exp（-x）和exp（+ x）之间的泰勒级数差

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源。

就像您自己说的那样，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，然后在处对其进行求值x=-100。

您实际上期望这种方法的准确性如何？为什么？

在较高的水平上，您应该只希望您的方法在附近的狭窄区域内是准确的x=0。泰勒的逼近定理告诉您，例如，如果您N围绕进行级数运算x=0，则您的逼近度O(|x|)^(N+1)至少将是准确的。因此，如果您使用200个字词，则您应该准确到例如10^(-60)范围内的[-0.5, 0.5]。但是在x=100泰勒定理中，这仅给您一个非常可怕的界限。

从概念上讲，您知道e^{-x}随着x负无穷大，趋向于零。但是您的逼近函数是一个固定次数的多项式，任何非常数多项式都趋于渐近于无穷大或负无穷大。因此，如果考虑的可能值的整个范围，则相对误差必须不受限制x。

简而言之，我认为您应该重新考虑您的方法。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x大于0.5f，请尝试除以x2并递归调用该函数，然后对结果求平方。或类似的东西。

为了获得最佳结果，您可能应该使用既定方法。

编辑：

我决定编写一些代码来查看我的想法实际如何运作。这似乎与整个范围的C库实现完美匹配x = -10000，以x = 10000至少为精度位数为我展示。:)

还要注意，即使对于x大于100的值，我的方法也是准确的，泰勒级数方法实际上在正端也会失去精度。

#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：

注释中提出了一些有关原始程序中浮点错误有多重要的问题。我最初的假设是它们可以忽略不计-我进行了测试以查看是否正确。事实证明这是不正确的，并且存在很大的浮点错误，但是即使没有浮点错误，仅taylor系列也会引入很大的错误。泰勒级数在200项时的真实价值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，这是建立在其任意精度整数类型上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);
}