A community for students.
Here's the question you clicked on:
 0 viewing
xartaan
 3 years ago
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)
xartaan
 3 years ago
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)

This Question is Closed

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

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2You 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)

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1as you can see every term will be undefined you can expand at 1 say

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1what @TuringTest said

xartaan
 3 years ago
Best ResponseYou've already chosen the best response.0Ok 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.

xartaan
 3 years ago
Best ResponseYou've already chosen the best response.0Aha! That worked perfectly! Substituting x=1. Thanks guys!
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.