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Taylor series generated at x=a

  1. Dec 16, 2014 #1
    I have just started learning about series and I don't see the benefit of shifting the series by using some "a" other than 0?

    My textbook doesn't really tell the benefits it just says "it is very useful"'
     
  2. jcsd
  3. Dec 16, 2014 #2

    Mark44

    Staff: Mentor

    You can't write a Maclaurin series (i.e., a Taylor series with a = 0) for f(x) = ln(x), since the function and all of its derivatives are not defined at x = 0. You can, however, write a Taylor series in powers of, say, x - 1, though.
     
  4. Dec 16, 2014 #3
    Would you be able to write a power series to estimate any value of lnx? I cannot think of a way to write a series to estimate ln9 for example... Is it possible? Every example I see is estimating ln2 and I feel like there must be some way to estimate other values but I cannot think of a way since the series diverges after 1 in each direction (from what ive seen)
     
  5. Dec 17, 2014 #4

    Mark44

    Staff: Mentor

    Off the top of my head I don't know what the radius of convergence is for this series. You could write it in powers of x - e2, with e2 being about 7.39, which might be close enough to 9.
     
  6. Dec 17, 2014 #5
    Oh I see! The point of having a Taylor Series centered around some arbitrary a is to move your radius of convergence in a sense, is that correct?
     
  7. Dec 17, 2014 #6

    Mark44

    Staff: Mentor

    Something like that. You move the interval of convergence. The radius doesn't change.
     
  8. Dec 17, 2014 #7

    mathman

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    Gold Member

    The power series for ln((1+x)/(1-x)) converges for -1<x<1, which can (in principal) be used for any y = (1+x)/(1-x) > 0.
     
  9. Dec 17, 2014 #8

    FactChecker

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    There are several reasons. Including:
    1) You can get more accurate answers from fewer terms if you can expand around point 'a' close to the x value you are interested in.
    2) Others have mentioned the radius of convergence, which can not go beyond any singularity. So you may need to expand in several areas.
    3) You may need to cancel one function's singularity at z=a by expanding another multiplying function that has a zero at z=a.
     
  10. Dec 25, 2014 #9

    HallsofIvy

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    Staff Emeritus
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    Another point: if you have a linear differential equation of the form [tex]y''+ f(x)y'+ g(x)y[/tex], with "initial values" y(a)= b, y'(a)= c; that is, with values of y and its derivative give at x= a, it is simplest to write the solution as a power series in powers of x-a. That is, in the form [tex]y= b+ c(x- a)+ p_2(x- a)^2+ p_3(x- a)^3+ \cdot\cdot\cdot[/tex] so that the first two coefficients are the given values, b and c. You can then write the functions f and g in Taylor series about a.
     
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