To solve sin series in c language?

Answer 1

# include <conio.h>

# include <stdio.h>

unsigned int factorial(int);

void main()

{

unsigned int f;

int a,i,g=0,c=1,j=-1;

clrscr();

scanf("%d",&a);

printf("sine series up to %d terms",a);

for(i=1;i<a;i++)

{

f=factorial(c);

j=j*(-1);

printf(" [(%d(x^%d))/(%d)] ",j,g,f);

if(i!=a)

printf("+");

c=c+2;

g=g+2;

}

getch();

}

unsigned int factorial(int x)

{

int fact=1,i;

for(i=1;i<=x;i++)

fact=fact*i;

return fact;

}

Answer 2

#include<iostream.h>

#include<conio.h>

#include<math.h>

void main()

{

int sum=1,i,k,x,n;

cout<<"enter the values for x,n";

cin>>n>>x;

for(i=2;i<=n;i+=2)

{

double fact=1;

for(k=1;k<=i;k++)

{

fact*=k;

}

int l=i/2;

if(l%2==0)

{

sum=sum+(pow(x,i)/fact);

}

else

sum=sum-(pow(x,i)/fact);

}

cout<<sum;

}

Answer 3

#include<stdio.h>

/*main start*/

main()

{

int inum=5,i,j;

for(i=inum;i>=1;i--)

{

for(j=1;j<=i;j++)

{

printf("%d",inum);

}

printf("\n");

inum--;

}

}

/*main end*/

Answer 4

//COSINE SERIES

#include<stdio.h>

#include<conio.h>

#include<math.h>

main()

{

int i,n,x;

float t,s;

clrscr();

printf("enter the value of x,n \n");

scanf("%d%d",&x,&n);

t=1;

s=1;

for(i=1;i<n;i++)

{

t=(-1)*((pow(x,2)*t*i))/((2*i)*((2*i-1)*i));

s=s+t;

printf("s=%6.2f & t=%6.2f\n",s,t);

}

printf("\n sum of the cos series=%5.2f\n",s);

}

Answer 5

The C and C++ implementation of sin(x) and cos(x) using Taylor's Series is straightforward, but care has to be taken in some of the details...

Start with Taylor's Series.

sin(x) is x1/1! - x3/3! + x5/5! - ... repeating for all (+/-) pairs of odd factors.

cos(x) is x0/0! - x2/2! + x4/4! - ... repeating for all (+/-) pairs of even factors.

A naive implementation could just iterate this series until a desired resolution is obtained. Several issues arise...

The series converges more rapidly near x=0 then it does elsewhere. It is very, very helpful to take advantage of the fact that sin(x) and cos(x) repeat every 2pi. The first step should be to add or subtract 2*pi from x until x is in the interval -pi to +pi. To narrow that further, you can take advantage of the fact that the functions are fully defined in the interval -pi/2 to +pi/2. If the final x is outside of the interval -pi/2 to +pi/2, you can add or subtract pi to get it in that interval, and then remember to flip the sign of the result.

In cases where the magnitude of x is very large, the problem of truncation rears its ugly head. In floating point form, a large number has poor resolution near zero. If it is large enough, a step of just one bit in the low order end of the mantissa can be more than 2pi. That makes the problem unsolvable in this context. This answer will not address that.

At this point, a series of 6 terms would be sufficient to get resolution within 6 digits, if you are using doubles with 18 digit resolution.

To go further, you need to consider that factorials get very large very fast, and you will need an arbitrary length math library to get good results. Theoretically, you could get whatever resolution you want, but truncation and round-off will cause accumulated error unless your math library has none. That is a bigger issue than the solution using Taylor's Series to start with.

Answer 6

#include<stdio.h>

#include<math.h>

void main()

{

int i = 2, n, s = 1, x, pwr = 1, dr;

float nr = 1, x1, sum;

clrscr();

printf("\n\n\t ENTER THE ANGLE…: ");

scanf("%d", &x);

x1 = 3.142 * (x / 180.0);

sum = x1;

printf("\n\t ENTER THE NUMBER OF TERMS…: ");

scanf("%d", &n);

while(i <= n)

{

pwr = pwr + 2;

dr = dr * pwr * (pwr - 1);

sum = sum + (nr / dr) * s;

s = s * (-1);

nr = nr * x1 * x1;

i+= 2;

}

printf("\n\t THE SUM OF THE SINE SERIES IS..: %0.3f",sum);

getch();

}

Answer 7

#include<stdio.h>

#include<math.h>

void main()

{

int i = 2, n, s = 1, x, pwr = 1, dr;

float nr = 1, x1, sum;

clrscr();

printf("\n\n\t ENTER THE ANGLE...: ");

scanf("%d", &x);

x1 = 3.142 * (x / 180.0);

sum = x1;

printf("\n\t ENTER THE NUMBER OF TERMS...: ");

scanf("%d", &n);

while(i <= n)

{

pwr = pwr + 2;

dr = dr * pwr * (pwr - 1);

sum = sum + (nr / dr) * s;

s = s * (-1);

nr = nr * x1 * x1;

i+= 2;

}

printf("\n\t THE SUM OF THE SINE SERIES IS..: %0.3f",sum);

getch();

}

Answer 8

Sure, do write. The attached wikipedia-entry will help.

Answer 9

∑ i+∑ (i+2)