C program for the approximation of ln(x+1) = x-(1/2)x^2 + (1/3)x^3 - (1/4)x^4 +...+ (((-1)^(n+1))*x)/n

Question

 Write a program, powerseries.c, to check the accuracy of this formula. Your program must
prompt the user for the value of x (where |x| < 1) and the degree, n, of the polynomial to use.
Your program should then write the values of x, n, the approximation, the actual value and
the percent relative error to the output file powerseriesout.txt.

Answer 1

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


void Go()
{

int N=-1;
double x;
char sign_flag=1;
double curTerm;
double sum=0;
int iLoop;
double powX;
FILE * fptr;
double percentError;
double actualResult;

while ((N<0) || (N>32767))
{
printf(" Please input the number of terms in the series :>");
scanf("%d",&N);
}

x = -2;
while ((x<-1) || (x>1))
{
printf(" Please input the value of x between -1 and 1 :>");
scanf("%lf",&x);
}
powX = x;

for (iLoop=0; iLoop {
curTerm = powX/(iLoop+1) ;
printf(" iLoop = %12ld \n",iLoop);
printf( " powX = %12.12lf \n",powX);
printf(" curTerm = %12.12lf \n",curTerm);
printf(" sum thus far is %12.12lf \n ",sum);

if (sign_flag==1)
{
sum = sum + curTerm;
sign_flag = 0;
}
else
{
sum = sum - curTerm;
sign_flag = 1;
}
powX = powX * x;


}

actualResult = log(1+x);

if (actualResult!=0)
{
percentError = (actualResult - sum)/actualResult;
}
else
{
percentError = sum;
}

printf(" N = %d \n", N);
printf( " x = %12.6lf \n",x);
printf(" Approximation is %12.12f \n",sum);
printf(" Actual value is %12.12f \n",actualResult);
printf(" error %12.12lf \n ",percentError);

fptr = fopen("powerseriesout.txt","w") ;
if (fptr != NULL)
{
fprintf(fptr," N = %d \n", N);
fprintf(fptr, " x = %12.6lf \n",x);
fprintf(fptr," Approximation is %12.12f \n",sum);
fprintf(fptr," Actual value is %12.12lf \n",actualResult);
fprintf(fptr," error %12.12lf \n ",percentError);
fclose(fptr);
}
}


void main()
{

Go();

}