/************************************************
Treibergs                            Jan 24, 2006 
HW 2a   Newton's series for pi
sum8.c 
************************************************/
/*   Newton's series may be given by

S = a_1 + a_2 + a_3 + a_4 + a_5 + ...

where a_1 = 3
      a_2 = 3/(3*2^3) 
      a_3 = 3*3/(4*5*2^5)
      a_4 = 3*3*5/(4*6*7*2^7)
      a_5 = 3*3*5*7/(4*6*8*9*2^9)
      ...
To see the pattern, it is easier to look at b_i
where  a_i = b_i/(2*i-1) so that

b_1 = 3       
b_2 = 3/(2^3)             = 1*[3]/8
b_3 = 3*3/(4*2^5)         = 3*[3/(2^3)]/(2*8)
b_4 = 3*3*5/(4*6*2^7)     = 5*[3*3/(4*2^5)]/(3*8)
b_5 = 3*3*5*7/(4*6*8*2^9) = 7*[3*3*5/(4*6*2^7)]/(4*8)

Thus  b_i  satisfies the recursion
b_1 = 3  and  b_(i+1) = (2*i-1)*b_i/(i*8)  
for i=1,2,3,....  Thus  a  and  b  are  b_i and a_i at 
the beginning of the do-while loop, as the 
index  i  runs from 1 to 20.                        */
      

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

int 
main (void) 
{
    int i, n;
    double a, b, sum;

    i = 1;
    n = 20;
    sum = 0.0;
    b = 3.0;

    printf ("   n          Term                 Sum\n");
   
    do
    {
         a = b/(2*i-1);
         sum = sum+a;
         printf ("%4d%20.15lf%20.15lf\n", i, a, sum);
         b = (2*i-1)*b/(8*i);
         i = i+1;
    } 
    while ( i <= n );
     
    printf ("The actual value of pi is%19.15lf\n", M_PI);
   
    return EXIT_SUCCESS;
}

/************************************************
Treibergs                            Jan 24, 2006 
HW 2b       Power series for cosine
sum8.c 
************************************************/
/*   Maclaurin's series for cosine is given by

S = a_0 + a_1 + a_2 + a_3 + a_4 + ...

where 
a_0 = 1
a_1 = -x^2/2              = -x^2*[1]/(1*2)
a_2 = x^4/(2*3*4)         = -x^2*[-x^2/2]/(3*4)
a_3 = -x^6/(2*3*4*5*6)    = -x^2*[x^4/(2*3*4)]/(5*6)
a_4 = x^8/(2*3*4*5*6*7*8) = -x^2*[-x^6/(2*3*4*5*6)]/(7*8)
      ...

Thus  a_i  satisfies the recursion
a_0 = 1  and  a_i = -x^2*a_(i-1)/((2*i-1)*(2*i))  
for i=1,2,3,....  

The partial sums S_n = a_1 + ... + a_n  corresponds to the 
Taylor polynomials  p_{2n} = p_{2n+1} = S_n  of degrees 
2n and  2n+1  since the series consists only of even terms.  
Thus the remainder
| cos(x) - S_n | = | cos(x) - p_{2n+1} | = | R_{2n+1} |.
For a general function  f(x),  the error term (the Lagrange
formula for the remainder) is
f(x) - p_{2n+1}(x) = R_{2n+1} = f^{2n+2}(c) x^{2n+2}/(2n+2)!
where  c  is a number strictly between  0  and  x.
But if  f(x)=cos(x),  | f^{2n+2}(c) |=| cos(c) | <= 1  so 
|R_{2n+1}| <= |x|^{2n+2}/(2n+2)! = |p_{2n+2}| = |a_{n+1}|.
Thus in this instance, the error has the magnitude of the 
next term neglected. Thus the term on the next line is 
printed first as the error on the current line.        */
               
      

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

int 
main(void) 
{

    int i=1, n=20;
    double  x, y, a, sum;
    
    
    a = 1.0;
    sum = a;
    
    
    printf ("What is your angle in radians?");
    scanf ("%lf", &x);
    
    y = -x*x/2.0;

    
    printf ("\n   n\t\t Term\t\t   Sum\t\t\t   Error\n");
    printf ("%4d%22.15lf%22.15lf", 0, a, sum);
    do
    {  
        a = y*a/(i*(2*i-1));
        sum = sum+a;
        printf ("%22.15lf\n%4d%22.15lf%22.15lf",fabs(a),i,a,sum);
        i = i+1;
     } 
     while ( i <= n);
     printf ("%22.15lf\n", fabs( y*a/(i*(2*i-1))) );

     printf (" Actual  cos(%10.6lf)  is%20.15lf\n", x, cos(x) );
   

    return EXIT_SUCCESS;
}