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Prove some properties of the function e-1/x2

Consider the function

    \[ f(x) = e^{-\frac{1}{x^2}} \qquad \text{when} \quad x \neq 0 \]

and f(0) = 0.

  1. Prove that for every positive number m we have

        \[ \lim_{x \to 0} \frac{f(x)}{x^m} = 0. \]

  2. Prove that if x \neq 0 then

        \[ f^{(n)}(x) = f(x) P \left( \frac{1}{x} \right) \]

    where P(t) is a polynomial in t.

  3. Prove that

        \[ f^{(n)}(0) = 0 \qquad \text{for all } n \geq 1. \]


  1. Proof. (A specific case of this general theorem is actually the first problem of this section, here. Maybe it’s worth taking a look since this proof is just generalizing that particular case.) We make the substitution t = \frac{1}{x^2}, so that t \to +\infty as x \to 0 and we have

        \begin{align*}  \lim_{x \to 0} \frac{e^{-\frac{1}{x^2}}}{x^m} &= \lim_{t \to +\infty} \frac{t^{\frac{m}{2}}}{e^t} \\[9pt]  &= 0 \end{align*}

    by Theorem 7.11 (page 301 of Apostol) since m > 0 implies \frac{m}{2} > 0 as well. \qquad \blacksquare

  2. Proof. The proof is by induction on n. In the case n = 1 we have

        \begin{align*}  f'(x) &= \left( \frac{2}{x^3} \right) e^{-\frac{1}{x^2}} \\[9pt]  &= 2 \left( \frac{1}{x} \right)^3 e^{-\frac{1}{x^2}} \\[9pt]  &= P \left(\frac{1}{x} \right) e^{-\frac{1}{x^2}}. \end{align*}

    So, indeed the formula is valid in the case n = 1. Assume then that the formula holds for some positive integer k. We want to show this implies the formula holds for the case k + 1.

        \begin{align*}  f^{(k+1)}(x) = \left( f^{(k)}(x) \right)' &= \left( P\left( \frac{1}{x} \right) e^{-\frac{1}{x^2}} \right)' \\[9pt]  &= P' \left( \frac{1}{x} \right) e^{-\frac{1}{x^2}} + \frac{2}{x^3} P \left( \frac{1}{x} \right) e^{-\frac{1}{x^2}} \\[9pt]  &= \left( P' \left( \frac{1}{x} \right) + \frac{2}{x^3} P \left( \frac{1}{x} \right) \right)e^{-\frac{1}{x^2}}. \end{align*}

    But then the leading term

        \[ P' \left( \frac{1}{x} \right) + 2 \left( \frac{1}{x} \right)^3 P \left( \frac{1}{x} \right) \]

    is still a polynomial in \frac{1}{x} since the derivative of a polynomial in \frac{1}{x} is still a polynomial in \frac{1}{x}, and so is the sum of two polynomials in \frac{1}{x}. Therefore, we have that the formula holds for the case k+1; hence, it holds for all positive integers n. \qquad \blacksquare

  3. Proof. The proof is by induction on n. If n = 1 then we use the limit definition of the derivative to compute the derivative at 0,

        \begin{align*}  f'(0) &= \lim_{h \to 0} \frac{f(0+h) - f(0)}{h} \\[9pt]  &= \lim_{h \to 0} \frac{f(h)}{h} &(f(0) = 0 \text{ by def of } f)\\[9pt]  &= 0 & \text{by part (a)}. \end{align*}

    So, indeed f'(0) = 0 and the statement is true for the case n = 1. Assume then that f^{(k)} (0) = 0 for some positive integer k. Then, we use the limit definition of the derivative again to compute the derivative f^{(k+1)}(0),

        \begin{align*}  f^{(k+1)}(0) &= \lim_{h \to 0} \frac{f^{(k)}(0+h) - f^{(k)}(0)}{h} \\[9pt]  &= \lim_{h \to 0} \frac{f^{(k)}(h)}{h} &(f^{(k)}(0) = 0 \text{ by Ind. Hyp.}) \\[9pt]  &= \lim_{h \to 0} f(h) P \left( \frac{1}{h} \right) \cdot \frac{1}{h} &(\text{part (b)}) \\[9pt]  &= \lim_{h \to 0} P \left( \frac{1}{h} \right) e^{-\frac{1}{h^2}}. \end{align*}

    This follow since \frac{1}{h} \cdot P \left( \frac{1}{h} \right) is still a polynomial in \frac{1}{h}, and by the definition of f(x) for x \neq 0. But then, by part (a) we know

        \[ \lim_{h \to 0} \frac{f(h)}{h^m} = 0 \qquad \text{for all } m \in \mathbb{Z}^+. \]

    Therefore,

        \[ f^{(k+1)}(0) = \lim_{h \to 0} P \left( \frac{1}{h} \right) e^{-\frac{1}{h^2}} = 0. \]

    Thus, the formula holds for the case k+1, and hence, for all positive integers n. \qquad \blacksquare

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