Writing a C program to solve an equation numerically - linux

Question

I am trying to solve x-cos(x)=0 numerically.
I need the program to accept one argument on the command line that becomes the desired accuracy of the solution.
The solution should yield an answer within some +/- the specified accuracy (epsilon) of 0 when the equation x-cos(x) is evaluated.
The maximum number of iterations should be set to 100.
The program should start with a first guess value of x=0.
The desired accuracy should accept both floating and scientific notation formats.
There should be a warning message if too few or too many arguments are supplied, and therefore exit the program.
If a solution is found in the max iterations, it should print the solution, accuracy and number of iterations.
If no solution is achieved in the max iterations, the program should print a message to indicate as such and then close.
Find the smallest accuracy that can be achieved in max iterations in powers of 10.
I know that there are loops involved. I've started it as such:`
#include<stdio.h>
#include<math.h>
int
main(void)
{
int MAX_ITERATIONS[100],x=0;
float epsilon;
double epsilon;
x=cos[x];
for (x=0; x<MAX_ITERATIONS; ++x)
if (MAX_ITERATIONS < x)
x=MAX_ITERATIONS[100];
}
I am not sure where to go from here or if I am even on the right track.

Answer 1

Here is some code to help you get started. My philosophy is to make only small changes and always keep a copy of what was working before. The way when I break something, I know exactly where I broke it. This code does not do everything you want, but you can make those changes yourself. To compile the code, I used cc -lm progname.c. To execute it, I used ./a.out 0.002.
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
int main(int argc, char *argv[]) {
double delta, x;
double epsilon=0.001;
double previous = 1.0;
if (argc > 1)
epsilon = strtof(argv[1], NULL);
printf("Using epsilon = %12.8f\n", epsilon);
for (x=0.1; x<1.0; x+=epsilon) {
delta = fabs(x-cos(x));
if (delta < previous)
previous = delta;
else
break;
}
printf("%12.8f %12.8f %12.8f\n", x, cos(x), delta);
}