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I Taylor series

  1. Jul 21, 2017 #1
    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?
     
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  3. Jul 21, 2017 #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.
     
  4. Jul 21, 2017 #3
    So why does it have log(x+1), and not log(x+1/2) or log(x+2), for example?
     
  5. Jul 22, 2017 #4

    Mark44

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    Mostly convenience, I suppose. Each of the other two expressions could also be expanded as Maclaurin series.
     
  6. Jul 22, 2017 #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.
     
  7. Jul 22, 2017 #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)
     
  8. Jul 22, 2017 #7

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    Expanding the Maclaurin series at x=0 would be trying to evaluate the log at negative numbers.
     
  9. Jul 23, 2017 #8

    mathman

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    MacLaurin series at x = 0 has minus infinity as the constant term.
     
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