Taylor series representation of lnx at a=3

[petrol.veem]

I'm find this question a little bit hard:

Find the Taylor series representation of the function lnx centered at a=3

I started by computing the first few derivatives of the function:

f^0(x) = lnx
f^1(x) = 1/x
f^2(x) = -1/x^2
f^3(x) = 2/x^3
f^4(x) = -6/x^4

So writing this now in the form of a Taylor series:

Sigma(1,inf) [ (-1)^(n+1) (n-1)! (x-a)^n / (a)^n n! ]

Then to write it as a Taylor series for a=3 do I simply substitute 3 in for a and I am done? I am somewhat concerned for n=0 and what to do with f^0(x) = lnx as well.

I find these Taylor and Maclaurin series somewhat confusing.

 

[Dr. Flim-Flam]

P^0(x) = ln(x) P^0(3) = ln(3)
P^1(x) = 0!/x P^1(3) = 1/3
P^2(x) = -1!/x^2 P^2(3) = -1/3^2
P^3(x) = 2!/x^3 P^3(3) = 2!/3^3
P^4(x) = -3!/x^4 P^4(3) = -3!/3^4
P^5(x) = 4!/x^5 P^4(3) = 4!/3^5
. .
.
. .
P^n(x) = (n-1)!/x^n P^n(3) = (-1)^(n+1)*(n-1)!/3^n n>=1

Hence the Taylor series = ln(3) + Sum[(-1)^(n+1)(n-1)!(x-3)^n]/[3^n*n!] Sum goes from 1 to infinity

equals ln(3)+Sum[(-1)^(n+1)(x-3)^n]/(n*3^n) = ln(3) +{ (x-3)/3 - [(x-3)^2]/(2*3^2) + [(x-3)^3]/(3*3^3) -[(x-3)^4]/(4*3^4) + [(x-3)^5/(5*3^5) - [(x-3)^6]/(6*3^6) + ... + [(x-3)^n]/(n*3^n)}, n>=1 to infinity.



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