Approximating the curve of a row of pascal's triangle using a bell curve equation

Question

When pascal's triangle extends to all real numbers (i.e., each factorial in the formula $\frac{n!}{k!(n-k)!}$ is extended to all real numbers, making the formula continuous), graphing the formula $\frac{n!}{k!(n-k)!}$ with any given positive $n$ yields a curve that resembles a bell curve. For instance, this is what it looks like when $n$ is 4. So my question is, how could I construct a bell curve formula that could approximate (or fit exactly, if it exists) any curve yielded by $n$?

Answer 1

You asked

So my question is, how could I construct a bell curve formula that could approximate (or fit exactly, if it exists) any curve yielded by n?

Part of the answer is given in the Wikipedia article on Binomial distribution. This distribution is defined by $\,\Pr(k;n,p) = \Pr(X = k) = \binom{n}{k}p^k(1-p)^{n-k}.\,$ When $\,p=\frac12\,$ this is a constant multiple of the binomial coefficients. The section on normal approximation states

If $\,n\,$ is large enough, then the skew of the distribution is not too great. In this case a reasonable approximation to $\,B(n, p)\,$ is given by the normal distribution $${\mathcal {N}}(np,\,np(1-p)),$$

which yields the bell curve shape approximation in a precise way.

Use the Gamma function formula $$\frac1{\Gamma(1+z)\Gamma(1-z)} = \frac{\sin \pi z}{\pi z}, $$ to get $\,{0 \choose x}=\frac{\sin \pi x}{\pi x},\,$ ${1 \choose x}=\frac{1!}{(1-x)}\frac{\sin \pi x}{\pi x},\,$ ${2 \choose x}=\frac{2!}{(1-x)(2-x)}\frac{\sin \pi x}{\pi x},\,$ ${3 \choose x}=\frac{3!}{(1-x)(2-x)(3-x)}\frac{\sin \pi x}{\pi x},\,$ etc. The graph of these functions is similar to a bell curve except that they have an infinite number of zero values and have negative values between zero pairs. However, this is a natural behavior required by the fact that they agree with the binomial coefficients exactly at all integer values of $\,x.$

Note that $\,\frac{\sin \pi x}{\pi x}\,$ is the normalized sinc function.

Note the division by zero in the analytical formula denominators for $\,{n \choose z}\,$ when $\,z=k=0,1,\dots,n\,$ but the limit exists and is equal to $\,{n \choose k}=\frac{n!}{k!(n-k)!}\,$ as expected.

Answer 2

The discrete case at least can be regarded as an instance of the central limit theorem: the binomial($n,1/2$) distribution for large $n$ has its PMF given by ${n \choose k} (1/2)^n$ and the normal approximation of that has a PDF given by $\sqrt{\frac{2}{n\pi}} e^{-2\frac{(x-n/2)^2}{n}}$. Therefore ${n \choose k}$ as a function of $k$ behaves roughly like $2^n \sqrt{\frac{2}{n\pi}} e^{-2\frac{(k-n/2)^2}{n}}$ if $n$ is large enough. The approximation is improved if you use $\int_{k-1/2}^{k+1/2} 2^n \sqrt{\frac{2}{n\pi}} e^{-2\frac{(x-n/2)^2}{n}} \, dx$ instead, especially if $n$ is not extremely huge.

The fact that the interaction with the Gamma function as the interpolant of the factorial plays nice with this is not totally obvious; it is probably easier to replace the factorials with Gamma functions directly instead.

Answer 3

Starting from @Ian's answer, we have two approximations $$f(n,k)=2^n \sqrt{\frac{2}{n\pi}} e^{-2\frac{ \left(k-\frac{n}{2}\right)^2}{n}}$$ $$g(n,k)=2^n \sqrt{\frac{2}{n\pi}}\int_{k-\frac 12}^{k+\frac 12}e^{-2\frac{ \left(x-\frac{n}{2}\right)^2}{n}}\,dx=2^{n-1} \Bigg[\text{erf}\left(\frac{n-2 k+1}{ \sqrt{2n}}\right)-\text{erf}\left(\frac{n-2 k-1}{ \sqrt{2n}}\right)\Bigg]$$

Comparing at $k=\frac n2$, some values $$\left( \begin{array}{cccc} n & f\left(n,\frac{n}{2}\right) & g\left(n,\frac{n}{2}\right) & \binom{n}{\frac{n}{2}}\\ 2 & 2.25676 & 2.08200 & 2 \\ 4 & 6.38308 & 6.12680 & 6 \\ 6 & 20.8470 & 20.2822 & 20 \\ 8 & 72.2163 & 70.7396 & 70 \\ 10 & 258.369 & 254.126 & 252 \\ 12 & 943.429 & 930.488 & 924 \\ 14 & 3493.78 & 3452.63 & 3432 \\ 16 & 13072.5 & 12937.6 & 12870 \\ 18 & 49299.6 & 48846.9 & 48620 \\ 20 & 187079.0 & 185532.0 & 184756 \end{array} \right)$$

If you cannot use the error function, use some quite good approximations (have a look at my naswer here).