Euler and Bernoulli Polynomial Identity Proof Given that the Euler Polynomials $E_n(z)$ are defined in terms of the generating function
$$\frac{2e^{xz}}{e^x+1}=\sum_{n=0}^\infty E_n(z)\frac{x^n}{n!}$$
and that the Bernoulli Polynomials $B_n(z)$ are defined in terms of a similar generating function
$$\frac{xe^{xz}}{e^x-1}=\sum_{n=0}^\infty B_n(z)\frac{x^n}{n!}$$
I am trying to find proofs on line of the following identity between the two polynomials:
$$E_{n-1}(z)=\frac{2^n}{n}\left[B_n\left(\frac{z+1}{2}\right)-B_n\left(\frac{z}{2}\right)\right]$$
I am looking for multiple proofs if possible but am having trouble locating the specific proofs.  Are there any individuals that can help me find this particular proof(s)?
 A: Here we use for convenience only the following notation for the generating functions of Bernoulli polynomials and Euler polynomials 
\begin{align*}
  \sum_{n=0}^\infty E_n(x)\frac{t^n}{n!}&=\frac{2e^{xt}}{e^t+1}\\
  \sum_{n=0}^{\infty}B_n(x)\frac{t^n}{n!}&=\frac{te^{xt}}{e^t-1}
  \end{align*}

We show the following is valid
  \begin{align*}
  E_{n-1}(x)=\frac{2^n}{n}\left[B_n\left(\frac{x+1}{2}\right)-B_n\left(\frac{x}{2}\right)\right]\qquad\qquad n\geq 1\tag{1}
  \end{align*}

$$ $$

We multiply both sides of (1) with $\frac{n}{2^n}$ and consider the generating function for the LHS
\begin{align*}
\sum_{n=1}^{\infty}&E_{n-1}(x)\frac{n}{2^n}\frac{t^n}{n!}\\
&=\frac{t}{2}\sum_{n=1}^{\infty}E_{n-1}(x)\frac{t^{n-1}}{2^{n-1}(n-1)!}\tag{2}\\
&=\frac{t}{2}\sum_{n=0}^{\infty}E_{n}(x)\frac{t^{n}}{2^{n}n!}\tag{3}\\
&=\frac{t}{2}\frac{2e^{\frac{1}{2}xt}}{e^{\frac{1}{2}t}+1}\\
&=\frac{te^{\frac{1}{2}xt}}{e^{\frac{1}{2}t}+1}\\
  \end{align*}
We obtain as generating function for the RHS of (1)
\begin{align*}
  \sum_{n=1}^{\infty}&\left[B_n\left(\frac{x+1}{2}\right)-B_n\left(\frac{x}{2}\right)\right]\frac{t^n}{n!}\\
  &=  \sum_{n=1}^{\infty}B_n\left(\frac{x+1}{2}\right)\frac{t^n}{n!}
  -\sum_{n=1}^{\infty}B_n\left(\frac{x}{2}\right)\frac{t^n}{n!}\\
&=\frac{te^{\frac{1}{2}(x+1)t}}{e^t-1}-\frac{te^{\frac{1}{2}xt}}{e^t-1}\tag{4}\\
  &=\frac{te^{\frac{1}{2}xt}}{e^t-1}\left(e^{\frac{t}{2}}-1\right)\\
  &=\frac{te^{\frac{1}{2}xt}}{e^{\frac{1}{2}t}+1}\tag{5}
  \end{align*}
and the claim follows.

Comment:


*

*In (2) we simplify and rearrange the sum

*In (3) we shift the index by one and observe that we get the generating function of Euler polynomials with argument $\frac{t}{2}$.

*In (4) we use the generating functions for the Bernoulli polynomials with arguments $\frac{x+1}{2}$ and $\frac{x}{2}$. We also use implicitly $B_0\left(\frac{x+1}{2}\right)=B_0\left(\frac{x}{2}\right)=1$.

*In (5) we use $e^t-1=\left(e^\frac{t}{2}+1\right)\left(e^\frac{t}{2}-1\right)$

Addendum:  According to OPs comment some links with related info
  
  
*
  
*In Eulerian Numbers and Polynomials by L. Carlitz we find near formula (5.4) the following generalisation for $m$ even
  \begin{align*}
m^n\sum_{s=0}^{m-1}(-1)^sB_n\left(x+\frac{s}{m}\right)=-\frac{n}{2}E_{n-1}(mx)
\end{align*}
  
*In Generating functions for generalized Stirling type numbers, Array type polynomials, Eulerian type polynomials and their applications by Yilmaz Simsek the same formula in (29) is provided.
  
*In Binomial Identities (vol.8) by H.W. Gould we find the formula given as (3.3).
  
*In The Umbral Calculus by S. Roman the formula is stated besides many other related formulas in section 2.5 Appell sequences.

Hint: Besides these references I suppose practically each section in Wiki's Bernoulli polynomials and Euler polynomials could be used as starting point for a proof of the stated formula.
