Question

what is the co-efficient of (x-2)^2 in the taylor series for f(x) = lnx about x=2? As far as I can see if x=2, then the answer will be zero as (2-2)^2 is zero? The answer is -1/8. Could someone please explain

Answer 1

taylot series \[\sum_{k=0}^{\infty}\frac{ f ^{(k)} (c)}{ k! }(x-c)^k\]

\[f(x)=lnx\] f(2)=ln2
f'(x)=1/x f'(2)=1/2 ______} 0!/2^1
f"(x)=-1/x^2 f"(2)=-1/4 ______} -1!/2^2
f"'(x)=2/x^3 f"'(2)=2/8 ______} 2!/2^3
\[f ^{(4)}(x)=-6/x^4\]
\[f ^{(4)}(2)=-6/16\] ______}-3!/2^4
it seems like ((-1)^(k+1)(k-1)!/2^k)
taylor series about x=2
\[\sum_{k=0}^{\infty}\frac{ f ^{(k)} (2)}{ k! }(x-2)^k=\ln2+\frac{ 1 }{ 2 }(x-2)-\frac{ 1 }{ 4}(x-2)^2+\frac{ 1 }{ 4 }(x-2)^3-\frac{ 3 }{ 8 }(x-2)^4\] ..............
\[lnx =\ln2 +\sum_{k=1}^{\infty}(-1)^{(k+1)}\frac{ (k-1)! }{ 2^k }(x-2)^k\]