$x^n/n!$ and Stirling's Formula

I saw this nice problem on Mathstodon a few months ago and decided to take a look today. It turns out that the curves of $x^n/n!$ have very similar shapes but shifted. What is the size of the shift between the curves?

Figure 1: The curves of $x^n/n!$ for $n=1,...,20.$ They show a characteristic parallel shape among each other.

The easiest way to do it is by using Stirling approximations, on which I did some rudimentary speed analysis sometime before. Assume that the value we’re trying to find the spacing at is $t.$ Then we need to find the corresponding value of $x$ for two successive curves. $x^n / n! = t.$ Recall Stirling’s formula and substitute it into this:

Two things happen here when $n$ is large and $t$ is small, we can ignore the effects of $\sqrt{n}$ in the product with $n^{n}$ firstly and notice that $t^{1/n}$ equals one. So we get

and the resulting spacing is simply $1/e.$ I thought this was very neat, and a good example of street-fighting calculations that are seemingly too complex. I would say that the hardest part in this entire thing was plotting the graphs, they needed some programming acrobatics to get the axes right.

def comp(n: int) -> list:
    
    x = []
    y = []
    for i in np.linspace(0, 10, 1000):
        if ((i ** n) / math.factorial(n)) <= 2:
            x.append(i)
            y.append((i ** n) / math.factorial(n))
    
    return x, y

for i in range(2, 20):
    plt.plot(comp(i)[0], comp(i)[1])
plt.xlabel("x")
plt.ylabel("x^n/n!")
plt.show()

An interesting fact here is that every curve intersects with the next one only at an integer. This is fairly easy to show,

This means that the $n^{\text{th}}$ curve and the $(n + 1)^{\text{th}}$ curve will intersect at $x = n + 1.$ Cute.