Solving a taylor serie with matlab. Homework question

Taylor serie of ln x:

ln x = (x − 1) − ((x − 1)^2)/ 2 + ((x − 1)^3)/ 3 − . . .

I need to find an approximately value of ln 3. The question suggested that we should replace x with 1/3 and then take the sum of all terms with absolute value bigger than 1e-8.

The questions suggested solution:

tol=1e-8; s=0; i=0;

term=;

while abs(term) > tol
s=s+term;
i=i+1;
term=...;
end

disp(-s)

What i've tried

tol=1e-8; s=0; i=1;

term = -2./3;


while abs(term) > tol
s = s + term;
i=i+1;
term = ((-2./3).^i)./i;
end

disp(-s)

My code gives the answear 0.51. The answear to ln 3 is 1.098...

How do i solve this?

Answer 1

[Alan Stevens]

There's a multiplier of (-1)^(i-1) that you need. Try

tol=1e-8; s=0; i=1;
k = -2/3;
s = k;
err = 1;
while err > tol
      sold = s;
      i=i+1;
      term = (-1)^(i-1)*(k.^i)./i;
      s = sold + term;
      err = abs(s-sold);
 end

disp(-s)