I need help with a taylor series approximation of ln(1+x)!

I am trying to do a taylor series approximation of ln(1+x) with a while loop and can't seem to get it right. this is what I have so far:

x = 3;
terms = 3;  
n = 0;  
lnx = 0;
arg = x + 1;    % argument for ln function
lnx_plot = zeros(terms,1);   % vector holding taylor series approximation values
real_lnx = zeros(terms,1);   % vector of same length holding true ln evaluation
iter = zeros(terms,1);

% loop for taylor series approxiation
while n < terms   
    n = n + 1;
    lnx = lnx + (-1)^(n-1)*x^n/n;
    lnx_plot(n) = lnx;      % puts current value at position 'n'
    real_lnx(n) = log(arg); % puts true ln value at position 'n'
    iter(n) = n;
end

Answer 1

[James Tursa]

This series only converges for abs(x) < 1. You are trying to use it with x = 3, which will diverge.

[Thomas MacDowell]

The problem this is solving asks to find the approximation evaluiated at x = 3. How can I write this program so that the approximation of ln(1+x) converges at x=3

[Thomas MacDowell]

This is the actual problem, I am trying to evaluate the function before i go ahead and add text to find the error associated with the approximation.

• Compute and determine the exact error when using only the first 3 and first 5 terms of the series
when compared to the actual solution when the function is evaluated at ? = 3.

[James Tursa]

You cannot change the convergence interval of this series. So if you use it with x=3 it will diverge. Maybe the intent of the assignment is to simply show this divergence.

[Thomas MacDowell]

How will it not converge if ln(1+3) is not equal to inf?

[James Tursa]

The fact that a function value itself is finite does not guarantee that a Taylor Series expansion of the function will converge for all values. Just look at the trend of the first few terms of your series:

log(1+x) = x - x^2/2 + x^3/3 - x^4/4 + x^5/5 - ...

Now plug in 3:

log(4) = 3 - 3^2/2 + 3^3/3 - 3^4/4 + 3^5/5 - ...
       = 3 - 9/2 + 27/3 - 81/4 + 243/5 - ...

This has larger and larger positive & negative swings in the terms ... does it really look like that will converge? No, of course not. This series simply will not converge if abs(x) > 1 even though the function value itself at x=3 is finite. This behavior, where the interval of convergence is finite, is not uncommon in Taylor Series expansions.