Taylor Series question here: So I am learning about Taylor series/polynomials and I am deriving some of the common functions just to get comfortable with the process.

My problem is with ln(x).
My process for finding taylor polynomials has been to take a few derivatives so in this case $$f(x)=\ln(x)....f'(x)=\frac{ 1 }{ x }...f''(x)=\frac{ -1 }{ x^2 }....f'''(x)=\frac{ 2 }{ x^3 } $$ etc...

But for the next step I have been evaluating each term at x=0 to find the general term. How can I do this for ln(x) since as far as I know, ln(0) isn't valid? (at least at my level of maths)

Question

Taylor Series question here: So I am learning about Taylor series/polynomials and I am deriving some of the common functions just to get comfortable with the process.  

My problem is with ln(x). 
My process for finding taylor polynomials has been to take a few derivatives so in this case $$f(x)=\ln(x)....f'(x)=\frac{ 1 }{ x }...f''(x)=\frac{ -1 }{ x^2 }....f'''(x)=\frac{ 2 }{ x^3 } $$  etc... 

But for the next step I have been evaluating each term at x=0 to find the general term. How can I do this for ln(x) since as far as I know, ln(0) isn't valid? (at least at my level of maths)

anonymous · Answer

$$\ln(x)$$ therefore has no taylor series expansion at $x=0$

turingtest · Answer

You don't take the taylor series of lnx about x=0 for the reason you discovered. It's much more common to take it around x=-1, or expand ln(x+1)

anonymous · Answer

as you can see every term will be undefined
you can expand at 1 say

anonymous · Answer

what @TuringTest  said

anonymous · Answer

Ok I will try from a different x then. I guess my confusion was just about finding a general term. I will expand at a different point and see if I can get it.

anonymous · Answer

Aha! That worked perfectly!  Substituting x=1.   Thanks guys!