How Can I Correct My Fortran Program for Newton's Method?

[jjohnson8]

I'm working on a program for Newton's method for solving equations.

This is my code:

=======================================================
program Newton
implicit real(a-h,o-z)
F(x) = x**2 - 4
!...&---1---------2---------3---------4---------5---------6---------7---

!----All values to be read in/set values----!
PRINT*, 'Please inter initial guess for root: '
READ*, x0
maxIT = 20
TOL = 1.e-14
EPS = 1.e-14
xn = 0
!----Print titles for output----!
PRINT 20, 'x0', 'maxIT', 'TOL', 'EPS', 'n', 'xn', 'F(xn)'

!----Do Loop to go from 0 to max iterations----!
DO n = 0,maxIT
CALL FCN(xn, Fn, DFn)
xn = x0 - (Fn/DFn)
Dx = xn - x0
x0 = xn
IF( ABS(Dx) .LE. TOL) THEN
IF( ABS(Fn) .LE. TOL) THEN
PRINT*, 'DONE: root=',xn,' found in',n,'iterations'
PRINT*, ' residual =', Fn
ELSE
PRINT*, 'Stuck at iteration:', n
PRINT*, ' relDX < EPS but residual=',Fn,' > TOL, exiting'
STOP
ENDIF
ENDIF
PRINT 10, x0, maxIT, TOL, EPS, n, xn, F(xn)
ENDDO
!----Formats for headings----!
10 FORMAT(F10.7,4x,I2,4x,F18.14,13x,F18.14,8x,I2,3x,F10.7,3x,F10.7)
20 FORMAT(A2,10x,A5,8x,A3,35x,A3,10x,A1,6x,A2,11x,A5)
END

subroutine FCN(xn, Fn, DFn)
Fn = (x0**2) - 4
DFn = 2*x0
return
END


Here is my output:

x0 maxIT TOL EPS n xn F(xn)
+Infinity 20 0.00000000000001 0.00000000000001 0 +Infinity +Infinity
+Infinity 20 0.00000000000001 0.00000000000001 1 +Infinity +Infinity
+Infinity 20 0.00000000000001 0.00000000000001 2 +Infinity +Infinity
+Infinity 20 0.00000000000001 0.00000000000001 3 +Infinity +Infinity
+Infinity 20 0.00000000000001 0.00000000000001 4 +Infinity +Infinity
+Infinity 20 0.00000000000001 0.00000000000001 5 +Infinity +Infinity
+Infinity 20 0.00000000000001 0.00000000000001 6 +Infinity +Infinity
+Infinity 20 0.00000000000001 0.00000000000001 7 +Infinity +Infinity
+Infinity 20 0.00000000000001 0.00000000000001 8 +Infinity +Infinity
+Infinity 20 0.00000000000001 0.00000000000001 9 +Infinity +Infinity
+Infinity 20 0.00000000000001 0.00000000000001 10 +Infinity +Infinity
+Infinity 20 0.00000000000001 0.00000000000001 11 +Infinity +Infinity
+Infinity 20 0.00000000000001 0.00000000000001 12 +Infinity +Infinity
+Infinity 20 0.00000000000001 0.00000000000001 13 +Infinity +Infinity
+Infinity 20 0.00000000000001 0.00000000000001 14 +Infinity +Infinity
+Infinity 20 0.00000000000001 0.00000000000001 15 +Infinity +Infinity
+Infinity 20 0.00000000000001 0.00000000000001 16 +Infinity +Infinity
+Infinity 20 0.00000000000001 0.00000000000001 17 +Infinity +Infinity
+Infinity 20 0.00000000000001 0.00000000000001 18 +Infinity +Infinity
+Infinity 20 0.00000000000001 0.00000000000001 19 +Infinity +Infinity
+Infinity 20 0.00000000000001 0.00000000000001 20 +Infinity +Infinity


I've tinkered with this all day, and I can't figure out how to get anything different than infinity for the values. Anyone out there that can correct me?

 

[Dr Transport]

Looks like you're not passing xn, the function should be

subroutine FCN(xn, Fn, DFn)
Fn = (xn**2) - 4
DFn = 2*xn
return

also try making tol and eps larger...

 

[quicknote]

I have to say that it is impressive that you are working on a program for Newton's method in Fortran. This method is a powerful tool for solving equations and is widely used in various fields of science and engineering.

Looking at your code and output, it seems that you are having trouble with the initial guess for the root. The program is not able to converge to a solution because the initial guess is too far from the actual root. This is why your values for x0, xn, and F(xn) are all showing as infinity.

To correct this, you can try changing your initial guess to a value closer to the actual root. You can also try tweaking the TOL and EPS values to see if that helps with convergence.

Additionally, it is important to carefully check your code and ensure that all variables are correctly defined and used in the calculations. Small errors in coding can also lead to incorrect results.

Overall, keep experimenting and fine-tuning your code. Newton's method is a valuable tool and with some adjustments, you will be able to successfully solve equations using your program. Good luck!