# Big O notation for e^(-1/x²)

• Jan 20th 2013, 12:01 PM
Big O notation for e^(-1/x²)
The following statement confuses me:

$\displaystyle e^{-\frac 1{x^2}} \text { is in } \text{O}(x^n) \text{ for all }n \, .$

In order to try to verify this, I used the Taylor expansion for $\displaystyle e^x$ about $\displaystyle x = 0$ yielding

$\displaystyle e^{-\frac 1{x^2}} = 1 + \left( -\frac 1{x^2} \right) + \frac {\left( -\frac 1{x^2} \right)^2}{2!} + \frac {\left( -\frac 1{x^2} \right)^3}{3!} + \, \ldots \, ,$

which can be simplified to

$\displaystyle e^{-\frac 1{x^2}} = 1 - x^{-2} + \frac 1{2!}x^{-4} - \frac 1{3!}x^{-6} + \, \ldots$

How can we now claim that this function is in $\displaystyle \text{O}(x^n)$ for all $\displaystyle n$? Correct me if I am wrong but being in $\displaystyle \text{O}(x^n)$ when $\displaystyle x \rightarrow 0$ means that the function approaches $\displaystyle 0$ at least as fast as $\displaystyle x^n$ (faster sure, but no less than $\displaystyle x^n$!). Let's suppose $\displaystyle n=1$; the function then apparently is in $\displaystyle \text{O}(x)$ i.e. the function approaches $\displaystyle 0$ at least as fast as x. Does this imply that the terms with negative exponents (who have a greater absolute value than $\displaystyle 1$) reach $\displaystyle 0$ faster than $\displaystyle x$? How is that possible when we know that negative exponents mean division by $\displaystyle x$ and if now $\displaystyle x \rightarrow 0$ then the quotient must approach infinity?

I am guessing that terms who have negative exponents approach zero faster than $\displaystyle x$ just as $\displaystyle x^2$ approach zero faster than $\displaystyle x$. Is this analogy mathematically valid? I am not sure at all since $\displaystyle x^{-2} \rightarrow \pm \infty$ when $\displaystyle x \rightarrow 0$ whereas $\displaystyle x^2 \rightarrow 0$ when $\displaystyle x \rightarrow 0$.