# CLUELESS about a Maclaurin series

1. Nov 10, 2004

I'm supposed to obtain a Maclaurin series for the function defined by

$$f(x) = \left\{ \begin{array}{lc} e^{-1/x^2} & \mbox{ if } x \neq 0 \\ 0 & \mbox{ if } x = 0 \end{array} \right.$$

I get immediately stuck as I find:

$$f(0) = 0$$
$$f^{\prime}(0) = \mbox{ undefined }$$
$$f^{\prime \prime}(0) = \mbox{ undefined }$$
$$f^{\prime \prime \prime}(0) = \mbox{ undefined }$$
$$f^{(4)}(0) = \mbox{ undefined }$$

So,

$$f^{(n)}(0) = \mbox{ undefined } \qquad n > 0$$

Thus, we may write

$$f(x) = f(0) + \frac{f^{\prime}(0)}{1!}x + \frac{f^{\prime \prime}(0)}{2!}x^2 + \frac{f^{\prime \prime \prime}(0)}{3!}x^3 + \cdots$$

Question

How do I handle the undefined results? (I'm supposed to show that the Maclaurin series is not equal to the given function)

Thank you very much.

Last edited: Nov 10, 2004
2. Nov 10, 2004

### Corneo

Not sure if this is right or not. We know that $$f'(x) = \left\{ \begin{array}{lc} e^{1/x^3} & \mbox{ if } x \neq 0 \\ 0 & \mbox{ if } x = 0 \end{array} \right.$$ So $f'(0)=0$, which is well defined.

3. Nov 10, 2004

### Hurkyl

Staff Emeritus
You messed up the chain rule, Corneo.

Thiago: are you sure f'(0) is undefined?

4. Nov 10, 2004

### Justin Lazear

0 is a constant.

What's the derivative of any constant?

--J

5. Nov 10, 2004

Justin, I don't think I should take the derivative of f'(0) = 0 (and then take the derivative of this constant), but instead find each derivative and later evaluate at x = 0. This is how I got my results.

Hurkyl, here you go:

$$f^{\prime} (x) = \frac{d}{dx} \left( e^{-1/x^2} \right) = \frac{2e^{-1/x^2}}{x^3} \Longrightarrow f^{\prime} (0) = \mbox{undefined}$$

I've just evaluated that over again in TI, and this time I also calculated

$$f(x) = 0 + \mbox{undefined} +\mbox{undefined} + \mbox{undefined} + \cdots = \mbox{undefined} \neq \left\{ \begin{array}{lc} e^{-1/x^2} & \mbox{ if } x \neq 0 \\ 0 & \mbox{ if } x = 0 \end{array} \right.$$

which sounds reasonable, since I need to show they're not equal.

6. Nov 10, 2004

### James R

Looks to me like you've done that.

7. Nov 11, 2004

### Hurkyl

Staff Emeritus
Ah, but near zero, the function is not defined as $f(x) = e^{-1/x^2}$!

8. Nov 11, 2004

### Duarh

that is, at zero, you mean. but yeah, that fact makes the function differentiable at zero (some tinkering indicates the lim of f(x)-f(0)/(x-0) as x goes to zero exists and is zero now that f(0) is defined; should doublecheck, though), so you can write a Maclaurin expansion that'll fail.

thiago_j - i think you're a bit confused about this, the derivative of 0 is not, in general, undefined (though it can be when a function is not differentiable at 0).