/*
The bissection method is used to determine the roots
of a function, generally not evident to find; such as
the function related to the third degree equation.
In this method,
- we first take a large range of
numbers (in the following example l = 0 - u = 50),
- we make sure that the function has a root by
writing the test : Is the product f(l) x f(u) negative ?
If yes we continue, otherwise no roots for this function .
-- If f(l) x f(u) < 0, We start then to divide the
range l - u in half and attribute the middle point (l + u)/2 to u
and repeat the test (using the loop for), until the the product
becomes > 0. The last half is fixed as c[i]
(for each step me put the middle point in the array c[101]
-- At this moment, we do the similar process (now at left) for
"l" until the root is reached .
*/
2. Example: function f(x) = x/16 - 1/8
// The bisection.c program
//. The function to for which we are to find its roots
float f(float x)
{
// Example:
return (x/16.0 - 1/8.0);
// any function
}
// Main program
// --------------------
int main()
{
printf("The middle points are:\n");
float l = 0, u = + 50, c[101];
int i, n = 30;
if(f(l)*f(u)<0)
{
for(i=1;i<=n;i++)
{
c[i]=(l+u)/2;
if(f(l)*f(c[i])<0)
{
u=c[i];
}
if(f(l)*f(c[i])>0)
{
l=c[i];
}
printf("\t - %1.6f\n",c[i]);
}
// c[n] is the last middle point: It is then the solution ( the searched root)
printf("The solution is : %1.2f\n",c[n]);
// To calculate the error
printf("The error: %1.6f\n",100*(c[n] - c[n-1])/c[n]);
}
else
// That is else if (f(l)*f(u)>0), there is no solution
{
printf("There is no roots in this range .\n");
}
return 0;
}
// ---------- end --------------------
The bisection.c program:
I. Compiled, gives:
--------------------
>gcc -msse2 -O3 -march=pentium4 -malign-double -funroll-loops
-pipe -fomit-frame-pointer -W -Wall -o "cable.exe" "bisection.c"
>Exit code: 0
Executed, gives:
----------------
C:\CLanguage>bisection
The middle points are:
- 25.000000
- 12.500000
- 6.250000
- 3.125000
- 1.562500
- 2.343750
- 1.953125
- 2.148438
- 2.050781
- 2.001953
- 1.977539
- 1.989746
- 1.995850
- 1.998901
- 2.000427
- 1.999664
- 2.000046
- 1.999855
- 1.999950
- 1.999998
- 2.000022
- 2.000010
- 2.000004
- 2.000001
- 2.000000
- 2.000000
- 2.000000
- 2.000000
- 2.000000
- 2.000000
The solution is : 2.00
The error: 0.000000
C:\CLanguage>
3. Example: Computing the tension T for a cable strung
/*This program "cable_strung.c" calls the function f(T) and
uses the bisection method to determine the tension in the cable.
Once this tension is computed, it is fixed as a parameter, like w and
ymin, to compute the values of the height "y" with respect to the
position of the right hand pole "x". The related dvalues are
written in a file of type .csv that a Ms Excel spreadsheet
can open and graph.
The function f(T) is: y - ymin + T/w - (T/w) * cosh(w*x/T). It is
is derived from the double differential equation of the cable strung:
d2y/dx2 = (w/T)[1 + (dy/dx)2]1/2
For more info, link to : cable strung .
*/
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
float x, y, ymin, w;
/*x is the distance from O to the right pole, y is the height
of the cable, ymin is the minimum height, and w is the uniform
weight w (in Newtons/meter) per unit length
y, x, and ymin : n meters. w in Newtons per meter
*/
// 1. Defining the fuction
//-----------------------------
// 1. 1. The fuction to compute T (in Newtons)
//----------------------------------------------
float f(float T)
{
double zz = w*x/T;
return y - ymin + T/w - (T/w) * cosh(zz) ;
//That is f(T) = 0
}
// 1. 1. The fuction to y with respect to x, once T is fixed
//-----------------------------------------------------------
float height(float x, float T)
{
double zz = w*x/T;
return (T/w) * cosh(zz) + ymin - T/w ;
//That is y
}
// 2. Main program:
//------------------
int main()
{
// 2 . 1. Input values:
//----------------------
printf("\n Input a value for y ");
printf("\n and for x in meters : --> ");
scanf("%f%f", &y, &x);
printf("\n Input values for w and ymin");
printf("\n w and ymin in meters : --> ");
scanf("%f%f", &w, &ymin);
/*Example:
------------
The inputs y=15,x = 50, ymin = 8 and w =8 are the
convenient values
*/
// 2. 2. Bisection method:
//----------------------------
// Finding the middle point ..
float l = 1, u = 3000, c[100];
// The tension T (in Newtons) is positive, start with 1,
//because zero leads to infinity
int i,n = 100;
float ss =0;
if(f(l)*f(u)<0)
{
for(i=1;i<=n;i++)
{
c[i]=(l+u)/2;
if(f(l)*f(c[i])<0)
{
u=c[i];
}
if(f(l)*f(c[i])>0)
{
l=c[i];
}
}
// 2. 3. The results ( value of T):
//---------------------------------
printf("\nThe solution is T = %1.2f Newtons.\n",c[n]);
//Fixing the result T :
ss= c[n];
}
else
{
printf("There is no roots in this range .\n");
}
// 2. 4. Computing "y" with respect to "x"
//--------------------------------------------
/*
Filling in the two array "array_cable_x", and "array_cable_y"
The values of "x" and "y" respectively
*/
printf("\n");
/*printf("The heights, y, of the cable at each x:\n");*/
int count = -1;
float k, array_cable_x[101], array_cable_y[101];
for(k= -50; k<= +50; k++)
{
count = count +1;
//printf("\t %1.2f\n",height(k,ss));
array_cable_x [count]= k;
array_cable_y[count] = height(k,ss);
// Here "k" is x, and height(k,ss); is "y"
}
FILE *fp;
/* Writes the set of data values from array_cale to the
file "cable_array.csv":*/
fp = fopen("cable_arrays.csv", "w");
int j;
for (j=0;j<=100;j++)
{
fprintf(fp,"%1.2f,%1.2f\n", array_cable_x[j], array_cable_y[j]);
}
fclose (fp);
//The graph "y" vs "x" is given by MS Excel
return 0;
}
//---------- end ----------------------------------
--------------------------
Compiled with SciTE gives:
---------------------------
>gcc -msse2 -O3 -march=pentium4 -malign-double
-funroll-loops -pipe -fomit-frame-pointer -W -Wall
-o "cable_strung.exe" "cable_strung.c"
>Exit code: 0
---------------
Executed gives:
----------------
C:\CLanguage>cable_strung
Input a value for y
and for x in meters : --> 15 50
Input values for w and ymin
w and ymin in meters : --> 8 8
scanf("%f%f", &w, &ymin);
The solution is T = 1437.81 Newtons.
C:\CLanguage>
------------------------
Graph given by MS Excel:
------------------------
The file "cable_array.csv" is immediately created within
the same directory: C:\CLanguage>
The graph is the following where the increment is 5:
2. Example: function f(x) = x/16 - 1/8 // The bisection.c program //. The function to for which we are to find its roots float f(float x) { // Example: return (x/16.0 - 1/8.0); // any function } // Main program // -------------------- int main() { printf("The middle points are:\n"); float l = 0, u = + 50, c[101]; int i, n = 30; if(f(l)*f(u)<0) { for(i=1;i<=n;i++) { c[i]=(l+u)/2; if(f(l)*f(c[i])<0) { u=c[i]; } if(f(l)*f(c[i])>0) { l=c[i]; } printf("\t - %1.6f\n",c[i]); } // c[n] is the last middle point: It is then the solution ( the searched root) printf("The solution is : %1.2f\n",c[n]); // To calculate the error printf("The error: %1.6f\n",100*(c[n] - c[n-1])/c[n]); } else // That is else if (f(l)*f(u)>0), there is no solution { printf("There is no roots in this range .\n"); } return 0; } // ---------- end -------------------- The bisection.c program: I. Compiled, gives: -------------------- >gcc -msse2 -O3 -march=pentium4 -malign-double -funroll-loops -pipe -fomit-frame-pointer -W -Wall -o "cable.exe" "bisection.c" >Exit code: 0 Executed, gives: ---------------- C:\CLanguage>bisection The middle points are: - 25.000000 - 12.500000 - 6.250000 - 3.125000 - 1.562500 - 2.343750 - 1.953125 - 2.148438 - 2.050781 - 2.001953 - 1.977539 - 1.989746 - 1.995850 - 1.998901 - 2.000427 - 1.999664 - 2.000046 - 1.999855 - 1.999950 - 1.999998 - 2.000022 - 2.000010 - 2.000004 - 2.000001 - 2.000000 - 2.000000 - 2.000000 - 2.000000 - 2.000000 - 2.000000 The solution is : 2.00 The error: 0.000000 C:\CLanguage> 3. Example: Computing the tension T for a cable strung /*This program "cable_strung.c" calls the function f(T) and uses the bisection method to determine the tension in the cable. Once this tension is computed, it is fixed as a parameter, like w and ymin, to compute the values of the height "y" with respect to the position of the right hand pole "x". The related dvalues are written in a file of type .csv that a Ms Excel spreadsheet can open and graph. The function f(T) is: y - ymin + T/w - (T/w) * cosh(w*x/T). It is is derived from the double differential equation of the cable strung: d2y/dx2 = (w/T)[1 + (dy/dx)2]1/2 For more info, link to : cable strung . */ #include <stdio.h> #include <stdlib.h> #include <math.h> float x, y, ymin, w; /*x is the distance from O to the right pole, y is the height of the cable, ymin is the minimum height, and w is the uniform weight w (in Newtons/meter) per unit length y, x, and ymin : n meters. w in Newtons per meter */ // 1. Defining the fuction //----------------------------- // 1. 1. The fuction to compute T (in Newtons) //---------------------------------------------- float f(float T) { double zz = w*x/T; return y - ymin + T/w - (T/w) * cosh(zz) ; //That is f(T) = 0 } // 1. 1. The fuction to y with respect to x, once T is fixed //----------------------------------------------------------- float height(float x, float T) { double zz = w*x/T; return (T/w) * cosh(zz) + ymin - T/w ; //That is y } // 2. Main program: //------------------ int main() { // 2 . 1. Input values: //---------------------- printf("\n Input a value for y "); printf("\n and for x in meters : --> "); scanf("%f%f", &y, &x); printf("\n Input values for w and ymin"); printf("\n w and ymin in meters : --> "); scanf("%f%f", &w, &ymin); /*Example: ------------ The inputs y=15,x = 50, ymin = 8 and w =8 are the convenient values */ // 2. 2. Bisection method: //---------------------------- // Finding the middle point .. float l = 1, u = 3000, c[100]; // The tension T (in Newtons) is positive, start with 1, //because zero leads to infinity int i,n = 100; float ss =0; if(f(l)*f(u)<0) { for(i=1;i<=n;i++) { c[i]=(l+u)/2; if(f(l)*f(c[i])<0) { u=c[i]; } if(f(l)*f(c[i])>0) { l=c[i]; } } // 2. 3. The results ( value of T): //--------------------------------- printf("\nThe solution is T = %1.2f Newtons.\n",c[n]); //Fixing the result T : ss= c[n]; } else { printf("There is no roots in this range .\n"); } // 2. 4. Computing "y" with respect to "x" //-------------------------------------------- /* Filling in the two array "array_cable_x", and "array_cable_y" The values of "x" and "y" respectively */ printf("\n"); /*printf("The heights, y, of the cable at each x:\n");*/ int count = -1; float k, array_cable_x[101], array_cable_y[101]; for(k= -50; k<= +50; k++) { count = count +1; //printf("\t %1.2f\n",height(k,ss)); array_cable_x [count]= k; array_cable_y[count] = height(k,ss); // Here "k" is x, and height(k,ss); is "y" } FILE *fp; /* Writes the set of data values from array_cale to the file "cable_array.csv":*/ fp = fopen("cable_arrays.csv", "w"); int j; for (j=0;j<=100;j++) { fprintf(fp,"%1.2f,%1.2f\n", array_cable_x[j], array_cable_y[j]); } fclose (fp); //The graph "y" vs "x" is given by MS Excel return 0; } //---------- end ---------------------------------- -------------------------- Compiled with SciTE gives: --------------------------- >gcc -msse2 -O3 -march=pentium4 -malign-double -funroll-loops -pipe -fomit-frame-pointer -W -Wall -o "cable_strung.exe" "cable_strung.c" >Exit code: 0 --------------- Executed gives: ---------------- C:\CLanguage>cable_strung Input a value for y and for x in meters : --> 15 50 Input values for w and ymin w and ymin in meters : --> 8 8 scanf("%f%f", &w, &ymin); The solution is T = 1437.81 Newtons. C:\CLanguage> ------------------------ Graph given by MS Excel: ------------------------ The file "cable_array.csv" is immediately created within the same directory: C:\CLanguage> The graph is the following where the increment is 5:
