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68 lines
2.2 KiB
C++
68 lines
2.2 KiB
C++
5 years ago
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/*--------------- SECANT METHOD TO FIND ROOT OF AN EQUATION ----------*/
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/* THE EXPRESSION FOR AN EQUATION IS DEFINED IN function fx
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YOU CAN WRITE DIFFERENT EQUATION IN function fx.
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HERE,
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f(x) = x*x*x - 5*x - 7
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INPUTS : 1) Initial interval [x0,x1] in which root is to
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be found.
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2) Number of iterations for given interval and
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permissible error.
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OUTPUTS : Value of the root in given interval. */
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/*---------------Program starts here ----------------------------------*/
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#include<stdio.h>
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#include<math.h>
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#include<stdlib.h>
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#include<conio.h>
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void main()
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{
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double fx ( double x); /* DECLARATION OF FUNCTION */
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double x0,x1,x2,f0,f1,f2,err;
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int n,i;
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clrscr();
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printf("\n SECANT METHOD TO FIND ROOT OF AN EQUATION");
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printf("\n\n f(x) = x*x*x - 5*x - 7");
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printf("\n\nEnter an interval [x0,x1] in "
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"which root is to be found");
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printf("\nx0 = ");
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scanf("%lf",&x0); /* INTERVAL [x0,x1] IS TO BE ENTERED HERE */
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printf("x1 = ");
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scanf("%lf",&x1);
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printf("\nEnter the number of iterations = ");
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scanf("%d",&n);
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printf("\npress any key for display of iterations...\n");
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getch();
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i = 0;
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while(n-- > 0)
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{
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f0 = fx(x0); /* CALCULATE f(x) AT x = x0 */
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f1 = fx(x1); /* CALCULATE f(x) AT x = x1 */
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x2 = x1 - ((x1 - x0)/(f1 - f0)) * f1;
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/* CALCULATION OF NEXT APPROXIMATION */
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i++;
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printf("\n%d x[%d] = %lf x[%d] = %lf",i,i-1,x0,i,x1);
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printf("\n f[%d] = %lf f[%d] = %lf",i-1,f0,i,f1);
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printf("\n x[%d] = %lf",i+1,x2);
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x0 = x1;
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x1 = x2;
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getch();
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}
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printf("\n\nThe value of root is = %20.15lf",x2); /* ROOT */
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}
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/*---------- FUNCTION PROCEDURE TO CALCULATE VALUE OF EQUATION --------*/
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double fx ( double x)
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{
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double f;
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f = x*x*x - 5*x - 7; /* FUNCTION f(x) */
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return(f);
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}
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/*-------------------------- END OF PROGRAM ---------------------------*/
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