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

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Homework Help Overview

The discussion revolves around finding the power series representation for a specific function, with a focus on understanding the process of deriving the series and determining the radius of convergence.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to use the nth derivative method for finding coefficients but encounters difficulties with higher derivatives. Some participants suggest starting with a known power series expansion for log(1+x) and modifying it for the function in question. Others question how to derive the terms from the suggested approaches.

Discussion Status

The discussion is ongoing, with participants exploring different methods for obtaining the power series. Some guidance has been offered regarding using a simpler function as a starting point, but there is no explicit consensus on the best approach yet.

Contextual Notes

Participants are navigating the complexities of power series and the implications of using derivatives, with some uncertainty about the correct application of methods and the specific function involved.

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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