Can someone help me find the power series representation for this function?

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SUMMARY

The discussion focuses on finding the power series representation for the function log(1+x). The user initially attempts to derive the series using the nth derivative method but encounters difficulties. A more efficient approach is suggested, involving the power series expansion of log(1+x) followed by substituting 2x for x and multiplying by 2x. This method simplifies the process and reduces the potential for errors compared to calculating higher-order derivatives.

PREREQUISITES
  • Understanding of power series and their representations
  • Familiarity with the function log(1+x)
  • Knowledge of derivatives and their applications in series expansion
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the power series expansion of log(1+x) in detail
  • Learn how to substitute variables in power series
  • Explore the concept of radius of convergence for power series
  • Practice calculating derivatives of functions for series representation
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Students and educators in calculus, mathematicians working with series expansions, and anyone interested in understanding power series representations and their applications.

cmantzioros
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I'm trying to do the question attached. I got the first three answers correct knowing that the nth derivative of a function evaluated at 0 divided by n! = c_n. However, I did the same for the others and the answer is incorrect. I know that I need the power series representation of that function in order to get the radius of convergence but I don't know how to get it. Can someone help please?
 

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Save yourself some labor and work out the power series expansion of log(1+x) first. It has a simple form. Then substitute 2x for x in that and multiply the whole thing by 2x. This is easier and less error prone than taking high order derivatives which get more and more complicated.
 
ok thanks but how do I get the terms?
 
Take higher derivatives again. But this is a much easier function to deal with. Try it.
 

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