Voofie | An Explicit Formula for the Euler zigzag numbers (Up/down numbers) from power series

Content

1 Introduction
2 Integrand of sec(x) + tan(x)
3 Power Series Expansion of f(x)
4 Simplification
5 Explicit Formula for Euler number
6 Explicit Formula for Bernoulli Numbers
7 Conclusion

Introduction

Euler zigzag numbers, $\f$A_n\f$$ , is the number of alternating permutation of the set {1,2,...,n}. And it is well known that:

$\f$\begin{align*} \sec x+\tan x= \sum _{n=0}^{\infty } \frac{A_n}{n!}x^n \end{align*}\f$$

In the following section, I am going to derive an explicit formula of $\f$A_n\f$$ by using power series expansion.

Integrand of sec(x) + tan(x)

Let's consider the integrand of sec(x) + tan(x):

$\f$\begin{align*} \sum _{n=0}^{\infty } \frac{A_n}{n!}\int_0^x y^n \, dy&=\int_0^x (\sec y+\tan y) \, dy \\ \sum _{n=1}^{\infty } \frac{A_{n-1}}{n!}x^n&= \int_0^x \frac{1+\sin y}{\cos y} \, dy\\ &= \int_0^x \frac{(1+\sin y)(1-\sin y)}{\cos y(1-\sin y)} \, dy\\ &= \int _0^x\frac{\cos ydy}{1- \sin y}\\ &=-\ln (1-\sin x) \end{align*}\f$$

Let:

$\f$f(x) = -\ln (1-\sin x)\f$$

I will do a power expansion of the function f(x), and a finite sum explicit formula for $\f$A_n\f$$ can be found by some simplification.

Power Series Expansion of f(x)

$\f$\begin{align*} f(x) &=-\ln (1-\sin y) \\ &= \sum _{n=1}^{\infty } \frac{\sin^n y}{n}\\ &= \sum _{n=1}^{\infty } \frac{\left(e^{i y}-e^{-i y}\right)^n}{2^ni^nn}\\ &= \sum _{n=1}^{\infty } \sum _{k=0}^n \frac{C_k^ne^{(n-k)i y }e^{-i k y}(-1)^k}{2^ni^nn}\\ &= \sum _{n=1}^{\infty } \sum _{k=0}^n \frac{C_k^ne^{(n-2k)i y }(-1)^k}{2^ni^nn}\\ &= \sum _{n=1}^{\infty } \sum _{k=0}^n \frac{C_k^n(-1)^k}{2^ni^nn}\sum _{j=0}^{\infty } \frac{(n-2k)^ji^jy^j}{j!}\\ &= \sum _{j=0}^{\infty } \sum _{n=1}^{\infty } \sum _{k=0}^n \frac{C_k^n(n-2k)^ji^j(-1)^k}{2^ni^n j!n}y^j \end{align*}\f$$

Therefore, by equating coefficients, we have:

$\f$\begin{align*} A_{j-1}&=i^j\sum _{n=1}^{\infty } \sum _{k=0}^n \frac{C_k^n(n-2k)^j(-1)^k}{2^ni^n n} \end{align*}\f$$

$\f$\begin{align*} A_j=i^{j+1}\sum _{n=1}^{\infty } \sum _{k=0}^n \frac{C_k^n(n-2k)^{j+1}(-1)^k}{2^ni^nn} \end{align*}\f$$

Now, I am going to simplify it, and show that it is actually equals to:

$\f$\begin{align*} A_j=i^{j+1}\sum _{n=1}^{j+1} \sum _{k=0}^n \frac{C_k^n(n-2k)^{j+1}(-1)^k}{2^ni^nn} \end{align*}\f$$

Simplification

In order to reduce the infinite sum to a finite sum, I first let:

$\f$B_n^j = \sum _{k=0}^n C_k^n(n-2k)^{j}(-1)^k \f$$

So that:

$\f$\begin{align*} A_j=i^{j+1}\sum _{n=1}^{\infty } \frac{B_n^{j+1}}{2^ni^nn} \end{align*}\f$$

I observed that:

$\f$\begin{align*} B_n^j = 0 \text{ if } n > j \end{align*} \f$$

To show that, let's define the translational operator $\f$D\f$$ such that:

$\f$\begin{cases} D a_n = a_{n+1}& \\ D^{-1} a_n = a_{n-1}& \end{cases}\f$$

Then:

$\f$\begin{align*} B_n^j&=\left(\sum _{k=0}^n C_k^n(-1)^kD^{-2k}\right)n^j \\ &= \left(1-D^{-2}\right)^nn^j\\ &= \left(1+D^{-1}\right)^n\left(1-D^{-1}\right)^nn^j \end{align*}\f$$

Here, $\f$\nabla = 1-D^{-1}\f$$ is the backward difference operator.

Now, consider:

$\f$\begin{align*} \nabla n^j &= n^j - (n-1)^j\\ &= n^j-\sum _{k=0}^j C_k^j(-1)^{j-k}n^k\\ &= -\sum _{k=0}^{j-1} C_k^j(-1)^{j-k}n^k \end{align*}\f$$

The result is a polynomial of degree $\f$j-1\f$$ . We can see that, if we apply backward difference operator to a polynomial, its degree decreases by 1. Therefore, if we apply the backward difference operator for a number of time which is larger than the degree of a polynomial, the result is zero.

As a result, we have $\f$\begin{align*} B_n^j = 0 \text{ if } n > j \end{align*} \f$$ . Therefore:

$\f$\begin{align*} A_j=i^{j+1}\sum _{n=1}^{j+1} \sum _{k=0}^n \frac{C_k^n(n-2k)^{j+1}(-1)^k}{2^ni^nn} \end{align*}\f$$

Explicit Formula for Euler number

Euler number $\f$E_n\f$$ is given by the generating function:

$\f$\begin{align*} \frac{1}{\cosh t}&=\frac{2}{e^t+e^{-t}}\\ &=\sum _{n=0}^{\infty } \frac{E_n}{n!}t^n \end{align*} \f$$

And it is given by:

$\f[\begin{cases} E_{2n}&=i\sum _{k=1}^{2n+1} \sum _{j=0}^k {k\choose{j}}\frac{(-1)^j(k-2j)^{2n+1}}{2^ki^kk} \\ E_{2n+1}&=0 \end{cases}\f]$

Explicit Formula for Bernoulli Numbers

From Wikipedia, we know:

$\f[B_{2n}=\frac{(-1)^{n-1}2n}{4^{2n}-2^{2n}}A_{2n-1}\f]$

Therefore:

$\f[\begin{cases} B_0 = 1 \\ B_1 = -\frac{1}{2} \\ B_{2n}=\frac{2n}{2^{2n}-4^{2n}} \sum_{k=1}^{2n} \sum_{j=0}^k \binom{k}{j} \frac{(-1)^j(k-2j)^{2n}}{2^ki^kk} &\text{ for } n > 0\\ B_{2n+1}=0 &\text{ for } n>0 \\ \end{cases}\f]$

Conclusion

I have found out an simple formula for the Euler zigzag number:

$\f$\begin{align*}A_n = i^{n+1}\sum _{k=1}^{n+1} \sum _{j=0}^k {k\choose{j}} \frac{(-1)^j(k-2j)^{n+1}}{2^ki^kk}\end{align*}\f$$

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An Explicit Formula for the Euler zigzag numbers (Up/down numbers) from power series