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Help with numerical precision issue

Help with numerical precision issue

Help with numerical precision issue

Hello. I need help to understand why my code is not giving me double numerical precision.
In this piece of code, I create a function that is analytically normalized, but when I calculate the numerical normalization factor, it seems to be with single precision. Here it is:

CODE --> fortran

program dyna
implicit none
integer i
integer, parameter :: Nq1=61
real(kind=8), parameter :: saw2au=1822.889950851334d0 
real(kind=8), parameter :: nitro=14.0067d0*saw2au 
real(kind=8), parameter :: mredu=nitro**2.0d0/(2.0d0*nitro)
real(kind=8) :: e0,pi,ch,x,x0,stepX,w,expo,c0,c1
complex(kind=8) :: soma,vec0(Nq1)
pi = 3.141592653589793d0
c0 = ((mredu*w)/pi)**(1.d0/4.d0)
c1 = (4.d0/pi*(mredu*w)**3.d0)**(1.d0/4.d0)
do i=1,Nq1
  expo = dexp(-(x-x0)**2.d0*mredu*w/2.d0)
  vec0(i) = c0 * expo
end do
!normalizing                             !
soma=0.0d0                               !
do i=1,Nq1                               !
  soma=soma+conjg(vec0(i))*vec0(i)*stepX !
end do                                   !
vec0=vec0/sqrt(soma)                     !
write(*,'(a24,2(f23.15))')'normalization constant =',soma
end program 

I get 0.999998932390816, and it should be 1.0d0.
I use complex vectors because later in the code they will become complex.
I don't understand where is the problem. Can someone help me, please?
Thanks in advance,

RE: Help with numerical precision issue

I think the problem is understanding what a precision of 15 digits means. This actually means 15 significant figures: not 15 decimal places. The bigger your number gets, the less decimal places you can expect.

For instance, if the biggest number in your calculations is 1000000 (7 digits) then you can only expect 8 decimal places of accuracy. That plus decimal/binary rounding will reduce the accuracy to about 6 decimal places.

RE: Help with numerical precision issue

Note how a simple operation can produce loss of digits:

I changed

normalization constant = 1.000000000000002 0.000000000000000

OPH. 2019-04-04 06:27

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