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In Newton's method, to solve a nonlinear system of equations we need to find the Jacobian matrix and the determinant of the inverse of the Jacobian matrix.
Here are my component functions,
real function f1(x,y)
parameter (pi = 3.141592653589793)
f1 = log(abs(x-y**2)) - sin(x*y) - sin(pi)
end function f1
real function f2(x,y)
f2 = exp(x*y) + cos(x-y) - 2
end function f2
For the 2x2 case I am computing the Jacobian matrix and determinant of the inverse of Jacobian matrix like this,
x = [2,2]
h = 0.00001
.
.
! calculate approximate partial derivative
! you can make it more accurate by reducing the
! value of h
j11 = (f1(x(1)+h,x(2))-f1(x(1),x(2)))/h
j12 = (f1(x(1),x(2)+h)-f1(x(1),x(2)))/h
j21 = (f2(x(1)+h,x(2))-f2(x(1),x(2)))/h
j22 = (f2(x(1),x(2)+h)-f2(x(1),x(2)))/h
! calculate the Jacobian
J(1,:) = [j11,j12]
J(2,:) = [j21,j22]
! calculate inverse Jacobian
inv_J(1,:) = [J(2,2),-J(1,2)]
inv_J(2,:) = [-J(2,1),J(1,1)]
DET=J(1,1)*J(2,2) - J(1,2)*J(2,1)
inv_J = inv_J/DET
.
.
How do I in Fortran extend this to evaluate a Jacobian for m functions evaluated at n points?
Here is a more flexible jacobian calculator.
The results with the 2×2 test case are what you expect
arguments (x)
2.00000000000000
2.00000000000000
values (y)
1.44994967586787
53.5981500331442
Jacobian
0.807287239448229 3.30728724371454
109.196300248300 109.196300248300
I check the results against a symbolic calculation for the given inputs of
Console.f90
program Console1
use ISO_FORTRAN_ENV
implicit none
! Variables
integer, parameter :: wp = real64
real(wp), parameter :: pi = 3.141592653589793d0
! Interfaces
interface
function fun(x,n,m) result(y)
import
integer, intent(in) :: n,m
real(wp), intent(in) :: x(m)
real(wp) :: y(n)
end function
end interface
real(wp) :: h
real(wp), allocatable :: x(:), y(:), J(:,:)
! Body of Console1
x = [2d0, 2d0]
h = 0.0001d0
print *, "arguments"
print *, x(1)
print *, x(2)
y = test(x,2,2)
print *, "values"
print *, y(1)
print *, y(2)
J = jacobian(test,x,2,h)
print *, "Jacobian"
print *, J(1,:)
print *, J(2,:)
contains
function test(x,n,m) result(y)
! Test case per original question
integer, intent(in) :: n,m
real(wp), intent(in) :: x(m)
real(wp) :: y(n)
y(1) = log(abs(x(1)-x(2)**2)) - sin(x(1)*x(2)) - sin(pi)
y(2) = exp(x(1)*x(2)) + cos(x(1)-x(2)) - 2
end function
function jacobian(f,x,n,h) result(u)
procedure(fun), pointer, intent(in) :: f
real(wp), allocatable, intent(in) :: x(:)
integer, intent(in) :: n
real(wp), intent(in) :: h
real(wp), allocatable :: u(:,:)
integer :: j, m
real(wp), allocatable :: y1(:), y2(:), e(:)
m = size(x)
allocate(u(n,m))
do j=1, m
e = element(j, m) ! Get kronecker delta for j-th value
y1 = f(x-e*h/2,n,m)
y2 = f(x+e*h/2,n,m)
u(:,j) = (y2-y1)/h ! Finite difference for each column
end do
end function
function element(i,n) result(e)
! Kronecker delta vector. All zeros, except the i-th value.
integer, intent(in) :: i, n
real(wp) :: e(n)
e(:) = 0d0
e(i) = 1d0
end function
end program Console1
I will answer about evaluation in different points. This is quite simple. You just need an array of points, and if the points are in some regular grid, you may not even need that.
You may have an array of xs and array of ys or you can have an array of derived datatype with x and y components.
For the former:
real, allocatable :: x(:), y(:)
x = [... !probably read from some data file
y = [...
do i = 1, size(x)
J(i) = Jacobian(f, x(i), y(i))
end do
If you want to have many functions at the same time, the problem is always in representing functions. Even if you have an array of function pointers, you need to code them manually. A different approach is to have a full algebra module, where you enter some string representing a function and you can evaluate such function and even compute derivatives symbolically. That requires a parser, an evaluator, it is a large task. There are libraries for this. Evaluation of such a derivative will be slow unless further optimizing steps (compiling to machine code) are undertaken.
Numerical evaluation of the derivative is certainly possible. It will slow the convergence somewhat, depending on the order of the approximation of the derivative. You do a difference of two points for the numerical derivative. You can make an interpolating polynomial from values in multiple points to get a higher-order approximation (finite difference approximations), but that costs machine cycles.
Normally we can use auto difference tools as #John Alexiou mentioned. However in practise I prefer using MATLAB to analytically solve out the Jacobian and then use its build-in function fortran() to convert the result to a f90 file. Take your function as an example. Just type these into MATLAB
syms x y
Fval=sym(zeros(2,1));
Fval(1)=log(abs(x-y^2)) - sin(x*y) - sin(pi);
Fval(2)=exp(x*y) + cos(x-y) - 2;
X=[x;y];
Fjac=jacobian(Fval,X);
fortran(Fjac)
which will yield
Fjac(1,1) = -y*cos(x*y)-((-(x-y**2)/abs(-x+y**2)))/abs(-x+y**2)
Fjac(1,2) = -x*cos(x*y)+(y*((-(x-y**2)/abs(-x+y**2)))*2.0D0)/abs(-
&x+y**2)
Fjac(2,1) = -sin(x-y)+y*exp(x*y)
Fjac(2,2) = sin(x-y)+x*exp(x*y)
to you. You just get an analytical Jacobian fortran function.
Meanwhile, it is impossible to solve the inverse of a mxn matrix because of rank mismatching. You should simplify the system of equations to get a nxn Jacobin.
Additionally, when we use Newton-Raphson's method we do not solve the inverse of the Jacobin which is time-consuming and inaccurate for a large system. An easy way is to use dgesv in LAPACK for dense Jacobin. As we only need to solve the vector x from system of linear equations
Jx=-F
dgesv use LU decomposition and Gaussian elimination to solve above system of equations which is extremely faster than solving inverse matrix.
If the system of equations is large, you can use UMFPACK and its fortran interface module mUMFPACK to solve the system of equations in which J is a sparse matrix. Or use subroutine ILUD and LUSOL in a wide-spread sparse matrix library SPARSEKIT2.
In addition to these, there are tons of other methods which try to solve the Jx=-F faster and more accurate such as Generalized Minimal Residual (GMRES) and Stabilized Bi-Conjugate Gradient (BICGSTAB) which is a strand of literature.
I am trying to write a FORTRAN code to evaluate the fast Fourier transform of the Gaussian function f(r)=exp(-(r^2)) using FFTW3 library. As everyone knows, the Fourier transform of the Gaussian function is another Gaussian function.
I consider evaluating the Fourier-transform integral of the Gaussian function in the spherical coordinate.
Hence the resulting integral can be simplified to be integral of [r*exp(-(r^2))*sin(kr)]dr.
I wrote the following FORTRAN code to evaluate the discrete SINE transform DST which is the discrete Fourier transform DFT using a PURELY real input array. DST is performed by C_FFTW_RODFT00 existing in FFTW3, taking into account that the discrete values in position space are r=i*delta (i=1,2,...,1024), and the input array for DST is the function r*exp(-(r^2)) NOT the Gaussian. The sine function in the integral of [r*exp(-(r^2))*sin(kr)]dr resulting from the INTEGRATION over the SPHERICAL coordinates, and it is NOT the imaginary part of exp(ik.r) that appears when taking the analytic Fourier transform in general.
However, the result is not a Gaussian function in the momentum space.
Module FFTW3
use, intrinsic :: iso_c_binding
include 'fftw3.f03'
end module
program sine_FFT_transform
use FFTW3
implicit none
integer, parameter :: dp=selected_real_kind(8)
real(kind=dp), parameter :: pi=acos(-1.0_dp)
integer, parameter :: n=1024
real(kind=dp) :: delta, k
real(kind=dp) :: numerical_F_transform
integer :: i
type(C_PTR) :: my_plan
real(C_DOUBLE), dimension(1024) :: y
real(C_DOUBLE), dimension(1024) :: yy, yk
integer(C_FFTW_R2R_KIND) :: C_FFTW_RODFT00
my_plan= fftw_plan_r2r_1d(1024,y,yy,FFTW_FORWARD, FFTW_ESTIMATE)
delta=0.0125_dp
do i=1, n !inserting the input one-dimension position function
y(i)= 2*(delta)*(i-1)*exp(-((i-1)*delta)**2)
! I multiplied by 2 due to the definition of C_FFTW_RODFT00 in FFTW3
end do
call fftw_execute_r2r(my_plan, y,yy)
do i=2, n
k = (i-1)*pi/n/delta
yk(i) = 4*pi*delta*yy(i)/2 !I divide by 2 due to the definition of
!C_FFTW_RODFT00
numerical_F_transform=yk(i)/k
write(11,*) i,k,numerical_F_transform
end do
call fftw_destroy_plan(my_plan)
end program
Executing the previous code gives the following plot which is not for Gaussian function.
Can anyone help me understand what the problem is? I guess the problem is mainly due to FFTW3. Maybe I did not use it properly especially concerning the boundary conditions.
Looking at the related pages in the FFTW site (Real-to-Real Transforms, transform kinds, Real-odd DFT (DST)) and the header file for Fortran, it seems that FFTW expects FFTW_RODFT00 etc rather than FFTW_FORWARD for specifying the kind of
real-to-real transform. For example,
! my_plan= fftw_plan_r2r_1d( n, y, yy, FFTW_FORWARD, FFTW_ESTIMATE )
my_plan= fftw_plan_r2r_1d( n, y, yy, FFTW_RODFT00, FFTW_ESTIMATE )
performs the "type-I" discrete sine transform (DST-I) shown in the above page. This modification seems to fix the problem (i.e., makes the Fourier transform a Gaussian with positive values).
The following is a slightly modified version of OP's code to experiment the above modification:
! ... only the modified part is shown...
real(dp) :: delta, k, r, fftw, num, ana
integer :: i, j, n
type(C_PTR) :: my_plan
real(C_DOUBLE), allocatable :: y(:), yy(:)
delta = 0.0125_dp ; n = 1024 ! rmax = 12.8
! delta = 0.1_dp ; n = 128 ! rmax = 12.8
! delta = 0.2_dp ; n = 64 ! rmax = 12.8
! delta = 0.4_dp ; n = 32 ! rmax = 12.8
allocate( y( n ), yy( n ) )
! my_plan= fftw_plan_r2r_1d( n, y, yy, FFTW_FORWARD, FFTW_ESTIMATE )
my_plan= fftw_plan_r2r_1d( n, y, yy, FFTW_RODFT00, FFTW_ESTIMATE )
! Loop over r-grid
do i = 1, n
r = i * delta ! (2-a)
y( i )= r * exp( -r**2 )
end do
call fftw_execute_r2r( my_plan, y, yy )
! Loop over k-grid
do i = 1, n
! Result of FFTW
k = i * pi / ((n + 1) * delta) ! (2-b)
fftw = 4 * pi * delta * yy( i ) / k / 2 ! the last 2 due to RODFT00
! Numerical result via quadrature
num = 0
do j = 1, n
r = j * delta
num = num + r * exp( -r**2 ) * sin( k * r )
enddo
num = num * 4 * pi * delta / k
! Analytical result
ana = sqrt( pi )**3 * exp( -k**2 / 4 )
! Output
write(10,*) k, fftw
write(20,*) k, num
write(30,*) k, ana
end do
Compile (with gfortran-8.2 + FFTW3.3.8 + OSX10.11):
$ gfortran -fcheck=all -Wall sine.f90 -I/usr/local/Cellar/fftw/3.3.8/include -L/usr/local/Cellar/fftw/3.3.8/lib -lfftw3
If we use FFTW_FORWARD as in the original code, we get
which has a negative lobe (where fort.10, fort.20, and fort.30 correspond to FFTW, quadrature, and analytical results). Modifying the code to use FFTW_RODFT00 changes the result as below, so the modification seems to be working (but please see below for the grid definition).
Additional notes
I have slightly modified the grid definition for r and k in my code (Lines (2-a) and (2-b)), which is found to improve the accuracy. But I'm still not sure whether the above definition matches the definition used by FFTW, so please read the manual for details...
The fftw3.f03 header file gives the interface for fftw_plan_r2r_1d
type(C_PTR) function fftw_plan_r2r_1d(n,in,out,kind,flags) bind(C, name='fftw_plan_r2r_1d')
import
integer(C_INT), value :: n
real(C_DOUBLE), dimension(*), intent(out) :: in
real(C_DOUBLE), dimension(*), intent(out) :: out
integer(C_FFTW_R2R_KIND), value :: kind
integer(C_INT), value :: flags
end function fftw_plan_r2r_1d
(Because of no Tex support, this part is very ugly...) The integral of 4 pi r^2 * exp(-r^2) * sin(kr)/(kr) for r = 0 -> infinite is pi^(3/2) * exp(-k^2 / 4) (obtained from Wolfram Alpha or by noting that this is actually a 3-D Fourier transform of exp(-(x^2 + y^2 + z^2)) by exp(-i*(k1 x + k2 y + k3 z)) with k =(k1,k2,k3)). So, although a bit counter-intuitive, the result becomes a positive Gaussian.
I guess the r-grid can be chosen much coarser (e.g. delta up to 0.4), which gives almost the same accuracy as long as it covers the frequency domain of the transformed function (here exp(-r^2)).
Of course there are negative components of the real part to the FFT of a limited Gaussian spectrum. You are just using the real part of the transform. So your plot is absolutely correct.
You seem to be mistaking the real part with the magnitude, which of course would not be negative. For that you would need to fftw_plan_dft_r2c_1d and then calculate the absolute values of the complex coefficients. Or you might be mistaking the Fourier transform with a limited DFT.
You might want to check here to convince yourself of the correctness of you calculation above:
http://docs.mantidproject.org/nightly/algorithms/FFT-v1.html
Please do keep in mind that the plots on the above page are shifted, so that the 0 frequency is in the middle of the spectrum.
Citing yourself, the nummeric integration of [r*exp(-(r^2))*sin(kr)]dr would have negative components for all k>1 if normalised to 0 for highest frequency.
TLDR: Your plot is absolute state of the art and inline with discrete and limited functional analysis.
I'm attempting to test a program that calculates the discrete Fourier transform of a signal, namely a sine wave. To test it, I need to plot my results. However, the result is an array of size N (currently at 400) and is filled with complex numbers of the form z = x + iy. So I know that to test it I need to plot these results, and that to do this I need to plot |z|. Here's my program:
program DFT
implicit none
integer :: k, N, x, y, j, r, l, istat, p
integer, parameter :: dp = selected_real_kind(15,300)
real, allocatable,dimension(:) :: h
complex, allocatable, dimension(:) :: rst
complex, dimension(:,:), allocatable :: W
real(kind=dp) :: pi
p = 2*pi
!open file to write results to
open(unit=100, file="dft.dat", status='replace')
N = 400
!allocate arrays as length N, apart from W (NxN)
allocate(h(N))
allocate(rst(N))
allocate(W(-N/2:N/2,1:N))
pi = 3.14159265359
!loop to create the sample containing array
do k=1,N
h(k) = sin((2*pi*k)/N)
end do
!loop to fill the product matrix with values
do j = -N/2,N/2
do k = 1, N
W(j,k) = EXP((2.0_dp*pi*cmplx(0.0_dp,1.0_dp)*j*k)/N)
end do
end do
!use of matmul command to multiply matrices
rst = matmul(W,h)
print *, h, w
write(100,*) rst
end program
So my question is how do I take the magnitude of all the individual complex numbers in the array?
The ABS intrinsic function returns the magnitude of a complex number in Fortran. It is an elemental function as well, so for an array of type complex simply ABS( array ) will return a real array with the same kind as the original containing the results you want.
I am struggling with reading a text string in. Am using gfortran 4.9.2.
Below I have written a little subroutine in which I would like to submit the write format as argument.
Ideally I'd like to be able to call it with
call printarray(mat1, "F8.3")
to print out a matrix mat1 in that format for example. The numbers of columns should be determined automatically inside the subroutine.
subroutine printarray(x, udf_temp)
implicit none
real, dimension(:,:), intent(in) :: x ! array to be printed
integer, dimension(2) :: dims ! array for shape of x
integer :: i, j
character(len=10) :: udf_temp ! user defined format, eg "F8.3, ...
character(len = :), allocatable :: udf ! trimmed udf_temp
character(len = 10) :: udf2
character(len = 10) :: txt1, txt2
integer :: ncols ! no. of columns of array
integer :: udf_temp_length
udf_temp_length = len_trim(udf_temp)
allocate(character(len=udf_temp_length) :: udf)
dims = shape(x)
ncols = dims(2)
write (txt1, '(I5)') ncols
udf2 = trim(txt1)//adjustl(udf)
txt2 = "("//trim(udf2)//")"
do i = 1, dims(1)
write (*, txt2) (x(i, j), j = 1, dims(2)) ! this is line 38
end do
end suroutine printarray
when I set len = 10:
character(len=10) :: udf_temp
I get compile error:
call printarray(mat1, "F8.3")
1
Warning: Character length of actual argument shorter than of dummy argument 'udf_temp' (4/10) at (1)
When I set len = *
character(len=*) :: udf_temp
it compiles but at runtime:
At line 38 of file where2.f95 (unit = 6, file = 'stdout')
Fortran runtime error: Unexpected element '( 8
What am I doing wrong?
Is there a neater way to do this?
Here's a summary of your question that I will try to address: You want to have a subroutine that will print a specified two-dimensional array with a specified format, such that each row is printed on a single line. For example, assume we have the real array:
real, dimension(2,8) :: x
x = reshape([1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16], shape=[2,8], order=[2,1])
! Then the array is:
! 1.000 2.000 3.000 4.000 5.000 6.000 7.000 8.000
! 9.000 10.000 11.000 12.000 13.000 14.000 15.000 16.000
We want to use the format "F8.3", which prints floating point values (reals) with a field width of 8 and 3 decimal places.
Now, you are making a couple of mistakes when creating the format within your subroutine. First, you try to use udf to create the udf2 string. This is a problem because although you have allocated the size of udf, nothing has been assigned to it (pointed out in a comment by #francescalus). Thus, you see the error message you reported: Fortran runtime error: Unexpected element '( 8.
In the following, I make a couple of simplifying changes and demonstrate a few (slightly) different techniques. As shown, I suggest the use of * to indicate that the format can be applied an unlimited number of times, until all elements of the output list have been visited. Of course, explicitly stating the number of times to apply the format (ie, "(8F8.3)" instead of "(*(F8.3))") is fine, but the latter is slightly less work.
program main
implicit none
real, dimension(2,8) :: x
character(len=:), allocatable :: udf_in
x = reshape([1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16], shape=[2,8], order=[2,1])
udf_in = "F8.3"
call printarray(x, udf_in)
contains
subroutine printarray(x, udf_in)
implicit none
real, dimension(:,:), intent(in) :: x
character(len=*), intent(in) :: udf_in
integer :: ncols ! size(x,dim=2)
character(len=10) :: ncols_str ! ncols, stringified
integer, dimension(2) :: dims ! shape of x
character(len=:), allocatable :: udf0, udf1 ! format codes
integer :: i, j ! index counters
dims = shape(x) ! or just use: ncols = size(x, dim=2)
ncols = dims(2)
write (ncols_str, '(i0)') ncols ! use 'i0' for min. size
udf0 = "(" // ncols_str // udf_in // ")" ! create string: "(8F8.3)"
udf1 = "(*(" // udf_in // "))" ! create string: "(*(F8.3))"
print *, "Version 1:"
do i = 1, dims(1)
write (*, udf0) (x(i, j), j = 1,ncols) ! implied do-loop over j.
end do
print *, "Version 2:"
do i = 1, dims(1)
! udf1: "(*(F8.3))"
write (*, udf1) (x(i, j), j = 1,ncols) ! implied do-loop over j
end do
print *, "Version 3:"
do i = 1, size(x,dim=1) ! no need to create nrows/ncols vars.
write(*, udf1) x(i,:) ! let the compiler handle the extents.
enddo
end subroutine printarray
end program main
Observe: the final do-loop ("Version 3") is very simple. It does not need an explicit count of ncols because the * takes care of it automatically. Due to its simplicity, there is really no need for a subroutine at all.
besides the actual error (not using the input argument), this whole thing can be done much more simply:
subroutine printarray(m,f)
implicit none
character(len=*)f
real m(:,:)
character*10 n
write(n,'(i0)')size(m(1,:))
write(*,'('//n//f//')')transpose(m)
end subroutine
end
note no need for the loop constructs as fortran will automatically write the whole array , line wrapping as you reach the length of data specified by your format.
alternately you can use a loop construct, then you can use a '*' repeat count in the format and obviate the need for the internal write to construct the format string.
subroutine printarray(m,f)
implicit none
character(len=*)f
real m(:,:)
integer :: i
do i=1,size(m(:,1))
write(*,'(*('//f//'))')m(i,:)
enddo
end subroutine
end
I'm trying to use a complex matrix with the dimensions (n x n) with the fftw_mpi subroutines (version 3.3.2): A 1D FFT (complex to complex) on all the rows and after a 1D FFT (complex to complex) for all columns. First I have made a serial code in Fortran, and everything is working properly, but when I'm trying the fftw subroutines with mpi the results are not as I expected. Is something wrong in how is created the plan for the 1D FFT transform? Below it is an example (only the first 1D FFT on the rows):
! get local data size and allocate (dimension reversal)
alloc_local = fftw_mpi_local_size_2d(M, L, MPI_COMM_WORLD, &
local_M, local_j_offset)
cdata = fftw_alloc_complex(alloc_local)
call c_f_pointer(cdata, ldata, [L,local_M])
call c_f_pointer(cdata, data, [L,local_M])
! create MPI plan
plan = fftw_mpi_plan_many_dft(1, L*local_M, local_M, FFTW_MPI_DEFAULT_BLOCK, &
FFTW_MPI_DEFAULT_BLOCK, ldata, data, MPI_COMM_WORLD, FFTW_FORWARD, FFTW_ESTIMATE)
do j = 1, local_M
do i = 1, L
ldata(i,j)= my_function
enddo
enddo
! compute transform
call fftw_mpi_execute_dft(plan, ldata, data)
call fftw_destroy_plan(plan)
call fftw_free(cdata)
call MPI_FINALIZE(mpi_err)