Implementation of Exercise_08_c

[miladsade96]

Implementation of Exercise_08_c

[LKedward]

Sorry, I should have explained that a bit more.

x = z*(2^m)

therefore

z = x/(2^m)

we can choose any m as long as -1 < z < 1.

Questions asks for x=3, for which m=2 and z=0.75 works.

Possible pseudo-code for program:

set ln2 = 0.6931472
ask user for value of x
ask user for value of m
calculate z = x/(2^m)
check that -1 < z < 1

calculate xi= (1-z)/(1+z)
calculate lnx = m*ln2 - 2*sum( (xi^(1:N))/(1:N) )

compare with exact value of ln(x)

Can try different values of m - I think smaller m will give a larger z which will give a smaller xi which will lead to fewer terms necessary for a certain precision.

Hope that helps a bit more.

edit: this repository is a really nice collection of Fortran snippets - great work!

[miladsade96]

Thank you very much @LKedward

[LKedward]

@FortranFan, the exercises appear to be from: Bose, S., 2019. Numerical Methods Of Mathematics Implemented In Fortran. Singapore. Springer. See here on google.

The text does not provide a derivation for the relation; my summary conclusion is that it is an error since z <= 0 implies x <= 0 which would be problematic for then calculating ln(x).
As you point out, 0 < z <= 1 appear to be the correct bounds for z here.

[LKedward]

I think m=2 and z=0.75 perhaps?

[miladsade96]

I think m=2 and z=0.75 perhaps?

How?

[FortranFan]

@everlookneversee ,

Can you provide a reference to a derivation for the relationship you show in the original post?

A quick glance suggests the series is divergent for z <= 0, so a numerical approach is unclear for -1 <= z <= 0.

When z > 0, xi < 1, so one can compute the series numerically to determine lnx.

[RonShepard]

One derivation of the relation in the original post follows from the identities

 y(z) = (1-z)/(1+z)

z(y) = (1-y)/(1+y)

From this, it follows that

ln(z) = ln(1-y) - ln(1+y)

Taylor expansion at y=0 on the right then gives the result shown in the original post where all of the even terms vanish. The radius of convergence of ln() on the right is -1<y<1. However, convergence is slow for large |y|, so the only useful values for computation are when y is small, which is when z is near +1. Thus in the computation where x is given, m should be chosen to make z<=1 as close to 1 as possible.