Show that the function defined by
$\displaystyle f(x)=$
$\displaystyle e^{-\frac{1}{x^{2}}}$ if $\displaystyle x\neq0$
0 if $\displaystyle x=0$
is not equal to its Maclaurin series.
Thanks!!
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?...
Actually calculate the derivative. The derivative of $\displaystyle e^{-1/x^2}$ is $\displaystyle e^{-1/x^2}$ times the derivative of $\displaystyle 1/x^2= x^{-2}$: $\displaystyle -2x^{-3}e^{-1/x^2}$ 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 $\displaystyle 6x^-4e^{-1/x^2}+ 4x^{-6}e^{-1/x^2}= (4x^{-6}+ 6x^{-4})e^{-1/x^2}$. Again, that is a polynomial in $\displaystyle x^{-1}$ times $\displaystyle e^{-1/x^2}$. 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 $\displaystyle x^{-1}$ times $\displaystyle e^{-1/x^2}$ 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 $\displaystyle x^{-n}$ times $\displaystyle e^{-1/x^2}$.
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.