C程序——泰勒系列_长公式 [英] C program - taylor series_long formula

问题描述

这个公式来自我的一个朋友——我为他修正了它.但我似乎无法弄清楚如何获得每个角度的正确正弦计算.有人可以帮我在罪恶部分获得正确的命令吗?

This formula is from a friend of mine --- and I fixed it up for him. But I can't seem to figure out of how to get the right sine calculations per angle. Can someone please help me in getting the right commands in the sin part?

代码:

 #include<stdio.h>
#define PI 3.141592653589
#define NUMBER_OF_TERMS 10


double factorial(double x)
 {
  double counter, total;
  counter=x;
  total=x;
  while(counter>1)
  {
     counter--;
     total = total * counter;
  }
  return total;
  }
  double power(double x, double y)
  {
  double counter, j;
  counter=0;
  j = x;
  while (counter<(y-1))
  {
        counter++;
        x = x * j;
  }
  return x;
  }
  double cosine_func(double radians)
  {
  int counter, x;
  double cosine;
  x=0;
  counter=0;
  cosine = 0;
  while(counter<NUMBER_OF_TERMS-1)
  {
             counter++;
             x=x+2;
             if(counter%2 == 0)
              {
                  cosine = cosine - (power(radians, x)/factorial(x));
              }
             else if(counter%2 != 0)
             {
                  cosine = cosine + (power(radians, x)/factorial(x));
             }

  }
  return cosine;
  }
  double sine_func(double radians)
  {
  int counter, x;
  double sin;
  x=0;
  counter=0;
  sin = 0;
  while(counter<NUMBER_OF_TERMS-2)
  {
             counter++;
             x=x+3;
             if(counter%2 != 0)
              {
                  sin= sin -(power(radians, x)/factorial(x));
              }
             else if(counter%2 != 0)
             {
                sin= sin + (power(radians, x)/factorial(x));
             }
    }
  return sin;

    } 
   main()
 { 
  double counter;       
  double x_angles;         
  double x_radians;  
  double cosine;   
  double sin;      


  printf("11526769\n\n");
  printf("x(degrees)\t   x(radians)\t\t     cos x\t\t  sin x\t\t");
  counter = 0;
  x_angles = -185;
  while (x_angles<180)
  {
        counter++;
        x_angles+=5;
        x_radians=x_angles*PI/180.0;
        cosine=1-cosine_func(x_radians);
        sin=x_radians-sine_func(x_radians);
        printf("%.2lf\t\t%.12lf\t\t%.12lf\t\t%.12lf\n", x_angles, x_radians, cosine, sin);
  }
  return 0;
  } 

推荐答案

允许任意数量术语的代码

从https://stackoverflow.com/a/28227419/3088138

避免重新计算递归函数(例如相同参数的幂和阶乘)始终是一个好主意.因此，简化为(最少量的)基本算术运算如下所示:

Avoiding the recomputation of recursive functions like powers of the same argument and factorials is always a good idea. Thus a reduction to (a minimal amount of) elementary arithmetic operations looks like this:

public static double sine(int terms, double x) {
    double result = 1;
    double mxx = -x*x;
    double addens = 1;
    double temp = 2;
    for(int n = 2; n <= terms; n++) {
        addens *= mxx/temp++/temp++;
        result += addens; 
    }
    return x*result;

}

<小时>

要研究生产质量实现，您可以访问基本计算器 bc 的数学库 libmath，gnu 版本可以在 http://code.metager.de/source/xref/gnu/bc/1.06/bc/libmath.b

它规范了论点wrt.pi，然后使用罪的泰勒级数.

It normalizes the argument wrt. pi and then uses the Taylor series of sin.

如果您被限制在幂级数中的固定项数，您可以使用类似 Horner 的代码进行计算.例如，取 4 个术语，然后

If you are restricted to a fixed number of terms in the power series, you can use a Horner like code for its evaluation. As example, take 4 terms, then

x-x^3/3!+x^5/5!-x^7/7! = x*(1-x²/6*(1-x²/20*(1-x²/42)))

这可以放入一个循环中

res = 1
for( k=num_terms; k-->1;) {
    res = 1-x*x*res/(2*k*(2*k+1))
}
res = res*x;

或者你可以展开循环

res = 1-x*x/42;
res = 1-x*x/20*res;
res = 1-x*x/6*res;
res = x*res;

您可以自由组合 x2=x*x 或重命名 x=angle_radians.

You are free to combine x2=x*x or to rename x=angle_radians.

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