0
$\begingroup$

Let $$f(x)=\left\{ {\matrix{ {{e^{ - {1 \over {{x^2}}}}},x \ne 0} \cr {0,x = 0} \cr } } \right.$$

Show that $f^{[n]}(0)=0$ for all $n=0,1,2,\cdots$

The proof:
First note that for $x\ne 0$: $f'(x)=\frac{2}{x^3} e^{-\frac{1}{x^2}}$

Consider the polynomial function $P_1(x) = 2x^3$, then we have:

$$f'(x) = P_1\left(\frac{1}{x}\right)e^{-\frac{1}{x^2}}$$

Since,

$$\lim\limits_{x\to 0} \frac{1}{x^n} e^{-\frac{1}{x^2}} = 0 $$

for any integer $n\ge 0$:

$$\lim\limits_{x\to 0} P\left(\frac{1}{x}\right) e^{-\frac{1}{x^2}} = 0 $$

for any polynomial $P(x)$.

Then, by induction we prove there exists $P_n(x)$ such that: $f^{(n)}(x) = P_n\left(\frac{1}{x}\right)e^{-\frac{1}{x^2}}$

My Question:
I don't understand the base case: Why are the two limits approaches $0$? (And where the $\frac{1}{x^n}$ came from?). It looks like black-magic :)

Thanks in advance.

$\endgroup$
1
  • 1
    $\begingroup$ With a change of variable: $$\lim_{x\to 0}\frac{1}{x^n}e^{-\frac{1}{x^2}}=\lim_{y\to +\infty} y^n e^{-y^2} = 0.$$ The same applies if a polynomial of the variable $1/x$ is took in place of $\frac{1}{x^n}$. $\endgroup$ Aug 27, 2014 at 19:52

1 Answer 1

2
$\begingroup$

To make sure consider $u=\frac{1}{x}$. Then 1st limit $$\lim\limits_{u\to \infty} \frac{e^{-u^2}}{u^{n}} \quad \forall n \in \mathbb{N}$$ Using many well-known methods ($\varepsilon-\delta$, for example) you can prove that it is $0$. For better understanding it means that exponential decays faster than any power function. The second is a corollary from the 1st: it's no difference between polynomial or its greater power summand in denominator

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .