firstly we look at the formula for the Taylor series, which is:
f(x) = sum_(n=0)^oo f^((n))(a)/(n!)(x-a)^n
which equals:
f(a) + f'(a)(x-a) + (f''(a)(x-a)^2)/(2!) + (f'''(a)(x-a)^3)/(3!) + ...
So you would like to solve for f(x) = ln(x) at x=1 which I assume mean centered at 1 of which you would make a=1
To solve:
f(x) = ln(x) and f(1) = ln(1) = 0
f'(x) = 1/x and f'(1) = 1/1 = 1
f''(x) = -1/x^2 and f''(1) = -1/(1)^2 = -1
f^((3))(x) = 2/x^3 and f^((3))(1) = 2/(1)^3 = 2
f^((4))(x) = -((2)(3))/x^4 and f^((4))(1) = -((2)(3))/(1)^4 = -(2)(3)
Where now we can already start to see a pattern forming, so we starting using our formula(2):
0 + 1(x-1) - (1(x-1)^2)/(2!) + (2(x-1)^3)/(3!) - ((2)(3)(x-1)^4)/(4!) .....
and now try try see how we can write this as a series, which we get: (we start will n=1 as our first term is 0)
f(x) = ln(x) = sum_(n=1)^oo (-1)^(n-1) (((n-1)!)(x-1)^n)/(n!)
Which can then simplify to:
f(x) = ln(x) = sum_(n=1)^oo (-1)^(n-1) (x-1)^n/n
