It is well - known (and famous) that the identity \begin{align} E_{n - 1} \left( x \right) = \frac{2}{n}\left[ {B_n \left( x \right) - 2^n B_n \left( {\frac{1}{2}x} \right)} \right] \\ \end{align} holds for all $n \ge 1$ and $0 \le x \le 1$. Recall that $B_1=-1/2$, $ E_{2n - 1} \left( 0 \right) = B_{2n + 1} \left( 0 \right) = 0,\,\,\forall n \ge 1$ and $ E_{2n} \left( 0 \right) \ne0\ne B_{2n} \left( 0 \right) $, for all $n \ge 1$
Now, for $x = 0$, we have \begin{align} E_{n - 1} \left( 0 \right) = \frac{2}{n}\left( {1 - 2^n } \right)B_n \left( 0 \right) \end{align} Thus, \begin{align} E_1 \left( 0 \right) &= - 3B_2 \left( 0 \right),\,\,{\rm{but}}\,\,E_1 \left( 0 \right) = 0 \ne B_2 \left( 0 \right)?? \\ E_2 \left( 0 \right) &= - \frac{{\left( 2 \right)\left( 7 \right)}}{3}B_3 \left( 0 \right),\,\,{\rm{but}}\,\,E_2 \left( 0 \right) \ne 0 = B_3 \left( 0 \right)?? \\ &\vdots \end{align} How this come? Any explanation?
Edit:
It is amazing to note that the absolute term in even Euler polynomials are zero. for example $E_2(x)=x^2-x$; so how $E_2(0)\ne0$?.
However $E_3(x)=x^3-\frac{3}{2}x^2+\frac{1}{4}$; so how $E_3(0)=0$? .... (Big Question Mark)
I'm so confusing because on mathworld (http://mathworld.wolfram.com/EulerPolynomial.html)
The even numbers are zeros while the odd numbers are not (and this my thought)
However, in several tables book the wrote $E_{2n-1}=0$ for all $n=1,2,\cdots$ and $E_{2n}\ne0$ for example see
1- Table of Integrals, Series, and Products, by I.S. Gradshteyn and I.M. Ryzhik
page 1043.
2- Handbook of Mathematical Functions by Milton Abramowitz and Irene A. Stegun page 805.
and others?
It is OK. I just found the answer, the Euler numbers are simply not $E_n(0)$. It is only the value of $\frac{1}{2}$ multiplied by $2^n$. Explicitly, $E_n=2^nE_n(\frac{1}{2})$. I thought that these numbers like Bernoulli numbers so the misunderstanding comes from this point. No one in his book say $E_n=E_n(0)$.
Sorry for confusing ...