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Taylor series question

  1. Jan 11, 2013 #1
    What does it mean to calculate the Taylor series ABOUT a particular point?

    I understand the formula for the Taylor series but how do you solve it about a particular point for a function? It's the about the particular point that confuses me.

    Could someone please explain this and provide examples?

    Thanks!
     
  2. jcsd
  3. Jan 11, 2013 #2

    jbunniii

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    A Taylor series of a function [itex]f[/itex], about the point [itex]x = 0[/itex] is a representation of [itex]f[/itex] as a sum of polynomials centered at [itex]x = 0[/itex], namely [itex]x, x^2, x^3, x^4, \ldots[/itex]

    Thus it will have the form

    [tex]f(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4 + \ldots[/tex]

    More generally, we can express [itex]f[/itex] as a sum of polynomials centered at some other point, say [itex]x = c[/itex]. Then it will look like

    [tex]f(x) = a_0 + a_1 (x - c) + a_2(x - c)^2 + a_3 (x - c)^3 + a_4 (x - c)^4 + \ldots[/tex]

    This is a more general form: if I set [itex]c = 0[/itex] then it reduces to the previous equation. Note that the values of the coefficients generally change if I change [itex]c[/itex]. Here is the formula for the k'th coefficient:

    [tex]a_k = \frac{f^{(k)}(c)}{k!}[/tex]

    where [itex]f^{(0)}[/itex] is taken to mean [itex]f[/itex], and [itex]f^{(k)}[/itex] means the k'th derivative of [itex]f[/itex]. So to calculate [itex]a_k[/itex], you find the k'th derivative of [itex]f[/itex] and evaluate it at [itex]x = c[/itex].
     
  4. Jan 11, 2013 #3

    jbunniii

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    As a simple example, take [itex]f(x) = e^x[/itex]. This is a nice example because all of the derivatives also equal [itex]e^x[/itex]. Thus, if I center the Taylor series at [itex]c[/itex], the coefficients will be
    [tex]a_k = \frac{f^{(k)}(c)}{k!} = \frac{e^c}{k!}[/tex]
    and the series will therefore be
    [tex]e^x = \frac{e^c}{0!} + \frac{e^c}{1!}(x - c) + \frac{e^c}{2!}(x - c)^2 + \ldots[/tex]
    For the special case [itex]c = 0[/itex], we have [itex]e^c = e^0 = 1[/itex] so the coefficient simplifies to
    [tex]a_k = \frac{1}{k!}[/tex]
    and the series will therefore be
    [tex]e^x = \frac{1}{0!} + \frac{1}{1!}x + \frac{1}{2!}x^2 + \ldots[/tex]
     
  5. Jan 11, 2013 #4

    Simon Bridge

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    The Taylor series is only defined to be about a particular point. The Taylor series expansion of y(x) about point x=a would be: $$y(x)_a = \sum_{n=0}^\infty \frac{f^{(n)}}{n!}(x-a)^n$$ It will be exact at that point - see:

    220px-Exp_series.gif

    ... this is the Taylor series expansion (red line), taken about x=0, for the exponential function. See how after each term is added in, the red line has the same value as the blue at x=0? As an exercise, try doing the expansion for ##y(x)=e^x## about ##x=2##, and compare.
     
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