I am working on a research project on Bernoulli polynomials, which are defined as $$B_n(x) = \sum_{j} \binom{n}{j}B_j x^{n-j}$$ where $B_j$ are the Bernoulli numbers. There is also the "generating function" $$\frac{ze^{zx}}{e^z-1} = \sum_n B_n(x)\frac{z^n}{n!}$$ Many papers I have found cite the following property and use it extensively in their proofs: $$\Delta B_n(x) = n x^{n-1}$$ where $\Delta$ is the forward-difference operator from discrete calculus. However, this fact is always stated without proof, and I am wondering what the justification is.


4 Answers 4


One way is to use the generating function: \begin{align} \sum_n \Delta B_n(x) \frac{z^n}{n!} &= \sum_n(B_n(x+1) - B_n(x)) \frac{z^n}{n!} \\ &= \frac{z e^{z(x+1)}}{e^z-1} - \frac{z e^{zx}}{e^z-1} \\ &= z e^{zx} \\ &= \sum_m x^m \frac{z^{m+1}}{m!} \\ &= \sum_n x^{n-1} \frac{z^n}{(n-1)!} \end{align} Then compare coefficients.

  • $\begingroup$ Nice! I wonder why many authors choose to omit this space-efficient calculation. $\endgroup$
    – Joe Bob
    Dec 14, 2020 at 0:07

You can think of the space of polynomials as the vector space generated by the monomials $1$, $x$, $x^2$, $x^3$, $\ldots$. The Bernoulli polynomials are a different basis for the same vector space and can be defined by the property that the linear map $L$ that transforms each monomial into the corresponding Bernoulli polynomial, $$ Lx^n=B_n(x), $$ transforms the forward difference of a polynomial into the derivative of that polynomial: $$ L\Delta f(x)=\frac{d}{dx}f(x). $$ (Writing down the system of linear equations relating the forward difference of $f$ to the derivative of $f$ and then solving that system is one way to obtain $L$ and hence the Bernoulli polynomials. Your summation formula for the Bernoulli polynomials and the recurrence for the Bernoulli numbers, $\sum_{k=0}^n\binom{n+1}{k}B_k=\delta_{n,0}$, naturally emerge from the solution.)

Now the property you want to prove has $L$ and $\Delta$ in the opposite order: you want to prove $$ \Delta B_n(x)=\Delta L x^n=\frac{d}{dx} x^n $$ or, more generally, $\Delta L f(x)=\frac{d}{dx}f(x)$.

In fact, the three operators $L$, $\Delta$, and $\frac{d}{dx}$, which all act on the space of polynomials, are mutually commuting, $$ \frac{d}{dx}\Delta f(x)=\Delta\frac{d}{dx}f(x),\quad \frac{d}{dx}Lf(x)=L\frac{d}{dx}f(x),\quad \Delta Lf(x)=L\Delta f(x). $$ The first of these follows from the chain rule; the second can be checked by verifying from your definition that $B'_n(x)=nB_{n-1}(x)$, which is the rule in the case $f(x)=x^n$, and using linearity of $L$ and $\frac{d}{dx}$. The third one, which is the one you want to prove, then follows from the second: apply $L\Delta f(x)=\frac{d}{dx}f(x)$ to the polynomial $Lf(x)$ to obtain $L\Delta Lf(x)=\frac{d}{dx}Lf(x)=L\frac{d}{dx}f(x)$. Since $L$ is an invertible operator, it follows that $\Delta Lf(x)=\frac{d}{dx}f(x)$.

By the way, the operators $\frac{d}{dx}$ and $\Delta$ are not invertible, a fact that amounts to the observation that the antiderivative and the anti-forward-difference are not uniquely defined.

Added: This expands on the parenthetical remark in the first paragraph. Let $f(x)=\sum_{i=0}^n f_ix^i$ be a polynomial of degree $n$ and let $f'(x)=\sum_{i=0}^{n-1}g_ix^i$, $\Delta f(x)=\sum_{i=0}^{n-1}h_ix^i$. Comparing the first and last lines of \begin{align} \sum_{i=0}^{n-1}h_i x^i &= \Delta f(x)\\ &=f(x+1)-f(x)\\ &=\sum_{i=0}^nf_i((x+1)^i-x^i)\\ &=\sum_{i=1}^nf_i\sum_{j=0}^{i-1}\binom{i}{j}x^j\\ &=\sum_{i=1}^n if_i\sum_{j=0}^{i-1}\frac{1}{i}\binom{i}{j}x^j\\ &=\sum_{i=0}^{n-1} g_i\sum_{j=0}^{i}\frac{1}{i+1}\binom{i+1}{j}x^j \end{align} gives a system of linear equations relating the $h_i$ to the $g_i$. The coefficient matrix is upper triangular, so its inverse is found by straightforward iteration. Carrying out the inversion gives $$ \sum_{i=0}^{n-1}g_ix^i=\sum_{i=0}^{n-1}h_iB_i(x). $$ This is where the prescription "replace every power $x^i$ in $\Delta f(x)$ with $B_i(x)$ to obtain $f'(x)$" comes from. Taking this linear algebra problem as the definition of the Bernoulli polynomials allows a great many things to be easily explained, in my view.


To calculate

$$ \sum_{k=0}^{m-1} f(x+k) $$

one finds a function $g(x)$ such that

$$\Delta g(x) \colon = g(x+1) - g(x) = f(x)$$

and then

$$g(x+m) - g(x) = \sum_{k=0}^{m-1} f(x+k)$$

There is a formal procedure to find $g$ that provides a solution when $f$ is a polynomial. Say we want to solve

$$\Delta g(x) = f(x) = D h(x)$$

where $D$ is the usual derivative. Note that

$$\Delta g = (e^D-1) g$$

and so we have to solve

$$(e^D-1) g = D h$$ with a solution

$$g = \frac{D}{e^D-1} h$$

The Bernoulli polynomials are solutions of

$$B_n(x+1) - B_n(x) = n x^{n-1} = (x^n)'$$

so we take

$$B_n(x) = \frac{D}{e^D-1} x^n$$

Define the Bernoulli numbers to be the coefficients in the expansion

$$\frac{D}{e^D-1} = \sum_{k=0}^{\infty} \frac{B_k}{k!} D^k$$

and we get

$$B_n(x) = \sum_{k\ge 0} \frac{B_k}{k!} D^k(x^n)= \sum_{k=0 }^n\binom{n}{k} B_k x^{n-k}$$

From the above we conclude that

$$\frac{t}{e^t-1} \cdot e^{x t} = \sum_{n \ge 0} \frac{B_n(x)}{n!} t^n$$

$\bf{Added:}$ From $B_n(x) = \frac{D}{e^D-1} x^n$ we conclude that

$$B_n'(x) = n B_{n-1}(x)$$ for $n \ge 0$


Perhaps the most useful general proof is to show that the alleged identity actually implies the e.g.f. for the Appell-Sheffer sequence of Bernoulli polynomials. An operational definition of the Bernoulli polynomials is the umbral compositional/substitutional action on functions $f$ analytic at $x$ as

$$f(B.(x+1)) - f(B.(x)) = f'(x) = \partial_x f(x),$$


$$(B.(x))^n := (b.+x)^n := \sum_{k=0}^n \binom{n}{k} x^{n-k}(b.)^k = \sum_{k=0}^n \binom{n}{k} x^{n-k}b_k = B_n(x).$$

With $f(x) = x^n$, we have the alleged specific identity

$$(B.(x+1))^n - (B.(x))^n = B_n(x+1) - B_n(x) = (e^{\partial_x}-1)B_n(x) = \partial_x x^n = nx^{n-1}.$$

Then, with $f(x) = e^{xt}$, we have the identity ($\dagger$)

$$e^{B.(x+1)t} - e^{B.(x)t} = te^{xt}$$

$$ = e^{(b.+x+1)} - e^{(b.+x)t} = e^{(b.+x)t}(e^t-1) = e^{B.(x)t}(e^t-1),$$

so the e.g.f. for the Bernoulli polynomials is

$$e^{B.(x)t} = \frac{t}{e^t-1}e^{xt} = B(t) e^{xt}$$

with the e.g.f. for the Bernoulli numbers

$$e^{B.(0)t} = e^{b.t} = B(t) = \frac{t}{e^t-1}.$$

This all works in the reverse direction as well, and evaluating the action of $\partial_{t=0}^n$ on $(\dagger)$ gives the alleged specific finite difference identity, so the operational definition and the e.g.f. are equivalent definitions of the Bernoulli polynomials.

Note then the operational relations connected to finite differences (this comports with the umbral, finite operator calculus of Appell Sheffer polynomial sequences);

$$B(\partial_x)x^n = \frac{\partial_x}{e^{\partial_x}-1}x^n = e^{b.\partial_x}x^n = (b.+x)^n = B_n(x),$$


$$\partial_x x^n = \frac{\partial_x}{e^{\partial_x}-1}(e^{\partial_x}-1) x^n$$

$$ = \frac{\partial_x}{e^{\partial_x}-1}[(x+1)^n-x^n] = e^{b.\partial_x}[(x+1)^n-x^n]$$

$$ = (x+1+b.)^n-(x+b.)^n = (B.(x+1))^n-(B.(x))^n = B_n(x+1) - B_n(x).$$

These results are easily extended by interpolating and extrapolating (several ways are available) the Bernoulli polynomials to the Bernoulli function (or simply by defining)

$$B_{-s}(z) = s\zeta(1+s,z),$$

where $s$ is complex and $\zeta(s,x)$ is the Hurwitz zeta function.


$$B_{s}(z+1) - B_{s}(z) = (e^{\partial_z} -1)B_s(z) = sz^{s-1}$$

since by analytic continuation

$$ B_{-s}(z+1) - B_{-s}(z) = s\zeta(s+1,x+1) - s\zeta(s+1,x)$$

$$ = s[\sum_{n\geq 0}\frac{1}{(n+x+1)^{s+1}}-\sum_{n\geq 0}\frac{1}{(n+x)^{s+1}}] = -sz^{-s-1},$$

and I've shown above that for $s = -n$ this defines the Bernoulli polynomials.

Extended Faulhaber formulas fall directly from this extension of the Bernoulli polynomials to the Bernoulli function.


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