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Is there any benefit to using Taylor series centered at nonzero value

  1. Mar 28, 2013 #1
    over a Maclaurin series?

    Also, how do I calculate e^0 using Maclaurin series? I'm getting 0^0.
  2. jcsd
  3. Mar 28, 2013 #2


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    Staff: Mentor

    Try to express ln(x) as series around x=0 ;)
    Of course, instead of using 1 as expansion point, you can expand ln(x-1) around x=0, but that just gives the same result.

    If you want to approximate the function value somewhere, it can be useful to consider an expansion point close to that.

    0^0 has to be defined as 1 here to get a correct series, but you need the value of e^0 anyway to calculate the series of e^x around x=0.
  4. Mar 28, 2013 #3
    Why? Can't you use any Taylor series centered at any point?

    Also, when I calculate e^2 using Taylor series centered at 2, I get 0 since x-a=2-2=0. I just noticed that in order to calculate any power of e, you need to know that power of e first. Doesn't that make it kind of pointless?
  5. Mar 28, 2013 #4


    Staff: Mentor

    No - the function that is represented by the Taylor series has to be defined and infinitely differentiable at that point. So you can't write a Taylor series centered around a = 0 for ln(x).
    The idea is that if you know the exact value of your function at some point a, you can use a Taylor series to find approximations at points near a. For example, we know that cos(60°) = cos(##\pi/3##) = .5, exactly. We can use a Taylor series expansion about a = ##\pi/3## to get reasonable approximations to angles such as 60.5°, and so on.
  6. Mar 28, 2013 #5


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    I would like to point out that the function [itex]f(x)= e^{-1/x^2}[/itex] if [itex]x\ne 0[/itex], f(0)= 0, is infinitely differentiable so you can write a Taylor series centered around x= 0, just as Mark44 says but that Taylor series is not equal to f(x) anywhere except at x= 0.

    Another reason to use a Taylor series around a value other than 0 is to find a series solution to a linear differential equation with "initial values" given at some non-zero point.
  7. Mar 28, 2013 #6


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    ex = 1 + x + x2/2! + ....
    therefore e0 = 1.
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