Answer:
Here we just want to find the Taylor series for f(x) = ln(x), centered at the value of a (which we do not know).
Remember that the general Taylor expansion is:
[tex]f(x) = f(a) + f'(a)*(x - a) + \frac{1}{2!}*f''(a)(x -a)^2 + ...[/tex]
for our function we have:
f'(x) = 1/x
f''(x) = -1/x^2
f'''(x) = (1/2)*(1/x^3)
this is enough, now just let's write the series:
[tex]f(x) = ln(a) + \frac{1}{a} *(x - a) - \frac{1}{2!} *\frac{1}{a^2} *(x - a)^2 + \frac{1}{3!} *\frac{1}{2*a^3} *(x - a)^3 + ....[/tex]
This is the Taylor series to 3rd degree, you just need to change the value of a for the required value.
