Question

我刚开始使用Java，作为第一个项目，我正在编写一个程序来查找给定数字的根（在本例中为立方根）。现在我正在尝试Newton-Ralphson来实现这一目标。这是代码 -

import java.util.Scanner;

import static java.lang.Math.abs;

public class newClass {
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        System.out.println("Number whose cube root u wanna find:");
        Double number = input.nextDouble();
        Double epsilon = 0.0001;
        Double ans = number/2.00;
        while (abs((abs(number) - abs(Math.pow(ans,3))))>epsilon){
            System.out.println("in loop");
            ans = ans - ((Math.pow(ans,3) - number)/(3*Math.pow(ans,2)));
            System.out.println(ans);
            if ((number - ans)<=epsilon){
                System.out.println(ans);
            }
        }
        //System.out.println(Math.pow(number,1.0/3.0));


    }
}

这只能达到11位数，因为它太大而无法处理IDE。但是，如果我只使用Math.pow(number,1.0/3.0)，它可以用于更大的数字，并立即计算它。

那么，Math.pow()使用哪种算法可以立即给出答案？ 我明白我的方法依赖于猜测，我猜math.pow()实际上可能正在计算答案，但是如何？

Answer 1

这是一个有趣的问题。如果你查看Java的Math类的源代码，你会发现它调用StrictMath.pow（double1，double2），而StrictMath的签名是public static native double pow(double a, double b);

因此，最终，它是一个真正的本机调用，可能因平台而异。但是，某处有一个实现，并且它不容易看到。以下是函数的描述和函数本身的代码：

注意

通过数学计算，试图理解它可能不可避免地导致更多问题。但是，通过在Java Math Function Source Code上搜索这个Github并浏览数学摘要，你可以更好地理解本机函数。快乐探索：）

方法说明

Method: Let x = 2 * (1+f) 1. Compute and return log2(x) in two pieces: log2(x) = w1 + w2, where w1 has 53-24 = 29 bit trailing zeros. 2. Perform y*log2(x) = n+y' by simulating muti-precision arithmetic, where |y'|<=0.5. 3. Return x**y = 2**n*exp(y'*log2)

特殊情况

1. (anything) ** 0 is 1 2. (anything) ** 1 is itself 3. (anything) ** NAN is NAN 4. NAN ** (anything except 0) is NAN 5. +-(|x| > 1) ** +INF is +INF 6. +-(|x| > 1) ** -INF is +0 7. +-(|x| < 1) ** +INF is +0 8. +-(|x| < 1) ** -INF is +INF 9. +-1 ** +-INF is NAN 10. +0 ** (+anything except 0, NAN) is +0 11. -0 ** (+anything except 0, NAN, odd integer) is +0 12. +0 ** (-anything except 0, NAN) is +INF 13. -0 ** (-anything except 0, NAN, odd integer) is +INF 14. -0 ** (odd integer) = -( +0 ** (odd integer) ) 15. +INF ** (+anything except 0,NAN) is +INF 16. +INF ** (-anything except 0,NAN) is +0 17. -INF ** (anything) = -0 ** (-anything) 18. (-anything) ** (integer) is (-1)**(integer)*(+anything**integer) 19. (-anything except 0 and inf) ** (non-integer) is NAN

<强>精度

pow(x,y) returns x**y nearly rounded. In particular pow(integer,integer) always returns the correct integer provided it is representable.

源代码

#ifdef __STDC__ double __ieee754_pow(double x, double y) #else double __ieee754_pow(x,y) double x, y; #endif { double z,ax,z_h,z_l,p_h,p_l; double y1,t1,t2,r,s,t,u,v,w; int i0,i1,i,j,k,yisint,n; int hx,hy,ix,iy; unsigned lx,ly; i0 = ((*(int*)&one)>>29)^1; i1=1-i0; hx = __HI(x); lx = __LO(x); hy = __HI(y); ly = __LO(y); ix = hx&0x7fffffff; iy = hy&0x7fffffff; /* y==zero: x**0 = 1 */ if((iy|ly)==0) return one; /* +-NaN return x+y */ if(ix > 0x7ff00000 || ((ix==0x7ff00000)&&(lx!=0)) || iy > 0x7ff00000 || ((iy==0x7ff00000)&&(ly!=0))) return x+y; /* determine if y is an odd int when x < 0 * yisint = 0 ... y is not an integer * yisint = 1 ... y is an odd int * yisint = 2 ... y is an even int */ yisint = 0; if(hx<0) { if(iy>=0x43400000) yisint = 2; /* even integer y */ else if(iy>=0x3ff00000) { k = (iy>>20)-0x3ff; /* exponent */ if(k>20) { j = ly>>(52-k); if((j<<(52-k))==ly) yisint = 2-(j&1); } else if(ly==0) { j = iy>>(20-k); if((j<<(20-k))==iy) yisint = 2-(j&1); } } } /* special value of y */ if(ly==0) { if (iy==0x7ff00000) { /* y is +-inf */ if(((ix-0x3ff00000)|lx)==0) return y - y; /* inf**+-1 is NaN */ else if (ix >= 0x3ff00000)/* (|x|>1)**+-inf = inf,0 */ return (hy>=0)? y: zero; else /* (|x|<1)**-,+inf = inf,0 */ return (hy<0)?-y: zero; } if(iy==0x3ff00000) { /* y is +-1 */ if(hy<0) return one/x; else return x; } if(hy==0x40000000) return x*x; /* y is 2 */ if(hy==0x3fe00000) { /* y is 0.5 */ if(hx>=0) /* x >= +0 */ return sqrt(x); } } ax = fabs(x); /* special value of x */ if(lx==0) { if(ix==0x7ff00000||ix==0||ix==0x3ff00000){ z = ax; /*x is +-0,+-inf,+-1*/ if(hy<0) z = one/z; /* z = (1/|x|) */ if(hx<0) { if(((ix-0x3ff00000)|yisint)==0) { z = (z-z)/(z-z); /* (-1)**non-int is NaN */ } else if(yisint==1) z = -1.0*z; /* (x<0)**odd = -(|x|**odd) */ } return z; } } n = (hx>>31)+1; /* (x<0)**(non-int) is NaN */ if((n|yisint)==0) return (x-x)/(x-x); s = one; /* s (sign of result -ve**odd) = -1 else = 1 */ if((n|(yisint-1))==0) s = -one;/* (-ve)**(odd int) */ /* |y| is huge */ if(iy>0x41e00000) { /* if |y| > 2**31 */ if(iy>0x43f00000){ /* if |y| > 2**64, must o/uflow */ if(ix<=0x3fefffff) return (hy<0)? huge*huge:tiny*tiny; if(ix>=0x3ff00000) return (hy>0)? huge*huge:tiny*tiny; } /* over/underflow if x is not close to one */ if(ix<0x3fefffff) return (hy<0)? s*huge*huge:s*tiny*tiny; if(ix>0x3ff00000) return (hy>0)? s*huge*huge:s*tiny*tiny; /* now |1-x| is tiny <= 2**-20, suffice to compute log(x) by x-x^2/2+x^3/3-x^4/4 */ t = ax-one; /* t has 20 trailing zeros */ w = (t*t)*(0.5-t*(0.3333333333333333333333-t*0.25)); u = ivln2_h*t; /* ivln2_h has 21 sig. bits */ v = t*ivln2_l-w*ivln2; t1 = u+v; __LO(t1) = 0; t2 = v-(t1-u); } else { double ss,s2,s_h,s_l,t_h,t_l; n = 0; /* take care subnormal number */ if(ix<0x00100000) {ax *= two53; n -= 53; ix = __HI(ax); } n += ((ix)>>20)-0x3ff; j = ix&0x000fffff; /* determine interval */ ix = j|0x3ff00000; /* normalize ix */ if(j<=0x3988E) k=0; /* |x|<sqrt(3/2) */ else if(j<0xBB67A) k=1; /* |x|<sqrt(3) */ else {k=0;n+=1;ix -= 0x00100000;} __HI(ax) = ix; /* compute ss = s_h+s_l = (x-1)/(x+1) or (x-1.5)/(x+1.5) */ u = ax-bp[k]; /* bp[0]=1.0, bp[1]=1.5 */ v = one/(ax+bp[k]); ss = u*v; s_h = ss; __LO(s_h) = 0; /* t_h=ax+bp[k] High */ t_h = zero; __HI(t_h)=((ix>>1)|0x20000000)+0x00080000+(k<<18); t_l = ax - (t_h-bp[k]); s_l = v*((u-s_h*t_h)-s_h*t_l); /* compute log(ax) */ s2 = ss*ss; r = s2*s2*(L1+s2*(L2+s2*(L3+s2*(L4+s2*(L5+s2*L6))))); r += s_l*(s_h+ss); s2 = s_h*s_h; t_h = 3.0+s2+r; __LO(t_h) = 0; t_l = r-((t_h-3.0)-s2); /* u+v = ss*(1+...) */ u = s_h*t_h; v = s_l*t_h+t_l*ss; /* 2/(3log2)*(ss+...) */ p_h = u+v; __LO(p_h) = 0; p_l = v-(p_h-u); z_h = cp_h*p_h; /* cp_h+cp_l = 2/(3*log2) */ z_l = cp_l*p_h+p_l*cp+dp_l[k]; /* log2(ax) = (ss+..)*2/(3*log2) = n + dp_h + z_h + z_l */ t = (double)n; t1 = (((z_h+z_l)+dp_h[k])+t); __LO(t1) = 0; t2 = z_l-(((t1-t)-dp_h[k])-z_h); } /* split up y into y1+y2 and compute (y1+y2)*(t1+t2) */ y1 = y; __LO(y1) = 0; p_l = (y-y1)*t1+y*t2; p_h = y1*t1; z = p_l+p_h; j = __HI(z); i = __LO(z); if (j>=0x40900000) { /* z >= 1024 */ if(((j-0x40900000)|i)!=0) /* if z > 1024 */ return s*huge*huge; /* overflow */ else { if(p_l+ovt>z-p_h) return s*huge*huge; /* overflow */ } } else if((j&0x7fffffff)>=0x4090cc00 ) { /* z <= -1075 */ if(((j-0xc090cc00)|i)!=0) /* z < -1075 */ return s*tiny*tiny; /* underflow */ else { if(p_l<=z-p_h) return s*tiny*tiny; /* underflow */ } } /* * compute 2**(p_h+p_l) */ i = j&0x7fffffff; k = (i>>20)-0x3ff; n = 0; if(i>0x3fe00000) { /* if |z| > 0.5, set n = [z+0.5] */ n = j+(0x00100000>>(k+1)); k = ((n&0x7fffffff)>>20)-0x3ff; /* new k for n */ t = zero; __HI(t) = (n&~(0x000fffff>>k)); n = ((n&0x000fffff)|0x00100000)>>(20-k); if(j<0) n = -n; p_h -= t; } t = p_l+p_h; __LO(t) = 0; u = t*lg2_h; v = (p_l-(t-p_h))*lg2+t*lg2_l; z = u+v; w = v-(z-u); t = z*z; t1 = z - t*(P1+t*(P2+t*(P3+t*(P4+t*P5)))); r = (z*t1)/(t1-two)-(w+z*w); z = one-(r-z); j = __HI(z); j += (n<<20); if((j>>20)<=0) z = scalbn(z,n); /* subnormal output */ else __HI(z) += (n<<20); return s*z; }

Java中

1 个答案: