Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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)

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SIGN UP FOR FREE
\[\ln(x)\] therefore has no taylor series expansion at \(x=0\)
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)
as you can see every term will be undefined you can expand at 1 say

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

what @TuringTest said
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.
Aha! That worked perfectly! Substituting x=1. Thanks guys!

Not the answer you are looking for?

Search for more explanations.

Ask your own question