IEEE浮点异常 - 为什么？

Question

我正在制作一个简单的函数来重现Secant方法,我认为我有一些精确的问题.

首先,这是函数(以及调用它的main方法,以及与它一起使用的测试函数):

double secant_method(double(*f)(double), double a, double b){
    double c;
    for (int i = 0; i < 10; i++){
        c = a - f(a) * (a - b) / (f(a) - f(b));
        b = a; a = c;
    }
    return c;
}

typedef double (*func)(double);

//test function - x^3 + 4x - 10
double test(double x){
    return (x*x*x) + (4*x) - 10;
}

int main(){
    func f = &test;
    double ans;

    ans = secant_method(f, 0, 2);

    printf("\nRoot found:\t%.*g\n", DECIMAL_DIG * 2, ans);

    return 0;
}

注意:for函数中的secant_method()循环仅循环10次.这就是我的问题所在.

当按原样打印时,一切正常.它将正确的输出提供给16位小数:

但是,当我向for循环中添加迭代时secant_method(),会发生以下情况:

为什么会这样？我是否达到了C可以处理的最大代表性？

---

我从另一篇文章中读到了这个伟大的答案,但在其中,关于我收到的异常(-1.#IND),它只是说我的结果不是数字,或者我正在做某种非法操作.

编辑:用(x*x*x) + (4*x) - 10 + sin(x)我的测试功能给出了正确的答案-但只有当我循环时i < 9,而不是i < 10为(x*x*x) + (4*x) - 10

Answer 1

-1.#IND是微软输出不确定值的方式,特别是NaN.

一的会发生这种情况的办法是0 / 0,但我会检查所有的操作,看看那里的问题所在:

double secant_method(double(*f)(double), double a, double b){
    double c;
    printf("DBG =====\n");
    for (int i = 0; i < 10; i++){
        printf("\nDBG -----\n");
        printf("DBG i: %d\n",i);
        printf("DBG a: %30f\n",a);
        printf("DBG b: %30f\n",b);
        printf("DBG c: %30f\n",c);
        printf("DBG f(a): %30f\n",f(a));
        printf("DBG a-b: %30f\n",a-b);
        printf("DBG f(b): %30f\n",f(b));
        printf("DBG f(a)-f(b): %30f\n",f(a)-f(b));
        printf("DBG f(a)*(a-b): %30f\n",f(a)*(a-b));
        printf("DBG f(a)*(a-b)/(f(a)-f(b)): %30f\n",f(a)*(a-b)/(f(a)-f(b)));

        c = a - f(a) * (a - b) / (f(a) - f(b));
        b = a; a = c;
    }
    return c;
}

一旦你有了调试输出,那么你就可以弄清楚实际问题是什么,并采取策略来避免它.

当我这样做时,我看到(最后):

DBG -----
DBG i: 8
DBG a: 1.556773264394211375716281509085
DBG b: 1.556773264393484179635152031551
DBG c: 1.556773264394211375716281509085
DBG f(a): -0.000000000000000987057657830803
DBG a-b: 0.000000000000727196081129477534
DBG f(b): -0.000000000008196943991622962500
DBG f(a)-f(b): 0.000000000008195956933965131697
DBG f(a)*(a-b): -0.000000000000000000000000000718
DBG f(a)*(a-b)/(f(a)-f(b)): -0.000000000000000087577871187781

DBG -----
DBG i: 9
DBG a: 1.556773264394211375716281509085
DBG b: 1.556773264394211375716281509085
DBG c: 1.556773264394211375716281509085
DBG f(a): -0.000000000000000987057657830803
DBG a-b: 0.000000000000000000000000000000
DBG f(b): -0.000000000000000987057657830803
DBG f(a)-f(b): 0.000000000000000000000000000000
DBG f(a)*(a-b): -0.000000000000000000000000000000
DBG f(a)*(a-b)/(f(a)-f(b)): nan

Root found:     nan

因此,你可以看到,在第十次迭代,a并b已变得平等,因此有f(a)和f(b).所以你得到的表达方式是:

something * 0 / 0

如上所述,它将给你0 / 0或NaN.

就如何修复它而言,你只需要避免除以零,因为这会给你带来NaN无穷或无穷.所以你可以改用以下函数:

double secant_method(double(*f)(double), double a, double b){
    double c;
    for (int i = 0; i < 1000; i++) {
        if (f(a) == f(b)) break;
        c = a - f(a) * (a - b) / (f(a) - f(b));
        b = a; a = c;
    }
    return c;
}

一千个循环应该足以得到一个合适的答案,如果你要分零,它会提前选择退出.

如果您想要更高的精度,您可以查看long double类型或切换到使用任意精度算术库之一,如GMP或MPIR.

这通常是更多的工作,但你可以取得一些令人印象深刻的结果.这个程序建立在MPIR上:

#include <stdio.h>
#include <mpir.h>

void secant_method(mpf_t result, void(*f)(mpf_t, mpf_t), mpf_t a, mpf_t b){
    mpf_t c, fa, fb, temp1, temp2;

    mpf_init (fa);
    mpf_init (fb);
    mpf_init (temp1);
    mpf_init (temp2);

    for (int i = 0; i < 1000; i++){
        printf("DBG i: %d\n",i);

        f (fa, a);
        f (fb, b);
        if (mpf_cmp (fa, fb) == 0) break;

        mpf_set (temp1, a);
        mpf_sub (temp1, temp1, b);

        mpf_set (temp2, fa);
        mpf_sub (temp2, temp2, fb);

        mpf_set (result, fa);
        mpf_mul (result, result, temp1);
        mpf_div (result, result, temp2);
        mpf_sub (result, result, a);
        mpf_neg (result, result);

        mpf_set (b, a);
        mpf_set (a, result);
    }
}

void test (mpf_t result, mpf_t x){
    mpf_t temp;

    mpf_set (result, x);
    mpf_pow_ui (result, result, 3);

    mpf_init_set (temp, x);
    mpf_mul_ui (temp, temp, 4);

    mpf_add (result, result, temp);

    mpf_set_ui (temp, 10);
    mpf_sub (result, result, temp);

    mpf_clear (temp);
}

int main(){
    mpf_t ans, a, b;

    mpf_set_default_prec (8000);

    mpf_init_set_ui (ans, 0);
    mpf_init_set_ui (a, 0);
    mpf_init_set_ui (b, 2);

    secant_method (ans, &test, a, b);

    mpf_out_str (stdout, 10, 0, ans);

    return 0;
}

输出更精确,大约两千五百个数字:

DBG i: 1
:
DBG i: 19
0.155677326439421146326886324730853302634853266143
22856485101283627988036767055520913212330822780959
93349183787687346999781239000417393618333668026011
02048595843228945228507966189601958673920851932189
20626590635658264390975889008832048255537650792123
54916373054888140164770654992918100928227714960414
65208113116379497717707745267800989233875981344305
90022883167106124203999713536673991376957068731244
91919087980169395013246250812213656324598765244218
15974098310512802880727074335472786858740154287363
31949470951650710072488856623955478366217474755111
76368234254761541647442609230138418167182918204711
66713459423756284737546964906061587903876515793884
14091165347411853670752820576131460960421137744435
73729141652832258144582021037373967987171478026002
48487515446248979731517957120705447608265161099693
33098235693813752370774508652788986557620510981156
19907950657355934071535840759135251701581523712307
00051674680667972152582339710574822560693109306285
91240827697915787078746087225027856691436076089912
35551789799825731841345891629028445554314717823386
07885164744100235567602875364878328805811271289098
87558119684442289199181352023304600847178256323082
57317198584882656089836229208443415369358460418542
84083408696290686178971039756668669303212658278679
39542421457300944206839268283788585029652481323614
65995074020560963212330914882733926627309382310653
39023265929195094492468196461296569155421718696631
73798097369621805062145075113127308161572398104766
37356504104570136778437926442139603916930640425421
15655156674699552536588332891562053247342008145504
44336211031437923307615880759201695011419324719812
46482293928341901673056596202744639074280785106031
90197472588293352508389295101867514582271001202777
85575614897203080940643669476500979934666490279524
88486176409290187337498631681392563044899541391612
88438904336237873504970887963071622208868799638373
42186338496601471274609131141920820263780493617795
89714798662834913192777810386631915415021934333441
01797098172897161215116673422762953435902633516501
73788202968876596925999628999004575114529754782488
59959395407324243559011982543407738505315960009874
36510513519775603567237051670918870105777288994910
85524037720122749091827520695838000086150188462000
63190624219373460624686216781527327604063990319908
56547016812115842640285111265677758613385414834511
69237199199725030839166586376374587900611430229333
87296847315023767826706323911923435564643861604120
017381909481e1

并且,如果您将该数字传递回test()函数,则会得到一个相当接近零的数字.所以我认为使用它是一个相当大的改进.-1.15 x 10-2408double

并且,对于它的价值而言,它只需要大约十分之一秒的CPU时间,因此使用任意精度算法进行此操作至少是可行的.

要获得更高的精度,只需更改MPIR的默认精度设置,当前设置为:

mpf_set_default_prec (8000);

最高可达100,000的答案可以得到超过30,000个有效数字的答案,以及最后的"接近于零"答案.-5 x 10-30103