# formula for the $n$th derivative of $e^{-1/x^2}$

$$f(x) = \begin{cases} e^{-1/x^2} & \text{ if } x \ne 0 \\ 0 & \text{ if } x = 0 \end{cases}$$

so

$$\displaystyle f'(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x - 0} = \lim_{x \to 0} \frac {e^{-1/x^2}}x = \lim_{x \to 0} \frac {1/x}{e^{1/x^2}} = \lim_{x \to 0} \frac x {2e^{1/x^2}} = 0$$

(using l'Hospital's Rule and simplifying in the penultimate step).

Similarly, we can use the definition of the derivative and l'Hospital's Rule to show that $$f''(0) = 0, f^{(3)}(0) = 0, \ldots, f^{(n)}(0) = 0$$, so that the Maclaurin series for $$f$$ consists entirely of zero terms. But since $$f(x) \ne 0$$ except for $$x = 0$$, we can see that $$f$$ cannot equal its Maclaurin series except at $$x = 0$$.

This is part of a proof question. I don't think the answer sufficiently proves that any $$n$$th derivative of $$f(x)$$ is $$0$$. Would anyone please expand on the answer?

ps: I promise this is not my homework :)

• It doesn't matter if it is your homework. For the proof that the limit is zero it is enough to prove more: Prove that $x^{-n}e^{-1/x^2}$ tends to zero. This can be done by L'Hospital and induction. This way we don't need to know the exact formula for the derivative. The derivative is equal to $x^ne^{-1/x^2}$ times some rational function that doesn't have a pole at zero, and that is as much as we need to know to complete the proof. Commented Apr 29, 2015 at 21:29
• What is it that you do not understand? The Maclaurin series of the given function is equal to the constant zero function, yet the function you are given is non-zero for all but $x=0$. Thus the conclusion. (I am aware to have only rewritten what stands in the question, but I don't see how this could be better rephrased) Commented Apr 29, 2015 at 21:30
• @GFauxPas: My eyes send you many thanks.
– mvw
Commented Apr 29, 2015 at 21:34
• For more details (but not all) of what Alamos discussed, see my answer to Examples of applying L'Hôpitals rule ( correctly ) leading back to the same state?. Commented Apr 29, 2015 at 21:36
• @Alamos Perhaps jxhyc is having difficulty generalizing the method of finding the derivative at zero to higher orders. If that is the difficulty, he would appreciate an explanation as to why $f^{(n)}(x) = x^{-n} e^{-1/x^2} r(x)$ Commented Apr 29, 2015 at 21:39

\begin{align} f'(x) &= \left\{ \begin{array}{ll} e^{-1/x^2} (2/x^3) & \mbox {for } x \ne 0 \\ 0 & \mbox{for } x = 0 \end{array} \right. \\ f''(x) &= \left\{ \begin{array}{ll} e^{-1/x^2} (4/x^6-6/x^4) & \mbox {for } x \ne 0 \\ 0 & \mbox{for } x = 0 \end{array} \right. \\ & \vdots \\ \\ f^{(n)}(x) &= \left\{ \begin{array}{ll} e^{-1/x^2} P_n(x) & \mbox {for } x \ne 0 \\ 0 & \mbox{for } x = 0 \end{array} \right. \end{align} where $P_n(x)$ fulfills the recursive definition \begin{align} P_0(x) & = 1 \\ P_n(x) & = (2/x^3) P_{n-1}(x)+P_{n-1}'(x) \end{align}

Just to add some additional details to Alex M.'s response...this question can also be found in Spivak's Calculus, in Chapter 18 problem 40 (Ed. 4). There is an important lemma that is derived earlier in the exposition of the chapter, which reads as follows:

Theorem 6:For any natural number $$n$$: $$\displaystyle \lim_{x \to \infty} \frac{e^x}{x^n}=\infty \quad (*_1)$$.

We will need this for our proof:

Firstly, we will assume that we have already shown that for any $$x \neq 0$$ and $$\displaystyle k \in \mathbb N: f^{(k)}(x)=e^{-\frac{1}{x^2}}\times\left(\sum_{i=1}^{3k}\frac{a_i}{x^i}\right) \quad (\dagger_1)$$, where the $$i$$ used in $$a_i$$ is just meant as an index, but the $$i$$ used with $$x^i$$ refers to $$x$$ raised to some natural number power $$i$$. Note that many values of $$a_i$$ will be equal to $$0$$. Observe that $$\sum_{i=1}^{3k}\frac{a_i}{x^i}$$ is just some polynomial $$P$$ with $$(\frac{1}{x})$$ as its argument.

Although sort of clunky, this can be proven by induction in a straightforward manner by just taking the derivative of the above expression and showing that $$\displaystyle f^{(k+1)}(x)=e^{-\frac{1}{x^2}}\times\left(\sum_{i=1}^{3(k+1)}\frac{b_i}{x^i}\right)$$.

We will now show that for any arbitrary $$k$$, $$f^{(k)}(0)=0$$. We will also use induction to prove this. As the base case, recalling the definition of the derivative, we know that $$f^{(1)}(0)=\displaystyle \lim_{h \to 0}\frac{f(h)-f(0)}{h}= \lim_{h \to 0}\frac{f(h)}{h}=\cdots \text{refer to the OPs work}\cdots=0$$.

Next, assume that for an arbitrary $$n$$, we know that $$f^{(n)}(0)=0$$. We will now prove that $$f^{(n+1)}(0)=0:$$

By definition and $$(\dagger_1)$$, $$f^{(n+1)}(0)=\displaystyle \lim_{h \to 0}\frac{f^{(n)}(h)-f^{(n)}(0)}{h}=\lim_{h \to 0}\frac{e^{-\frac{1}{h^2}}\times\left(\sum_{i=1}^{3k}\frac{a_i}{h^i}\right)-0}{h}$$.

How exactly should we deal with the numerator? Well, $$e^{-\frac{1}{h^2}}$$ will be multiplied to every member of the summation...so let us just consider the $$j$$th arbitrary member of this expression: $$e^{-\frac{1}{h^2}}\times \frac{a_j}{h^j}$$. If we can show that $$\displaystyle \lim_{h \to 0}\frac{e^{-\frac{1}{h^2}}\times \frac{a_j}{h^j}}{h}=0$$, then the additive law of limits will let us conclude that $$\displaystyle\lim_{h \to 0}\frac{e^{-\frac{1}{h^2}}\times\left(\sum_{i=1}^{3k}\frac{a_i}{h^i}\right)}{h}=0$$.

It will be useful to rewrite $$\displaystyle \lim_{h \to 0}\frac{e^{-\frac{1}{h^2}}\times \frac{a_j}{h^j}}{h}$$ in the following way:

$$\frac{e^{-\frac{1}{h^2}}\times \frac{a_j}{h^j}}{h}=\frac{\frac{a_j}{h^{j+1}}}{e^{\frac{1}{h^2}}}=a_j \times \left(\frac{\frac{1}{h^{j+1}}}{e^{\frac{1}{h^2}}}\right)$$

To address this, let us prove the more general statement:

For any $$\displaystyle n \in \mathbb N: \lim_{h \to 0}\frac{\frac{1}{h^n}}{e^{\frac{1}{h^2}}}=0$$

To prove this, we will break down the above statement into two different cases:

1. $$\displaystyle \lim_{h \to 0^+}\frac{\frac{1}{h^n}}{e^{\frac{1}{h^2}}}=0$$

2. $$\displaystyle \lim_{h \to 0^-}\frac{\frac{1}{h^n}}{e^{\frac{1}{h^2}}}=0$$

Note that limit laws let us evaluate the following two equivalent cases:

1. $$\displaystyle \lim_{h \to \infty}\frac{h^n}{e^{h^2}}=0$$

2. $$\displaystyle \lim_{h \to -\infty}\frac{h^n}{e^{h^2}}=0$$

Recall from $$(*_1)$$ that $$\displaystyle \lim_{x \to \infty}\frac{e^x}{x^n}=\infty$$. Because $$\exp$$ is a strictly increasing function, for $$x \gt 1$$, we note that $$e^{x} \lt e^{x^2}$$. Therefore, we most certainly have that $$\displaystyle \lim_{x \to \infty}\frac{e^x}{x^n}=\infty \implies \displaystyle \lim_{x \to \infty}\frac{e^{x^2}}{x^n}=\infty$$. Of course, then, the inverse of such an expression must equal $$0$$...i.e $$\displaystyle \lim_{x \to \infty}\frac{e^{x^2}}{x^n}=\infty \implies \displaystyle \lim_{x \to \infty}\frac{x^n}{e^{x^2}}=0$$. This deals with case 1.

Because $$e^{x^2}$$ is an even function, the relevant consideration for case 2 is whether or not $$n$$ is even or odd. If $$n$$ is even, then $$\displaystyle \lim_{x \to -\infty}\frac{x^n}{e^{x^2}} = \lim_{x \to \infty}\frac{x^n}{e^{x^2}}=0$$. But if $$n$$ is odd, then $$\displaystyle \lim_{x \to -\infty}\frac{x^n}{e^{x^2}}=-\displaystyle \lim_{x \to \infty}\frac{x^n}{e^{x^2}}=-1 \times 0=0$$. With the two cases established, we conclude that for any $$\displaystyle n \in \mathbb N: \lim_{h \to 0}\frac{\frac{1}{h^n}}{e^{\frac{1}{h^2}}}=0$$.

Of course, then, we have that $$\displaystyle \lim_{h \to 0} a_j \times \left(\frac{\frac{1}{h^{j+1}}}{e^{\frac{1}{h^2}}}\right)=a_j \lim_{h \to 0}\left(\frac{\frac{1}{h^{j+1}}}{e^{\frac{1}{h^2}}}\right)=a_j \times 0=0$$

Show by induction that $$f^{(n)}(x)=P_n(\frac 1 x) \mathbb{e} ^{-\frac 1 {x^2}}$$ with $$P_n$$ a polynomial function of degree $$3n$$, and then compute (again by induction if you want) $$\lim \limits _{x \to 0^+} \space f^{(n)}(x)$$. You'll have to use l'Hospital's thorem.