Question

How do you find the Maclaurin series for `f(x)= e^(-1/(x^2))` centered at 0?

Answer 1

The Maclaurin series (i.e. Taylor series centred at `0`) for `f(x)` is `0`, which only matches `f(x)` at `x=0`.

In other words, there is no useful Maclaurin series for `f(x)`

Explanation:

Note first that:

`e^(-1/x^2)`

is undefined when `x = 0`, but:

`lim_(x->0) e^(-1/x^2) = 0`

So we could patch up the definition of `f(x)` as:

`f(x) = { (0, " if " x = 0), (e^(-1/x^2), " otherwise") :}`

In general, if `r(x)` is any rational function in `x` then:

`d/(dx) (r(x) e^(-1/x^2)) = r'(x) e^(-1/x^2) + r(x) 2/x^3 e^(-1/x^2)`

`color(white)(d/(dx) (r(x) e^(-1/x^2))) = (r'(x) + (2r(x))/x^3) e^(-1/x^2)`

and `r'(x) + (2r(x))/x^3` is also a rational function in `x`

So `f(x)` and all of its derivatives are of the form `r(x) e^(-1/x^2)` for some rational function `r(x)` and all satisfy:

`lim_(x->0) f^((n))(x) = 0`

Hence all of the terms in the Maclaurin series are `0`, so the series only matches the value of `f(x)` at `x = 0`.

Here's what the graph of `f(x)` looks like:

graph{e^(-1/x^2) [-2.5, 2.5, -1.25, 1.25]}

So why does Taylor's theorem fail?

If we examine `e^(-1/x^2)` for Complex values of `x`, then we find that it has an essential singularity at `x=0`: In any open neighbourhood of `0`, no matter how small, you can find a value of `x` such that `f(x)` is arbitrarily close to any chosen Complex value.

So as a Complex function `f(x) = e^(-1/x^2)` is not even close to being differentiable at `x = 0`.