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 Post subject: Euler numbers as determinantsPosted: Sat Jun 25, 2016 6:33 am
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Joined: Mon Nov 09, 2015 1:36 am
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Location: Ioannina, Greece
It is known that $\displaystyle\frac{1}{\cos{x}}=\sum_{n=0}^{+\infty}{\frac{(-1)^{n}{\rm{E}}_{2n}x^{2n}}{(2n)!}}\,,\quad |x|<\tfrac{\pi}{2}$ where ${\rm{E}}_{2n}\,,\; n=0,1,2,\ldots$ are the even-indexed Euler numbers. Prove for $n=0^{(*)},1,2,\ldots\,,$ that ${\rm{E}}_{2n}=(-1)^n(2n)!\left|{\begin{array}{ccccccc} \frac{1}{2!} & 1 & 0 & 0 & \cdots & 0 & 0\\ \\ \frac{1}{4!} & \frac{1}{2!} & 1 & 0 & \cdots & 0 & 0\\ \\ \frac{1}{6!} & \frac{1}{4!} & \frac{1}{2!} & 1 & \cdots & 0 & 0\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ \\ \frac{1}{(2n-2)!} & \frac{1}{(2n-4)!} & \frac{1}{(2n-6)!} & \frac{1}{(2n-8)!} & \cdots & \frac{1}{2!} & 1\\\\ \frac{1}{(2n)!} &\frac{1}{(2n-2)!} & \frac{1}{(2n-4)!} & \frac{1}{(2n-6)!} & \cdots & \frac{1}{4!} & \frac{1}{2!}\end{array}}\right|\,.$

Note: The determinant of the $0\times0$-matrix is equal to $1$.

HINT

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Grigorios Kostakos

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 Post subject: Re: Euler numbers as determinantsPosted: Sat Jun 25, 2016 6:35 am

Joined: Sat Nov 07, 2015 6:12 pm
Posts: 828
Location: Larisa
Hello Grigoris,

\begin{aligned} 1 &=\cos x \sec x \\ &= \sum_{n=0}^{\infty}\frac{(-1)^n{\rm E}_{2n}x^{2n}}{(2n)!}\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{\left ( 2n \right )!}\\ &= \sum_{n=0}^{\infty}\left ( \sum_{k=0}^{n}\frac{1}{\left ( 2n-2k \right )!}\frac{{\rm E}_{2k}}{(2k)!} \right )(-1)^n x^{2n}\\ &=\sum_{n=0}^{\infty}\left ( \sum_{k=0}^{n}\binom{2n}{2k}{\rm E}_{2k} \right )\frac{(-1)^n x^{2n}}{(2n)!} \\ \end{aligned}

Hence by equating coefficients we get that:

$${\rm E}_0=1, \;\; {\rm E}_{2n}=-\sum_{k=0}^{n-1}\binom{2n}{2k}{\rm E}_{2k}, \; n\geq 1$$

which is the recursive formula of the Euler even indexed numbers.

We , now prove by induction that the relation given and the recursive are equal.

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