A sequence of polynomials related to the evaluation of the Riemann zeta function Javier Duoandikoetxea Departamento de Matem´aticas Universidad del Pa´ıs Vasco/Euskal Herriko Unibertsitatea Apartado 644, 48080, Bilbao (Spain)
[email protected] The Riemann zeta function ζ(s) is defined for s > 1 as ∞ X 1 . ζ(s) = ns n=1
Euler was the first to evaluate the sum of the series for even values of s. Before 1740 he had obtained the values of ζ(2) and ζ(4), and in his Introductio in analysin infinitorum (1748) he explicitly gave the values for even s up to s = 26. Later he discovered the well-known formula valid for all even s in terms of the so-called Bernoulli numbers. (See [2] for an account of Euler’s work.) After Euler’s original work many other methods have been given to compute ζ(2k) for integer k. Several of them have in common the use of moments of trigonometric functions (definite integrals of xk cos nx or xk sin nx). We will modify that approach by defining a sequence of interpolating polynomials adapted in a natural way to the evaluation and express the sum ζ(2k) in terms of the values of the polynomials at a point. There is a recursive formula giving these values without using the polynomials themselves. Using the derivatives of these polynomials we obtain integral formulae for ζ(2k + 1). This is achieved using only elementary calculus: trigonometric identities, integration by parts and the differentiability of some particular functions. Writing the Fourier series expansion of our sequence of polynomials we will recognize them as being essentially the Euler polynomials.
1
The basic ingredients We will need two elementary trigonometric identities: 1 sin(N + 1/2)πx + cos πx + cos 2πx + . . . + cos N πx = , 2 2 sin πx/2 1 − cos 2N πx , sin πx + sin 3πx + . . . + sin(2N − 1)πx = 2 sin πx
(1) (2)
and the following lemma. Lemma 1. Let g be a function of class C 1 on [a, b]. Then Z b Z b lim g(x) sin λx dx = lim g(x) cos λx dx = 0. λ→∞
a
λ→∞
(3)
a
To prove the trigonometric identities multiply the left-hand side of (1) and (2) by the denominator of the right-hand side and use the trigonometric formulae for products of sines and cosines to get telescoping sums. To prove the lemma use integration by parts. The reader familiar with the theory of Fourier series will recognize in (1) the Dirichlet kernel and in Lemma 1 the simplest form of the RiemannLebesgue lemma. P −s The evaluations will give the value of ∞ n=0 (2n + 1) . This is enough to compute ζ(s) since we have ∞ ∞ ∞ X X X 1 1 1 1 = − = 1 − s ζ(s). (4) (2n + 1)s ns n=1 (2n)s 2 n=0 n=1
Evaluation of ζ(2k) An elementary way to evaluate ζ(2) is based on the following fact: the integral of x cos kπx on [0, 1] is −2(kπ)−2 for odd k ≥ 1 and is 0 for even k ≥ 2. Thus, multiplying both sides of (1) by x, integrating on [0, 1], passing to the limit, and using (3) and (4) we get ζ(2) = π 2 /6. This evaluation is essentially the same as the one given by E. L. Stark in [10], who used the second mean value theorem for integrals instead of (3). 2
Integrating x2k−1 cos(2n + 1)πx instead of x cos(2n + 1)πx lets us evaluate ζ(2k) recursively using the values of ζ(2l) for l < k. We will modify this approach and look for a polynomial of degree 2k − 1, denoted by p2k−1 , such that the integral of p2k−1 (x) cos(2n + 1)πx on [0, 1] reduces to just one term, a constant (in n) multiple of (2n + 1)−2k . Integration by parts shows the conditions required on p2k−1 ; we add the condition p2k−1 (0) = 0 to apply (3). Altogether we make the following definition. Definition 2. Let p1 (x) = x. For each integer k ≥ 2 define p2k−1 as the unique polynomial of degree 2k − 1 with leading coefficient 1 such that p2k−1 (0) = 0 and (2j−1)
(2j−1)
p2k−1 (0) = p2k−1 (1) = 0 for j = 1, 2, ..., k − 1. Define also p2k (x) =
(5)
p02k+1 (x) for k ≥ 1. 2k + 1
The polynomials pk are uniquely defined. Since p2k−1 has leading coefficient 1 and vanishes at 0, there are 2k − 2 coefficients to be determined. Half of them are 0 (those corresponding to odd powers of x) because the derivatives of odd order vanish at 0. The remainder can be obtained one by one from the condition (5) at x = 1, starting with j = k − 1. We will get an explicit formula later (equation (14)). By integration by parts, the conditions on the derivatives of p2k−1 give exactly Z 1 (−1)k (2k − 1)![1 − (−1)n ] p2k−1 (x) cos nπx dx = . (6) (nπ)2k 0 We are now ready to establish the first relationship between ζ(2k) and pk . Let N = 2M + 1. Multiply both sides of (1) by p2k−1 (x), integrate on [0, 1] and use (6) to get 1 2
Z
1
p2k−1 (x) dx + (−1) 0
M X
− 1)! π 2k
k (2k
n=0 1
Z =
0
3
2 (2n + 1)2k p2k−1 (x) sin(2M + 3/2)πx dx. 2 sin πx/2
Taking the limit as M goes to infinity and applying (3) with g(x) = we obtain
∞ X n=0
(−1)k+1 π 2k 1 = (2n + 1)2k 4(2k − 1)!
Z
p2k−1 (x) 2 sin πx/2
1
p2k−1 (x) dx;
(7)
0
and using (4) we conclude that (−1)k+1 22k−2 π 2k ζ(2k) = 2k (2 − 1)(2k − 1)!
Z
1
p2k−1 (x) dx.
(8)
0
Some properties of the polynomials pk The polynomial p2k−1 (x) satisfies p2k−1 (1) − p2k−1 (1 − x) = p2k−1 (x).
(9)
To see this, note that the left-hand side fulfills all the requirements of Definition 2 and the interpolating polynomial is unique. From (9) we deduce several useful properties. Z 1 1 Proposition 3. (i) p2k−1 (x) dx = p2k−1 (1). 2 0 (ii) p2k−1 (1/2) = p2k−1 (1)/2. (j)
(j)
(iii) p2k−1 (x) = (−1)j+1 p2k−1 (1 − x), j = 1, . . . , 2k − 1. (2j)
(iv) p2k−1 (1/2) = 0, j = 1, . . . , k − 1. (v) p2k−1 (2) = 2. For (i) integrate both sides of (9) from 0 to 1 and make the change of variables u = 1−x; (ii) and (iii) are easy consequences of (9); and (iv) follows from (iii). To prove (v) note that by Taylor’s theorem, p2k−1 (2) = p2k−1 (1) − p2k−1 (−1) =
2k−1 X j=0
4
(j)
p2k−1 (0) (1 − (−1)j ), j!
and all the terms in the sum are zero except the one corresponding to j = 2k − 1. Equations (iii) and (iv) also give information about the polynomials p2k ; for instance, p02k (1/2) = 0, a result that we will use later. Using (i) in (8) we can state the following theorem. Theorem 4. Let p2k−1 be the interpolating polynomial introduced in Definition 2 and denote Ak = p2k−1 (1). Then ζ(2k) =
(−1)k+1 22k−3 π 2k Ak . (22k − 1)(2k − 1)!
(10)
From the above computation of the coefficients of p2k−1 it should be clear that they are rational, so Ak is a rational number. As a consequence (10) shows that ζ(2k) is a rational multiple of π 2k .
Explicit calculations For k = 1, we have A1 = 1 and ζ(2) = π 2 /6. (This is the evaluation of ζ(2) mentioned above.) There are many proofs of this result: see, for instance, [5, Section 11.3] or the paper [7], and the references in both of them. For higher values of k we use induction. The polynomial p002k−1 (x) has leading coefficient (2k − 1)(2k − 2) and does not vanish at the origin; apart from these things it satisfies all the requirements of p2k−3 . Therefore p002k−1 (x) = (2k − 1)(2k − 2)(p2k−3 (x) − c)
(11)
for some constant c. Evaluating at x = 1/2 and using properties (iv) and (ii) in Proposition 3 we deduce c = p2k−1 (1/2) = Ak−1 /2. Bringing this value to (11) we obtain 1 p002k−1 (x) = (2k − 1)(2k − 2)(p2k−3 (x) − Ak−1 ), 2
(12)
which yields the following rule: to get p2k−1 from p2k−3 , integrate term by term twice, add −Ak−1 x2 /4 and normalize the leading coefficient. 5
From (12) and Proposition 3 we also obtain p002k−1 (0) = −p002k−1 (1) = −
(2k − 1)(2k − 2) Ak−1 , 2
and so by induction (2j)
(2j)
p2k−1 (0) = −p2k−1 (1) = −
(2k − 1)! Ak−j . 2(2k − 2j − 1)!
(13)
This gives an explicit formula for p2k−1 (x), namely p2k−1 (x) = x
2k−1
k−1 1 X 2k − 1 Aj x2k−2j . − 2 j=1 2k − 2j
(14)
Evaluating it at x = 1 gives a recurrence formula for Ak : k−1 1 X 2k − 1 Ak = 1 − Aj . 2 j=1 2k − 2j
(15)
Thus we can fill the following table. k
p2k−1
Ak
1
x
1
2 3 4 5
3 x3 − x2 2 5 5 x5 − x4 + x2 2 2 7 35 21 x7 − x6 + x4 − x2 2 4 2 9 153 2 x9 − x8 + 21x6 − 63x4 + x 2 2
−
1 2
1 −
17 4
31
ζ(2k) π2 6 π4 90 π6 945 π8 9450 π 10 93555
Other recurrence formulae for Ak can be found by evaluating (14) at x = 1/2 and x = 2 and using Proposition 3 to replace p2k−1 (1/2) and p2k−1 (2). More interesting for us is the formula obtained by differentiating (14) and
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evaluating at x = 1; doing this after replacing 2k − 1 by 2k + 1 in (14) we obtain k X 2k + 1 jAj = 2k + 1, (16) 2j j=1 which can be rewritten as k X 2k Aj = 2. 2j − 1 j=1
In search of the old formula The classical formula for ζ(2k) uses Bernoulli numbers of even order and is ζ(2k) =
(−1)k+1 22k−1 2k π B2k . (2k)!
(17)
Bernoulli numbers can be obtained from the Taylor expansion of x/(ex − 1), namely ∞ X x xn B = , n ex − 1 n=1 n! or can be computed using the recurrence relation k X 2k + 1 j=0
2j
B2j =
2k + 1 2
(k ≥ 1),
(18)
starting with B0 = 1. To check that (10) is the same as (17) we need to show that kAk = 2(22k − 1)B2k . (19) This will follow if we prove that both sides coincide for k = 1 and satisfy the same recurrence relation. Equality for k = 1 holds because B2 = 1/6; on the other hand, according to Lemma 2 in [1, p. 429] the Bernoulli numbers satisfy k X 2k + 1 2j 2 B2j = 2k + 1. (20) 2j j=0 7
Subtracting (18) from (20) and multiplying by 2 we obtain for 2(22j − 1)B2j the recurrence relation given by (16) for jAj . Apart from Euler’s method, there are many other ways of evaluating ζ(2k); see, for instance, [1], [3], [4], [6] and [9]. The proofs in [3] and [9] also use sequences of polynomials.
Integral forms of ζ(2k + 1) It is well-known that no explicit evaluations of ζ(2k + 1) have yet been obtained. Instead, there are equivalent expressions given by other series or definite integrals. Here we present a formula obtained using the sequence p2k . Take (6) with k + 1 instead of k and integrate by parts to deduce Z 1 (−1)k (2k)!2 . p2k (x) sin(2n + 1)πx dx = [(2n + 1)π]2k+1 0 Multiply both sides of (2) by p2k (x), integrate from 0 to 1 and take the limit as N goes to infinity to get ∞
(−1)k (2k)!2 X 1 = 2k+1 2k+1 π (2n + 1) n=0
1
Z 0
p2k (x) dx = 2 sin πx
Z 0
1/2
p2k (x) dx. sin πx
(21)
(The last equality is obtained by changing x into 1 − x in the integral from 1/2 to 1 and using p2k (x) = p2k (1 − x) from Proposition 3, (iv).) This yields the following theorem. Theorem 5. Let {pk } be the sequence of polynomials introduced in Definition 2. Then Z 1/2 (−1)k 22k π 2k+1 p2k (x) ζ(2k + 1) = dx (22k+1 − 1)(2k)! 0 sin πx Z 1/2 (−1)k 22k π 2k Ak πx = (p2k−1 (x) − ) log cot dx. 2k+1 (2 − 1)(2k − 1)! 0 2 2
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The second formula is obtained from the first one by integrating by parts and using (12), since πx π d log cot =− . dx 2 sin πx (For the evaluation at 0 remember that p2k (0) = 0 and log cot πx/2 ∼ log 1/x.) For k = 1 and k = 2 we get 2π 3 ζ(3) = 7
1/2
x − x2 2π 2 dx = sin πx 7
Z
x4 − 2x3 + x 2π 4 dx = sin πx 93
Z
Z 0
1/2
(1 − 2x) log cot 0
πx dx, 2
and 2π 5 ζ(5) = 93
Z 0
1/2
1/2
(4x3 − 6x2 + 1) log cot
0
πx dx. 2
Using Fourier series to recognize our polynomials For a function f defined in [0, 1], its Fourier cosine series is ∞
a0 X + aj cos jπx, 2 j=1 where Z aj = 2
1
f (x) cos jπx dx. 0
This corresponds to the Fourier series of the (unique) even function of period 2 whose restriction to [0, 1] is f . One of the simplest convergence theorems for Fourier series states that if f is continuous and piecewise differentiable, then the Fourier series of f converges (uniformly) to f . See, for instance, [11] for this and the results below on Fourier series. The coefficients of the Fourier cosine series of p2k−1 (x) are a0 = A k ,
a2n+2 = 0,
a2n+1 =
(−1)k 4(2k − 1)! (n = 0, 1, 2, . . .). ((2n + 1)π)2k
The value of a0 is given by Proposition 3 (i); the other coefficients are in (6).
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Using the convergence theorem we get the equality ∞
Ak (−1)k 4(2k − 1)! X cos(2n + 1)πx p2k−1 (x) = + , 2 π 2k (2n + 1)2k n=0
0 ≤ x ≤ 1.
(22)
Particular values of x in (22) provide the sum of the corresponding numerical series. For instance, choosing x = 0 (or x = 1) in (22) leads to the evaluation of ζ(2k). The derivative of the even extension of p2k+1 (x) to [−1, 1] is odd and coincides with (2k + 1)p2k (x) (see Definition 2) on [0, 1]. The corresponding periodic function of period 2 is continuous and has piecewise continuous derivative. From the relation between the Fourier coefficients of a function and those of its derivative, and from the convergence theorem for Fourier series we deduce ∞ (−1)k 4(2k)! X sin(2n + 1)πx , 0 ≤ x ≤ 1. (23) p2k (x) = π 2k+1 (2n + 1)2k+1 n=1 Again, particular values of x in (23) allow us evaluate numerical series. An interesting case is x = 1/2: this yields the value of the so-called Dirichlet P n −s beta function, β(s) = ∞ n=0 (−1) (2n + 1) , for odd s. Comparing (22) with the Fourier series expansion of the Euler polynomials we see that p2k−1 (x) − Ak /2 = E2k−1 . The sequence {En } of Euler polynomials is defined by ∞
X 2ext tn E (x) = . n 1 + et n! n=0 See [8, Chapter 20] for the definition and properties of the Euler polynomials including their Fourier series expansion. The property En0 = nEn−1 shows that E2k coincides with p2k . This suggests that Euler polynomials could be defined as interpolating polynomials using some variant of Definition 2. Indeed, it would be enough to replace the condition p2k−1 (0) = 0 (introduced because it was suited to our aim of using Lemma 3) with p2k−1 (0) = −p2k−1 (1) or, alternatively, with p2k−1 (1/2) = 0. 10
Acknowledgment The author is very grateful to both referees. Their comments and suggestions served to improve the first version of this paper, and moreover, one of them provided valuable additional information. He is also grateful to David Cruz-Uribe for his help with the formal aspects of the paper.
References [1] T. M. Apostol, Another elementary proof of Euler’s formula for ζ(2n), Amer. Math. Monthly 80 (1973), 425-431. [2] R. Ayoub, Euler and the zeta function, Amer. Math. Monthly 81 (1974), 1067-1086. [3] E. Balanzario, M´etodo elemental para la evaluaci´ on de la funci´ on zeta de Riemann en los enteros pares, Miscel´anea Mat. 33 (2001), 31-41. [4] B. Berndt, Elementary Evaluation of ζ(2n), Math. Magazine 48 (1975), 148-154. [5] G. Boros and V. H. Moll, Irresistible Integrals, Cambridge University Press, Cambridge, 2004. [6] Ji Chungang and Chen Yonggao, Euler’s formula for ζ(2k) proved by induction on k, Math. Magazine 73 (2000), 154-155. [7] J. Hofbauer, A simple proof of 1+ 212 + 312 +· · · = Amer. Math. Monthly 109 (2002), 196-200.
π2 6
and related identities,
[8] J. Spanier and K. B. Oldham, An Atlas of Functions, Hemisphere, Washington DC, 1987. P −s [9] E. L. Stark, The series ∞ k=1 k , s = 2, 3, 4, . . . once more, Math. Magazine 47 (1974), 197-202.
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[10] E. L. Stark, Application of a mean value theorem for integrals to series summation, Amer. Math. Monthly 85 (1978), 481-483. [11] G. P. Tolstov, Fourier Series, Dover, New York, 1976.
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