Quandaries and Queries


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*2n))*(x-2)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/x2)*1/2!

f---(x)= (2/x3)*1/3!

f----(x)= (-6/x4)* 1/4!

so the pattern shows me that f(n)= ((-1)(n+1))/xn *n)

so f(2)= sigma ((-1)(n+1))/2n *n) *(x-2)n

so as you see i don`t get ln 2




The Taylor series expression for f(x) at x = a is

where f(n)(a) is the n-th 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! (x-2)0

But 0! = 1 and (x-2)0 = 1 and hence the the first term is

f(0)(2) = f(2) = ln(2)



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