!Name: Nicholas Farkash
!Purpose: To gain experience using iterative do loops for summation and
!         products, as well as to think about algorithms.
!Date: October 4th, 2021
!Due: October 8th, 2021 at 6:00PM

program HW_7
  implicit none

  integer :: n     ! Integer number and upper bound of summation
  real :: x        ! Variable of summation formula
  integer :: i     ! Loop index
  real :: sum      ! Value of calculation
  real :: denom    ! Denominator of summation; for easier calculation

  
  write(*,*) "Enter an integer value, n, and a real, positive value, x:"
  read(*,*) n, x

  sum = 0.0e0      ! Initialize variable "sum"
  denom = 0.0e0    ! Initialize variable "denom"

  do i=0,n,1       ! Begin iterative loop, from 0 to n

     denom = denom * (2.0e0*i) * ( (2.0e0*i) - 1.0e0)

     ! Calculate the denominator of the summation function

     
     if(i==0) then      ! Handle the case where i==0 to avoid
        denom = 1.0e0   ! division by 0.0e0 in the denom calculation.
     endif              ! Directly inserts the value of 0! as 1.0e0.

     
     ! According to the assignment, it is possible to calculate this
     ! without using an if statement. I cannot figure out how to do this.
     ! Grader, please comment how this is accomplished.


     sum = sum + ( (-1.0e0)**i ) * ( (x**(2.0e0*i)) / (denom) )

     ! Calculates the summation function by adding the newly calculated
     ! sum to the previous sum.

     
  enddo            ! Ends iterative loop, returns to top of loop.

  write(*,*) "The value of the summation function is",sum

  
  stop 0
end program HW_7