How to program natural log in matlab

I'm trying to write a Matlab function that will compute the natural log for any input within limits I have set. I'm considering the Taylor series expansion of f(x)=ln(1+x). Where the series is only converges for -1< x <=1. I'm also using E<10^-6 for convergence criterion. I wrote a program but its not running correctly. Any suggestions? Thank you in an advance for your help!

[Walter Roberson]

What behavior do you observe instead of it "running correctly" ?

[Josh]

 The values it gives me are inaccurate.

Answer 1

[John D'Errico]

You can learn a vast amount by reading what I did in my HPF toolbox, although there is no need for you to write a complete class to do the work. HPF uses series approximations, along with various range reduction schemes for all of the special functions I wrote. There I show how I compute logs that are accurate to many thousands of digits if desired.

Section 4.2 of HPFMod2.pdf describes the tricks I used in computation of the natural log. Find HPF here:

HPF

As an example...

log(hpf(2,100))
ans =
0.6931471805599453094172321214581765680755001343602552541206800094933936219696947156058633269964186875

And for comparison to the symbolic toolbox:

vpa(log(sym(2)),100)
ans =
0.6931471805599453094172321214581765680755001343602552541206800094933936219696947156058633269964186875

Some points to note:

You don't need to use a series to compute the log for all values of x, but only a restricted range, since you have various identities. For example, here is a simple one.

log(e*x) = log(e) + log(x) = 1 + log(x)
log(x/e) = -log(e) + log(x) = -1 + log(x)

So, if x is greater than sqrt(e), you can always divide by e repeatedly until x is not. Likewise, if x is less than 1/sqrt(e), you can always multiply by e until it is in the interval [1/sqrt(e),sqrt(e)].

So you might use a while loop to deal with x outside of the interval of interest.

exp(1/2)
ans =
          1.64872127070013

exp(-1/2)
ans =
         0.606530659712633

This is a reasonably small range to deal with. Once you get x into that interval, then a series approximation will do reasonably well. You can do a bit more of course, as I show in the pdf.