1
$\begingroup$

Let $g(x) = e^{-1/x^2}$ for $x$ not equal to zero, and $g(0) = 0$.

a) Please Show that $g^{(n)}(0) = 0$, for all $n = 0,1,2,3,4, \ldots$

Can someone please elaborate on the comments below for this one?

b) Please Show that the Taylor Series for $g$ about 0 agrees with $g$ only at $x = 0$.

I think this would be easy once I have part a, all I have to do is plug in n = 0?

Can someone please show how to do this?

  • 0
    @GEdgar, can you please write out a solution, I am not fully sure I get it2011-04-09

2 Answers 2

4

A good way to do this is to prove something stronger. Let $P$ be any polynomial. Then show that the function $h$ defined by $h(x) = P(1/x)e^{-1/x^2}$ for $x \ne 0$ and $h(x)=0$ has your property $h^{(n)}(0) = 0$ for all $n$. Do it by showing: (1) $h$ is continuous; and (2) the derivative of such a function $h$ is another function of the same kind.

-1

All you have to do is prove that $\dfrac{\rm{d}g(x)}{\rm{d}x} = (\text{something})\,g(x)$ so that the limit of any derivative when $x\rightarrow 0$ is zero, since g(0)=0.

I see that

$\dfrac{\rm{d}g(x)}{\rm{d}x} = \dfrac{2}{x^3}\,g(x) $

  • 0
    This does no good at all if the (something) blows up as $x$ approaches 0 - which in fact the first derivative does, as per the formula you give. You need the fact that $g(x)$ approaches 0 'exponentially' as $x$ approaches 0, and this essentially boils down to GEdgar's method.2011-04-10