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Outline of the method: In (10), we replace the unknown value s (1=n) by some PadÃƒ -approximant

e

to s (x), at the point x = 1=n. We get the following approximation:

n

1 1

(s) â‰ˆ + [p=q] s (x = 1=n): (11)

ks (s)

k=1

We only consider the particular case p = q.

240 M. PrÃƒ vost / Journal of Computational and Applied Mathematics 122 (2000) 231â€“250

e

Case (2): If s = 2 then (10) becomes

n

1

(2) = + 2 (1=n);

k2

k=1

and its approximation (11):

n

1

(2) â‰ˆ + [p=p] 2 (x = 1=n); (12)

k2

k=1

where

âˆž

Bk xk+1 = B0 x + B1 x2 + B2 x3 + Â· Â· Â·

2 (x) = (asymptotic expansion): (13)

k=0

The asymptotic expansion (13) is Borel-summable and its sum is

âˆž

eâˆ’u=x

2 (x) = uu du:

e âˆ’1

0

Computation of [p=p] 2 (x)=x : We apply Section 1.1, where function f(x) = 2 (x)=x. The PadÃƒ ap- e

proximants [p=p]f are linked with the orthogonal polynomial with respect to the sequence B0 ; B1 ; B2 : : :.

As in Section 1, we deÃ¿ne the linear functional B acting on the space of polynomials by

B : Pâ†’R

xi â†’ B; xi = Bi ; i = 0; 1; 2; : : : :

The orthogonal polynomials satisfy

p

B; xi p (x) = 0; i = 0; 1; : : : ; p âˆ’ 1: (14)

These polynomials have been studied by Touchard ([31,9,28,29]) and generalized by Carlitz ([12,13]).

The following expressions

2

2x + p âˆ’ 2r x

p (x) =

p âˆ’ 2r r

2r6p

p p+k x+k p p+k x

p p

p k

= (âˆ’1) (âˆ’1) = (15)

k k k k k k

k=0 k=0

hold (see [34,12]).

Note that the p â€™s are orthogonal polynomials and thus satisfy a three-term recurrence relation.

The associated polynomials p of degree p âˆ’ 1 are deÃ¿ned as

p (x) âˆ’ p (t)

p (t) = B; ;

xâˆ’t

where B acts on x.

From expression (15) for p , we get the following formula for p :

x t

âˆ’

p p+k

p

k k

p (t) = B; :

xâˆ’t

k k

k=0

M. PrÃƒ vost / Journal of Computational and Applied Mathematics 122 (2000) 231â€“250

e 241

The recurrence relation between the Bernoulli numbers Bi implies that

x (âˆ’1)k

B; = :

k +1

k

x t

Using the expression of the polynomial (( k ) âˆ’ ( k ))=(x âˆ’ t) on the Newton basis on 0; 1; : : : ; k âˆ’ 1;

x t x

âˆ’

t k

k k iâˆ’1

= ;

xâˆ’t t

k i=1

i

i

we can write a compact formula for p:

p p+k t

p k

(âˆ’1)iâˆ’1

p (t) = âˆˆ Ppâˆ’1 :

t

k k k i=1

k=1

i2

i

Approximation (12) for (2) becomes

Ëœ p (t)

n n

1 1 p (n)

(2) â‰ˆ +t = + :

Ëœ p (t)

k2 k2 p (n)

k=1 k=1

t=1=n

n

Using partial decomposition of 1= with respect to the variable n, it is easy to prove that

i

dn

âˆˆ N; âˆ€i âˆˆ {1; 2; : : : ; n} (16)

n

i

i

with dn :=LCM(1; 2; : : : ; n).

A consequence of the above result is

d2 p (n) âˆˆ N; âˆ€p âˆˆ N

n

and

d2 (2) âˆ’ d2 (Sn

p (n) p (n) + p (n)) (17)

n n

is a Diophantine approximation of (2), for all values of integer p, where Sn denotes the partial

sums Sn = n 1=k 2 . It remains to estimate the error for the PadÃƒ approximation:

e

k=1

2 (t) âˆ’ [p=p] 2 (t) = 2 (t) âˆ’ [p âˆ’ 1=p] (t):

2 =t

Touchard found the integral representation for the linear functional B:

+iâˆž

dx

k

xk

B; x :=Bk = âˆ’i ; âˆ’1 Â¡ Â¡ 0:

sin2 ( x)

2 âˆ’iâˆž

242 M. PrÃƒ vost / Journal of Computational and Applied Mathematics 122 (2000) 231â€“250

e

Thus, formula (1) becomes

2

p (x)

t 2p +iâˆž

dx

âˆ’1

t 2 (t) âˆ’ [p âˆ’ 1=p] 2 =t (t) = âˆ’i ;

2 Ëœ 2 (t) 1 âˆ’ xt sin2 ( x)

âˆ’iâˆž

p

and we obtain the error for the PadÃƒ approximant to

e 2:

2

p (x)

+iâˆž

t dx

2 (t) âˆ’ [p=p] 2 (t) = âˆ’i

1 âˆ’ xt sin2 ( x)

2 p (t âˆ’1 )

2

âˆ’iâˆž

and the error for formula (17):

2

p (x)

+iâˆž

1 dx

d2 d2 (Sn âˆ’d2 i

p (n) (2) âˆ’ p (n) + p (n)) = : (18)

n n n

1 âˆ’ x=n sin2 ( x)

2n p (n) âˆ’iâˆž

If p = n, we get ApÃƒ ryâ€™s numbers [4]:

e

2

n n+k

n

bn = n (n) =

k k

k=0

and

2

n n+k

n n k

(âˆ’1)iâˆ’1

1

an = Sn n (n) + n (n) = bn + :

k2 n

k k i=1

k=1 k=1

i2

i

The error in formula (18) becomes

+iâˆž 2

1 n (x)

dx

d2 bn d2 an âˆ’d2 i

(2) âˆ’ = (19)

n n n

1 âˆ’ x=n sin2 x

2n bn âˆ’iâˆž

In order to prove the irrationality of (2), we have to show that the right-hand side of (19) tends

to 0 when n tends to inÃ¿nity, and is di erent from 0, for each integer n.

We have

1

2

n (âˆ’ 2

+ iu) du

âˆ’1=2+iâˆž 2 +âˆž

n (x) dx 1 2

6 6 B; n (x)

1 âˆ’ x=n sin2 x 1 + 1=2n cosh2 u 1 + 1=2n

âˆ’1=2âˆ’iâˆž âˆ’âˆž

since cosh2 u is positive for u âˆˆ R and n (âˆ’ 1 + iu) real positive for u real ( n has all its roots on

2

2

the line âˆ’ 2 + iR; because n (âˆ’ 2 + iu) is orthogonal with respect to the positive weight 1=cosh2 u

1 1

2

on R). The quantity B; n (x) can be computed from the three term recurrence relation between

the n s [31]:

(âˆ’1)n

2

B; n (x) = :

2n + 1

The Diophantine approximation (19) satisÃ¿es

1

|d2 bn (2) âˆ’ d2 an |6d2 Ã— :

n n n

(2n + 1)2 bn

M. PrÃƒ vost / Journal of Computational and Applied Mathematics 122 (2000) 231â€“250

e 243

âˆš

In [14], it is proved that bn âˆ¼ A ((1 + 5)=2)5n nâˆ’1 when n â†’ âˆž, for some constant A . From a

result concerning dn = LCM(1; 2; : : : ; n): (dn = e(n(1+o(1)) ), we get

lim |d2 bn (2) âˆ’ d2 an | = 0; (20)

n n

nâ†’âˆž

where d2 bn and d2 an are integers.

n n

Relation (20) proves that (2) is not rational.

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