The expression $x^n / n!$ appears in the infinite sum defining $e^x$ and similar terms in the sums defining $\cos(x)$, $\sin(x)$, etc. I would like to know if there is some combinatorial/probabilistic meaning or analogy to the term $x^n / n!$ and appropriately an example of a scenario of selection or decision making that could be represented by a consecutive sum of these terms similar to the infinite sums defining $e^x$ or $\cos(x)/\sin(x)$.

In other words: some probabalistic/combinatorical scenario whose calculation would converge to one of these functions ($e^x/\cos(x)/\sin(x)$).

  • $\begingroup$ There are extremely many. One of the best known is a derangement. $\endgroup$
    – whuber
    Jul 5, 2023 at 14:31
  • 2
    $\begingroup$ @whuber I do not understand what $x^n/n!$ has to do with derangements, can you clarify? $\endgroup$ Jul 5, 2023 at 15:54

2 Answers 2


If you divide every term by $e^{x}$ so the sum is $1$

then you get the Poisson distribution probability $e^{-x} \dfrac{x^n}{n!}$

of seeing $n$ items when the expected number of items is $x$.

  • $\begingroup$ this @Henry still has the term e^−x . could you give some intuitive example of a set of calculations that converge to e^x? $\endgroup$
    – Gilad
    Jul 8, 2023 at 21:48

If $X_1,X_2,\ldots,X_n$ are i.i.d $\text{Uniform}(0,1)$ random variables, then (check this for details)

$$P\left(\sum_{i=1}^n X_i \le x\right)=\frac{x^n}{n!}\quad,\, x\in [0,1]$$

In other words, $x^n/n!$ is the volume of the region $$\left\{(x_1,\ldots,x_n)\in \mathbb R^n: x_1,\ldots,x_n\ge 0, \sum_{i=1}^n x_i \le x\right\}$$ whenever $x\in [0,1]$. When $x=1$, this is related to the volume of a simplex.

We can also define a stopping rule associated with the $X_i$'s as

$$N=\min\left\{n: \sum_{i=1}^n X_i>x\right\} \quad,\, x\in [0,1]$$

Then, using the result above,

$$E(N)=\sum_{n=0}^\infty P(N>n)=\sum_{n=0}^\infty P\left(\sum_{i=1}^n X_i \le x\right)=e^x$$

This is in fact a popular exercise discussed here previously.

  • $\begingroup$ What I like about this is the result is simple and yet the interpretation cannot be more manifesting. +1. $\endgroup$ Jul 6, 2023 at 6:22
  • $\begingroup$ @StubbornAtom what is the meaning of E(N)? $\endgroup$
    – Gilad
    Jul 8, 2023 at 21:46
  • $\begingroup$ @Gilad It is the expected number of $X_i$s when their cumulative sum first exceeds the given $x$ $\endgroup$
    – Henry
    Apr 2 at 10:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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