Prove a Bernoulli polynomial equation: $B_s(x+y)=\sum_{j=0}^s \binom{s}{j} B_{j}(x)\cdot y^{s-j}$

Question

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?

Answer 1

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.

Answer 2

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 $$ B_n(x+y) = L_B( (B+x+y)^n ) = L_B(((B+x)+y)^n) =\\ L_B\left(\sum_{k=0}^n {n\choose k} (B+x)^k y^{n-k} \right) = \sum_{k=0}^n {n\choose k} B_k(x) y^{n-k}. $$