我观察到数字越大，获得正确平方根所需的循环次数就越多。如果我确定循环次数和初始数，当计算出的数字变大时，根变得不太准确，但如果我能确保初始值不比平方根大很多，我想我总能得到一个准确的平方根。那么，如何使这个初始数字大致相同？
答案 0 :(得分：2)
double frexp(double x, int *p)
返回范围为[0.5 .. 1.0]的0或分数f。将*p设置为2的幂，使 f * 2 *p 等于x。
double ldexp(double x, int p)
返回 x * 2 p
因此，可以通过以下方式找到x的平方根的合适的第一近似值：
int p;
double f = frexp(x, &p);
double r = ldexp(f, p / 2);
正如此代码所示，您需要最多5个Newton-Raphson'除法和平均'周期才能产生完全准确的答案：
#include <stdio.h>
#include <math.h>
static double test_sqrt(double x)
{
if (x <= 0.0)
return 0.0;
int p;
double frac = frexp(x, &p);
printf(" x = %22.15e; f = %22.15e; p = %d\n", x, frac, p);
double x0 = ldexp(frac, p / 2);
printf("x0 = %22.15e; x0*x0 = %22.15e\n", x0, x0 * x0);
for (int i = 0; i < 5; i++)
{
double x1 = (x / x0 + x0) / 2.0;
printf("x1 = %22.15e; x1*x1 = %22.15e\n", x1, x1 * x1);
x0 = x1;
}
printf(" r = %22.15e; r*r = %22.15e; x = %22.15e\n", x0, x0 * x0, x);
return x0;
}
int main(void)
{
double x = 1.23456E-125;
while (x < 1E+125)
{
test_sqrt(x);
x *= 987.654;
}
return 0;
}
从输出中提取：
x = 1.234560000000000e-125; f = 5.223124843710012e-01; p = -414
x0 = 2.539342632858085e-63; x0*x0 = 6.448261007050633e-126
x1 = 3.700536659343551e-63; x1*x1 = 1.369397156714553e-125
x1 = 3.518350710213579e-63; x1*x1 = 1.237879172006039e-125
x1 = 3.513633767135102e-63; x1*x1 = 1.234562224955201e-125
x1 = 3.513630600961067e-63; x1*x1 = 1.234560000001003e-125
x1 = 3.513630600959640e-63; x1*x1 = 1.234560000000000e-125
r = 3.513630600959640e-63; r*r = 1.234560000000000e-125; x = 1.234560000000000e-125
x = 1.219318122240000e-122; f = 5.037734516005438e-01; p = -404
x0 = 7.837474714727561e-62; x0*x0 = 6.142600990399386e-123
x1 = 1.169750645469021e-61; x1*x1 = 1.368316572575191e-122
x1 = 1.106062520605688e-61; x1*x1 = 1.223374299488608e-122
x1 = 1.104228909406935e-61; x1*x1 = 1.219321484370028e-122
x1 = 1.104227387018778e-61; x1*x1 = 1.219318122242317e-122
x1 = 1.104227387017728e-61; x1*x1 = 1.219318122240000e-122
r = 1.104227387017728e-61; r*r = 1.219318122240000e-122; x = 1.219318122240000e-122
x = 8.193538364768438e-27; f = 6.339443253229819e-01; p = -86
x0 = 7.207112563753321e-14; x0*x0 = 5.194247150661096e-27
x1 = 9.287898167957462e-14; x1*x1 = 8.626505237834758e-27
x1 = 9.054816977123554e-14; x1*x1 = 8.198971048920493e-27
x1 = 9.051817090894058e-14; x1*x1 = 8.193539264700178e-27
x1 = 9.051816593794011e-14; x1*x1 = 8.193538364768461e-27
x1 = 9.051816593793999e-14; x1*x1 = 8.193538364768438e-27
r = 9.051816593793999e-14; r*r = 8.193538364768438e-27; x = 8.193538364768438e-27
x = 8.092380940117008e-24; f = 6.114430162915473e-01; p = -76
x0 = 2.224416735012882e-12; x0*x0 = 4.948029811005370e-24
x1 = 2.931197771052297e-12; x1*x1 = 8.591920373021955e-24
x1 = 2.845986967837607e-12; x1*x1 = 8.099641821101499e-24
x1 = 2.844711332870451e-12; x1*x1 = 8.092382567361578e-24
x1 = 2.844711046858202e-12; x1*x1 = 8.092380940117088e-24
x1 = 2.844711046858188e-12; x1*x1 = 8.092380940117008e-24
r = 2.844711046858188e-12; r*r = 8.092380940117008e-24; x = 8.092380940117008e-24
x = 7.511130028860641e-06; f = 9.844988351428219e-01; p = -17
x0 = 3.845698574776648e-03; x0*x0 = 1.478939752803914e-05
x1 = 2.899411787388324e-03; x1*x1 = 8.406588712846356e-06
x1 = 2.744991037655443e-03; x1*x1 = 7.534975796808706e-06
x1 = 2.740647532044505e-03; x1*x1 = 7.511148894901633e-06
x1 = 2.740644090149702e-03; x1*x1 = 7.511130028872487e-06
x1 = 2.740644090147541e-03; x1*x1 = 7.511130028860642e-06
r = 2.740644090147541e-03; r*r = 7.511130028860642e-06; x = 7.511130028860641e-06
x = 7.418397617524327e-03; f = 9.495548950431139e-01; p = -7
x0 = 1.186943618803892e-01; x0*x0 = 1.408835154219280e-02
x1 = 9.059718094019462e-02; x1*x1 = 8.207849194310363e-03
x1 = 8.624024859090236e-02; x1*x1 = 7.437380477020637e-03
x1 = 8.613019058546711e-02; x1*x1 = 7.418409730288888e-03
x1 = 8.613012026886571e-02; x1*x1 = 7.418397617529272e-03
x1 = 8.613012026883701e-02; x1*x1 = 7.418397617524329e-03
r = 8.613012026883701e-02; r*r = 7.418397617524329e-03; x = 7.418397617524327e-03
x = 7.326810080538372e+00; f = 9.158512600672964e-01; p = 3
x0 = 1.831702520134593e+00; x0*x0 = 3.355134122267418e+00
x1 = 2.915851260067297e+00; x1*x1 = 8.502188570836042e+00
x1 = 2.714301457717203e+00; x1*x1 = 7.367432403365734e+00
x1 = 2.706818441652081e+00; x1*x1 = 7.326866076067802e+00
x1 = 2.706808098230342e+00; x1*x1 = 7.326810080645360e+00
x1 = 2.706808098210579e+00; x1*x1 = 7.326810080538371e+00
r = 2.706808098210579e+00; r*r = 7.326810080538371e+00; x = 7.326810080538372e+00
x = 7.236353283284045e+03; f = 8.833439066508844e-01; p = 13
x0 = 5.653401002565660e+01; x0*x0 = 3.196094289581041e+03
x1 = 9.226700501282829e+01; x1*x1 = 8.513200214037281e+03
x1 = 8.534770092045144e+01; x1*x1 = 7.284230052406829e+03
x1 = 8.506722020095668e+01; x1*x1 = 7.236431952718052e+03
x1 = 8.506675780525467e+01; x1*x1 = 7.236353283497856e+03
x1 = 8.506675780399794e+01; x1*x1 = 7.236353283284046e+03
r = 8.506675780399794e+01; r*r = 7.236353283284046e+03; x = 7.236353283284045e+03
x = 7.147013265648620e+06; f = 8.519903738079810e-01; p = 23
x0 = 1.744876285558745e+03; x0*x0 = 3.044593251905283e+06
x1 = 2.920438142779372e+03; x1*x1 = 8.528958945800630e+06
x1 = 2.683839109930689e+03; x1*x1 = 7.202992367993553e+06
x1 = 2.673410187023612e+03; x1*x1 = 7.147122028081623e+06
x1 = 2.673389845507460e+03; x1*x1 = 7.147013266062398e+06
x1 = 2.673389845430072e+03; x1*x1 = 7.147013265648622e+06
r = 2.673389845430072e+03; r*r = 7.147013265648622e+06; x = 7.147013265648620e+06
x = 5.714937347533526e+60; f = 8.891035606430090e-01; p = 202
x0 = 2.254145324628333e+30; x0*x0 = 5.081171144543771e+60
x1 = 2.394723262542396e+30; x1*x1 = 5.734699504161695e+60
x1 = 2.390597074573772e+30; x1*x1 = 5.714954372960678e+60
x1 = 2.390593513658524e+30; x1*x1 = 5.714937347546205e+60
x1 = 2.390593513655872e+30; x1*x1 = 5.714937347533526e+60
x1 = 2.390593513655872e+30; x1*x1 = 5.714937347533526e+60
r = 2.390593513655872e+30; r*r = 5.714937347533526e+60; x = 5.714937347533526e+60
x = 5.644380731040877e+63; f = 8.575455938313579e-01; p = 212
x0 = 6.957236395157723e+31; x0*x0 = 4.840313825810723e+63
x1 = 7.535100118309196e+31; x1*x1 = 5.677773379294325e+63
x1 = 7.512942052902534e+31; x1*x1 = 5.644429829027134e+63
x1 = 7.512909377296952e+31; x1*x1 = 5.644380731147647e+63
x1 = 7.512909377225895e+31; x1*x1 = 5.644380731040878e+63
x1 = 7.512909377225895e+31; x1*x1 = 5.644380731040878e+63
r = 7.512909377225895e+31; r*r = 5.644380731040878e+63; x = 5.644380731040877e+63
x = 4.457671009241398e+120; f = 8.631370354723219e-01; p = 401
x0 = 1.387007739709396e+60; x0*x0 = 1.923790470013767e+120
x1 = 2.300441914113688e+60; x1*x1 = 5.292033000211049e+120
x1 = 2.119093716219477e+60; x1*x1 = 4.490558178120875e+120
x1 = 2.111333991241823e+60; x1*x1 = 4.457731222573127e+120
x1 = 2.111319731695020e+60; x1*x1 = 4.457671009444733e+120
x1 = 2.111319731646867e+60; x1*x1 = 4.457671009241398e+120
r = 2.111319731646867e+60; r*r = 4.457671009241398e+120; x = 4.457671009241398e+120
x = 4.402636602961303e+123; f = 8.325007281566217e-01; p = 411
x0 = 4.280886694234198e+61; x0*x0 = 1.832599088887140e+123
x1 = 7.282645088745868e+61; x1*x1 = 5.303691948863432e+123
x1 = 6.664013166606369e+61; x1*x1 = 4.440907148470304e+123
x1 = 6.635298828449913e+61; x1*x1 = 4.402719054282879e+123
x1 = 6.635236697622269e+61; x1*x1 = 4.402636603347328e+123
x1 = 6.635236697331380e+61; x1*x1 = 4.402636602961305e+123
r = 6.635236697331380e+61; r*r = 4.402636602961305e+123; x = 4.402636602961303e+123
如果您愿意，可以应用更复杂的技术（例如，改进分数的估计值），但frexp()和ldexp()函数可能很有用。
您可以在Plauger的The Standard C Library以及Bentley的More Programming Pearls中找到对此的讨论。
答案 1 :(得分：1)
那么，如何使这个初始数字大致相同？
这是一个问题22。如果你能找到一个大致相同的数字，为什么你需要解决它呢？
我认为没有任何好的公式可以用于所有数字的第一次猜测。
您可以根据数字的范围使用初始值。
例如：
1.0的所有数字，请使用2*x作为初始值。1.0和1.0E4之间的所有数字，请使用x/1.0E2作为初始值。1.0E4和1.0E8之间的所有数字，请使用x/1.0E6作为初始值。等