从函数的平方根返回首选根

Question

I am trying to solve the following equation for a special case, where l=b=0; and it should return a square root of a perfect square function eg sqrt((xd)^2). This can have two solutions, (xd) or (dx). I would like to obtain (xd) as my final solution, but the program by default returns (dx) solution. I tried changing positions of d and x, but nothing seems to work. Here is my program :

float y(float x) {
  float l=0., b=0., d=8.5, r_0=3., z_0=0.1;
  return exp(-pow(x*x*cos(b*PI/180.)*cos(b*PI/180.)+d*d-2*d*x*cos(b*PI/180.)*cos(l*PI/180.), 0.5)/r_0)*exp(-x*pow(1-cos(b*PI/180.)*cos(b*PI/180.),0.5)/z_0) ;
}
int main(){
  FILE* fp =NULL;
  float x0,xn,step,s,int_val, tau; /* s = distance to the star from the sun*/

  int i,n, j ;
  scanf("%f%f%d",&x0,&xn,&n);
  step = (xn-x0)/n;
  s = y(x0) + y(xn);
  fp = fopen("trap.txt", "w");
  for(i = 1; i < n; i++) {
    s += 2*y(x0+i*step);
    fprintf(fp,"%e\n",s*step/2);
  }
  fclose(fp);

Answer 1

I am assuming you are talking for the part:

-pow( x*x*cos( b*PI / 180. )*cos( b*PI / 180. ) + d*d - 2 * d*x*cos( b*PI / 180. )*cos( l*PI / 180. ), 0.5 )

First things first, there actually is a function double sqrt( double x ) which is for calculating square roots.

The second thing is, hand in hand with the mathematics, sqrt( square( anything ) ) will return absolutevalue( anything ) . In your example case, sqrt( (xd)^2 ) will be equivalent to absolutevalue( xd ) . Since absolutevalue( xd ) is equal to absolutevalue( dx ) , changing the places of the values won't change anything...

If x > d , then it will evaluate to x - d ; otherwise to d - x , that's what mathematics says.

Not with changing places, but you can simply put a minus sign before the whole sqrt( square( ) ) thing to have their places changed. You already have one minus there, you can simply remove that.

With the knowledge that square root of a square evaluates to the absolute value, you also can replace that specific extract I wrote above with fabs( x * cos( b * PI / 180. ) - d ) , where fabs is the function that takes absolute value of a double , and is defined in math.h .

Answer 2

Use substitutions to replace repetitive terms. With select constants, simplify.

#include <math.h>
#define PI 3.1415926535897932384626433832795

float y(float x) {
  float l=0., b=0., d=8.5, r_0=3., z_0=0.1;
  double y;
  // y = exp(-pow(x*x*cos(b*PI/180.)*cos(b*PI/180.)+d*d-2*d*x*cos(b*PI/180.)*cos(l*PI/180.), 0.5)/r_0)
  // y *= exp(-x*pow(1-cos(b*PI/180.)*cos(b*PI/180.),0.5)/z_0);
  if (l != 0.0 || b != 0.0) {
    double xcosb = x*cos(b*PI/180.);
    double xsinb = x*sin(b*PI/180.);
    double  cosl =   cos(l*PI/180.);
    // general solution
    y = exp(-sqrt(xcosb*xcosb - 2*xcosb*d*cosl + d*d)/r_0);
    y *= exp(-xsinb/z_0);  // @abiessu
  } else {
    // y = exp(-sqrt(x*x - 2*x*d*1.0 + d*d)/r_0);
    // y *= exp(-0/z_0);
    // y = exp(-sqrt((x-d)*(x-d))/r_0);
    // y *= 1.0;
    y = exp(-fabs(x - d))/r_0;
  }
  return y;
}

This does obtain (xd) as the final solution. Suggest OP review the function for correctness.