The even-numbered coefficients of the Maclaurin series of $ \frac{1}{\cos(x)} $ are odd integers.

Question

Let’s consider $ G(z) \stackrel{\text{df}}{=} \dfrac{1}{\cos(z)} $ as the exponential generating function of the sequence of Euler numbers. How can one prove that in the Maclaurin series of $ G $, $$ G(z) = \sum_{k = 0}^{\infty} a_{2 k} \cdot \frac{z^{2 k}}{(2 k)!}, $$ the coefficients $ a_{2 k} $ are all odd integers?

For instance, it’s reasonable to use the fact that $$ \forall n \in \mathbb{N}_{0}: \quad a_{2 n} = \frac{{G^{(2 n)}}(0)}{(2 n)!}, $$ but this way of bringing it into life seems too much complicated.

Any help would be appreciated.

Answer 1

To expand my comment into a solution: consider the (formal) product $1=\cos(z)\cdot\frac1{\cos(z)}$. Expanding this out (using the already-established fact that $\frac1{\cos(z)}$ is even) and comparing powers of $x^2$, we get $a_0=1$ and $\displaystyle\sum_{i=0}^n\dfrac{a_{2i}}{(2i)!}\dfrac{(-1)^i}{(2n-2i)!}=0$; multiplying the latter by $(2n)!$ gives $\displaystyle\sum_{i=0}^n(-1)^ia_{2i}{2n\choose 2i}=0$. But we already know that $a_0=1$, and since ${2n\choose 2i}={2n\choose2n-2i}$, all of the terms for $i\in\{1\ldots n-1\}$ pair off (by induction, using the fact that $a_{2i}$ is odd for $i\lt n$) to give even values. (The exception is the middle value when $n$ is even; but in this case, $2n\choose n$ is even, by well-established properties of the binomial coefficents — see, e.g., Kummer's Theorem.) This thus implies that $a_{2n}$ must be odd.

Answer 2

Note: We know, that $\cos$ is an even function, i.e. $\cos(z)=\cos(-z)$ and the powers of $z$ with non-zero coefficients of the power series representation of $\cos$ are even.

\begin{align*} \cos(z)=\sum_{n=0}^{\infty}(-1)^n\frac{z^{2n}}{(2n)!} \end{align*}

Let's consider $f(z)=\frac{1}{\cos(z)}$. Since $\cos$ is even we obtain \begin{align*} f(z)=\frac{1}{\cos (z)}=\frac{1}{\cos (-z)}=f(-z) \end{align*} So, $f(z)$ is also an even function and we get the representation

\begin{align*} f(z)=\sum_{n=0}^{\infty}a_n\frac{z^{n}}{n!}=\sum_{n=0}^{\infty}a_n\frac{(-z)^{n}}{n!}=f(-z) \end{align*}

Comparing coefficients of $z^n$ implies that $a_{2n+1}=0$ for $n\geq 0$ and we get

\begin{align*} f(z)=\sum_{n=0}^{\infty}a_{2n}\frac{z^{2n}}{(2n)!} \end{align*}

Answer 3

This is not a full solution, but I hope that it might help you in some way.

There is a nice characterization of the Maclaurin series of $ \sec $ by the Frenchman Désiré André, which he discovered while studying Bertrand’s Ballot Problem.

Define a sequence $ (A_{n})_{n \in \mathbb{N}} $ of natural numbers by $$ \forall n \in \mathbb{N}: \quad A_{n} \stackrel{\text{df}}{=} \text{The number of alternating permutations of $ \{ 1,\ldots,n \} $}. $$ An alternating permutation of $ \{ 1,\ldots,n \} $ is simply a permutation $ \sigma $ of $ \{ 1,\ldots,n \} $ satisfying $$ \sigma(1) > \sigma(2) < \sigma(3) > \ldots. $$ André showed that $$ \forall x \in \left( - \frac{\pi}{2},\frac{\pi}{2} \right): \quad \sec(x) = \sum_{n = 0}^{\infty} A_{2 n} \cdot \frac{x^{2 n}}{(2 n)!}, $$ where $ A_{0} \stackrel{\text{df}}{=} 1 $.

Hence, your problem boils down to showing that $ A_{2 n} \equiv 1 ~ (\text{mod} ~ 2) $ for all $ n \in \mathbb{N}_{0} $. This is as far as I can go.

Answer 4

Now that I've had a chance to read Steven Stadnicki's answer, I see that my answer below is essentially the same, with some added detail.

Since $\frac1{\cos(x)}$ is even, all the odd order terms are $0$. So we will look only at the even order terms. Since $$ 1=\overbrace{\sum_{k=0}^\infty\frac{a_{2k}}{(2k)!}x^{2k}}^{1/\cos(x)}\overbrace{\sum_{j=0}^\infty\frac{(-1)^j}{(2j)!}x^{2j}}^{\cos(x)}\tag{1} $$ we have that $a_0=1$, and for $k\gt0$, $$ \begin{align} 0 &=(2k)!\sum_{j=0}^k(-1)^{k-j}\frac{a_{2j}}{(2j)!(2k-2j)!}\\ &=\sum_{j=0}^k(-1)^{k-j}\binom{2k}{2j}a_{2j}\\ &=(-1)^ka_0+a_{2k}+\sum_{j=1}^{k-1}(-1)^{k-j}\binom{2k}{2j}a_{2j}\\ &=\left\{\begin{array}{} \small\displaystyle(-1)^k+a_{2k}+\sum_{j=1}^{(k-1)/2}(-1)^{k-j}\binom{2k}{2j}(a_{2j}-a_{2k-2j})&\text{if $k$ is odd}\\ \small\displaystyle(-1)^k+a_{2k}+\sum_{j=1}^{(k-2)/2}(-1)^{k-j}\binom{2k}{2j}(a_{2j}+a_{2k-2j})+\binom{2k}{k}a_k&\text{if $k$ is even} \end{array}\right.\tag{2} \end{align} $$ Now we proceed inductively. Assume that for all $j\lt k$, $a_{2j}$ is odd.

If $k$ is odd, then each term of the sum in $(2)$ is even since the difference of two odds is even.

If $k$ is even, then each term of the sum in $(2)$ is even since the sum of two odds is even. For $k\gt0$, $\binom{2k}{k}=2(k-1)\binom{2k-2}{k-1}$ is even.

Therefore, whether $k$ is even or odd, we have that $(-1)^k+a_{2k}$ is even. Thus, $a_{2k}$ is odd.