Fortran Program For Secant Method Vs Newton

$UWHPSC/codes/fortran/newton/newton.f90 module newton! Module parameters: implicit none integer, parameter:: maxiter = 20 real ( kind = 8 ), parameter:: tol = 1. D - 14 contains subroutine solve ( f, fp, x0, x, iters, debug )! Estimate the zero of f(x) using Newton's method.! F: the function to find a root of! Fp: function returning the derivative f'! X0: the initial guess!

Debug: logical, prints iterations if debug=.true.! The estimate x satisfying f(x)=0 (assumes Newton converged!)! The number of iterations iters implicit none real ( kind = 8 ), intent ( in ):: x0 real ( kind = 8 ), external:: f, fp logical, intent ( in ):: debug real ( kind = 8 ), intent ( out ):: x integer, intent ( out ):: iters! Declare any local variables: real ( kind = 8 ):: deltax, fx, fxprime integer:: k!

Initial guess x = x0 if ( debug ) then print 11, x 11 format ( 'Initial guess: x = ', e22.15 ) endif! Newton iteration to find a zero of f(x) do k = 1, maxiter! Evaluate function and its derivative: fx = f ( x ) fxprime = fp ( x ) if ( abs ( fx ). $UWHPSC/codes/fortran/newton/test1.f90 program test1 use newton, only: solve use functions, only: fsqrt, fprimesqrt implicit none real ( kind = 8 ):: x, x0, fx real ( kind = 8 ):: x0vals ( 3 ) integer:: iters, itest logical:: debug! Print., 'Test routine for computing zero of f' debug =.true.! Values to test as x0: x0vals = ( / 1. D0 / ) do itest = 1, 3 x0 = x0vals ( itest ) print., ' '!

Secant Method Geeks For Geeks

Blank line call solve ( fsqrt, fprimesqrt, x0, x, iters, debug ) print 11, x, iters 11 format ( 'solver returns x = ', e22.15, ' after', i3, ' iterations' ) fx = fsqrt ( x ) print 12, fx 12 format ( 'the value of f(x) is ', e22.15 ) if ( abs ( x - 2. D0 ) 1 d - 14 ) then print 13, x 13 format ( '. Unexpected result: x = ', e22.15 ) endif enddo end program test1This makes use of a module functions.f90 that includes the definition ofthe functions used here. Since (f(x) = x^2 - 4), we are attempting tocompute the square root of 4.

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