How to compute $\sum_{n=1}^N e^{-( n-c)^2}$

Question

I have to compute or at least find good upper and lower bounds on

\begin{align*} \sum_{n=1}^N e^{-( n-c)^2/b} \end{align*}

and

\begin{align*} \sum_{n=1}^N ne^{-( n-c)^2/b} \end{align*}

where $c$ and $b\ge 0$ are constants.

What I tried

-I tried to rewrite the above two using hyperbolic function but this approach led to nothing.

-this reminds of the functions \begin{align*} \int_{0}^a e^{-( n-c)^2/b} dn= 0.5 \sqrt{\pi b} ({\rm erf(c/\sqrt{b})}-{\rm erf((c-a)/\sqrt{b})} \end{align*}

and

\begin{align*} \int_{0}^a ne^{-( n)^2/} dn =1/2 -e^{-a}/2 \end{align*}

but how to relate those?

Thank you for any help

Answer 1

A possible upper bound may be obtained by considering geometric sums as follows $$\sum_{n=1}^{N}e^{-(n-c)^{2}}= \sum_{n=1}^{N}e^{2cn-c^{2}-n^{2}}\leq \sum_{n=1}^{N}e^{2cn-c^{2}}=e^{-c^{2}}\sum_{n=1}^{N}\left( e^{2c}\right) ^{n}=e^{-c^{2}}e^{2c}\frac{1-e^{N(2c)}}{1-e^{2c}}(\leq e^{2c}\frac{1-e^{2Nc}% }{1-e^{2c}})$$ The last inequality is optional (it depends on what you want)

Answer 2

You can see the sum as the integral of the step function $$e^{-(\lfloor x\rfloor-c)^2/b}.$$

In the increasing area ($x<c$),

$$e^{-(x-1-c)^2/b}\le e^{-(\lfloor x\rfloor-c)^2/b}\le e^{-(x-c)^2/b},$$ and conversely ($x>c$), $$e^{-(x-c)^2/b}\le e^{-(\lfloor x\rfloor-c)^2/b}\le e^{-(x-1-c)^2/b}.$$

Integrating these bracketings with the relevant bounds, you will obtain an approximation in terms of the error function.

The bounds are less tight in the vicinity of $x=c$, but you can handle these by computing a number of discrete $n$ terms and using the $\text{erf}$ approximation for the tails.