# Evaluating the “limit function” $f_n$ with $n \to \infty$

I want to find the following "limit function" for $n$ to infinity with $x \in [0;1]$

$$f_n (x) = \frac{ \sum_{i = 1}^{n -1} \eta(nx-i) \left ( \left\{nx\right\} \binom{n}{i} + (1 - \left\{nx\right\}) \binom{n}{i + 1} \right ) }{\binom{n}{\left \lfloor n/2 \right \rfloor}}$$

with

$$\left\{x\right\} = x - \left \lfloor x \right \rfloor$$

and

$$\eta (x) =\begin{cases}0 & x \notin [0;1] \\ 1 & x \in [0;1]\end{cases}$$

I pieced together that functions because I was curious about the sequence of binomial coefficients. So I took that sequence, linearly interpolated between the terms and scaled the function down to the unit interval.

Now I want to know what happens if I let $n$ towards infinity. In a sense, I want to know what $f_\infty$ is.

But as I'm only grade ten, it is hard to proceed further. Can you help me?

EDIT: As Ian pointed out in his comment, all but one terms in my sum are equal to 0, which reduces my functional equation to:

$$f_n (x) = \frac{\left\{nx\right\} \binom{n}{\lfloor nx \rfloor} + (1 - \left\{nx\right\}) \binom{n}{\lfloor nx \rfloor + 1} }{\binom{n}{\left \lfloor n/2 \right \rfloor}}$$

• Are your binomial coefficients backwards? Also, have you noticed that the only i that contributes is $\lfloor nx \rfloor$? – Ian Jul 18 '17 at 20:10
• $\mathrm{Frac}\left(\,x\,\right)$ is usually denoted as $\left\{\,x\,\right\}$. The code is $\texttt{\left\\{x\right\\}}$. – Felix Marin Jul 18 '17 at 20:17
• $\eta$ will be zero in your sum for all i except $\lfloor nx \rfloor$. – Ian Jul 18 '17 at 20:19
• I guess the coefficient of the linear interpolation is interchanged; the correct formulation would be $$f_n(x) = \frac{(1-\{nx\}) \binom{n}{[nx]} + \{nx\}\binom{n}{[nx]+1}}{\binom{n}{[n/2]}}.$$ Anyway, due the the concentration this function will converge pointwise to $$\lim_{n\to\infty} f_n(x) = \begin{cases} 1, & x = 1/2 \\ 0, & \text{otherwise} \end{cases}.$$ – Sangchul Lee Jul 18 '17 at 20:54
• The more interesting question is really about the rate of decay away from $x=1/2$ for large finite $n$. If I made no calculation error, it should turn out that ${n \choose k}$ is roughly $2^n f \left ( \frac{2k-n}{\sqrt{n}} \right )$ where $f(x)$ is the standard normal PDF $\frac{1}{\sqrt{2 \pi}} e^{-x^2/2}$. This "roughly" is actually quite rough, but it at least gives the right order of magnitude. – Ian Jul 18 '17 at 22:42