Taylor series

  • #1
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Main Question or Discussion Point

I am looking at examples of Maclaurin expansions for different functions, such as e^x, and sinx. But there is no expansion for log(x), only log(x+1). Why is that?
 

Answers and Replies

  • #2
FactChecker
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The Maclaurin series uses the values of the function and it's derivatives at x=0. But log(x) is not defined at x=0.
 
  • #3
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The Maclaurin series uses the values of the function and it's derivatives at x=0. But log(x) is not defined at x=0.
So why does it have log(x+1), and not log(x+1/2) or log(x+2), for example?
 
  • #4
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So why does it have log(x+1), and not log(x+1/2) or log(x+2), for example?
Mostly convenience, I suppose. Each of the other two expressions could also be expanded as Maclaurin series.
 
  • #5
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Issues to consider are the radius of convergence and the speed of convergence of the series. An expansion of log(x+1) around x=0 will give values for log(y), y=x+1 from y=0 to y=2.
 
  • #6
mathman
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Log((x+1)/(x-1)) gives a series that can be used for any y=(x+1)/(x-1)
 
  • #7
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Log((x+1)/(x-1)) gives a series that can be used for any y=(x+1)/(x-1)
Expanding the Maclaurin series at x=0 would be trying to evaluate the log at negative numbers.
 
  • #8
mathman
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MacLaurin series at x = 0 has minus infinity as the constant term.
 

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