Question

ln2 with Taylor series

Answer 1

Try expanding the series about 1
\[f(x) = \sum_{k=0}^{\infty} \frac{ f^{(k)}(a) }{k! }(x-a)^k\]
i.e. let a=1.
\[f(x) = \sum_{k=0}^{\infty} \frac{ f^{(k)}(1) }{k! }(x-1)^k\]
if you want f(2), you have
\[ f(2) = \sum_{k=0}^{\infty} \frac{ f^{(k)}(1) }{k! }(2-1)^k \\
= \sum_{k=0}^{\infty} \frac{ f^{(k)}(1) }{k! }1^k
= \sum_{k=0}^{\infty} \frac{ f^{(k)}(1) }{k! }
\]

now you need the first and higher order derivatives of ln(x)
evaluated at x=a i.e. x=1
f' = 1/x = 1
f'' = -x^(-2) = -1
f'''= +2x^(-3) = +1 \cdot 2!
f''''= -3*2 x^(-4)= -1 3!
f''''' = +4*3*2 x^(-5) = +1 / 4!
or in general
\[ f^{(k)}(1) = (-1)^{k+1} \cdot (k-1)!\cdot x^{-k} = (-1)^{k+1} \cdot (k-1)! \]