Taylor Expansion where the derivatives are undefined?

Click For Summary

Homework Help Overview

The discussion revolves around expanding the function x/(x-1) at the point a=1. The original poster notes that the book provides an expansion but lacks an explanation of the process, leading to confusion regarding the validity of the expansion at a point where the function is undefined.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to understand the expansion process and questions the validity of the Taylor expansion given the undefined nature of the function at x=1. Some participants suggest considering the Laurent expansion instead, indicating a potential shift in approach.

Discussion Status

The discussion is ongoing, with participants exploring the distinction between Taylor and Laurent series. The original poster expresses confusion about the reasoning behind the provided expansion and the implications of the function being undefined at the point of expansion.

Contextual Notes

There is a noted concern regarding the undefined nature of the function x/(x-1) at x=1, which raises questions about the appropriateness of using a Taylor expansion in this context.

pandaBee
Messages
23
Reaction score
0

Homework Statement


Expand x/(x-1) at a=1
The book already gives the expansion but it doesn't explain the process. The expansion it gives is:
x/(x-1) = (1+x-1)/(x-1) = (x-1)^(-1) + 1

Homework Equations


The Attempt at a Solution


I've already solved for the Mclaurin expansion for the same polynomial which gives me -x - x^2 - x^3 - x^4 - ...
The derivatives all have a power of (x-1) as its denominator therefore all of its derivatives are undefined. I don't know any other way to expand polynomials other than term by term, I've been staring at

x/(x-1) = (1+x-1)/(x-1) = (x-1)^(-1) + 1

But it doesn't make sense. All they did was just split the fraction up into two parts.
 
Physics news on Phys.org
Are you sure you don't want the Laurent expansion instead of the Taylor expansion?
 
micromass said:
Are you sure you don't want the Laurent expansion instead of the Taylor expansion?

Sorry about that, yes I believe that is the case.
 
Are you confused about how (x-1)-1 + 1 is a Laurent series?
 
Office_Shredder said:
Are you confused about how (x-1)-1 + 1 is a Laurent series?

I'm confused about the method used to derive it. As in I don't understand the method of the reasoning behind it.

Also I don't see how they can come up with that answer when the function x/(x-1) is undefined around x=1.
 

Similar threads

Taylor Expansion for very small and very big arguments
  • · Replies 1 ·
Replies
1
Views
2K
What is the Taylor expansion of x/sin(ax)?
  • · Replies 5 ·
Replies
5
Views
3K
Using a Taylor expansion to prove equality
  • · Replies 1 ·
Replies
1
Views
2K
How should I show that this root is given approximately by this?
  • · Replies 2 ·
Replies
2
Views
2K
How can the Taylor expansion of x^x at x=1 be simplified to make solving easier?
  • · Replies 2 ·
Replies
2
Views
3K
Limit of the remainder of Taylor polynomial of composite functions
  • · Replies 2 ·
Replies
2
Views
2K
First order term in the taylor expansion of ln(x) abut 1
  • · Replies 3 ·
Replies
3
Views
3K
Coefficient for a Term in a Taylor Expansion for Cosine
  • · Replies 10 ·
Replies
10
Views
2K
Interval of convergence for Taylor series exp of 1/x^2
  • · Replies 10 ·
Replies
10
Views
4K
Determining domain for C^1 function
  • · Replies 4 ·
Replies
4
Views
1K