# Relationship between factorial and derivatives

I was wondering if there is any relationship between factorials and derivatives because I notice that if we had $x^n$ and we take the $n$-th derivative of this function it will be equal to the factorial of $n$: $$\frac{d^n}{dx^n}(x^n)=n!$$

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I'd love to find an explanation that uses the geometric fact that $\frac{x^n}{n!}$ is the (hyper-)volume of the region $\{(x_1,x_2,\dots,x_n)\mid 0\leq x_1\leq x_2\leq \cdots x_n\leq x\}$. Or the set $\{(x_i)\mid x_i\geq 0, \text{ and }\sum x_i\leq x\}$. – Thomas Andrews Jan 29 at 15:18
can you post an image for the region of some visual data ? – Aram Rafeq Jan 29 at 15:27
Not really possible to post an image beyond $n=3$, for obvious reasons. It is the convex region with corners $(0,0,1)$, $(0,1,1)$, $(1,1,1)$ and $(0,0,0)$ in three dimensions. – Thomas Andrews Jan 29 at 15:54
thanks for your comment :) – Aram Rafeq Jan 29 at 15:57

Yes, and that's precisely why $n!$ appears in the denominator of the term of a Taylor series containing $x^n$ (for simplicity, I'll assume the series is centered at $x=0$).

That term is $\frac{f^{(n)}(0)}{n!}x^n$. When you take $n$ derivatives and plug in $x=0$, you get just $f^{(n)}(0)$ as desired. That's why the Taylor series has the correct derivatives of all orders at $x=0$; all other terms vanish (they have either been differentiated away or they still contain a real factor of $x$), leaving just exactly what you want.

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Glad to help ${}{}{}$ – MPW Jan 29 at 17:37

Let $U_{n,x}=\{(x_1,x_2,\dots,x_n)\mid 0<x_1<x_2<\cdots <x_n\leq x\}$. It is not hard to show, by a symmetry argument, that the hypervolume of this region is $\frac{x^n}{n!}$, because there are $n!$ ways to permute the $x_i$ to get a different order, and this covers "almost all" of the $n$-dimension hypercube with side $x$.

An interesting feature can be seen that the value:

$$|U_{x+y,n}| = \sum_{k=0}^{n} |U_{x,k}|\cdot|U_{y,n-k}|$$

This is the binomial theorem written geometrically. It's saying that we can break up $U_{x+y,n}$ into components $A_k$ where the first $k$ elements of $(x_1,\dots,x_n)$ are less than $x$ and the next $n-k$ elements are greater than $x$. (Where $|U_{z,0}|$ is considered to be $1$.)

This lets us calculate the derivative, because $y=|U_{y,1}|$ and you get:

$$\frac{U_{x+y,n}-U_{x,n}}{y} = |U_{x,n-1}| + \sum_{k=0}^{n-2} |U_{x,k}|\cdot \frac{|U_{y,n-k}|}{y}$$ and we get that $\frac{|U_{y,j}|}{y}\to 0$ as $y\to 0$ when $j>1$.

It feels like there is something more that can be done with this idea, perhaps using something like:

$$f^{(n)}(x)=\lim_{h_1\to 0} \lim_{h_2\to 0} \cdots \lim_{h_n\to 0}\text{ some horrible expression }$$

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Very interesting insights! – jeremy radcliff Jan 29 at 23:22

Look at the general expression $$y_i = \frac{{\rm d}^i x^n}{{\rm d}x^i}$$

This is the i-th derivative of $x^n$ for $i=1\ldots n$. This can be directly evaluated as

$$\boxed{ \frac{{\rm d}^i x^n}{{\rm d}x^i} = \frac{n!}{(n-i)!} x^{n-i} }$$

So the i-th derivative of a n-th order polynomial contains terms of $n!$, $(n-1)!$, $(n-2)!$ etc

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