Question

最近我使用Math.log()编写了一些程序代码，现在我问自己方法log()是如何工作的。计算机如何计算对数？谢谢你的解决方案。

Answer 1

openjdk中Math.log的定义是：

public static double log(double a) {
    return StrictMath.log(a); // default impl. delegates to StrictMath
}

这引导我查看StrictMath的来源，其中log被声明为：

public static native double log(double a);

原生关键字indicates JNI。换句话说，它是一个libc源来找到我们正在寻找的东西。由于我使用的是NetBSD，我将发布它的日志定义：

/* __ieee754_log(x)
 * Return the logrithm of x
 *
 * Method :
 *   1. Argument Reduction: find k and f such that
 *          x = 2^k * (1+f),
 *     where  sqrt(2)/2 < 1+f < sqrt(2) .
 *
 *   2. Approximation of log(1+f).
 *  Let s = f/(2+f) ; based on log(1+f) = log(1+s) - log(1-s)
 *       = 2s + 2/3 s**3 + 2/5 s**5 + .....,
 *           = 2s + s*R
 *      We use a special Reme algorithm on [0,0.1716] to generate
 *  a polynomial of degree 14 to approximate R The maximum error
 *  of this polynomial approximation is bounded by 2**-58.45. In
 *  other words,
 *              2      4      6      8      10      12      14
 *      R(z) ~ Lg1*s +Lg2*s +Lg3*s +Lg4*s +Lg5*s  +Lg6*s  +Lg7*s
 *      (the values of Lg1 to Lg7 are listed in the program)
 *  and
 *      |      2          14          |     -58.45
 *      | Lg1*s +...+Lg7*s    -  R(z) | <= 2
 *      |                             |
 *  Note that 2s = f - s*f = f - hfsq + s*hfsq, where hfsq = f*f/2.
 *  In order to guarantee error in log below 1ulp, we compute log
 *  by
 *      log(1+f) = f - s*(f - R)    (if f is not too large)
 *      log(1+f) = f - (hfsq - s*(hfsq+R)). (better accuracy)
 *
 *  3. Finally,  log(x) = k*ln2 + log(1+f).
 *              = k*ln2_hi+(f-(hfsq-(s*(hfsq+R)+k*ln2_lo)))
 *     Here ln2 is split into two floating point number:
 *          ln2_hi + ln2_lo,
 *     where n*ln2_hi is always exact for |n| < 2000.
 *
 * Special cases:
 *  log(x) is NaN with signal if x < 0 (including -INF) ;
 *  log(+INF) is +INF; log(0) is -INF with signal;
 *  log(NaN) is that NaN with no signal.
 *
 * Accuracy:
 *  according to an error analysis, the error is always less than
 *  1 ulp (unit in the last place).
 *
 * Constants:
 * The hexadecimal values are the intended ones for the following
 * constants. The decimal values may be used, provided that the
 * compiler will convert from decimal to binary accurately enough
 * to produce the hexadecimal values shown.
 */

#include "math.h"
#include "math_private.h"

static const double
ln2_hi  =  6.93147180369123816490e-01,  /* 3fe62e42 fee00000 */
ln2_lo  =  1.90821492927058770002e-10,  /* 3dea39ef 35793c76 */
two54   =  1.80143985094819840000e+16,  /* 43500000 00000000 */
Lg1 = 6.666666666666735130e-01,  /* 3FE55555 55555593 */
Lg2 = 3.999999999940941908e-01,  /* 3FD99999 9997FA04 */
Lg3 = 2.857142874366239149e-01,  /* 3FD24924 94229359 */
Lg4 = 2.222219843214978396e-01,  /* 3FCC71C5 1D8E78AF */
Lg5 = 1.818357216161805012e-01,  /* 3FC74664 96CB03DE */
Lg6 = 1.531383769920937332e-01,  /* 3FC39A09 D078C69F */
Lg7 = 1.479819860511658591e-01;  /* 3FC2F112 DF3E5244 */

static const double zero   =  0.0;

double
__ieee754_log(double x)
{
    double hfsq,f,s,z,R,w,t1,t2,dk;
    int32_t k,hx,i,j;
    u_int32_t lx;

    EXTRACT_WORDS(hx,lx,x);

    k=0;
    if (hx < 0x00100000) {          /* x < 2**-1022  */
        if (((hx&0x7fffffff)|lx)==0)
        return -two54/zero;     /* log(+-0)=-inf */
        if (hx<0) return (x-x)/zero;    /* log(-#) = NaN */
        k -= 54; x *= two54; /* subnormal number, scale up x */
        GET_HIGH_WORD(hx,x);
    }
    if (hx >= 0x7ff00000) return x+x;
    k += (hx>>20)-1023;
    hx &= 0x000fffff;
    i = (hx+0x95f64)&0x100000;
    SET_HIGH_WORD(x,hx|(i^0x3ff00000)); /* normalize x or x/2 */
    k += (i>>20);
    f = x-1.0;
    if((0x000fffff&(2+hx))<3) { /* |f| < 2**-20 */
        if(f==zero) { if(k==0) return zero;  else {dk=(double)k;
                   return dk*ln2_hi+dk*ln2_lo;}
        }
        R = f*f*(0.5-0.33333333333333333*f);
        if(k==0) return f-R; else {dk=(double)k;
                 return dk*ln2_hi-((R-dk*ln2_lo)-f);}
    }
    s = f/(2.0+f);
    dk = (double)k;
    z = s*s;
    i = hx-0x6147a;
    w = z*z;
    j = 0x6b851-hx;
    t1= w*(Lg2+w*(Lg4+w*Lg6));
    t2= z*(Lg1+w*(Lg3+w*(Lg5+w*Lg7)));
    i |= j;
    R = t2+t1;
    if(i>0) {
        hfsq=0.5*f*f;
        if(k==0) return f-(hfsq-s*(hfsq+R)); else
             return dk*ln2_hi-((hfsq-(s*(hfsq+R)+dk*ln2_lo))-f);
    } else {
        if(k==0) return f-s*(f-R); else
             return dk*ln2_hi-((s*(f-R)-dk*ln2_lo)-f);
    }
}

如果您需要更多帮助，请发表评论，我会详细说明。

Answer 2

如果怀疑，请查看source code（使用Java 8）。

public static double log(double a) {
    return StrictMath.log(a); // default impl. delegates to StrictMath
}

您看到它从StrictMath.java调用静态方法。

static native double log(double a);

native个关键字表示您可以阅读here。基本上，它说该方法是用Java之外的另一种语言实现的。我在GitHub上找到了带有数学函数的文件夹jdk/src/share/native/java/lang/fdlibm/src/。该算法在e_log.c中实现 - 其余部分对我来说仍然是一种魔力，因为计算时会咬人。

Answer 3

对数函数（以及许多其他函数，例如sin（x））可以用系列近似： https://en.wikipedia.org/wiki/Natural_logarithm#Series

如果需要足够好的逼近度（但不是全精度），则可以在系列的前几次迭代后停止。这就是在早期的游戏/隔代编程中采用三角函数的方式。

查找表也是一种很好的方法，这里的关键是找到合适的查找值分布（在这种情况下，线性可能不是理想的选择））。

Math.log背后的算法 - Java

3 个答案: