1. ## Maclaurin Series

Show that the function defined by

$f(x)=$
$e^{-\frac{1}{x^{2}}}$ if $x\neq0$
0 if $x=0$

is not equal to its Maclaurin series.

Thanks!!

2. Have you tried anything at all on this? What is the Maclaurin series? Is this function differentiable at x= 0? (It had better be- if it is not infinitely differentiable at 0, it won't even have a Maclaurin series!) What is the derivative at 0? What is the second derivative at 0?...

3. the function has a turning point at x=0,
so I have f(0)=0 and f'(0)=0. Therefore the first two terms in the Maclaurin series is zero
But what about f''(0), f'''(0) and the rest?
How to continue from here?

Thanks

4. Actually calculate the derivative. The derivative of $e^{-1/x^2}$ is $e^{-1/x^2}$ times the derivative of $1/x^2= x^{-2}$: $-2x^{-3}e^{-1/x^2}$ and, as x goes to 0, that goes to 0 (the exponential goes to 0 faster than any power of x so faster than any negative power of x increases). The second derivative, using the product law is $6x^-4e^{-1/x^2}+ 4x^{-6}e^{-1/x^2}= (4x^{-6}+ 6x^{-4})e^{-1/x^2}$. Again, that is a polynomial in $x^{-1}$ times $e^{-1/x^2}$. The limit, as x goes to 0, is 0. The third derivative is (-24e^{-7}- 24x^{-5})e^{-1/x^2}+ (4x^{-6}+ 6x^{-4})(-x^{-3})e^{-1/x^2}= (-4x^{-9}- 30x^{-7}- 24x^{-5})e^{-1/x^2}[/tex] and, again, that is a polynomial in $x^{-1}$ times $e^{-1/x^2}$ which goes to 0 as x goes to 0. It should be easy to show, perhaps by induction on n, that the nth derivative is a polynomial in $x^{-n}$ times $e^{-1/x^2}$.

By the way, while the dervivative of a function is not necessarily continuous, it must satisfy the 'intermediate value' property and so, as long as the derivative at 0 exists, the nth derivative at 0 is the same as the limit, as x goes to 0, of the nth derivative for n not 0.