Prove a Bernoulli polynomial equation: $B_s(x+y)=\sum_{j=0}^s \binom{s}{j} B_{j}(x)\cdot y^{s-j}$ I want to show the equation for Bernoulli polynomials $B_s(x+y)$:
$$B_s(x+y)=\sum_{j=0}^s \binom{s}{j} B_{j}(x)\cdot y^{s-j}$$
I begin with:
$$B_s(z)=\sum_{j=0}^s \binom{s}{j} B_{s-j} \cdot z^j$$
Replace $z$ with $x+y$
$$B_s(x+y)=\sum_{j=0}^s \binom{s}{j} B_{s-j} \cdot(x+y)^j$$
where $B_{s-j}$ is Bernoulli coefficient.
Next, I do the binomial expansion for $(x+y)^j=\sum_{k=0}^j \binom{j}{k} x^k \cdot y^{j-k}$
So we got:
$$B_s(x+y)=\sum_{j=0}^s \binom{s}{j} B_{s-j} \cdot \sum_{k=0}^j \binom{j}{k} x^k \cdot y^{j-k}$$
I guess next step is to do the variable transformation from $j,k$ to some other index, how should I define the variable transformation?
 A: The EGF of Bernoulli polynomials is
$$\frac{t\exp(xt)}{\exp(t)-1} =
\sum_{n\ge 0} B_n(x) \frac{t^n}{n!}.$$
We seek to show that
$$B_s(x+y) = \sum_{j=0}^s {s\choose j} B_j(x) y^{s-j}
= \sum_{j=0}^s {s\choose j} B_{s-j}(x) y^j.$$
The RHS is
$$\sum_{j=0}^s {s\choose j} y^j
(s-j)! [t^{s-j}]
\frac{t\exp(xt)}{\exp(t)-1}
\\ = s! [t^s]
\frac{t\exp(xt)}{\exp(t)-1}
\sum_{j\ge 0} \frac{1}{j!} y^j t^j.$$
Now here we have extended $j$ to infinity because the coefficient
extractor in $t$ enforces the upper limit through $t^j.$ Continuing,
$$s! [t^s]
\frac{t\exp(xt)}{\exp(t)-1} \exp(yt)
= s! [t^s] \frac{t\exp((x+y)t)}{\exp(t)-1}
= B_s(x+y).$$
This is the claim.
Remark. We can also start from
$$B_s(x+y) = \sum_{j=0}^s {s\choose j} B_{s-j}
\sum_{k=0}^j {j\choose k} x^{j-k} y^k
\\ = \sum_{k=0}^s x^{-k} y^k
\sum_{j=k}^s B_{s-j} {s\choose j} {j\choose k} x^j.$$
We have
$${s\choose j} {j\choose k}
= \frac{s!}{(s-j)! \times k! \times (j-k)!}
= {s\choose k} {s-k\choose j-k}$$
and obtain
$$\sum_{k=0}^s {s\choose k} x^{-k} y^k
\sum_{j=k}^s B_{s-j} {s-k\choose j-k} x^j
\\ = \sum_{k=0}^s {s\choose k} y^k
\sum_{j=0}^{s-k} B_{s-k-j} {s-k\choose j} x^j
\\ = \sum_{k=0}^s {s\choose k} y^k B_{s-k}(x)
= \sum_{k=0}^s {s\choose k} B_{k}(x) y^{s-k}.$$
Again we have the claim.
A: This method uses the Umbral calculus.
Define the linear map
$$ L_B( B^n ) := B_n $$
where $B_n$ is the Bernoulli numbers. The key result is
$$ B_n(z) = L_B( (B+z)^n ) = L_B\left(\sum_{k=0}^n {n\choose k} B^k z^{n-k} \right) = \sum_{k=0}^n {n\choose k} B_k z^{n-k}.$$
Apply this result to your case with
$$ B_n(x+y) = L_B( (B+x+y)^n ) = L_B(((B+x)+y)^n) =\\
L_B\left(\sum_{k=0}^n {n\choose k} (B+x)^k y^{n-k} \right) = \sum_{k=0}^n {n\choose k} B_k(x) y^{n-k}. $$
