The Bernoulli Numbers

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Mathematicians define sets of numbers through the criteria by which they include numbers in the set and/or exclude numbers from the set. As an example, consider the set of the square numbers: we include a number in that set if and only if we can produce it by multiplying one of the natural numbers by itself. By that criterion we know that we can include four (2x2) in the set but not five; we can include thirty-six (6x6) but not forty-eight (6x8); and so on. If we explore that set, we will discover further that we can include a number in the set if it equals the sum of all of the odd positive integers less than 2N (for any positive integer N), that sum equaling N2. Part of our motive for studying this particular set comes from our awareness of the fact that square numbers give us a measure of areas and, through the Pythagorean theorem, distances on a flat plane. In essence we use the set of the square numbers to make a connection between arithmetic and geometry. That connection, in turn, lies at the heart of modern mathematical physics.

In the case of the Bernoulli numbers, we don’t have an easy way to define the set or a clear motive for considering them as a set. That observation may mean that the Bernoulli numbers constitute a perfectly trivial and meaningless set. Or it may bespeak a profundity that obliges us to look further into the deep structure of mathematics. Not knowing a priori which case stands true to mathematics, we can only explore the properties of the set and see what we find. In this essay I will simply lay out the fundamentals of the set of the Bernoulli numbers.

As the elements of a distinct set, the Bernoulli numbers first emerged when mathematicians sought formulae for the Gaussian sums of the various integer powers of the natural numbers. The sum of the squares of the positive integers from one to N; the sum of the cubes of the positive integers from one to N; the sum of the fourth powers of the positive integers from one to N; and so on – each of those sums has its own formula. Discovery of the set began in 1631 when Johann Faulhaber (1580 May 05– 1635 Sep 10) published "Academia Algebra" and laid out the beginnings of the general equation,

(Eq’n 1)

in which we have the binomial coefficients

(Eq’n 2)

and the Bernoulli numbers Bk. Faulhaber didn’t actually devise the general equation for Σim himself; but rather he devised formulae for various values of m up to m=23. In "Ars Conjectandi", published posthumously in 1713, Jakob Bernoulli (1654 Dec 27 – 1705 Aug 16) referred to Faulhaber and his work and then extended that work. In that work Bernoulli described an equation that, when solved in an iterated procedure, yields the Bernoulli numbers. That equation takes the form, in modern notation,

(Eq’n 3)

The first five of those equations come out as

(Eq’ns 4)

Solving those equations and subsequent equations in the series, one after another, yields the sequence of the Bernoulli numbers, one by one. If we prime the pump by setting B0=1, then we get the first twenty-four Bernoulli numbers as

Note that, except for B1, all of the odd-indexed Bernoulli numbers equal zero.

We can derive Equation 3 from Equation 1 through a simple procedure. Set N=1 and multiply the resulting equation by m+1 to get

(Eq’n 5)

Setting B0=1 and solving that equation for subsequent values of m yields the same Bernoulli numbers above except that B1=+½. If we then set B1=-½, then the resulting right-hand side of Equation5 comes out m+1 smaller than the unmodified version, so we must subtract m+1 from the left-hand side, thereby getting Equation 3.

In addition to the Bernoulli numbers, we have another set – the set of the Bernoulli polynomials. To define the Bernoulli polynomials we have the equation

(Eq’n 6)

For the first seven elements of the set we have

(Eq’ns 7)

Compare the last five of those equations with Equations 4. If we make x=1, we can see, through that comparison, that Bn(1)=Bn. And, trivially, we also have Bn(0)=Bn.

Next I want to devise a generating function for the Bernoulli numbers and for the Bernoulli polynomials; specifically, I want to derive an equation that relates a closed-form formula to an infinite series containing the target entity in each term,

(Eq’n 8)

Applying to that equation an m-fold partial differentiation with respect to t followed by setting t=0 yields

(Eq’n 9)

To determine the algebraic form of f(x,t) begin by differentiating Equation 8 with respect to x to get

(Eq’n 10)

To understand the second step in that derivation, see the Appendix. The last step looks a bit fishy, but the differentiation merely adds a term to the infinite series. That term involves B-1(x), which doesn’t exist, so it adds nothing to the series in Equation 8. We know that whenever partial differentiation of some function with respect to some variable equals the same function multiplied by a constant or by some other variable, the function takes the form of an exponential. In the present case we have

(Eq’n 11)

in which T(t) represents a function of only t, which function responds to the operator ∂/∂x as a constant.

Next integrate Equation 11 with respect to x between the limits of x=0 and x=1. That integration yields

(Eq’n 12)

But we also have, via Equation 8,

(Eq’n 13)

Only the n=0 term exists in that integration because zero raised to the zeroth power equals one by definition, all other powers of zero equaling zero. Combining Equations 12 and 13 thus yields

(Eq’n 14)

which we can substitute into Equation 11 to get the generating function of the Bernoulli polynomials as

(Eq’n 15)

We have already determined that Bn(1)=Bn(0)=Bn, so the two forms of the generating function for the set of the Bernoulli numbers follows easily from Equation 15. For x=0 we get

(Eq’n 16)

and for x=1 we get

(Eq’n 17)

To extend that analysis we extend the variable t onto the complex plane by making the substitution z=t+iw. In making that substitution we merely pretend that the imaginary component remained invisible until now, so that we could work out our preliminary analysis on the real number line. Now that we have acknowledged the fact that our functions exist on the complex plane, we can bring the calculus of residues into play. To that end substitute the generating function of Equation 16 into Equation 9 and then replace the multiple differentiation with the equivalent loop integral in the complex plane. That move yields

(Eq’n 18)

In that integral the circular contour C, traversed in the counterclockwise direction, encloses the origin of the complex plane and has a radius less than 2π in order to avoid the poles at w=±2π. The contour encloses the singularity at z=0, which holds an (m+1)-order pole. The complex plane also holds (m+1)-order poles at each of the points z=±j2πi, in which j represents a natural number. Rather than try to evaluate the integral at z=0, I will exploit Cauchy’s integral theorem to come up with a simpler calculation.

First I reform the contour of the integration in Equation 18. Where the circle crosses the positive t-axis I break the circle and make a gap a minuscule distance wide. From the endpoints of the gap I extend two straight lines parallel to the t-axis and toward positive infinity. The far ends of the lines connect to the endpoints of a gap in a second circle, which circle has a pseudo-infinite radius and lies centered on the origin of the complex plane.

Integration of the function around the larger circle, carried out in the clockwise direction, must yield zero, in accordance with Cauchy’s integral theorem, as the radius goes to infinity. That fact necessitates that all of the function’s residues within the circle add up to zero. The integrations of the function along the straight lines, evaluated in opposite directions, cancel each other out of the calculation. And integration of the function around the smaller circle isolates the singularity at z=0 and extracts its residue. That residue equals the negative of the sum of the residues of the singularities that lie outside the smaller circle,

(Eq’n 19)

That equation lets me rewrite Equation 18 as

(Eq’n 20)

At any point z=j2πi where our function has a singularity we have a contour z’=j2πi+reiφ which encircles the singularity and enables us to carry out the loop integration that yields the singularity’s residue. The radius r of the contour takes a minuscule value that goes to zero as we take the relevant limit. Thus we have the integral in Equation 20 as

(Eq’n 21)

Expanding the exponential into its infinite series representation and then letting r go to zero transforms that integral into

(Eq’n 22)

That result makes Equation 20 take the form

(Eq’n 23)

If m represents an odd number, then the second term in the square bracket on the right side of that equation cancels the first term, so we have Bm=0 for m=3,5,7,etc. If m represents an even number, then the second term adds to the first term an amount equal to the first term, so we have in that case

(Eq’n 24)

in which ζ(m) represents the Riemann zeta function.

Now I need to make a comment on the above derivation; to wit, it doesn’t apply to the cases m=0 and m=1. To derive values for B0 and B1 we must evaluate the integral of Equation 18 directly around the point z=0.

For the case m=0 we have

(Eq’n 25)

Setting z’=0+reiφ gives us dz’=ireiφdφ=iz’dφ. Expanding the exponential into its infinite-series form and taking the limit of the integrand as r approaches zero then yields

(Eq’n 26)

For m=1 we have

(Eq’n 27)

To set up the integration we first expand the exponential in the generating function,

(Eq’n 28)

in which

(Eq’n 29)

Next we apply a binomial expansion with an exponent of minus one to the last expression in Equation 28. We get

(Eq’n 30)

Equation 27 then becomes

(Eq’n 31)

In the last step I exploited the fact that integration of the exponential around a circle equals a precise zero, leaving only th non-exponential term to contribute to the generating function’s residue.

Thus we have the fundamental mathematical facts concerning the set of the Bernoulli numbers. With those facts we can conduct further explorations of patterns among the elements of the set. Such explorations properly belong in other essays.

Appendix:

The Calculus of the Bernoulli Polynomials

Because we have the Bernoulli polynomials as smooth, continuous functions of a continuous variable, we expect that we can differentiate them or integrate them with respect to that variable. In order to work out the form of the derivatives and integrals of the Bernoulli polynomials we start with the definition of the n-th Bernoulli polynomial,

(Eq’n A-1)

in which Bk represents the k-th Bernoulli number and

(Eq’n A-2)

represents the binomial coefficients.

Differentiating Equation A-1 with respect to x seems straightforward enough; we get

(Eq’n A-3)

But now we want to simplify that result. We can see that

(Eq’n A-4)

so we have

(Eq’n A-5)

If you look at Equation 7 in the above essay, you will see that the differentiation shifts the range of k from (0 to n) to (-1 to n-1). Because B-1 does not exist, we have

(Eq’n A-6)

Integration of Equation A-1 simply reverses the differentiation, except for the fact that in the integral we have to specify limits between which the integral takes its value. Thus we have

(Eq’n A-7)

With that equation and the one above it we can explore the Bernoulli numbers, as I’ve shown in the essay above.

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