Natural logarithm taylor series

ELNS · June 6, 2020, 2:14pm

Hi all
I have created this issue on my repository and really need help.
Any ideas about how to solve and implement this exercise in fortran?

parekhvs · June 7, 2020, 3:47am

If you’re writing a simple program with user interaction, you can ask the user to enter a value of ‘z’ that is in [-1,1] range.

With the user input of ‘z’ and x=3, you can calculate ‘m’ - I think a simple Newton method should converge in a few iterations.

Then determine the summation series involving Greek letter ‘xi’ by adding terms and stopping when the increment is less than EPSILON of the floating-point precision being used.

‘lnx’ is yielded by the Taylor’s series expansion you show.

ELNS · June 7, 2020, 5:28am

SOLVED

ELNS · June 7, 2020, 7:28am

And this the implementation:

github.com

EverLookNeverSee/FCS/blob/master/src/numerical methods/Exercise_08_c.f90

! Created by EverLookNeverSee@GitHub on 6/5/20
! For more information see FCS/img/Exercise_08_c.png

! Solution:
!   Since x = z * (2 ** m) Hence z = x / (2 ** m)
!   We can choose any `m` as long as -1 <= z <= 1
!   Our x = 3 , for which m = 2 and z = 0.75 works.
!   Smaller `m` will give a larger `z` which will give
!   smaller `xi` which will lead to fewer terms necessary
!   for a certain precision.

! I could not have done it without @taless474, @diehlpk and @LKedward

program main
    implicit none
    ! xi_sum --> sum of xi elements
    ! previous --> sum of xi elements in previous step
    ! Ea --> approximation error
    ! declaring and initializing variables
    integer :: i = 1

This file has been truncated. show original

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Natural logarithm taylor series

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