Open Book Publishers logo Open Access logo
  • button
  • button
  • button
GO TO...
book cover

Problem 46:  Bernoulli polynomials ( ) 1987 Paper III

The Bernoulli polynomials, Bn(x) (where n = 0,1,2,), are defined by B0(x) = 1 and, for n 1,

dBn dx = nBn1(x) (1)


01B n(x)dx = 0. (2)
Show that B4(x) = x2(x 1)2 + c, where c is a constant (which you need not evaluate).
Show that, for n 2, Bn(1) Bn(0) = 0.
Show, by induction or otherwise, that
Bn(x + 1) Bn(x) = nxn1(n 1). (3)
Hence show that n m=0kmn1 = B n(k + 1) Bn(0),

and deduce that m=01000m3 = (500500)2.


The Swiss family Bernoulli included no fewer than eight mathematicians who were counted amongst the leading scholars of their day. They made major contributions to all branches of mathematics, especially differential calculus. There was great rivalry between some members of the family; between brothers Jakob (1654–1705) and Johann (1667–1748), in particular.

Johann once published an important result in the form of a Latin anagram, in order to retain the priority of discovery without giving the game away to his brother. The anagram was: 24a, 6b, 6c, 8d, 33e, 5f, 2g, 4h, 33i, 6l, 21m, 26n, 16o, 8p, 5q, 17r, 16s, 25t, 32u, 4x, 3y, +, , –, ±, =, 4, 2, 1, . The notation means that his important result contained, for instance, the letter a 24 times either in text or in equations. After waiting for a year for someone to solve it, Bernoulli weakened and published the solution himself. If you are trying to solve the anagram yourself, you might like to know that it is about the Riccati equation y = ay2 + bxn, which can be solved (very cunningly, as it turns out) when n is of the form 4m(2m ± 1) for any positive integer m. (Newton also published some work in the form of anagrams, during his conflict with Leibniz).

The polynomials described above were discovered by Jakob Bernoulli. They are defined recursively; that is to say, the zeroth polynomial is given an explicit value, and the nth is determined from the (n 1)th. Here, Bn1 has to be integrated to obtain Bn, which means that Bn is a polynomial of degree n. The constant of integration is determined by the condition (2), so Bn is uniquely determined. We have to do this explicitly for part (i).

Solution to problem 46

(i) First we find B1(x) by integrating 1 ×B0(x), using equation (1): B1(x) = x + k, where k is a constant. We find k by applying the condition 01B1(x)dx = 0, which gives k = 1 2. Next we find B2(x) by similar means, giving x2 x + 1 6, and similarly B3(x) = x3 3 2x2 + 1 2x and B4(x) is as given.

(ii) We are asked to prove a result involving Bn(x) evaluated at x = 1 and x = 0, i.e. at the limits of the integral (2). We therefore try the effect of integrating both sides of equation (1) between these limits:

Bn(1) Bn(0) 01dBn(x) dx dx = n01B n1(x)dx = 0,

using property (2) with n replaced by (n 1).

(iii) First the easy bit of the induction proof. For n = 1, we have B1(x) = x 1 2, so

B1(x + 1) B1(x) = (x + 1 1 2) (x 1 2) = 1 nxn1,

so the formula holds.

Now suppose that it holds for n = k:

Bk(x + 1) Bk(x) kxk1 = 0 (4)

and investigate

Bk+1(x + 1) Bk+1(x) (k + 1)xk, (5)

which we hope will also equal zero.

The only helpful thing we know about Bernoulli polynomials involves the derivatives. Therefore, let us see what happens when we differentiate the expression (5):

d dxBk+1(x + 1) d dxBk+1(x) (k + 1)kxk1.

Now using (1) gives

(k + 1)Bk(x + 1) (k + 1)Bk(x) (k + 1)kxk1.

Note that we have used the chain rule to differentiate Bk+1(x + 1) with respect to x rather than with respect to (x + 1). Note also that there is a pleasing overall factor of (k + 1), which suggests that we are on the right track. In fact, taking out this factor gives exactly the left hand side of equation (4), which is zero.

Of course, we are not finished yet: we have only shown that the derivative of equation (5) is equal to zero; the expression (5) is therefore constant:

Bk+1(x + 1) Bk+1(x) (k + 1)xk = A.

We must show that A = 0. Setting x = 0 gives Bk+1(1) Bk+1(0) = A, which implies that A = 0 by part (ii).

(iv) Summing (3) from x = 0 to x = k gives the first of these results immediately because nearly all the terms cancel in pairs. The evaluation of the sum follows by calculating B4(1001) B4(0) from the result given in part (i).