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Problem 8A (5 points) Write a C program to find the roots for the following poly

ID: 3779858 • Letter: P

Question

Problem 8A (5 points)
Write a C program to find the roots for the following polynomial using Newton-Raphson method. F(x) = Ax^3 + Bx^2 + Cx + D Where A, B, C and D are coefficients/constant of the polynomial. The program should ask the user to
enter the values for each and also the initial value of estimated root value. Use 10 decimal places for precision.

inding root for three degree polynomial Ax 3 Bx 2 Cx D using Newton-Raphson method nter value for the following coefficients nitial root estimate :1.0 1 1.0000000000 Estimate 1.1250000000 2 Estimate 3 Estimate 1.1149754501 4 Estimate 1.1149075446 5 Estimate 1.1149075415 Approximate solution 1.1149075415

Explanation / Answer


#include<stdio.h>
#include<stdlib.h>
#include<math.h>

int my_power,i=0,count=0,f=0;
int coefficient[10]={0};
float x1=0,x2=0,t=0;
float fx1=0,fdx1=0;

int main()
{

  

printf("Finding root for three degree polynomial Ax^3+Bx^2+Cx+D using Newton-Raphson method ");

my_power=3;int j=0;
printf("Enter the value for following coefficientficent ");
for(i=my_power;i>=0;i--)
{
printf(" %c::",'A'+j);
scanf("%d",&coefficient[i]);
j++;
}

printf(" ");


printf(" initial root estimate : ");
scanf("%f",&x1);

int cc=0;
do
{
count++;
fx1=fdx1=0;
for(i=my_power;i>=1;i--)
{
fx1+=coefficient[i] * (pow(x1,i)) ;
}
fx1+=coefficient[0];
for(i=my_power;i>=0;i--)
{
fdx1+=coefficient[i]* (i*pow(x1,(i-1)));
}
t=x2;
x2=(x1-(fx1/fdx1));

x1=x2;
cc++;

printf(" %d : Estimate : %.13f ",count,x2);

}while(cc<5);
printf(" Approximate solution = %.13f",x2);
  
}

===========================================================================

akshay@akshay-Inspiron-3537:~/Chegg$ gcc newton.c -lm
akshay@akshay-Inspiron-3537:~/Chegg$ ./a.out
Finding root for three degree polynomial Ax^3+Bx^2+Cx+D
using Newton-Raphson method
Enter the value for following coefficientficent

   A::1

   B::3

   C::-1

   D::-4

initial root estimate : 1.0

1 : Estimate : 1.1250000000000   
2 : Estimate : 1.1149754524231   
3 : Estimate : 1.1149076223373   
4 : Estimate : 1.1149076223373   
5 : Estimate : 1.1149076223373   
   Approximate solution = 1.1149076223373

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