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Riemann Zeta Function

 
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arctica



Joined: 10 Sep 2006
Posts: 105
Location: United Kingdom

PostPosted: Wed Dec 20, 2023 9:15 am    Post subject: Riemann Zeta Function Reply with quote

Hello,

I would like to convert the following Scilab code that I generated into Fortran 95, but not entirely sure of how to deal with it. Two issues, one how to deal with an equivalent to linspace and secondly the recursive function in zeta_3.

// Riemann Zeta Function (all domains)

// Re > 1
function s=zeta_1(z,n)
n=linspace(1,n,n);
//s=0.0;
if z == 0
s = -0.5
elseif z == 1
s = %inf
else
s=sum(n.^-z);
end

endfunction

// Re < 0
function zfn = zeta_2(s,n)
// Riemann's functional equation
// Analytic contiuation for negative values
zfn = 2.^s .* %pi.^(s - 1) .* sin(%pi.*s./2) .* gamma(1 - s) .* zeta_1((1 - s),n)
endfunction

// 0 < Re < 1
function zs1 = zeta_3(s,n)
// Vectorised version
zs1=0
k=linspace(1,n,n);
zs1 = sum((-1).^(k+ 1)./(k.^s ));
zs1 = 1./(1 - 2.^( 1-s )).*zs1;
endfunction

Any suggestions would be most helpful.
Thanks

Lester
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mecej4



Joined: 31 Oct 2006
Posts: 1896

PostPosted: Wed Dec 20, 2023 11:31 am    Post subject: Reply with quote

There are no zeta functions here, if n is finite.

Try this for your zeta_1:
Code:
program tstz1
   print *, zta_1(0.5,10),zta_1(-0.4,20)
contains
   function zta_1(z,n) result(s)
   integer,allocatable :: iv(:)
   real s, z
   integer i,n
   allocate (iv(n))
   iv = (/(i,i=1,n)/)
   s = sum(z**iv)
   end function zta_1
end program

There is no recursion in your zeta_2. A function name behaves as an ordinary variable when no argument list follows it.
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arctica



Joined: 10 Sep 2006
Posts: 105
Location: United Kingdom

PostPosted: Wed Dec 20, 2023 2:31 pm    Post subject: Re: Reply with quote

Thanks for the solution, I have coded and tested the three functions and all looks fine:

Code:

program Riemann_Zeta_test

   print *, zeta_1(2.0,1000), zeta_2(-2.0,1000), zeta_2(-3.0,1000), zeta_3(0.5,1000), zeta_3(0.6,1000)
   
contains
   
! Re > 1.0 (if z=0 then s=-0.5 and if z=1 then s=infinity)
   function zeta_1(z,n) result(s)
   integer,allocatable :: nn(:)
   real s, z
   integer i,n
   allocate (nn(n))
   nn = (/(i,i=1,n)/)
   s = sum(nn**(-1*z))
   end function zeta_1

! Re < 0.0
   function zeta_2(s,n) result(zfn)
   integer,allocatable :: nn(:)
   integer i,n
   real s
   real :: pi= 4.0*ATAN(1.0)
   allocate (nn(n))
   nn = (/(i,i=1,n)/)
   zfn=2**s*pi**(s-1)*sin(pi*s/2)*gamma(1-s)*zeta_1((1-s),n)
   end function zeta_2

! 0.0 < Re < 1.0
   function zeta_3(s,n) result(zs1)
   integer,allocatable :: nn(:)
   integer i,n
   real s
   allocate (nn(n))
   nn = (/(i,i=1,n)/)   
   zs1 = sum((-1)**(nn + 1)/(nn**s ))
   zs1 = 1/(1 - 2**( 1-s ))*zs1 
   end function zeta_3

end program Riemann_Zeta_test


The values at 0.0 and 1.0 are already defined as -0.5 and infinity respectively.

mecej4 wrote:
There are no zeta functions here, if n is finite.

Try this for your zeta_1:
[code]program tstz1
print *, zta_1(0.5,10),zta_1(-0.4,20)
contains
function zta_1(z,n) result(s)
integer,allocatable :: iv(Smile
real s, z
integer i,n
allocate (iv(n))
iv = (/(i,i=1,n)/)
s = sum(z**iv)
end function zta_1
end program

There is no recursion in your zeta_2. A function name behaves as an ordinary variable when no argument list follows it.
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arctica



Joined: 10 Sep 2006
Posts: 105
Location: United Kingdom

PostPosted: Wed Dec 20, 2023 6:17 pm    Post subject: Reply with quote

Given the function Zeta_3

Code:

   function zeta_3(s,n) result(zs1)
   integer,allocatable :: nn(:)
   integer i,n
   real s
   allocate (nn(n))
   nn = (/(i,i=1,n)/)   
   zs1 = sum((-1)**(nn + 1)/(nn**s ))
   zs1 = 1/(1 - 2**( 1-s ))*zs1 
   end function zeta_3


How can the variable (s) be passed as a complex number of the form s = 0.5 + it , where t would be a linear vector range and i is the imaginary number?

Thanks
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mecej4



Joined: 31 Oct 2006
Posts: 1896

PostPosted: Thu Dec 21, 2023 3:41 am    Post subject: Reply with quote

It is not clear to me what you want. If t is a real array, s = 0.5 + i.t is a complex array, not a complex number.

Will the following do, perhaps?

Code:
program tz3
  integer, parameter :: k = 7
  integer j
  complex s(k)
  do j = 1, k
     s(j) = cmplx(0.5,j*0.1)
  end do
  print *,zeta_3(s,5)
contains
elemental function zeta_3(s,n) result(zs1)
   integer,allocatable :: nn(:)
   integer, intent(in) :: n
   complex zs1
   complex, intent(in) :: s
   allocate (nn(n))
   nn = (/(i,i=1,n)/)
   zs1 = sum((-1)**(nn + 1)/(nn**s ))
   zs1 = 1/(1 - 2**( 1-s ))*zs1
end function zeta_3
end program
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arctica



Joined: 10 Sep 2006
Posts: 105
Location: United Kingdom

PostPosted: Thu Dec 21, 2023 1:42 pm    Post subject: Re: Reply with quote

That worked fine, many thanks for the guidance.

Had to stick the write statement in a do-loop to get the columns right:

do i = 1, k
write(25,'(f15.12,1x,f15.12)') real_part(i), imaginary_part(i) ! write data to file
end do


mecej4 wrote:
It is not clear to me what you want. If t is a real array, s = 0.5 + i.t is a complex array, not a complex number.

Will the following do, perhaps?

Code:
program tz3
  integer, parameter :: k = 7
  integer j
  complex s(k)
  do j = 1, k
     s(j) = cmplx(0.5,j*0.1)
  end do
  print *,zeta_3(s,5)
contains
elemental function zeta_3(s,n) result(zs1)
   integer,allocatable :: nn(:)
   integer, intent(in) :: n
   complex zs1
   complex, intent(in) :: s
   allocate (nn(n))
   nn = (/(i,i=1,n)/)
   zs1 = sum((-1)**(nn + 1)/(nn**s ))
   zs1 = 1/(1 - 2**( 1-s ))*zs1
end function zeta_3
end program
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mecej4



Joined: 31 Oct 2006
Posts: 1896

PostPosted: Thu Dec 21, 2023 4:42 pm    Post subject: Reply with quote

You could replace the explicit DO loop
Code:
do i = 1, k
write(25,'(f15.12,1x,f15.12)') real_part(i), imaginary_part(i) ! write data to file
end do


with the implied DO loop (available since Fortran 77)

Code:
write(25,'(f15.12,1x,f15.12)') (real_part(i), imaginary_part(i), i=1,k)
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