Taylor Expansion for ln(1+x)/(1-x) About x=0

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SUMMARY

The discussion focuses on obtaining the first three non-zero terms of the Taylor Expansion for the function \(\frac{\ln(1+x)}{1-x}\) around \(x=0\). Participants suggest two primary methods: multiplying the Taylor expansions of \(\frac{1}{1-x}\) and \(\ln(1+x)\), or computing the Taylor series for \(\ln(1+x)\) and then dividing by \(1-x\). Both methods aim to simplify the differentiation process, which some users found cumbersome.

PREREQUISITES
  • Understanding of Taylor series expansions
  • Familiarity with the functions \(\ln(1+x)\) and \(\frac{1}{1-x}\)
  • Basic calculus, particularly differentiation
  • Knowledge of series multiplication and coefficient gathering
NEXT STEPS
  • Study the Taylor series expansion of \(\ln(1+x)\)
  • Learn about the geometric series and its application in \(\frac{1}{1-x}\)
  • Explore methods for multiplying power series
  • Investigate the process of gathering coefficients in series expansions
USEFUL FOR

Mathematicians, students studying calculus, and anyone interested in series expansions and their applications in analysis.

phalanx123
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Hi I wonder if there is a simpler way to obtain the first three non-zero terms of Taylor Expansion for the function \frac{Ln(1+x)}{1-x} about x=0?

I differentiated it directly, but it was such a nightmare to do:mad: . So I am wondering if there is a simpler way to do it?
 
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You can take the Taylor expansions of 1/(1-x) and ln(1+x) and multiply those. Gather coefficients etc.
 
Or, you could compute the Taylor series for Ln(1+x), and then divide by 1-x.
 

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