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Math Help - Maclaurin Series

  1. #1
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    Maclaurin Series

    Show that the function defined by

    f(x)=
    e^{-\frac{1}{x^{2}}} if x\neq0
    0 if x=0

    is not equal to its Maclaurin series.

    Thanks!!
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  2. #2
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    Have you tried anything at all on this? What is the Maclaurin series? Is this function differentiable at x= 0? (It had better be- if it is not infinitely differentiable at 0, it won't even have a Maclaurin series!) What is the derivative at 0? What is the second derivative at 0?...
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  3. #3
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    the function has a turning point at x=0,
    so I have f(0)=0 and f'(0)=0. Therefore the first two terms in the Maclaurin series is zero
    But what about f''(0), f'''(0) and the rest?
    How to continue from here?

    Thanks
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  4. #4
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    Actually calculate the derivative. The derivative of e^{-1/x^2} is e^{-1/x^2} times the derivative of 1/x^2= x^{-2}: -2x^{-3}e^{-1/x^2} and, as x goes to 0, that goes to 0 (the exponential goes to 0 faster than any power of x so faster than any negative power of x increases). The second derivative, using the product law is 6x^-4e^{-1/x^2}+ 4x^{-6}e^{-1/x^2}= (4x^{-6}+ 6x^{-4})e^{-1/x^2}. Again, that is a polynomial in x^{-1} times e^{-1/x^2}. The limit, as x goes to 0, is 0. The third derivative is (-24e^{-7}- 24x^{-5})e^{-1/x^2}+ (4x^{-6}+ 6x^{-4})(-x^{-3})e^{-1/x^2}= (-4x^{-9}- 30x^{-7}- 24x^{-5})e^{-1/x^2}[/tex] and, again, that is a polynomial in x^{-1} times e^{-1/x^2} which goes to 0 as x goes to 0. It should be easy to show, perhaps by induction on n, that the nth derivative is a polynomial in x^{-n} times e^{-1/x^2}.

    By the way, while the dervivative of a function is not necessarily continuous, it must satisfy the 'intermediate value' property and so, as long as the derivative at 0 exists, the nth derivative at 0 is the same as the limit, as x goes to 0, of the nth derivative for n not 0.
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  5. #5
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    Great...that's really helpful.
    So the Maclaurin series is just a constant function, i.e. f(x)=0!!!
    so that is why they are not equal. Thanks!!!
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