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Homework Help: CLUELESS about a Maclaurin series

  1. Nov 10, 2004 #1
    CLUELESS about a Maclaurin series!!

    I'm supposed to obtain a Maclaurin series for the function defined by

    [tex] f(x) = \left\{ \begin{array}{lc} e^{-1/x^2} & \mbox{ if } x \neq 0 \\
    0 & \mbox{ if } x = 0 \end{array} \right. [/tex]

    I get immediately stuck as I find:

    [tex] f(0) = 0 [/tex]
    [tex] f^{\prime}(0) = \mbox{ undefined } [/tex]
    [tex] f^{\prime \prime}(0) = \mbox{ undefined } [/tex]
    [tex] f^{\prime \prime \prime}(0) = \mbox{ undefined } [/tex]
    [tex] f^{(4)}(0) = \mbox{ undefined } [/tex]

    So,

    [tex] f^{(n)}(0) = \mbox{ undefined } \qquad n > 0 [/tex]

    Thus, we may write

    [tex] f(x) = f(0) + \frac{f^{\prime}(0)}{1!}x + \frac{f^{\prime \prime}(0)}{2!}x^2 + \frac{f^{\prime \prime \prime}(0)}{3!}x^3 + \cdots [/tex]

    Question

    How do I handle the undefined results? (I'm supposed to show that the Maclaurin series is not equal to the given function)

    Thank you very much. :smile:
     
    Last edited: Nov 10, 2004
  2. jcsd
  3. Nov 10, 2004 #2
    Not sure if this is right or not. We know that [tex]f'(x) = \left\{ \begin{array}{lc} e^{1/x^3} & \mbox{ if } x \neq 0 \\
    0 & \mbox{ if } x = 0 \end{array} \right.[/tex] So [itex]f'(0)=0[/itex], which is well defined.
     
  4. Nov 10, 2004 #3

    Hurkyl

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    You messed up the chain rule, Corneo.


    Thiago: are you sure f'(0) is undefined?
     
  5. Nov 10, 2004 #4
    0 is a constant.

    What's the derivative of any constant?

    --J
     
  6. Nov 10, 2004 #5
    Thanks for your input, guys.

    Justin, I don't think I should take the derivative of f'(0) = 0 (and then take the derivative of this constant), but instead find each derivative and later evaluate at x = 0. This is how I got my results.

    Hurkyl, here you go:

    [tex] f^{\prime} (x) = \frac{d}{dx} \left( e^{-1/x^2} \right) = \frac{2e^{-1/x^2}}{x^3} \Longrightarrow f^{\prime} (0) = \mbox{undefined} [/tex]

    I've just evaluated that over again in TI, and this time I also calculated

    [tex] f(x) = 0 + \mbox{undefined} +\mbox{undefined} + \mbox{undefined} + \cdots = \mbox{undefined} \neq \left\{ \begin{array}{lc} e^{-1/x^2} & \mbox{ if } x \neq 0 \\
    0 & \mbox{ if } x = 0 \end{array} \right. [/tex]

    which sounds reasonable, since I need to show they're not equal.
     
  7. Nov 10, 2004 #6

    James R

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    Looks to me like you've done that.
     
  8. Nov 11, 2004 #7

    Hurkyl

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    Ah, but near zero, the function is not defined as [itex]f(x) = e^{-1/x^2}[/itex]!
     
  9. Nov 11, 2004 #8
    that is, at zero, you mean. but yeah, that fact makes the function differentiable at zero (some tinkering indicates the lim of f(x)-f(0)/(x-0) as x goes to zero exists and is zero now that f(0) is defined; should doublecheck, though), so you can write a Maclaurin expansion that'll fail.

    thiago_j - i think you're a bit confused about this, the derivative of 0 is not, in general, undefined (though it can be when a function is not differentiable at 0).
     
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