A program to calculate the sum of the series using function.
A program to calculate the sum of the series using function.
Codes
/*A program to calculate the sum of given series using function*/
/*Programed by Jitendra Kr Yadav
10116
2 Mar,2011*/
#include<stdio.h>
#include<conio.h>
#include<math.h>
void series1(int, int);
void series2(int, int);
void series3(int, int);
main()
{
int a,n,choice;
printf(“Series 1: y=1+x^2/2!+x^4/4!+x^6/6!+…\n”);
printf(“Series 2: y=x-x^3/3+x^5/5-x^7/7+x^9/9-…\n”);
printf(“Series 3: y=x^3/5-x^6/10+x^9/15-x^12/20+…\n”);
printf(“Enter your choice:”);
scanf(“%d”,&choice);
printf(“Enter the value of x: “);
scanf(“%d”,&a);
printf(“Enter the value of n: “);
scanf(“%d”,&n);
switch(choice)
{
case 1:
series1 (a,n);
break;
case 2:
series2 (a,n);
break;
case 3:
series3 (a,n);
break;
}
getch();
}
void series1 (int x,int n)
{
int i, j, fact;
float sum, term;
sum=0;
for(i=1;i<n;i++)
{
for(j=1,fact=1;j<=((2*i)-2);j++)
fact=fact*j;
term=(pow(x,((2*i)-2))*1.0/fact);
sum+=term;
}
printf(“Sum = %.2f.\n”,sum);
printf(“\n\nPress any key to continue…”);
}
void series2 (int x,int n)
{
int i;
float sum, term;
sum=0;
for(i=1;i<n+1;i++)
{
term=(pow(x,((2*i)-1))*(pow(-1,(i+1))))*1.0/((2*i)-1);
sum+=term;
}
printf(“Sum = %.2f.\n”,sum);
printf(“\n\nPress any key to continue…”);
}
void series3 (int x,int n)
{
int i;
float sum, term;
sum=0;
for(i=0;i<n;i++)
{
term=(pow(x,((3*i)+3))*(pow(-1,(i+2))))*1.0/((5*i)+5);
sum+=term;
printf(“%d\n”,term);
}
printf(“Sum = %.2f.\n”,sum);
printf(“\n\nPress any key to continue…”);
}
Output
Series 1: y=1+x^2/2!+x^4/4!+x^6/6!+…
Series 2: y=x-x^3/3+x^5/5-x^7/7+x^9/9-…
Series 3: y=x^3/5-x^6/10+x^9/15-x^12/20+…
Enter your choice: 2
Enter the value of x: 2
Enter the value of n: 3
Sum = 5.73.
Press any key to continue…
