Visual Calculus, Power Sums and Bernoulli Coefficients

Intro

My goal here is to motivate a natural construction of the general formula for sums of powers of integers through a geometrical understanding of derivation and integration of power functions. The core insight then lies in peeling an hypercube with the finite difference operator and decomposing layers in blocks of volumes of lower powers. Transitioning from that elementary induction formula to a general polynomial expression for the $n$-th power sum leads to the introduction of a well known family of coefficients: the Bernoulli numbers..

In the following sections $f$ denotes the n-th power function : $x\mapsto x^n$.

Differentiation

---

(Differentiation of Power Functions). $\forall x \in \mathbb R$, $f'(x)=n\cdot x^{n-1}$.

---

Proof/Geometrical Intuition. The rate of growth of an hypercube with respect to the length of one of its edges equals half the volume of its outer shell, i.e the volume of one of its walls times its number of walls, divided by 2. Consider the $n$-cube cornered at the origin so that half of its walls are leaning against our frame (see fig.1 for $n$=3), the derivative of $f$ corresponds to the rate at which our hypervolume grows/shrinks as $x$ grows/diminishes. A look at figure 1.1 should help one gain confidence in the fact that this rate is indeed equal to half the number of walls of the hypercube times the current volume of one of these walls. There are $2n$ walls ¹, each having an area of $x^{n-1}$.

With a bit more algebra :

The coloured volume is :

In particular :

And:

Integration

---

(Integration of Power Functions). $\forall x \in \mathbb R$, $\int_{0}^{x}f(t)dt=\frac{x^{n+1}}{n+1}$.

---

We use similar insights and a similar setting to get a grasp of the formula for a primitive of the n-th power function. For simplicity we consider the integral of $f$ to be from $0$ to $x$.

Proof/Geometrical Intuition. The integral of the $n$-th power is equal to the volume of the $(n+1)$-cube divided by half of its number of walls. To see this, look at what happens when we integrate our $n$-cubes : we start with a point (the $n$-cube of side-length $0$) at the origin of our frame and keep piling up $n$-cubes until we’ve reached one of the $n+1$ walls of the $(n+1)$-cube that don’t have our starting point as a vertex (figure 1.2). There are $n+1$ ways to choose the destination wall, hence the division by $n+1$ in the formula.

Power Sums

Let $S_m(n)$ stand for the power sum of power $m$ :

The formulas for the sums of low order are the following well known identities :

• $S_1(n)=1 + 2 + \ldots + (n-1)=\frac{n (n-1)}{2}$

• $S_2(n)=1^2 + 2^2 + \ldots + (n-1)^2=\frac{n (n-1) (2n+1)}{6}$

• $S_3(n)=1^3 + 2^3 + \ldots + (n-1)^3=\left[1 + 2 + \ldots + (n-1)\right]^2=\frac{n^2(n-1)^2}{4}$

These $n$-th power sums are Riemann sums for the $(n+1)$-th power function with a subdivision of length $1$. Hence the expression of the sum of power $m$ being of dimension $m+1$. One way to come up with a general formula for $S_m (n)$ would then be to check the gap (in blue in fig. 3) between the continuous sum (the integral) and the discrete sum on each interval of the subdivision and then sum for the whole subdivision (see fig. 3).

---

(Induction Formula for Power Sums).

---

Looking back, the main geometrical insight behind this induction formula is the fact that peeling the $n$-cube layer by layer with the finite difference operator $\Delta[f]$ yields a collection of $\binom{n}{p}$ blocks of volume $p$. We can thus reach the same formula without any mention of integration by looking at the decomposition of a single layer:

And then putting back all the layers together to re-assemble the $n$-cube :

Which is the exact same formula :

---

(Decomposition of the Hypercube).

---

Bernoulli Numbers

Because of this decomposition of the $n$-cube in blocks of volume $p < n$ it is easy to see that $S_m(n)$ is always a polynomial of degree $m+1$ in $n$ with leading coefficient $\frac{1}{m+1}$ and constant term zero. Now we want the expression of all its coefficients in order to write it down as a function of $n$. Since these coefficients depend on $m$, we label the $k$-th coefficient for $S_m(n)$ as $A_{k-1}(m)$ (for the term in $n^{m-k}$).

That is to say :

($A_0(m)$ is the coefficient of the term of highest order in $S_m(n)$ (i.e the term in $n^{m+1}$), $A_1(m)$ the coefficient of the term of second highest order in $S_m(n)$ (i.e the term in $n^{m}$), etc.)

From (I) we can deduce the following induction relation for the $A_k$ :

If we look at the first few coefficients we find that :

As one works his way through the successive computations it becomes clear that every $A_i(m)$ can be written under the following form:

It is not difficult to prove that result by induction. We’ve already verified it for $A_0(m)$, $A_1(m)$ and $A_2(m)$ and if we suppose that this is true for all $i<k$ then :

Which proves the result and gives us the expression for $B_k$ :

These coefficients are the Bernoulli numbers :

---

(Bernoulli Numbers). The members of the integer sequence $\left(B_k\right)_{k\in\mathbb N}$ defined inductively by $B_0=1$ and $B_k=-\frac{1}{k+1}\sum_{i=0}^{k-1}\binom{k+1}{i}B_i$ are called the Bernoulli numbers.

---

We then have the following identity :

---

(Induction Formula for Power Sums).

---

Now that we’ve reached this expression of $S_m(n)$ as a polynomial in $n$ we can make sense of a broader continuous version $S_m(x)$ of $S_m(n)$. This leads to the introduction of the Bernoulli polynomials :

---

(Bernoulli Polynomials).

---

Which lead to the identity :

---

(Formula for Power Sums).

---