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Moments of infinite convolutions of symmetric Bernoulli distributions
Journal of Computational and Applied Mathematics 153 (2003) 191 – 199
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Moments of in"nite convolutions of symmetric Bernoulli distributions C. Escribano∗ , M.A. Sastre, E. Torrano Departamento de Matem atica Aplicada, Facultad de Inform atica, Universidad Polit ecnica de Madrid, Campus de Montegancedo, Boadilla del Monte, 28660 Madrid, Spain Received 7 November 2001; received in revised form 15 January 2002
Abstract We study the in"nite convolution of symmetric Bernoulli distributions associated to a parameter r. We obtain an explicit formula for the moments as a function of Bernoulli numbers and conditioned partitions. Applying this formula we obtain the moments as a quotient of polynomials in the parameter r. The leading coe2cient of the numerator is related to the asymptotic behavior of the moments and, unexpectedly, this coe2cients are the absolute values of Euler numbers. c 2002 Elsevier Science B.V. All rights reserved. Keywords: In"nite Bernoulli convolution; Orthogonal polynomials; Exponential generating function; Euler numbers
1. Introduction Let S denote the set consisting of the two points r; −r and let (E) be the symmetric Bernoulli distribution function which is 0; 12 or 1 according to whether E contains neither, one or both of these points. For every r ∈ (0; 1), we consider the sets S0 = {1; −1}; S1 = {r; −r}; S2 = {r 2 ; −r 2 }; : : : , and r; 0 ; r; 1 ; r; 2 ; : : : , the corresponding Bernoulli distributions. Its characteristic function is (x) = 1 −irx (e +eirx )=cos rx. The in"nite convolution distribution r; 0 ∗r; 1 ∗r; 2 ∗· · · is absolutely convergent 2 to the distribution r , we call such r an in&nitely convolved symmetric Bernoulli measure (ICSBM). The spectrum S(r ) is a compact set, the point spectrum is empty and r is either singular or absolutely continuous. Its characteristic function is (x) = cos(rx) cos(r 2 x) · · · [8]. For every r ∈ (0; 1=2) the ICSBM r is singular and its spectrum is a Cantor set but there is not an explicit expression for its orthogonal polynomials. In the last years, several papers related to this ∗
Corresponding author. E-mail addresses: cescribano@".upm.es (C. Escribano), masastre@".upm.es (M.A. Sastre), emilio@".upm.es (E. Torrano). c 2002 Elsevier Science B.V. All rights reserved. 0377-0427/03/$ - see front matter PII: S 0 3 7 7 - 0 4 2 7 ( 0 2 ) 0 0 5 9 5 - 2
192
C. Escribano et al. / Journal of Computational and Applied Mathematics 153 (2003) 191 – 199
problem have been published [11,5,6]. For r = 12 we get the Lebesgue distribution and Legendre polynomials. It is well known that if the characteristic function does not tend to zero at in"nity then the distribution is singular. The converse is not true. In 1939 Erd˝os [3] showed that, when r is the reciprocal of a Pisot–Vijayaraghavan number 1 then the corresponding characteristic function does not tend to zero at in"nity. Salem in 1944 [13] proved that the characteristic function does not tend to zero if and only if r is the reciprocal of a P.V. number. Nowadays it is conjectured that the ICSBM r is singular for 1=2 ¡ r ¡ 1 if and only if r is the reciprocal of a P.V. number. It has been recently proved [12,14] that for 12 ¡ r ¡ 1 the ICSBM r is absolutely continuous almost everywhere. In this paper we deduce an explicit formula for the moments of this distributions for every r from their characteristic functions. Applying this formula we obtain the moments as a quotient of polynomials in the parameter r. From a recurrent formula for the moments we prove that the leading coe2cient of the numerator is the absolute value of an Euler number. 2. Moments of the innite convolution Bernoulli distribution Theorem 1. Let r ∈ ( 12 ; 1) be an arbitrary and &xed value and let r denote the corresponding in&nite Bernoulli convolution distribution de&ned above. The moments of the distribution are given in terms of Bernoulli Numbers and partitions of n by the formula n n (2n)! (1 − r)2k k 1 (−1)k 2k 2k n S2n = (−1) × : (1) 2 (2 − 1)B2k n 1 ! · · · nn ! (2k)! 2k 1 − r 2k 1·n +···+n·n =n 1
k=1
n
Proof. The moments of the distribution r are de"ned by xk dr (x) Sk = S(ˆr )
and can be expressed in terms of the derivatives of the characteristic function in x = 0 as (k) (0) Sk = ; k = 0; 1; 2; : : : : ik In order to develop a formula for the moments from these derivatives we take the auxiliary function ’(x) de"ned as ∞ ∞ ’(x) = ln (t) = ln cos xr k = ln(cos xr k ): k=1
k=1
Its derivative is given by ∞ r k tan(xr k ): ’ (x) = − k=1
To seethat this series converges near x = 0, we apply the Weierstrass M-test with the convergent 2k series ∞ k=1 r . The same argument works for all derivatives of ’. 1
An algebraic number is called a P.V. number when all its conjugates are in absolute value less than 1.
C. Escribano et al. / Journal of Computational and Applied Mathematics 153 (2003) 191 – 199
193
If we denote by yk the kth derivative of the tangent function, it is clear that ∞ r nk yn−1 (xr k ); n = 1; 2; : : : : ’(n) (x) = − k=1
Taking x = 0, ’(n) (0) = −yn−1 (0)
∞
r nk = −yn−1 (0)
k=1
rn ; 1 − rn
n = 1; 2; : : : :
The function tan(x), near x =0, can be written in terms of the Bernoulli numbers [1, Formula 4.3.67] by, ∞ ∞ ∞ tan(n) (0) n 4n (4n − 1)B2n 2n−1 yn (0) n x = x x : (−1)n+1 = tan x = n! (2n − 1)! · 2n n! n=1 n=1 n=1 We have y2n (0) = 0, for n = 0; 1; 2; : : : and for the odd terms 4n (4n − 1)B2n y2n−1 (0) = tan(2n−1) (0) = (−1)n+1 ; n = 1; 2; 3; : : : : 2n Then, the above derivatives are given by ’(2n−1) (0) = 0;
n = 1; 2; : : : ;
’(2n) (0) = (−1)n
4n (4n − 1)B2n r 2n ; 2n 1 − r 2n
n ¿ 1:
(2)
From the identity (x) = e’(x) , by using their expressions in power series, we obtain 1+
(0) 2 (0) [’(t)] [’(t)]2 [’(t)]3 t+ t + ··· = 1 + + + + ···; 1! 2! 1! 2! 3!
(3)
where ’ (0) ’ (0) 2 t+ t + ··· : 1! 2! We will use the following basic result: ’(t) =
j Lemma 1. Let ak ∈ R or C, ∀k ∈ N, and let ∞ j=1 xj t be a formal series. Then the coe9cient of ∞ ∞ m j n t in n=1 an ( j=1 xj t ) is given by m (n1 + n2 + · · · + nm )! an1 +n2 +···+nm (xk )nk : n 1 !n2 ! · · · nm ! 1·n +2·n +···+m·n =m 1
2
k=1
m
Remark 1. This formula is obtained from the following [15]: k ∞ ∞ xj t j = xp1 xp2 : : : xpk t j ; j=1
j=k
p1 +p2 +···+pk =j; pi ¿1
rearranging the sum in terms of partitions of m.
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C. Escribano et al. / Journal of Computational and Applied Mathematics 153 (2003) 191 – 199
Then, for all n ¿ 1 we have
(m) (0) = m!
1·n1 +2·n2 +···+m·nm
m 1 ’(k) (0) nk : nk ! k! =m
(4)
k=1
We can obtain any moment by substituting (2) in (4). Since ’(2n−1) (0) = 0, for n = 1; 2; : : :, in order to obtain a value diJerent from zero in the product above, the odd indexes n1 = n3 = n5 = · · · must be set equal to zero. Hence, the product is over 2 · n2 + 4 · n4 + · · · + m · nm = m; for every even m. So, setting m = 2n and simplifying, we have n 1 ’(2k) (0) nk (2n) (0) = (2n)! : n (2k)! k! 1·n +2·n +···+n·n =n 1
2
n
(5)
k=1
Using S2n = (2n) (0)=(−1)n , we have the expression for the moments n n 1 (−1)k 2k 2k (2n)! (1 − r)2k k n × 2 (2 − 1)B2k S2n = (−1) : n1 !n2 ! · · · nn ! (2k)! 2k 1 − r 2k 1·n +2·n +···+n·n =n 1
2
k=1
n
(6) The number of terms in the sum is the number of partitions of n. Example. Let us calculate S8 . The conditioned partition of 4 and the "rst even Bernoulli numbers will be
Ordinary
n1
n2
n3
n4
1+1+1+1 2+2 1+1+2 1+3 4
4 0 2 1 0
0 2 1 0 0
0 0 0 1 0
0 0 0 0 1
Hence, S8 =
1 (−1)4
8! n !n 1 2 !n3 !n4 ! 1·n +2·n +3·n +4n =4 1
2
3
4
n 4 1 (−1)k k k (1 − r)2k k 4 (4 − 1)B2k × (2k)! 2k 1 − r 2k k=1
B2
1 6
B4
−1 30
B6
1 42
B8
−1 30
B10
5 66
B12
−691 2730
C. Escribano et al. / Journal of Computational and Applied Mathematics 153 (2003) 191 – 199
=
(−1)4 28 (28 − 1) (−1)4 28 (24 − 1)2 2 (1 − r)8 (1 − r)8 8! 8! + B4 B8 0!0!0!1! (2 · 4)8! 1 − r8 0!2!0!0! (2 · 2)2 (4!)2 (1 − r 4 )2 +
(−1)1 22 (22 − 1) 8! (−1)3 26 (26 − 1) (1 − r)8 B2 B6 1!0!1!0! (2 · 1)2! (2 · 3)6! (1 − r 2 )(1 − r 6 )
+
(−1)2 24 (22 − 1)2 2 (−1)2 24 (24 − 1) 8! (1 − r)8 B B 4 2 2!1!0!0! (2!)2 (2 · 1)2 (2 · 2)4! (1 − r 2 )2 (1 − r 4 )
+
(−1)4 28 (22 − 1)4 4 (1 − r)8 8! B2 4!0!0!0! (2 · 1)4 (2!)4 (1 − r 2 )4
= −272
(1 − r)8 (1 − r)8 (1 − r)8 + 140 + 448 1 − r8 (1 − r 4 )2 (1 − r 2 )(1 − r 6 )
− 420 =
(1 − r)8 (1 − r)8 + 105 (1 − r 2 )2 (1 − r 4 ) (1 − r 2 )4
1−r 1+r
4
1385r 12 + 323r 10 + 365r 8 + 350r 6 + 69r 4 + 27r 2 + 1 : 1 · (1 + r 2 ) · (1 + r 2 + r 4 ) · (1 + r 2 + r 4 + r 6 )
In the same way we can calculate S0 = 1; 1−r ; r+1
1 − r 2 5r 2 + 1 S4 = ; 1 + r (1 + r 2 )
S2 =
S6 = S8 =
1−r 1+r 1−r 1+r
S10 =
3
4
1−r 1+r
:::
5
61r 6 + 14r 4 + 14r 2 + 1 ; (1 + r 2 )(1 + r 2 + r 4 ) 1385r 12 + 323r 10 + 365r 8 + 350r 6 + 69r 4 + 26r 2 + 1 ; (1 + r 2 )(1 + r 2 + r 4 )(1 + r 2 + r 4 + r 6 ) 50521r 20 + 11804r 18 + 13529r 16 + 14535r 14 + 14700r 12 + 3902r 10 (1 + r 2 )(1 + r 2 + r 4 )(1 + r 2 + r 4 + r 6 )
+2940r 8 + 1215r 6 + 209r 4 + 44r 2 + 1 : (1 + r 2 + r 4 + r 6 + r 8 )
195
196
C. Escribano et al. / Journal of Computational and Applied Mathematics 153 (2003) 191 – 199
Unexpectedly, the conductor coe2cient of the polynomial in r appearing in the numerator of the moment S2n is, taking the absolute value, the Euler number E2n . The "rst Euler numbers are E0 E2 E4 E6 E8 E10 E12
1 −1 5 −61 1385 −50521 2702765
To prove this conjecture we deduce a recurrence formula for the moments. 3. Recurrence formula for the moments and the Euler numbers 3.1. Self-similar measures Given a family of contractive maps{’j }nj=1 de"ned in a complete metric space, there exists a unique compactum K satisfying K = nj=1 ’j (K). This compactum is obtained as a limit in the metric space of compacta with the HausdorJ metric, iterating the maps, taking as initial set any compactum of the space. The family {’j }nj=1 is called an Iterated Functions System (IFS) [2]. If we assign a probability pj ¿ 0 to every ’j , with nj=1 pj = 1, there exists a unique probability measure invariant for the Markov operator de"ned as n 1 T = pj ’− j : j=1
We say that K and are self-similar when the maps ’j are contractive similarities. Moreover, if the ’j (K) are disjoints sets, then any of them is similar to K [7,10]. The same happens to the self-similar measure restricted to these sets. The ICSBM r is a remarkable example of invariant measure for the Markov operator 1 −1 1 T = 12 ’− 1 + 2 ’2 ;
where ’1 (x) = rx − 1 + r and ’2 (x) = rx + 1 − r, [9]. Using the “balance property” for invariant measures [4, p. 36]; [11], we have 1 1 1 1 1 Sn = xn dr = (rx − (1 − r))n d(x) + (rx + (1 − r))n d(x): 2 2 0 0 0 The moments of the ICSBM r are given by the recurrence formula
2k m− 1 (1 − r)2m 2m r S2k ; m = 1; 2; 3; : : : ; S2m = 1 − r 2m 1−r 2k k=0
where S0 = 1:
C. Escribano et al. / Journal of Computational and Applied Mathematics 153 (2003) 191 – 199
197
Proposition 1. The moment of the ICSBM satis&es S2m =
1−r 1+r
2
m Q m (r ) 2
r 2 (m + 1)
;
where r 2 (m+1)=1(1+r 2 )(1+r 2 +r 4 ) · · · (1+r 2 +r 4 +· · ·+r 2(m−1) ), and Q m (r 2 ) are polynomials 2
m
2
in r with integer coe9cients of degree 2 , satisfying the following recurrence formula: m− 1 m −1 2m 2 2k 2 Q m (r ) = (1 − r 2j ); r Q k (r ) 2k 2 2 k=0 j=k+1 with Q0 (r 2 ) = 1. Proof. We can write the formula r 2 (m + 1) in the following way. Since −1 (1 − r 2j ) (1 − r 2m ) mj=1 ; r 2 (m + 1) = (1 − r 2 )m we have, (1 − r)2m = 1 − r 2m
1−r 1+r
m m−1 j=1
(1 − r 2j )
r 2 (m + 1)
then, S2m = =
1−r 1+r 1−r 1+r
m
m− 1
1 r 2 (m + 1) k=0
m
1 r 2 (m + 1)
2m
;
2k
m− 1 2m k=0
2k
r
r 1−r
2k
2k
S2k
1+r 1−r
m −1
(1 − r 2j )
j=1
k
r 2 (k + 1)S2k
S2k =
1−r 1+r
2
k Q k (r ) 2
r 2 (k + 1)
;
k = 1; 2; : : : ; m − 1:
For k = m we substitute this into the formula for S2k given above
m− 1 m −1 2m 1 1−r m S2k = (1 − r 2j ) r 2k Q k (r 2 ) 1+r r 2 (m + 1) 2k 2 =
1−r 1+r
m
k=0
1 Q (r 2 ) r 2 (m + 1) m2
m −1 j=k+1
In order to apply induction with respect to m, assume
j=k+1
(1 − r 2j ):
198
C. Escribano et al. / Journal of Computational and Applied Mathematics 153 (2003) 191 – 199
and, 2
Q m (r ) = 2
m− 1
2m
r 2k Q k (r 2 )
2k
k=0
2
m −1
(1 − r 2j );
j=k+1
where Q0 (r 2 ) = 1. Note that the coe2cients are integer numbers. Proposition 2. If Q m (r ) = a m m r 2
2
2
;
2
m 2
2
a m m r 2
;
2
−1
2
m 2
−2
+ · · ·, then a m m = |E2n |. 2
;
2
Proof. Consider the reverse polynomial Q m (r 2 ), obtained with r = 1=s, and multiplying by s
2
2
m 2
s
2
: m 2
Q m 2
1 s2
=
k=0
consider H k (s ) = s 2
m− 1
k 2 2
2
m k m− 1 2 2 −2 2
k 2m 2 2 1 1 s 1 − 2j ; s Q k 2 2k s s s 2k 2 j=k+1
Q k (1=s2 ). Then 2
m− 1
m− 1 2m sm(m−1)−k(k −1) 2 2j H k (s2 ) (s − 1) H m (s ) = 2 )(k+1)+(k+2)+···+(k+(m−1)−k) s2k (s 2k 2 2 k=0 j=k+1 =
m− 1
k=0
2m 2k
m− 1 (s2j − 1) H k (s2 ): j=k+1
If we take s = 0 we get H k (0) = a k k ; 2 ;
2
therefore m k=0
2m 2k
k = 1; 2; : : :
2
(−1)m−k −1 a k k = 0: 2 ;
2
As (−1)−k = (−1)k , we have m 2m k = 0: (−1) a k k ; 2 2k 2 k=0
2
C. Escribano et al. / Journal of Computational and Applied Mathematics 153 (2003) 191 – 199
The Euler numbers satisfy m 2m E2k = 0; 2k k=0
199
with E0 = 1:
Consequently (−1)k a k k = E2k , and "nally a k k = |E2k |. 2 ;
2
2 ;
2
References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15]
M. Abramowitz, I.A. Stegun, Handbook of Mathematical Functions, Dover, New York, 1972. M. Barnsley, Fractals Everywhere, Academic Press, Boston, 1988. P. Erd˝os, On a family of symmetric Bernoulli convolutions, Amer. J. Math. 61 (1939) 974–976. K.J. Falconer, Techniques in Fractal Geometry, Wiley, Chichester, 1997. H.J. Fischer, On the paper Asymptotics for the moments of singular distributions, J. Approx. Theory 82 (1995) 362–374. W. Goh, J. Wimp, Asymptotics for the moments of singular distributions, J. Approx. Theory 74 (1993) 301–334. J. Hutchinson, Fractal and self-similarity, Indiana Univ. Math. J. 30 (1981) 713–747. B. Jessen, A. Wintner, Distribution functions and the Rieman zeta function, Trans. Amer. Math. Soc. 38 (1935) 48–88. K.S. Lau, Fractal measures and mean p-variations, J. Funct. Anal. 108 (1992) 427–457. B.B. Mandelbrot, The Fractal Geometry of Nature, Freeman, San Francisco, 1977. G. Mantica, A stable technique for computing orthogonal polynomials and Jacobi matrices associated with a class of singular measures, Control Approx. 12 (1996) 509–530. Y. Peres, B. Solomyak, Absolute continuity of Bernoulli convolutions, a simple proof, Math. Res. Lett. (1996) 231–239. R. Salem, A remarkable class of algebraic integers proof of a conjecture of Vijayaraghavan, Duke Math. J. 11 (1944) 103–108. B. Solomyak, Measure and dimension for some fractal families, Math. Proc. Cambridge Philos. Soc. 124 (1998) 531–546. H.S. Wilf, Generating Functionology, Academic Press Inc., New York, 1994.
www.elsevier.com/locate/cam
Moments of in"nite convolutions of symmetric Bernoulli distributions C. Escribano∗ , M.A. Sastre, E. Torrano Departamento de Matem atica Aplicada, Facultad de Inform atica, Universidad Polit ecnica de Madrid, Campus de Montegancedo, Boadilla del Monte, 28660 Madrid, Spain Received 7 November 2001; received in revised form 15 January 2002
Abstract We study the in"nite convolution of symmetric Bernoulli distributions associated to a parameter r. We obtain an explicit formula for the moments as a function of Bernoulli numbers and conditioned partitions. Applying this formula we obtain the moments as a quotient of polynomials in the parameter r. The leading coe2cient of the numerator is related to the asymptotic behavior of the moments and, unexpectedly, this coe2cients are the absolute values of Euler numbers. c 2002 Elsevier Science B.V. All rights reserved. Keywords: In"nite Bernoulli convolution; Orthogonal polynomials; Exponential generating function; Euler numbers
1. Introduction Let S denote the set consisting of the two points r; −r and let (E) be the symmetric Bernoulli distribution function which is 0; 12 or 1 according to whether E contains neither, one or both of these points. For every r ∈ (0; 1), we consider the sets S0 = {1; −1}; S1 = {r; −r}; S2 = {r 2 ; −r 2 }; : : : , and r; 0 ; r; 1 ; r; 2 ; : : : , the corresponding Bernoulli distributions. Its characteristic function is (x) = 1 −irx (e +eirx )=cos rx. The in"nite convolution distribution r; 0 ∗r; 1 ∗r; 2 ∗· · · is absolutely convergent 2 to the distribution r , we call such r an in&nitely convolved symmetric Bernoulli measure (ICSBM). The spectrum S(r ) is a compact set, the point spectrum is empty and r is either singular or absolutely continuous. Its characteristic function is (x) = cos(rx) cos(r 2 x) · · · [8]. For every r ∈ (0; 1=2) the ICSBM r is singular and its spectrum is a Cantor set but there is not an explicit expression for its orthogonal polynomials. In the last years, several papers related to this ∗
Corresponding author. E-mail addresses: cescribano@".upm.es (C. Escribano), masastre@".upm.es (M.A. Sastre), emilio@".upm.es (E. Torrano). c 2002 Elsevier Science B.V. All rights reserved. 0377-0427/03/$ - see front matter PII: S 0 3 7 7 - 0 4 2 7 ( 0 2 ) 0 0 5 9 5 - 2
192
C. Escribano et al. / Journal of Computational and Applied Mathematics 153 (2003) 191 – 199
problem have been published [11,5,6]. For r = 12 we get the Lebesgue distribution and Legendre polynomials. It is well known that if the characteristic function does not tend to zero at in"nity then the distribution is singular. The converse is not true. In 1939 Erd˝os [3] showed that, when r is the reciprocal of a Pisot–Vijayaraghavan number 1 then the corresponding characteristic function does not tend to zero at in"nity. Salem in 1944 [13] proved that the characteristic function does not tend to zero if and only if r is the reciprocal of a P.V. number. Nowadays it is conjectured that the ICSBM r is singular for 1=2 ¡ r ¡ 1 if and only if r is the reciprocal of a P.V. number. It has been recently proved [12,14] that for 12 ¡ r ¡ 1 the ICSBM r is absolutely continuous almost everywhere. In this paper we deduce an explicit formula for the moments of this distributions for every r from their characteristic functions. Applying this formula we obtain the moments as a quotient of polynomials in the parameter r. From a recurrent formula for the moments we prove that the leading coe2cient of the numerator is the absolute value of an Euler number. 2. Moments of the innite convolution Bernoulli distribution Theorem 1. Let r ∈ ( 12 ; 1) be an arbitrary and &xed value and let r denote the corresponding in&nite Bernoulli convolution distribution de&ned above. The moments of the distribution are given in terms of Bernoulli Numbers and partitions of n by the formula n n (2n)! (1 − r)2k k 1 (−1)k 2k 2k n S2n = (−1) × : (1) 2 (2 − 1)B2k n 1 ! · · · nn ! (2k)! 2k 1 − r 2k 1·n +···+n·n =n 1
k=1
n
Proof. The moments of the distribution r are de"ned by xk dr (x) Sk = S(ˆr )
and can be expressed in terms of the derivatives of the characteristic function in x = 0 as (k) (0) Sk = ; k = 0; 1; 2; : : : : ik In order to develop a formula for the moments from these derivatives we take the auxiliary function ’(x) de"ned as ∞ ∞ ’(x) = ln (t) = ln cos xr k = ln(cos xr k ): k=1
k=1
Its derivative is given by ∞ r k tan(xr k ): ’ (x) = − k=1
To seethat this series converges near x = 0, we apply the Weierstrass M-test with the convergent 2k series ∞ k=1 r . The same argument works for all derivatives of ’. 1
An algebraic number is called a P.V. number when all its conjugates are in absolute value less than 1.
C. Escribano et al. / Journal of Computational and Applied Mathematics 153 (2003) 191 – 199
193
If we denote by yk the kth derivative of the tangent function, it is clear that ∞ r nk yn−1 (xr k ); n = 1; 2; : : : : ’(n) (x) = − k=1
Taking x = 0, ’(n) (0) = −yn−1 (0)
∞
r nk = −yn−1 (0)
k=1
rn ; 1 − rn
n = 1; 2; : : : :
The function tan(x), near x =0, can be written in terms of the Bernoulli numbers [1, Formula 4.3.67] by, ∞ ∞ ∞ tan(n) (0) n 4n (4n − 1)B2n 2n−1 yn (0) n x = x x : (−1)n+1 = tan x = n! (2n − 1)! · 2n n! n=1 n=1 n=1 We have y2n (0) = 0, for n = 0; 1; 2; : : : and for the odd terms 4n (4n − 1)B2n y2n−1 (0) = tan(2n−1) (0) = (−1)n+1 ; n = 1; 2; 3; : : : : 2n Then, the above derivatives are given by ’(2n−1) (0) = 0;
n = 1; 2; : : : ;
’(2n) (0) = (−1)n
4n (4n − 1)B2n r 2n ; 2n 1 − r 2n
n ¿ 1:
(2)
From the identity (x) = e’(x) , by using their expressions in power series, we obtain 1+
(0) 2 (0) [’(t)] [’(t)]2 [’(t)]3 t+ t + ··· = 1 + + + + ···; 1! 2! 1! 2! 3!
(3)
where ’ (0) ’ (0) 2 t+ t + ··· : 1! 2! We will use the following basic result: ’(t) =
j Lemma 1. Let ak ∈ R or C, ∀k ∈ N, and let ∞ j=1 xj t be a formal series. Then the coe9cient of ∞ ∞ m j n t in n=1 an ( j=1 xj t ) is given by m (n1 + n2 + · · · + nm )! an1 +n2 +···+nm (xk )nk : n 1 !n2 ! · · · nm ! 1·n +2·n +···+m·n =m 1
2
k=1
m
Remark 1. This formula is obtained from the following [15]: k ∞ ∞ xj t j = xp1 xp2 : : : xpk t j ; j=1
j=k
p1 +p2 +···+pk =j; pi ¿1
rearranging the sum in terms of partitions of m.
194
C. Escribano et al. / Journal of Computational and Applied Mathematics 153 (2003) 191 – 199
Then, for all n ¿ 1 we have
(m) (0) = m!
1·n1 +2·n2 +···+m·nm
m 1 ’(k) (0) nk : nk ! k! =m
(4)
k=1
We can obtain any moment by substituting (2) in (4). Since ’(2n−1) (0) = 0, for n = 1; 2; : : :, in order to obtain a value diJerent from zero in the product above, the odd indexes n1 = n3 = n5 = · · · must be set equal to zero. Hence, the product is over 2 · n2 + 4 · n4 + · · · + m · nm = m; for every even m. So, setting m = 2n and simplifying, we have n 1 ’(2k) (0) nk (2n) (0) = (2n)! : n (2k)! k! 1·n +2·n +···+n·n =n 1
2
n
(5)
k=1
Using S2n = (2n) (0)=(−1)n , we have the expression for the moments n n 1 (−1)k 2k 2k (2n)! (1 − r)2k k n × 2 (2 − 1)B2k S2n = (−1) : n1 !n2 ! · · · nn ! (2k)! 2k 1 − r 2k 1·n +2·n +···+n·n =n 1
2
k=1
n
(6) The number of terms in the sum is the number of partitions of n. Example. Let us calculate S8 . The conditioned partition of 4 and the "rst even Bernoulli numbers will be
Ordinary
n1
n2
n3
n4
1+1+1+1 2+2 1+1+2 1+3 4
4 0 2 1 0
0 2 1 0 0
0 0 0 1 0
0 0 0 0 1
Hence, S8 =
1 (−1)4
8! n !n 1 2 !n3 !n4 ! 1·n +2·n +3·n +4n =4 1
2
3
4
n 4 1 (−1)k k k (1 − r)2k k 4 (4 − 1)B2k × (2k)! 2k 1 − r 2k k=1
B2
1 6
B4
−1 30
B6
1 42
B8
−1 30
B10
5 66
B12
−691 2730
C. Escribano et al. / Journal of Computational and Applied Mathematics 153 (2003) 191 – 199
=
(−1)4 28 (28 − 1) (−1)4 28 (24 − 1)2 2 (1 − r)8 (1 − r)8 8! 8! + B4 B8 0!0!0!1! (2 · 4)8! 1 − r8 0!2!0!0! (2 · 2)2 (4!)2 (1 − r 4 )2 +
(−1)1 22 (22 − 1) 8! (−1)3 26 (26 − 1) (1 − r)8 B2 B6 1!0!1!0! (2 · 1)2! (2 · 3)6! (1 − r 2 )(1 − r 6 )
+
(−1)2 24 (22 − 1)2 2 (−1)2 24 (24 − 1) 8! (1 − r)8 B B 4 2 2!1!0!0! (2!)2 (2 · 1)2 (2 · 2)4! (1 − r 2 )2 (1 − r 4 )
+
(−1)4 28 (22 − 1)4 4 (1 − r)8 8! B2 4!0!0!0! (2 · 1)4 (2!)4 (1 − r 2 )4
= −272
(1 − r)8 (1 − r)8 (1 − r)8 + 140 + 448 1 − r8 (1 − r 4 )2 (1 − r 2 )(1 − r 6 )
− 420 =
(1 − r)8 (1 − r)8 + 105 (1 − r 2 )2 (1 − r 4 ) (1 − r 2 )4
1−r 1+r
4
1385r 12 + 323r 10 + 365r 8 + 350r 6 + 69r 4 + 27r 2 + 1 : 1 · (1 + r 2 ) · (1 + r 2 + r 4 ) · (1 + r 2 + r 4 + r 6 )
In the same way we can calculate S0 = 1; 1−r ; r+1
1 − r 2 5r 2 + 1 S4 = ; 1 + r (1 + r 2 )
S2 =
S6 = S8 =
1−r 1+r 1−r 1+r
S10 =
3
4
1−r 1+r
:::
5
61r 6 + 14r 4 + 14r 2 + 1 ; (1 + r 2 )(1 + r 2 + r 4 ) 1385r 12 + 323r 10 + 365r 8 + 350r 6 + 69r 4 + 26r 2 + 1 ; (1 + r 2 )(1 + r 2 + r 4 )(1 + r 2 + r 4 + r 6 ) 50521r 20 + 11804r 18 + 13529r 16 + 14535r 14 + 14700r 12 + 3902r 10 (1 + r 2 )(1 + r 2 + r 4 )(1 + r 2 + r 4 + r 6 )
+2940r 8 + 1215r 6 + 209r 4 + 44r 2 + 1 : (1 + r 2 + r 4 + r 6 + r 8 )
195
196
C. Escribano et al. / Journal of Computational and Applied Mathematics 153 (2003) 191 – 199
Unexpectedly, the conductor coe2cient of the polynomial in r appearing in the numerator of the moment S2n is, taking the absolute value, the Euler number E2n . The "rst Euler numbers are E0 E2 E4 E6 E8 E10 E12
1 −1 5 −61 1385 −50521 2702765
To prove this conjecture we deduce a recurrence formula for the moments. 3. Recurrence formula for the moments and the Euler numbers 3.1. Self-similar measures Given a family of contractive maps{’j }nj=1 de"ned in a complete metric space, there exists a unique compactum K satisfying K = nj=1 ’j (K). This compactum is obtained as a limit in the metric space of compacta with the HausdorJ metric, iterating the maps, taking as initial set any compactum of the space. The family {’j }nj=1 is called an Iterated Functions System (IFS) [2]. If we assign a probability pj ¿ 0 to every ’j , with nj=1 pj = 1, there exists a unique probability measure invariant for the Markov operator de"ned as n 1 T = pj ’− j : j=1
We say that K and are self-similar when the maps ’j are contractive similarities. Moreover, if the ’j (K) are disjoints sets, then any of them is similar to K [7,10]. The same happens to the self-similar measure restricted to these sets. The ICSBM r is a remarkable example of invariant measure for the Markov operator 1 −1 1 T = 12 ’− 1 + 2 ’2 ;
where ’1 (x) = rx − 1 + r and ’2 (x) = rx + 1 − r, [9]. Using the “balance property” for invariant measures [4, p. 36]; [11], we have 1 1 1 1 1 Sn = xn dr = (rx − (1 − r))n d(x) + (rx + (1 − r))n d(x): 2 2 0 0 0 The moments of the ICSBM r are given by the recurrence formula
2k m− 1 (1 − r)2m 2m r S2k ; m = 1; 2; 3; : : : ; S2m = 1 − r 2m 1−r 2k k=0
where S0 = 1:
C. Escribano et al. / Journal of Computational and Applied Mathematics 153 (2003) 191 – 199
197
Proposition 1. The moment of the ICSBM satis&es S2m =
1−r 1+r
2
m Q m (r ) 2
r 2 (m + 1)
;
where r 2 (m+1)=1(1+r 2 )(1+r 2 +r 4 ) · · · (1+r 2 +r 4 +· · ·+r 2(m−1) ), and Q m (r 2 ) are polynomials 2
m
2
in r with integer coe9cients of degree 2 , satisfying the following recurrence formula: m− 1 m −1 2m 2 2k 2 Q m (r ) = (1 − r 2j ); r Q k (r ) 2k 2 2 k=0 j=k+1 with Q0 (r 2 ) = 1. Proof. We can write the formula r 2 (m + 1) in the following way. Since −1 (1 − r 2j ) (1 − r 2m ) mj=1 ; r 2 (m + 1) = (1 − r 2 )m we have, (1 − r)2m = 1 − r 2m
1−r 1+r
m m−1 j=1
(1 − r 2j )
r 2 (m + 1)
then, S2m = =
1−r 1+r 1−r 1+r
m
m− 1
1 r 2 (m + 1) k=0
m
1 r 2 (m + 1)
2m
;
2k
m− 1 2m k=0
2k
r
r 1−r
2k
2k
S2k
1+r 1−r
m −1
(1 − r 2j )
j=1
k
r 2 (k + 1)S2k
S2k =
1−r 1+r
2
k Q k (r ) 2
r 2 (k + 1)
;
k = 1; 2; : : : ; m − 1:
For k = m we substitute this into the formula for S2k given above
m− 1 m −1 2m 1 1−r m S2k = (1 − r 2j ) r 2k Q k (r 2 ) 1+r r 2 (m + 1) 2k 2 =
1−r 1+r
m
k=0
1 Q (r 2 ) r 2 (m + 1) m2
m −1 j=k+1
In order to apply induction with respect to m, assume
j=k+1
(1 − r 2j ):
198
C. Escribano et al. / Journal of Computational and Applied Mathematics 153 (2003) 191 – 199
and, 2
Q m (r ) = 2
m− 1
2m
r 2k Q k (r 2 )
2k
k=0
2
m −1
(1 − r 2j );
j=k+1
where Q0 (r 2 ) = 1. Note that the coe2cients are integer numbers. Proposition 2. If Q m (r ) = a m m r 2
2
2
;
2
m 2
2
a m m r 2
;
2
−1
2
m 2
−2
+ · · ·, then a m m = |E2n |. 2
;
2
Proof. Consider the reverse polynomial Q m (r 2 ), obtained with r = 1=s, and multiplying by s
2
2
m 2
s
2
: m 2
Q m 2
1 s2
=
k=0
consider H k (s ) = s 2
m− 1
k 2 2
2
m k m− 1 2 2 −2 2
k 2m 2 2 1 1 s 1 − 2j ; s Q k 2 2k s s s 2k 2 j=k+1
Q k (1=s2 ). Then 2
m− 1
m− 1 2m sm(m−1)−k(k −1) 2 2j H k (s2 ) (s − 1) H m (s ) = 2 )(k+1)+(k+2)+···+(k+(m−1)−k) s2k (s 2k 2 2 k=0 j=k+1 =
m− 1
k=0
2m 2k
m− 1 (s2j − 1) H k (s2 ): j=k+1
If we take s = 0 we get H k (0) = a k k ; 2 ;
2
therefore m k=0
2m 2k
k = 1; 2; : : :
2
(−1)m−k −1 a k k = 0: 2 ;
2
As (−1)−k = (−1)k , we have m 2m k = 0: (−1) a k k ; 2 2k 2 k=0
2
C. Escribano et al. / Journal of Computational and Applied Mathematics 153 (2003) 191 – 199
The Euler numbers satisfy m 2m E2k = 0; 2k k=0
199
with E0 = 1:
Consequently (−1)k a k k = E2k , and "nally a k k = |E2k |. 2 ;
2
2 ;
2
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