International Mathematical Forum, Vol. 8, 2013, no. 1, 9 - 14
Some Applications of the Euler-Maclaurin Summation Formula Rafael Jakimczuk Divisi´on Matem´atica, Universidad Nacional de Luj´an Buenos Aires, Argentina
[email protected] Abstract Let n be a fixed positive integer. In this article we obtain asymptotic N (j n ) n formulae for N . For example if n = 1 we j=1 j log j and i=1 j obtain the asymptotic formulae, N i=1
j log j =
N2 1 1 N2 log N − + N log N + log N + C1 + o(1), 2 4 2 12
and
N
j
C1
j ∼e
N
N2 +N 2 2
i=1
e
N2 4
1 + 12
,
where C1 is a constant.
Mathematics Subject Classification: 11A99 Keywords: Euler-Maclaurin summation formula, asymptotic formulae
1
Introduction
Bernoulli numbers are among the most distinguished and important numbers in all of mathematics. Indeed, they play a vital role in number theory. The Bernoulli numbers can be defined in the following way (see [1], chapter V) B0 = 1 n−1 i=0
n Bi = 0 i
(n ≥ 2)
Using this relation we obtain the first Bernoulli numbers, namely B0 = 1, B1 = −1/2, B2 = 1/6, B3 = 0, B4 = −1/30, B5 = 0.
R. Jakimczuk
10
It is well known (see [1], chapter V) that if i ≥ 3 is odd then Bi = 0 . The n-th Bernoulli polynomial is defined in the following way Bn (x) =
n
j=0
n Bj xn−j . j
The first few Bernoulli polynomials are 1 B1 (x) = x − , 2
B0 (x) = 1,
1 B2 (x) = x2 − x + , 6
3 1 B3 (x) = x3 − x2 + x 2 2
In the following theorem we establish the Euler-Maclaurin summation formula. Theorem 1.1 Let a < b be integers and let m a positive integer. If f (x) has m continuous derivatives on the interval [a, b], then b
f (j) =
j=a+1
b a
f (x) dx + m−1
+
(−1) m!
b a
m
(−1)i
i=1
Bi (i−1) f (b) − f (i−1) (a) i!
Bm (x − x) f (m) (x) dx.
(1)
Proof. See [1], chapter V. Let n be a fixed positive integer. We shall need the following well-known formula N
n N n+1 1 n 1 n+1 + N + i = Bi N n+1−i Pn (N) = n + 1 2 n + 1 i i=1 i=2 n
(N ≥ 1).
For example, we have N2 N + , P1 (N) = 1 + 2 + · · · + N = 2 2 N3 N2 N + + . P2 (N) = 1 + 2 + · · · + N = 3 2 6 4 3 N N2 N 3 3 3 + + . P3 (N) = 1 + 2 + · · · + N = 4 2 4 2
2
2
2
Preliminary Results
Let us consider the function f (x) = f (0) (x) = xn log x, where n is a positive integer. Let f (k) (x) be its k-th derivative. The following theorem holds.
(2)
Euler-Maclaurin summation formula
11
Theorem 2.1 We have f (k) (x) = ak xn−k log x + bk xn−k
(k = 1, . . . , n),
(3)
where ak = n(n − 1) · · · (n − (k − 1)), and bk =
k−1 i=0
(4)
n(n − 1) · · · (n − (k − 1)) . n−i
(5)
Besides
n! , (6) x n! f (n+2) (x) = − 2 . (7) x Proof. Clearly formulae (3), (4) and (5) are true if k = 1. This is the unique case if n = 1. If n ≥ 2, suppose that formulae (3), (4) and (5) are true for k such that 1 ≤ k ≤ n − 1. That is f (n+1) (x) =
f (k) (x) = n(n − 1) · · · (n − (k − 1))xn−k log x +
k−1 i=0
n(n − 1) · · · (n − (k − 1)) n−k x . n−i
If we derive then we find that f (k+1) (x) = n(n − 1) · · · (n − (k − 1))(n − k)xn−(k+1) log x + n(n − 1) · · · (n − (k − 1))xn−(k+1) +
k−1 i=0
n(n − 1) · · · (n − (k − 1))(n − k) n−(k+1) x n−i
= n(n − 1) · · · (n − k)xn−(k+1) log x +
k
n(n − 1) · · · (n − k) n−(k+1) x . n−i i=0
That is, formulae (3), (4) and (5) are true for k + 1. This proves formulae (3), (4) and (5 ). If k = n then equation (3) becomes (see (4) and (5)) f
(n)
(x) = n! log x +
n−1 i=0
n! . n−i
(8)
Equations (6) and (7) are an immediate consequence of equation (8). The theorem is proved. We shall need the following integral 1 xn+1 log x − xn log x dx = xn+1 + C. n+1 (n + 1)2
(9)
R. Jakimczuk
12
3
Main Results Lemma 3.1 The integral ∞ 1
Bn+2 (x − x)
1 dx x2
(n ≥ 1)
(10)
is convergent. Proof. Note that 0 ≤ x − x < 1. On the other hand (see (1), chapter V) Bn+2 (0) = Bn+2 (1). Consequently Bn+2 (x−x) is continuous and with period 1 on the interval (−∞, ∞). Therefore there exist A > 0 such that |Bn+2 (x − x)| ≤ A. Consequently we have 1 A Bn+2 (x − x) (x ≥ 1). ≤ 2 2 x x Now, the integral
∞ 1
A dx x2
is convergent. Therefore (comparison criterion) the integral ∞ 1 Bn+2 (x − x) x2 1
dx
(n ≥ 1)
is also convergent. Thus, the integral (10) converges absolutely and hence converges. The lemma is proved. The following theorem is our main theorem. Theorem 3.2 Let n be an arbitrary but fixed positive integer. The following asymptotic formula holds N j=1
j n log j = Dn (N) log N − Hn (N) + Cn + o(1)
(11)
where n+1
Dn (N) = Pn (N) + (−1)
Bn+1 , (n + 1)
n Bi N n+1 − Hn (N) = bi−1 N n−(i−1) . 2 (n + 1) i=2 i!
and Cn is a constant depending of n. Note that Dn (N) and Hn (N) are polynomials in N of degree n + 1.
Euler-Maclaurin summation formula
13
Proof. We have (see (1) with m = n + 2, (8), (6) , (7), (9) and lemma 3.1) N n
N n
j=1
j=2 n
j log j =
j log j =
N 1
xn log x dx − B1 f (0) (N) − f (0) (1)
Bi (i−1) Bn+1 f (N) − f (i−1) (1) + (−1)n+1 i! (n + 1)! i=2 Bn+2 (n! log N) + (−1)n+2 (n + 2)! n! (−1)n − n! + N (n + 1)(n + 2) N N n+1 1 N n+1 log N − Bn+2 (x − x) 2 dx = x n+1 (n + 1)2 1 n Bi Bn+1 − B1 f (0) (N) + (−1)i f (i−1) (N) + (−1)n+1 i! (n + 1) i=2 log N + Cn + o(1),
+
(−1)i
where Cn is a constant. That is N
n Bi N n+1 N n+1 (0) j log j = − B1 f (N) + (−1)i f (i−1) (N) log N − 2 n+1 (n + 1) i! j=1 i=2 n
+ (−1)n+1
Bn+1 log N + Cn + o(1). (n + 1)
(12)
Substituting (2) and (3) into (12) we obtain N
j n log j =
j=1
N n+1 N n+1 1 log N − + N n log N 2 n+1 (n + 1) 2
Bi ai−1 N n−(i−1) log N + bi−1 N n−(i−1) i! i=2 Bn+1 log N + Cn + o(1) + (−1)n+1 (n + 1)
+
n
(−1)i
(13)
Since B1 = −1/2 (see the introduction). Equation (13) can be written in the form N n j=1
n+1
j log j = An (N) + (−1)
where An (N) =
Bn+1 log N − Hn (N) + Cn + o(1) (14) (n + 1)
n Bi N n+1 1 n + N + ai−1 N n−(i−1) n+1 2 i! i=2
(15)
R. Jakimczuk
14 and Hn (N) =
n N n+1 Bi − bi−1 N n−(i−1) . 2 (n + 1) i! i=2
(16)
Since if i ≥ 2 then (−1)i Bi = Bi (see the introduction). Substituting (4) into (15) we find that (see the polynomial Pn (N) in the introduction) An (N ) = =
n Bi N n+1 1 n + N + (n(n − 1) · · · (n − (i − 2))N n−(i−1) n+1 2 i! i=2
n N n+1 1 n 1 (n + 1)n · · · (n − (i − 2)) n+1−i + N + N Bi n+1 2 n + 1 i=2 i!
n n+1 N n+1 1 n 1 + N + = Bi N n+1−i = Pn (N) n+1 2 n + 1 i=2 i
(17)
Equations (14), (15), (16) and (17) give equation (11). The theorem is proved. Theorem 3.3 Let n be an arbitrary but fixed positive integer. The following asymptotic formula holds N (j n )
j
j=1
∼ eCn
N Dn (N) , eHn (N)
(18)
Proof. It is an immediate consequence of equation (11). The theorem is proved. Remark 3.4 Note that equation (18) is a generalization of the Stirling’s formula. Namely N N N +1/2 j = N! ∼ eC0 , eN j=1 √
2π . where in this case C0 = log ACKNOWLEDGEMENTS. The author is very grateful to Universidad Nacional de Luj´an.
References [1] R. A. Mollin, Advanced Number Theory with Applications, Chapman and Hall/CRC, Taylor and Francis Group, Boca Raton, London, New York, 2010. Received: August, 2012
