Taylor Series of $e^{-1/x^2}$ about $0$

Question

So I worked out the Taylor series of $e^{-1/x^2}$ about $0$. However, all derivatives at $0$ came out to be $0$. So how can we write the Taylor series of the above function about $0$?

Answer 1

As you worked out, the Taylor series of this function is 0. It means that, around 0, this function and all of its derivatives evaluate to 0.

Answer 2

We're going to prove that the function $$f(x)=\left\{ \begin{array}{ll} e^{-1/x^2} & \text{if } x\neq 0 \\ 0 & \text{if } x=0 \end{array} \right.$$ is infinitely differentiable and all his derivatives at $x=0$ are zero.

By induction, it's easy to check that, if $x\neq 0$, for each $n\in \mathbb{N}$ we have $$f^{n)}(x)=\frac{e^{-1/x^2}P_{2n-2}(x)}{x^{3n}},$$ where $P_{2n-2}(x)$ is a polynomial with degree $2n-2$, and if $x=0$ $$f^{n)}(0)=0. $$

Then, we can apply Taylor's Theorem. However, as $f^{n)}(0)=0$ for each $n\in \mathbb{N}$, the Taylor Series is just $0$.

Answer 3

@user3650050 : I suppose that you expect a series which approximates the function $y(x)=\exp\left(-\frac{1}{x^2} \right)$ for $x$ close to $0$.

The pertinent comments and answer that you already received might disappoint you because whey don't give an approximate series as expected : As already pointed out, the Taylor series for $y(x)=\exp\left(-\frac{1}{x^2} \right)$ at $x=0$ doesn't fit the function for $x$ not exactly $0$.

I don't know if the next form of series can be useful for you, but I hope so :

Around $x=\epsilon\:\:$ close to $0$, the first terms of the series expansion are : $$\exp\left(-\frac{1}{x^2} \right)=\exp\left(-\frac{1}{\epsilon^2} \right)\left(1+\frac{2}{\epsilon^3}x +\frac{2-3\epsilon^2}{\epsilon^6}x^2 +\frac{\frac43 -6\epsilon^2+4\epsilon^4}{\epsilon^9}x^3 +... \right)$$