So I am using the taylor series to calculate sin(0.75) in fortran 90 up until a certain point, so I need to run it in a do while loop (until my condition is met). This means I will need to use a factorial, here's my code:
program taylor
implicit none
real :: x = 0.75
real :: y
integer :: i = 3
do while (abs(y - sin(0.75)) > 10.00**(-7))
i = i + 2
y = x - ((x**i)/fact(i))
print *, y
end do
end program taylor
Where i've written fact(i) is where i'll need the factorial. Unfortunately, Fortran doesn't have an intrinsic ! function. How would I implement the function in this program?
Thanks.
The following simple function answers your question. Note how it returns a real
, not an integer. If performance is not an issue, then this is fine for the Taylor series.
real function fact(n)
integer, intent(in) :: n
integer :: i
if (n < 0) error stop 'factorial is singular for negative integers'
fact = 1.0
do i = 2, n
fact = fact * i
enddo
end function fact
But the real answer is that Fortran 2008 does have an intrinsic function for the factorial: the Gamma function. For a positive integer n
, it is defined such that Gamma(n+1) == fact(n)
.
(I can imagine the Gamma function is unfamiliar. It's a generalization of the factorial function: Gamma(x)
is defined for all complex x
, except non-positive integers. The offset in the definition is for historical reasons and unnecessarily confusing it you ask me.)
In some cases you may want to convert the output of the Gamma function to an integer. If so, make sure you use "long integers" via INT(Gamma(n+1), kind=INT64)
with the USE, INTRINSIC :: ISO_Fortran_env
declaration. This is a precaution against factorials becoming quite large. And, as always, watch out for mixed-mode arithmetic!
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With