Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $n$: $$\frac{d^n}{dx^n}(x^n)=n!$$

share|cite|improve this question
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
up vote 7 down vote accepted

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.

share|cite|improve this answer
thank you for your answer it was helpful – Aram Rafeq Jan 29 at 15:28
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:

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

share|cite|improve this answer
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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.