Fitting a closed curve on the roots of ${x \choose k}-c$

Let $${n \choose k} = \frac1{(n+1) \operatorname{B}(n-k+1, k+1)}.$$ be the generalized binomial coefficient, and here $\operatorname{B}$ is the Beta function.

Let $f_{k,c}(x)$ be the following function

$$f_{k,c}(x) = {x \choose k}-c,$$

where $k$ and $c$ are nonzero scalars.

Are there a special general closed curve, what we can fit on the roots of $f_{k,c}$?

The roots on the complex plane for $(k,c)=(6,1)$, $(k,c)=(9,1)$, $(k,c)=(15,1)$ are the followings.

The results are similar for other $(k,c)$ pairs. First I thought that it is an ellipse, but actually after fitting ellipses on arbirtrary five roots of higher order $f_{k,c}$ functions it turned out that it is not. The motivation of the problem was this question.

Solution 1:

For convenience let me set $k = 2n$.

We can represent the binomial coefficient by

$$ \binom{x}{2n} = \frac{1}{(2n)!}\prod_{j=0}^{2n-1} (x-j), $$

so to make this an even polynomial we will replace $x$ by $y+n-\tfrac{1}{2}$. This will make the roots of the equation

$$ \binom{y+n-\tfrac{1}{2}}{2n} = c \tag{0} $$

symmetric about the real and imaginary axes (at least when $c$ is real), effectively centering the "oval" at the origin.

It appears that the roots tend to $\infty$ as $n \to \infty$. From the product representation we observe that

$$ \binom{y+n-\tfrac{1}{2}}{2n} \sim \frac{y^{2n}}{(2n)!} $$

for large $y$, so we might expect that equation $(0)$ behaves a little like

$$ \frac{y^{2n}}{(2n)!} \approx c $$

when $n$ is large. Multiplying both sides by $(2n)!$ and raising things to the $1/(2n)$ power the "equation" becomes, by Stirling's formula,

$$ y \approx \frac{2n}{e}. $$

This heuristic reasoning leads us to suspect that the roots grow approximately on the order of $n$. We'll make one more substitution to reflect this, replacing $y$ by $nz$.

In total we want to make the substitution

$$ x = nz + n - \tfrac{1}{2} $$

into the original equation $\binom{x}{2n} = c$ and hence study the new equation

$$ \binom{nz + n - \tfrac{1}{2}}{2n} = c. \tag{1} $$

Using the Beta function representation of the binomial coefficient we can rewrite it in terms of Gamma functions as

$$ \binom{nz + n - \tfrac{1}{2}}{2n} = \frac{\Gamma\left(nz+n+\tfrac{3}{2}\right)}{\left(nz+n+\tfrac{1}{2}\right) \Gamma\left(nz-n+\tfrac{1}{2}\right) \Gamma(2n+1)}. $$

We now replace the Gamma functions with their Stirling formula equivalents;

$$ \Gamma\left(nz+n+\tfrac{3}{2}\right) \sim \left(\frac{nz+n+\tfrac{3}{2}}{e}\right)^{nz+n+3/2}\sqrt{\frac{2\pi}{nz+n+\tfrac{3}{2}}}, $$

$$ \Gamma\left(nz-n+\tfrac{1}{2}\right) \sim \left(\frac{nz-n+\tfrac{1}{2}}{e}\right)^{nz-n+1/2}\sqrt{\frac{2\pi}{nz-n+\tfrac{1}{2}}}, $$

$$ \Gamma(2n+1) \sim \left(\frac{2n+1}{e}\right)^{2n+1} \sqrt{4\pi n}, $$

where the first two are valid for $\operatorname{Im} z \neq 0$ with a relative error of $O(1/n)$ as long as $z$ remains bounded away from the points $z=\pm 1$. With a little bit of algebra we find that

$$ \binom{nz + n - \tfrac{1}{2}}{2n} \sim \frac{(2n+1)^{-2n-1}}{\sqrt{4\pi n} \left(nz+n+\tfrac{1}{2}\right)} \sqrt{\frac{nz-n+\tfrac{1}{2}}{nz+n+\tfrac{3}{2}}} \frac{\left(nz+n+\tfrac{3}{2}\right)^{nz+n+3/2}}{\left(nz-n+\tfrac{1}{2}\right)^{nz-n+1/2}}. $$

A straightforward calculation will show that

$$ \frac{\left(nz+n+\tfrac{3}{2}\right)^{nz+n+3/2}}{\left(nz-n+\tfrac{1}{2}\right)^{nz-n+1/2}} \sim e n^{2n+1} \frac{(z+1)^{nz+n+3/2}}{(z-1)^{nz-n+1/2}} $$

as $n \to \infty$, so on taking absolute values and raising both sides to the $1/n$ power we obtain

$$ \left|\binom{nz + n - \tfrac{1}{2}}{2n}\right|^{1/n} = \frac{1}{4} \left|\frac{(z+1)^{z+1}}{(z-1)^{z-1}}\right|\left(1+O\left(\frac{\log n}{n}\right)\right), $$

as $n \to \infty$, where the error term holds uniformly with respect to $z$ as long as $z$ remains bounded away from the points $z = \pm 1$. So, after taking absolute values of both sides of $(1)$ followed by $n^\text{th}$ roots,

$$ \left|\binom{nz + n - \tfrac{1}{2}}{2n}\right|^{1/n} = |c|^{1/n}, $$

letting $n \to \infty$ yields

$$ \frac{1}{4} \left|\frac{(z+1)^{z+1}}{(z-1)^{z-1}}\right| = 1. $$

So...

The roots of equation $(1)$ tend to the limit curve $$ \left|\frac{(z+1)^{z+1}}{(z-1)^{z-1}}\right| = 4 \tag{2} $$ as $n \to \infty$.

Below is a plot of the roots of equation $(1)$ with $n=30$ and $c=1$ in $\color{blue}{\text{blue}}$ along with the limit curve $(2)$ in $\color{red}{\text{red}}$.

And here is a plot with $n=100$ and $c=1$.

Remark 1: These calculations indicate that the rate at which the zeros approach the limit curve is $O(\log n/n)$ away from the points $z = \pm 1$. Near these points the rate of approach should be different and a more detailed analysis is required to determine it.

Remark 2: The variable $c$ does not need to remain fixed; it may in fact depend on $n$. As long as

$$ |c|^{1/n} \to 1 $$

as $n \to \infty$, the roots of equation $(1)$ will still tend to the limit curve $(2)$, though perhaps at a different rate than if $c$ were held fixed.

Remark 3: With more work it is possible to obtain better approximations to the actual curve on which the roots lie. Some rough calculations seem to indicate that

$$ \left|\frac{(z+1)^{z+1}}{(z-1)^{z-1}}\right| = 4 + \frac{\log(16e^4|c|^4n^2)}{n} $$

is a much better one. Here's a plot with $n = 30$ and $c=5$, showing this 'adjusted' curve as a thin blue line.