Quandaries
and Queries 

Question: i have to represent ln(x) as a power series about 2 i`m not getting the final answer which is ln 2+ sigma (((1)^{(n+1)}/
(n*2^{n}))*(x2)^{n}). i don`t get the ln 2 part i show you my trial f(x)= ln x. f(x)=(1/x) . f(x)= (1/x^{2})*1/2! f(x)= (2/x^{3})*1/3! f(x)= (6/x^{4})* 1/4! so the pattern shows me that f(n)= ((1)^{(n+1)})/x^{n} *n) so f(2)= sigma ((1)^{(n+1)})/2^{n} *n) *(x2)^{n} so as you see i don`t get ln 2 

Anood, The Taylor series expression for f(x) at x = a is where f^{(n)}(a) is the nth derivative of f(x) at x=a if n ≥ 1 and f^{(0)}(a) is f(a). The series you developed for ln(x) at a=2 is correct for n ≥ 1 but what about the first term, that is n = 0? The first term (n = 0) is ^{f(0)(2)}/_{0!} (x2)^{0} But 0! = 1 and (x2)^{0} = 1 and hence the the first term is
Penny


