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Maclaurin series

  1. Mar 12, 2008 #1
    1. The problem statement, all variables and given/known data
    Show that the function defined by

    (stepwise)
    f(x)= e^(-1/x^2) if x =/ 0
    = 0 if x=0

    is NOT equal to its Maclauren series.

    Then graph the function and comment on its behavior near the origin.


    3. The attempt at a solution
    Well, I honestly don't know how to prove this. I graphed it and noticed that as it approaches the origin, it concaves up, so f''(x)>0 at (0,0).


    Am I supposed to first u-substitute in order to differentiate the e^(-1/x^2)?
     
    Last edited: Mar 13, 2008
  2. jcsd
  3. Mar 12, 2008 #2

    morphism

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    Show that the function is what?
     
  4. Mar 12, 2008 #3

    Dick

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    The function may 'concave up' as x->0, but if you look carefully at the limit of f''(x) as x->0 you may find f''(0) is not positive. You should note f(x)=x^4 also 'concaves up' and f''(0)=0 for that function. And as morphism asks, you must have a more specific idea of what you are supposed to comment on.
     
  5. Mar 13, 2008 #4

    HallsofIvy

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    As morphism pointed out, you didn't say what it is you want to prove! Because you title this "MacLaurin series", I assume you want to prove that the MacLaurin series of this function is a certain thing. I would think it would be obvious "how to prove" this: find the derivatives at x= 0 and construct the MacLaurin series. It shouldn't take more than a few derivatives to see what is happening.

    I'll start you off. If
    [tex]f(x)= e^{-\frac{1}{x^2}}[/tex]
    then, using the chain rule,
    [tex]\frac{df}{dx}= e^{-\frac{1}{x^2}}(-2x^{-3})= -\frac{e^{-\frac{1}{x^2}}}{2x^3}[/itex]
    as long as x is not 0. To find the derivative at 0, take the limit as x goes to 0. As x goes to 0, the argument of the exponential goes to negative infinity so the exponential goes to 0. Both numerator and denominator of this derivative, separately, go to 0. Remembering that an exponential will go to 0 (or infinity) faster than any polynomial, only the numerator is relevant and the limit is 0. Hint: every derivative will be [itex]e^{-\frac{1}{x^2}}[/itex] over some polynomial.

    This is an example showing that a function may have a MacLaurin (Or Taylor) series that converges everywhere but not to the value of the function!
     
  6. Mar 13, 2008 #5
    Sorry! I just edited in the rest of the question. It asks to show that the function does not equal the Maclauren series, but why wouldn't it?
     
  7. Mar 13, 2008 #6

    Dick

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    Figure out what the Maclauren series of the function is, and it will be pretty easy to figure out why it's not equal to the function. I promise.
     
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