Teylor expantion of ln(1-e^-x)^-1

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Discussion Overview

The discussion revolves around the Taylor expansion of the expression ln((1-e^(-x))^(-1)). Participants are exploring the appropriate point for expansion and the implications of the logarithmic transformation involved.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant asks for clarification on the Taylor expansion point, indicating that x=0 may not be suitable.
  • Another participant seeks to confirm whether the (-1) is inside or outside the logarithm, clarifying that it is indeed inside.
  • A suggestion is made to perform Taylor expansions of ln(.), e^-x, and (.)^-1 separately, followed by function composition and the use of the multinomial theorem for expansion.
  • There is a mention of using polynomial multiplication as a means to collect coefficients, with a note that it may become complex.
  • A later reply emphasizes simplifying the logarithmic expression using the identity log(A^(-1)) = -log(A).

Areas of Agreement / Disagreement

Participants express uncertainty regarding the expansion point and the structure of the logarithmic expression. There is no consensus on the best approach to take for the Taylor expansion.

Contextual Notes

Participants have not yet established the assumptions regarding the expansion point or the specific form of the logarithmic expression, which may affect the analysis.

Who May Find This Useful

Readers interested in Taylor series expansions, logarithmic functions, and polynomial multiplication techniques may find this discussion relevant.

zendani
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(ln(1-e^(-x))^(-1))^2=?
thankyou very much
http://www.uploadgeek.com/share-69EF_4A600E7F.html
 
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Taylor's expansion about what point?
 
and is the (-1) inside or outside the logarithm?
 
(-1) is inside ln
ln((1-e^-x)^-1)
thank u
 
Do, the taylor expansions, of ln(.), e^-x and (.)^-1separately, then compose (function composition) the results. Use the multinomial theorem to expand the terms in brackets. Then collect terms. As opposed to using the multinomial theorem you may be able to also use the fact that polynomial multiplication is the convolution of the coeficients. Have fun. It is going to be a mess. If you are skilled you may be able to write a computer algebra program which will give you each coefficient of the resulting polynomial.
 
He still didn't answer "about what point"... x=0 is NOT a candidate.

Simplify what Johns said by using the exponent identity for logarithms.
log(A^(-1)) = -log(A)
 

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