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Prove that: $${\rm{E}}_{2n}=(-1)^n(2n)!\left|{\begin{array}{ccccccc} \frac{1}{2!} & 1 & 0 & 0 & \cdots & 0 & 0\\ \\ \frac{1}{4!} & \frac{1}{2!} & 1 & 0 & \cdots & 0 & 0\\ \\ \frac{1}{6!} & \frac{1}{4!} & \frac{1}{2!} & 1 & \cdots & 0 & 0\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ \\ \frac{1}{(2n-2)!} & \frac{1}{(2n-4)!} & \frac{1}{(2n-6)!} & \frac{1}{(2n-8)!} & \cdots & \frac{1}{2!} & 1\\\\ \frac{1}{(2n)!} &\frac{1}{(2n-2)!} & \frac{1}{(2n-4)!} & \frac{1}{(2n-6)!} & \cdots & \frac{1}{4!} & \frac{1}{2!}\end{array}}\right|$$

where ${\rm E}_{2n}$ are the Euler numbers with even index. (given also http://en.wikipedia.org/wiki/Euler_number )

I started the problem using the fact that: $$\sec x =\sum_{n=0}^{\infty}\frac{(-1)^n {\rm E}_{2n}x^{2n}}{(2n)!}, \;\;\; \left | x \right |<\frac{\pi}{2}$$

Hence: $$\sec x \cdot \cos x =1 \implies \sum_{n=0}^{\infty}\frac{(-1)^n{\rm E}_{2n}x^{2n}}{\left ( 2n \right )!}\cdot \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}=1$$

Now I have to count , closely, in order to get what I want. Now I got stuck.. How can I proceed further? I cannot seem to combine the final result to get I want.

Thank you.

Tolaso
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1 Answers1

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By expanding these determinants with the Laplace rule, you can simply check that they satisfy the same recursion satisfied by the Euler numbers, and that the identity holds for $n=0$ and $n=1$.

See also this other question (line $(2)$ in robjohn's answer).

Community
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Jack D'Aurizio
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  • I got to (if I did the calculations right) : $$\sum_{n=0}^{\infty}\left ( \sum_{k=0}^{n}\binom{2n}{2k}{\rm E}_{2n} \right )\frac{(-1)^n x^{2n}}{(2n)!}$$

    Equating coefficients I get that ${\rm E}0=1$ and $\displaystyle {\rm E}{2n}=-\sum_{k=0}^{n-1}\binom{2n}{2k}{\rm E}_{2k} $. How can I proceed to get that det?

     – Tolaso Jan 10 '15 at 12:13
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    You mean expanding the det and find the recursive formula of the det? – Tolaso Jan 10 '15 at 12:27
  • @Tolaso: yes, exactly. Expand it with respect to the last column. – Jack D'Aurizio Jan 10 '15 at 12:29
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    Hmm... I'm not that good at doing that, but I'll give it a try... there. – Tolaso Jan 10 '15 at 12:38
  • I tried expanding the det but unsuccessfully. What I have also tried was to use induction... but I get stuck. Could you show me the way for the det? – Tolaso Feb 12 '15 at 03:26