Find an approximately value of ln3 with Taylor

I am going to find the approximately value of ln3 by putting x = 1/3 and sum all the terms with a bigger abs-value than 1e-8. Anyone who know how I should continue my code?

I got the taylorserie of lnx for x = 1 as help in the task.

Answer 1

[Raunak Gupta]

Hi,

You may find below code useful. I am using ln(1-x) expansion of Taylor Series as it is more intuitive to code. It is not be very different from ln(1+x).

tol = 1e-8; 
s = 0; 
x = 2/3; 
count = 1; 
while 1 
    update = (x.^count)./count; 
    s = s - update; 
    count = count + 1; 
    if ~(abs(update) > tol) 
        return; 
    end 
end
% s is the finalvalue of ln(1/3)
% reversing the sign of s will give ln(3)