# Showing $e^{-x^2}$ is fast decreasing

How can we show that $$e^{-x^2}$$ is a fast decreasing function? I am using the definition of fast decreasing as follows:

A smooth function $f$ is fast decreasing on a conical neighbourhood $V$ if $$\forall N\in \mathbb{Z} \ \ \ \exists\ C_{N} : |f(x)|\leq C_N(1+|x|)^{-N} \ \ \forall x\in V$$

• What can you say about $\lim e^{-x^2}(1+|x|)^N$ at $\pm \infty$?
– A.Γ.
May 19, 2018 at 10:57
• @A.Γ. The essence here is that the exponential is increasing faster than any polynomial, right? (Just to make sure I get it) May 19, 2018 at 11:09
• @A.Γ. it would be 0, but how would you go about defining the constants.
– Sav
May 19, 2018 at 11:15
• If $g$ is continuous on $\Bbb R$ and has limits $0$ at $\pm\infty$ then there is an interval $[-T,T]$ such that $|g(x)|\le 1$ outside the interval. Now what can you say about the values inside $[-T,T]$?
– A.Γ.
May 19, 2018 at 11:26
• @JustDroppedIn Yes, the limit is zero.
– A.Γ.
May 19, 2018 at 11:27

Well, one could go in three different ways to show this.

## First way: Using the definition of $e$.

Since we know that: $$e^x=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n$$ for every $x\in\mathbb{R}$, we also have that: $$e^{x^2}=\lim_{n\to\infty}\left(1+\frac{x^2}{n}\right)^n$$ and, the most important, that te sequence $\left(1+\frac{x^2}{n}\right)^n$ is increasing to $e^{x^2}$.

Now, let, at first, $N\in\mathbb{N}$. Then, we have that: $$e^{x^2}\geq\left(1+\frac{x^2}{N}\right)^N\geq\left(\frac{1+x^2}{N}\right)^N=\frac{1}{N^N}(1+x^2)^N.$$ What we would like to show is that: $$(1+x^2)^N\geq K_N(1+|x|)^N$$ for some constant $K_N$, independent of $x$. For $|x|\geq1$ the wanted inequality holds for $K_N=1$, but for $|x|<1$ we have that: $$(1+x^2)^N<(1+|x|)^N.$$ For $|x|<1$ what we want is, essentially, to find an upper bound for the quantity: $$\frac{1+|x|}{1+x^2}$$ which will be uniform for $x\in(-1,1)$. Note that $x^2\geq0$, so we get that $1+x^2\geq1$ and then: $$\frac{1+|x|}{1+x^2}\leq1+|x|<1+1=2$$ and, as a result: $$(1+x^2)^N\geq2^N(1+|x|)^N.$$ So, letting $K_N=\max\left\{1,2^N\right\}=2^N$ we have found the constant we wanted and our inequalites become: $$e^{x^2}\geq\frac{1}{N^N}(1+x^2)^N\geq\left(\frac{2}{N}\right)^N(1+|x|)^N$$ so, letting $C_N:=\left(\frac{N}{2}\right)^N$ completes the proof for $N\in\mathbb{N}$ since this is equivalent to the inequality: $$e^{-x^2}\leq C_N(1+|x|)^{-N}.$$ Also, for $N=0$, we have nothing to prove, since $|f(x)|\leq1$ for every $x\in\mathbb{R}$.

Now, let $N\in\mathbb{Z}\setminus\mathbb{N}$, $N\neq0$. Then $-N\in\mathbb{N}$ and, from the previous, there exists a $C_{-N}$ such that: $$e^{-x^2}\leq C_{-N}(1+|x|)^N\leq C_{-N}(1+|x|)^{-N}$$ since $N<-N$ and $1+|x|\geq1$, so the sequence $(1+|x|)^{k}$, $k\in\mathbb{Z}$ is increasing. So, letting $C_N=C_{-N}$ gives the wanted constant and the proof is complete.

## Second way: Using the series expansion of $e^x$.

Since we know that: $$e^x=\sum_{k=0}^\infty\frac{x^k}{k!}$$ and the convergence is uniform for $x\in\mathbb{R}$, we also have that: $$e^{x^2}=\sum_{k=0}^\infty\frac{x^{2k}}{k!}$$ uniformly on $\mathbb{R}$.

Let $N\in\mathbb{N}$. Then, we have that: $$e^{x^2}\geq\sum_{k=0}^N\frac{x^{2k}}{k!}=\sum_{k=0}^N\binom{N}{k}\frac{1}{\prod\limits_{i=n-k+1}^Ni}x^{2k}\geq\sum_{k=0}^N\binom{N}{k}\frac{1}{N^N}x^{2k}=\frac{1}{N^N}(1+x^2)^N$$ using the binomial formula in the last step. From this point and over the proof is essentially the same with the previous one.

## Third way: Using de L'Hospital rule.

Since both $e^{-x^2}$ and $(1+|x|)^N$ are even for every $N\in\mathbb{Z}$, we only need to examine their limits as $x\to+\infty$. Note that as $x\to+\infty$, $|x|=x$. We will prove by induction that: $$\lim_{x\to+\infty}\frac{(1+x)^N}{e^{x^2}}=0.$$ For $N=1$, it is evident, since $$0\leq\frac{1+x}{e^{x^2}}\leq\frac{1+x}{1+x^2}\to0$$ from the standard inequality $e^x\geq1+x$. Let us suppose that this is true for $N-1$, $N\geq2$. Then, we have, using de L'Hospital rule: $$\lim_{x\to+\infty}\frac{(1+x)^N}{e^{x^2}}\overset{\infty/\infty}{=}\lim_{x\to+\infty}\frac{N(1+x)^{N-1}}{2xe^{x^2}}= \lim_{x\to+\infty}\frac{N}{2x}\lim_{x\to+\infty}\frac{(1+x)^{N-1}}{e^{x^2}}=0\cdot0=0$$ since both of the latter limits do exist (the one from the inductive hypothesis, the other is trivial).

Also, it is evident that the same is true for every $N\in\mathbb{Z}$, as well. So, we finally have: $$\lim_{x\to+\infty}\frac{(1+|x|)^N}{e^{x^2}}=0,\ \forall N\in\mathbb{Z}.$$ Let, for our convenience, $$h(x)=\frac{(1+|x|)^N}{e^{x^2}}.$$ Since $\lim\limits_{x\to+\infty}h(x)=0$, there exists a $M\in\mathbb{N}$ such that, for every $x\geq M$, we have: $$0\leq h(x)<1.$$ Since $h$ is continuous on $[0,M]$, there exists a number $K>0$ such that, for every $x\in[0,M]$: $$0\leq h(x)\leq K$$ so, letting $C_N:=\max\{1,K\}$ and taking into consideration that $h$ is even, we have that: $$h(x)\leq C_N,\ \forall x\in\mathbb{R}.$$

## Comment

The main difference between the first and the third proof (the second is in general same to the first) is that the first one uses more elementary concepts andfind an exact value for the constant $C_N$ while the third one uses higher level concepts (differentiability etc) and proves only the existence of that constant.

$$-x^2-\log C_N+N\log(1+x)$$

is maximum when $$-2x+\frac N{1+x}=0.$$

Then with

$$\log C_N=-\left(\dfrac{\sqrt{2N+1}-1}2\right)^2+N\log\left(\dfrac{\sqrt{2N+1}+1}2\right)$$ the inequation is verified for all $x$.

This expression is asymptotic to $$-\frac{N\log N}2.$$