sqrt sin cos exp 函数泰勒公式和迈克劳林实现

sin cos exp 是用泰勒公式和麦克劳林公式来计算。

为防止幂运算指数过高，在计算较大输入参数的时候容易导致溢出，考虑到sin和cos都是以2*PI为周期的，所以在函数内设置一个阀值（可自 行修改，此处使用2*PI作为阀值），当实参大于阀值的时候，将其计算到-2*PI~2*PI象限，既防止了大数溢出问题，又提高了运算速度。

另外，误差范围也可自行控制。

sqrt的计算是采用一个逼近算法，此处设置了一个数组，作为逼近算法的起始基值。

exp的计算因子存在一个大数溢出的问题，此处采用步数来限制无穷次计算，步数值为经验值，不能全面计算该函数的所有值。

//mathtest.c
#include "stdio.h"
#include "math.h"

#define PI 3.14156
float sinx(float x);
float fun_sinx(float x, int m);
float expx(float x);
float fun_exp(float x,int n);
float sqrtx(float t);
float cosx(float x);
float fun_cos(float x, int m);
int main()
{
float x = PI/2;
printf("sin(%f)=%f\n",x,sin(x));
printf("sinx(%f)=%f\n",x,sinx(x));
printf("cos(%f)=%f\n",x,cos(x));
printf("cosx(%f)=%f\n",x,cosx(x));
x = PI/1.3;
printf("sin(%f)=%f\n",x,sin(x));
printf("sinx(%f)=%f\n",x,sinx(x));
printf("cos(%f)=%f\n",x,cos(x));
printf("cosx(%f)=%f\n",x,cosx(x));
x = PI/2.3;
printf("sin(%f)=%f\n",x,sin(x));
printf("sinx(%f)=%f\n",x,sinx(x));
printf("cos(%f)=%f\n",x,cos(x));
printf("cosx(%f)=%f\n",x,cosx(x));
x = PI/0.3;
printf("sin(%f)=%f\n",x,sin(x));
printf("sinx(%f)=%f\n",x,sinx(x));
printf("cos(%f)=%f\n",x,cos(x));
printf("cosx(%f)=%f\n",x,cosx(x));
x = 4.5*PI;
printf("sin(%f)=%f\n",x,sin(x));
printf("sinx(%f)=%f\n",x,sinx(x));
printf("cos(%f)=%f\n",x,cos(x));
printf("cosx(%f)=%f\n",x,cosx(x));
x = 1999;
printf("sin(%f)=%f\n",x,sin(x));
printf("sinx(%f)=%f\n",x,sinx(x));
printf("cos(%f)=%f\n",x,cos(x));
printf("cosx(%f)=%f\n",x,cosx(x));
x = 1.0;
printf("exp(%f) = %f\n",x,exp(x));
printf("expx(%f) = %f\n",x,expx(x));

x = 2.0;
printf("exp(%f) = %f\n",x,exp(x));
printf("expx(%f) = %f\n",x,expx(x));

x= .0014;
printf("sqrt(%f) = %f\n",x,sqrt(x));
printf("sqrtx(%f) = %f\n",x,sqrtx(x));
printf("hello, math!\n");
return 0;
}

float sinx(float x)
{
int m = 1;
float tempRet;
float retVal = 0.0;

float Pi = 3.1415926;

if (x > 2*Pi || x < -2*Pi)
{
   x -=((int)(x/(2*Pi)))*(2*Pi);
}

do
{
   tempRet = fun_sinx(x,m);
   retVal += tempRet;
   m++;
} while (tempRet<-.0000005||tempRet>0.0000005);

return retVal;
}

float fun_sinx(float x, int m)
{
float ret = 0.0;
int i = 0;

if (m%2 == 0)
{
   ret = -1.0;
}else
{
   ret = 1.0;
}

for (i=1;i<=2*m-1;i++)
{
   ret = ret * x/i;
}

return ret;
}

float expx(float x)
{
float retVal = 1.0;
float tempRet;
int n = 1;

int step;

if(x>10)

    step = 20;

else

   step = 40;

do
{
   tempRet = fun_exp(x,n);
   retVal += tempRet;
   n++;
} while ((tempRet<-0.0000005||tempRet>0.0000005)&&(n<step));

return retVal;
}

float fun_exp(float x,int n)
{
float ret = 1.0;
int i = 0;

for (i=1;i<=n;i++)
{
   ret = ret*x/i;
}

return ret;
}
float sqrt_array[] ={0.0,1.0,4.0,3.0*3.0,4.0*4.0,5.0*5.0,6.0*6.0,7.0*7.0,8.0*8.0,9.0*9.0,10.0*10.0,100.0*100.0,1000.0*1000.0};
float sqrtx(float t)
{
float sqrt_base =1.0;
int i = 0;
float sqrt_ret;
float temp;

while(i<=12&&t>sqrt_array[i++]) ;//遍历寻找一个基的范围

if (i>=1&&i<=11)//0.0 < t < 100.0*100.0
{
   sqrt_base *= (i-1.0);
}
else if (i==12)
{
   sqrt_base = 100.0;
}else if (i==13)
{
   sqrt_base = 1000.0;
}

sqrt_ret = sqrt_base;
temp = (sqrt_ret * sqrt_ret - t)/t;

while (temp>0.00005 || temp < -0.00005)
{
   sqrt_ret = (sqrt_base + t/sqrt_base)/2.0;
   sqrt_base = sqrt_ret;
   temp = (sqrt_ret * sqrt_ret - t)/t;
}

return sqrt_ret;

}

float fun_cos(float x, int m)
{
float ret_val;
int i;

if (m%2 == 0)
{
   ret_val = 1.0;
}else
{
   ret_val = -1.0;
}

for (i=1;i<=2*m;i++)
{
   ret_val = ret_val * x/i;
}

return ret_val;
}

float cosx(float x)
{
float ret_val = 1.0;
float temp_ret;
int m = 1;
float Pi = 3.1415926;

if (x > 2*Pi || x < -2*Pi)
{
   x = x-((int)(x/(2*Pi)))*(2*Pi);
}

do
{
   temp_ret = fun_cos(x,m++);
  
   ret_val += temp_ret;

} while (temp_ret>0.00005 || temp_ret<-0.00005);

return ret_val;
}
from：http://blog.edu.cn/user1/12168/archives/2007/1921714.shtml