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xartaan
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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)
 one year ago
 one year ago
xartaan Group Title
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)
 one year ago
 one year ago

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satellite73 Group TitleBest ResponseYou've already chosen the best response.1
\[\ln(x)\] therefore has no taylor series expansion at \(x=0\)
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
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)
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
as you can see every term will be undefined you can expand at 1 say
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
what @TuringTest said
 one year ago

xartaan Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

xartaan Group TitleBest ResponseYou've already chosen the best response.0
Aha! That worked perfectly! Substituting x=1. Thanks guys!
 one year ago
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