# Bounds on function $\exp(-\frac{1}{2}x^2)$

I have the following function : $$f(x)=\exp(-\frac{1}{2}x^2),$$ where $x >0$.

I am looking for some tight bounds (upper bound and lower bounds) on $f(x)$. Any idea ?

P.S.: The problem arises when I tried to find an upper bound on $x_2 \exp(-x_1^2/2)/ (x_1 \exp(-x_2^2/2))$. I prefer the bounds to be tight so that the error when bounding will be small.

• When $x=0$ $f(x)=1$ and it decreases from there to zero in the limit. So $0$ and $1$ are bounds. Commented Feb 10, 2016 at 22:17
• What sort of bounds are you looking for? The obvious bound is $f$ itself - this is an incredibly rapidly decaying function.
– user296602
Commented Feb 10, 2016 at 22:18
• To get a good answer, you should probably tell us what you want to use the bounds for. Showing convergence of some integral? Estimating the error function? Something else?
– mrf
Commented Feb 10, 2016 at 22:22
• I am trying to find an upper bound on this function $x_2 \exp(-x_1^2/2)/ (x_1 \exp(-x_2^2/2))$, and I prefer the bounds to be tight so that the error when bounding will be small.
– din
Commented Feb 10, 2016 at 22:27
• @din I would find the maxima by differentiating and setting equal to $f'(x)=0$. Commented Feb 10, 2016 at 22:43

We have $$|f(x)|=\left|\exp(-\frac{1}{2}x^2)\right|<1, \quad x>0,$$ and $$\lim_{x \to +\infty}|f(x)|=\lim_{x \to +\infty}\left|\exp(-\frac{1}{2}x^2)\right|=0.$$
When $x=0$ $f(x)=1$ and it decreases from there to zero in the limit. So $0$ and $1$ are bounds.