Question

Please write in Language c using only the files stdio.h and math.h

Suppose you wish to find the root r of a function f(x), that is, the value r where f(r)=0. One method is to make an initial guess, x0, compute the line tangent to f at x0, and find where the tangent line intercepts the x-axis, x1. Then, use x1 as the second guess and repeat this procedure n times until f(xn) approximately equals 0 and report xn as the root r. In the figure we see successive tangent lines of the function f(x) in red eventually intercept the xaxis at the root r.

The Taylor series of a function f(x) that is infinitely differentiable at a number x0, is approximately

Setting this equation to 0, we have

Write a program using an iteration structure to calculate the approximate root of the equation

Prompt the user to enter an initial guess x0. Example output:

Answer 1

Hi, According to your question, you need a C program that uses only 2 header files, i.e., 'main.h' and 'math.h' for finding the roots of the equation .

x3 += 22 +3

The given equation
x3 += 22 +3
can be re-written as
f(0) = 23 – 2.+ 1 - 3
.

The differentiation of given eqn is  .

f'(x) = 3.12 – 4.+1

The program for the same is given below:

*************************MAIN.C************************

#include<stdio.h>
#include<math.h>
long double f(long double x) //The given function f(x)
{
return x*x*x-2*x*x+x-3;
}
long double df (long double x) // Differentiation of given function df(x)
{
return 3*x*x-4*x+1;
}
int main() // The main function
{
int itr, maxmitr=20; // Variables for iteration and maximum allowed iterations
long double h, x0, x1, allerr=0.1e-50; // Variables for allowed errors, initial value of x as x0, next value of x as x1, h as the difference and allerr as allowed error
printf("?");
scanf("%Lf", &x0);
printf("i\t\t\t\t\t\t\t\tx\t\t\t\tf(x)\n"); // For formatting stuffs
printf("%02d\t%.30Lf\t%Le\n", 0, x0, f(x0));
for (itr=1; itr<=maxmitr; itr++)
{
h=f(x0)/df(x0);
x1=x0-h;
printf("%02d\t%.30Lf\t%Le\n", itr, x1, f(x1));
if (fabsl(h) < allerr)
{
printf("%02d\t%.30Lf\t%Le\n", itr, x1, f(x1));
return 0;
}
x0=x1;
}
printf("Solution does not converge or iterations are insufficient\n");
return 1;
}

*************************MAIN.C ENDs*******************

The images of the above program are:

2 #include<stdio.h> #include<math.h> long double f(long double x) //The given function f(x) vouw return x*x*x-2*x*x+X-3; long double df (long double x) // Differentiation of given function df(x) 8 return 3*x*x-4*x+1; 9 10 11 int main() // The main function 12 13 int itr, maxmitr=20; // Variables for iteration and maximum allowed iterations long double h, x0, x1, allerr=0.1e-50; // Variables for allowed errors, initial value of xas xo, next value of xas x1, h as the difference and allerr as allowed error printf("?"); scanf("%Lf", &xo); printf("i\t\t\t\t\t\t\t\tx\t\t\t\tf(x) \n"); // For formatting stuffs printf("%02d\t%.30Lf\t%Le\n", 0, xo, f(x0)); for (itr=1; itr<=maxmitr; itr++) h=f(x0)/df(x0); X1=x-h; printf("%02d\t%.30LF\t%Le\n", itr, x1, f(x1)); if (fabsl(h) < allerr) printf("%02d\t%.30Lf\t%Le\n", itr, x1, f(x1)); return 0; XO=x1; printf("solution does not converge or iterations are insufficient\n"); return 1;

The output of the above program are:

Run1:

3./main ?4.0 i 02 00 4.000000000000000000000000000000 01 3.000000000000000000000000000000 2.437500000000000000000000000000 03 2.213032716315109771831534657416 04 2.175554938721488266702500102845 05 2.174560100666445745971888658943 06 2.174559410293312394042361535007 07 2.174559410292980074150270830557 08 2.174559410292980074150270830557 08 2.174559410292980074150270830557 f(x) 3.300000e+01 9.000000e+00 2.036865e+00 2.563634e-01 6.463361e-03 4.479068e-06 2.156054e-12 0.000000e+00 0.000000e+00 0.000000e+00

Run2:

3./main 215.0 i 00 15.000000000000000000000000000000 01 10.232142857142857142981051676855 02 7.062076371831236789450292334891 03 4.965797461779154731500995900717 04 3.603176463485630989357771936099 05 2.764475070663096292322008507014 06 2.328791578255298578203555392996 07 2.189009445308850030937480868332 08 2.174703020889649505422921982323 09 2.174559424671094085826025810171 10 2.174559410292980218349159771130 11 2.174559410292980074150270830557 12 2.174559410292980074150270830557 12 2.174559410292980074150270830557 f(x) 2.937000e+03 8.691107e+02 2.565226e+02 7.509983e+01 2.141703e+01 5.606840e+00 1.111917e+00 9.469779e-02 9.318228e-04 9.328360e-08 9.352328e-16 0.000000e+00 0.000000e+00 0.000000e+00

Run3:

./main ?-200 f(x) 00 -200.000000000000000000000000000000 -8.080203e+06 01 -133.111456030993120916439131917741 -2.394130e+06 02 -88.519247679724349139873673664169 -7.093693e+05 03 -58.791317515468891563568121227945 -2.101821e+05 04 -38.972959426450921937429283659071 -6.227546e+04 05 -25.760998802751498691632203730251 -1.845177e+04 06 -16.953183655239736063805033872143 -5.467296e+03 07 -11.080963103492844898378955065255 -1.620263e+03 -7.164324242762062072871487306003 -4.805466e+02 09 -4.547537251403170445798740395560 -1.429512e+02 10 -2.787713714283297577632356190058 -4.299470e+01 11 -1.575396628561823317944343147978 -1.344908e+01 12 -0.663421931878106638171173348173 -4.835670e+00 13 0.308753098528772199386164404311 -2.852471e+00 14 56.269144279311781458446928283479 1.718811e+05 15 37.736632809359557223782832835468 5.092561e+04 16 25.382685608246783647359845481617 1.508740e+04 17 17.148607029293502912806190785489 4.468971e+03 18 11.662714729188597952147521397137 1.322975e+03 19 8.012181613863674221540089881444 3.909645e+02 20 5.591894801238047055941260410350 1.149079e+02 Solution does not converge or iterations are insufficient

I hope this will help you. I have added comments in assistance with the code to be understandable. If you have any query related to the above question feel free to ask.