100.0, sum = 1.0, last = 1.0; for(int i = 1; i < 200; i++) { last *= x / i; //mul
long double x = 100.0, 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
}
cout << "exp(+x) = " << sum << endl;
x = -100.0; //redo but now letting x<0
sum = 1.0;
last = 1.0;
for(int i = 1; i < 200; i++) {
last *= x / i;
sum += last;
}
cout << "exp(-x) = " << sum << endl;
exp(+x) = 2.68811714182e+43
exp(-x) = 3.72007597602e-44
提前谢谢 我认为这实际上与浮点近似误差无关，我认为还有另一个更重要的误差来源 正如你自己所说，你的方法非常简单。您将在
x=0x=-100x=0附近的狭窄区域内准确。泰勒近似定理告诉你，例如，如果你取N
关于x=0
的级数项，你的近似至少会精确到O（| x |）^（N+1）
。因此，如果使用200个术语，则应精确到10^（-60）
范围内的[-0.5,0.5]
左右。但在x=100时，泰勒定理只给出了一个非常糟糕的界
从概念上讲，当x
变为负无穷大时，e^{-x}
趋于零。但你的近似函数是一个固定次数的多项式，任何非常数多项式都趋向于渐近正无穷大或负无穷大。因此，如果考虑“代码> x的全部可能值范围，则相对误差必须是无限的。
总之，我认为你应该重新考虑你的方法。你可以考虑的是，只使用泰勒级数法来求< <代码> x >代码>满足<代码> -0.fp的泰勒多项式可能不是一个好主意。有关使用切比雪夫多项式进行函数逼近的文章，请参阅
rickandross指出了这种情况下的错误来源，即exp（-100）的泰勒展开式包含了大值的差异
对Taylor尝试进行了一个简单的修改，可以为我尝试的几个测试用例获得合理的答案，即使用exp（-x）=1/exp（x）这一事实。本方案：
#include <iostream>
#include <cmath>
double texp(double x)
{
double last=1.0;
double sum=1.0;
if(x<0)
return 1/texp(-x);
for(int i = 1; i < 200; i++) {
last *= x / i;
sum += last;
}
return sum;
}
void test_texp(double x)
{
double te=texp(x);
double e=std::exp(x);
double err=te-e;
double rerr=(te-e)/e;
std::cout << "x=" << x
<< "\ttexp(x)=" << te
<< "\texp(x)=" << e
<< "\terr=" << err
<< "\trerr=" << rerr
<< "\n";
}
int main()
{
test_texp(0);
test_texp(1);
test_texp(-1);
test_texp(100);
test_texp(-100);
}
这取决于你在“它对正计算很有效”中所说的“好”是什么意思。就绝对值而言，正指数的结果实际上比负指数的结果离真实答案更远|2.68811691354e+43-2.68811714182e+43 |=2.2828e+36表示e^100与|-8.42078025179e+24-3.72007597602e-44 |=8.42078e+24表示e^-100Ah是的，对不起，我指的是罚款，相对误差为正指数~0.00000001，负指数~10^68：（这是为了兴趣吗？或者你有一个应用程序，你可以使用双精度数字，但是不能使用标准库ExpEd）？@ RryYyKek这仅仅是为了兴趣，我对C++是相当新的，我正在尝试获得精确和舍入错误的处理。哇，非常感谢！这非常有帮助，我非常感谢这个非常周到/有洞察力的答案。我非常喜欢这种混合方法。这是一个很好的答案，除了第一段。一些求和的绝对值（远！）大于1，误差是由舍入误差造成的（如rickandross所述）@RoryWorker：谢谢你的评论，我已经编辑了我的答案。我对这里的浮点错误的严重程度做了一些调查，现在附在答案上。有一些严重的浮点错误正在发生，但我认为公平地说，它不是这里的主要错误源。
#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);
}
$ ./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
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
#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);
}
#include <iostream>
#include <cmath>
double texp(double x)
{
double last=1.0;
double sum=1.0;
if(x<0)
return 1/texp(-x);
for(int i = 1; i < 200; i++) {
last *= x / i;
sum += last;
}
return sum;
}
void test_texp(double x)
{
double te=texp(x);
double e=std::exp(x);
double err=te-e;
double rerr=(te-e)/e;
std::cout << "x=" << x
<< "\ttexp(x)=" << te
<< "\texp(x)=" << e
<< "\terr=" << err
<< "\trerr=" << rerr
<< "\n";
}
int main()
{
test_texp(0);
test_texp(1);
test_texp(-1);
test_texp(100);
test_texp(-100);
}
x=0 texp(x)=1 exp(x)=1 err=0 rerr=0
x=1 texp(x)=2.71828 exp(x)=2.71828 err=4.44089e-16 rerr=1.63371e-16
x=-1 texp(x)=0.367879 exp(x)=0.367879 err=-5.55112e-17 rerr=-1.50895e-16
x=100 texp(x)=2.68812e+43 exp(x)=2.68812e+43 err=1.48553e+28 rerr=5.52628e-16
x=-100 texp(x)=3.72008e-44 exp(x)=3.72008e-44 err=-2.48921e-59 rerr=-6.69128e-16