User:Michael Hardy/proof of André's theorem

The nth Euler number En is equal to the number of alternating permutations of the set { 1, 2, 3, ..., n}, i.e. arrangements of that set into a sequence a1, ..., an satisfying


 * $$ a_1 > a_2 < a_3 > a_4 < \cdots. \, $$

(There are differing conventions as to how the term "Euler number" is defined, but all are related concepts.) The Euler number is also equal to the number of reverse-alternating permutations, i.e. those that satisfy


 * $$ a_1 < a_2 > a_3 < a_4 > \cdots. \, $$

(A bijection between alternating and reverse-alternating permutations is given by


 * $$ a_i \mapsto n+1-a_i.) \, $$

The first several Euler numbers are


 * $$ E_0 = 1,\quad E_1=1,\quad E_2=1,\quad E_3=2,\quad E_4=5,\quad E_5=16,\quad E_6=61,\quad E_7=272,\quad \ldots $$

André's theorem states that the exponential generating function of the Euler numbers is


 * $$ f(x) = \sum_{n=0}^\infty E_n \frac{x^n}{n!} = \sec x + \tan x. $$

Here we prove André's theorem by means of a combinatorial argument showing that this generating function satisfies a certain differential equation, and we use the initial condition &fnof;(0) = 1. This proof appears is due to and also appears in. See also this preprint by Stanley.

The principal combinatorial result that we need is this identity:


 * $$ 2E_{n+1} = \sum_{k=0}^n \binom n k E_k E_{n-k}\text{ if }n\ge 1.\qquad\qquad(1) $$

The provision that n ≥ 1 is crucial.

A proof of this combinatorial identity runs as follows. To choose an alternating or reverse-alternating permutation of the set { 1, 2, 3, ..., n, n + 1 }, we


 * choose a subset of size k of the set { 1, ..., n }, then
 * choose a reverse-alternating permutation a1, ..., ak of the set { 1, ..., k }, then
 * choose a reverse-alternating permutation b1, ..., bn&minus;k of the set { k + 1, ..., n }.

Then the permutation


 * $$ a_k, a_{k-1}, a_{k-2},\ldots, a_1, n+1, b_1, b_2, b_3,\ldots b_{n-k} \, $$

is either alternating or reverse-alternating. The number of ways to choose a permutation of { 1, 2, 3, ..., n, n + 1 } that is either alternating or reverse-alternating is En+1, and the number of ways to complete the sequence of steps above is


 * $$ \binom n k E_k E_{n-k}. \, $$

This needs to be done for each possible value of k to get a complete list, hence we sum from k = 0 to k = n. This completes the proof of the identity (1).

Multiplication of both sides of (1) by xn+1/(n+1)! and summing over n ≥ 1, and then prepending the constant and first-degree terms, yields


 * $$ 2f(x) = 2E_0 + 2E_1 x + 2\sum_{n=1}^\infty E_{n+1}\frac{x^{n+1}}{(n+1)!} = 2E_0 + 2 E_1 x + \sum_{n=1}^\infty \sum_{k=0}^n \binom n k E_k E_{n-k} \frac{x^{n+1}}{(n+1)!}. $$

Differentiating both sides, we get


 * $$ 2f'(x) = 2E_1 + 2\sum_{n=1}^\infty E_{n+1}\frac{x^n}{n!} = 2E_1 + \sum_{n=1}^\infty \sum_{k=0}^n \binom n k E_k E_{n-k} \frac{x^n}{n!}. $$

In the last sum, the index n goes from 1 to &infin;, not from 0 to &infin;. If we change the lower bound of summation from 1 to 0, we simply add 1 to the sum, and compensate by changing the initial term, 2E1 = 2, to E1 = 1, getting


 * $$ E_1 + \sum_{n=0}^\infty \sum_{k=0}^n \binom n k E_k E_{n-k} \frac{x^n}{n!} = 1 + \sum_{n=0}^\infty \sum_{k=0}^n \left(E_k \frac{x^k}{k!}\right) \left( E_{n-k} \frac{x^{n-k}}{(n-k)!}\right). $$

The last sum is over all pairs of positive integers, so the expression above can be rearranged as


 * $$ 1 + \sum_{i=0}^\infty \sum_{j=0}^\infty E_i\frac{x^i}{i!}\cdot E_j \frac{x^j}{j!} $$

(see Cauchy product).

The expression $$ E_i\frac{x^i}{i!} $$ does not change as j goes from 0 to &infin;; hence it can be pulled out of the inside sum, getting


 * $$ 1 + \sum_{i=0}^\infty\left( E_i\frac{x^i}{i!} \sum_{j=0}^\infty E_j\frac{x^j}{j!} \right). $$

Now the second sum does not change as i goes from 0 to &infin;; hence it can be pulled out of the outer sum, getting


 * $$ 1 + \left(\sum_{j=0}^\infty E_j\frac{x^j}{j!} \right)\left(\sum_{i=0}^\infty E_i\frac{x^i}{i!} \right). $$

This is


 * $$ 1 + (f(x))^2. \, $$

Consequently we have a differential equation


 * $$ 2f'(x) = 1 + (f(x))^2. \, $$

This can be solved by separation of variables, getting


 * $$ f(x) = \tan\left( \frac x 2 + \text{constant} \right). $$

We have an initial condition &fnof;(0) = 1, so we have


 * $$ f(x) = \tan\left( \frac x 2 + \frac\pi4 \right). $$

Finally, a standard tangent half-angle trigonometric identity gives us


 * $$ f(x) = \sec x + \tan x. \, $$

This completes the proof of André's theorem.