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Math Help - Cauchy's function and differentiability

  1. #1
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    Cauchy's function and differentiability

    How can I show that <br />
f(x) = \{ \begin{array}{ll}<br />
         exp(\frac{-1}{x^2} & x>0\\<br />
        0 & x\leq 0 \end{array}
    is infinitely differentiable?
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by dannyboycurtis View Post
    How can I show that <br />
f(x) = \{ \begin{array}{ll}<br />
         exp(\frac{-1}{x^2}) & x>0\\<br />
        0 & x\leq 0 \end{array}
    is infinitely differentiable?
    f(x)=e^{-x^{-2}}

    The first derivative is x^{-3}(2)f(x)

    The second derivative is x^{-6}(4-6x^2)f(x)

    The third derivative is x^{-9}(8-36x^2+24x^4)f(x)

    The fourth derivative is x^{-12}(16-144x^2+300x^4-140x^6)f(x)

    So f^{(n)}(x)=x^{-3n}(\text{polynomial~with~degree}<3n)f(x)

    (You might want to prove this using induction, though it will likely be ugly.)

    Next you'd have to do another induction proof. Like assume that f^{(n)}(0)=0 and then prove that:

    f^{(n+1)}(0)=\lim_{x\to0^+}\frac{f^{(n)}(x)-f^{(n)}(0)}{x}=0

    Since you assumed f^{(n)}(0)=0, you need to prove that

    \lim_{x\to0^+}\frac{x^{-3n}(\text{polynomial})f(x)}{x}=\lim_{x\to0^+}x^{-3n-1}(\text{polynomial})e^{-x^{-2}}=0.

    There could be a cleaner way to do this using some theorems or something, but this should work.
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