Euler numbers as determinants
Posted: Sat Jun 25, 2016 6:33 am
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 \(1\).
Note: The determinant of the \(0\times0\)-matrix is equal to \(1\).
HINT