SIMULTANEOUS RATIONAL APPROXIMATION VIA RICKERT S INTEGRALS

Transcription

1 SIMULTAEOUS RATIOAL APPROXIMATIO VIA RICKERT S ITEGRALS SARADHA AD DIVYUM SHARMA Abstract Using Rickert s contour integrals, we give effective lower bounds for simultaneous rational approximations to numbers in the sets { a ) ν, + a ) ν } { + a ) ν, + 2a ) ν } Here > a are integers, 0 < ν < is a rational number at least one of the radicals is irrational in each set The result is valid for all q where q denotes the denominator of the approximating rational number Introduction The method of Padé approximation has been widely used to obtain effective lower bounds for simultaneous rational approximations to algebraic numbers See for example [9], [2], [7], [8], [0] [5] In these papers hypergeometric polynomials were used as Padé approximants to transcendental functions of the form + ax) ν with a Z \ {0} ν Q + \ Rickert [] used contour integral representation for the Padé approximants to these transcendental functions to show the following result Let ) θ = /), θ 2 = + /) with 2 Then 2) max { θ p q, θ 2 p } 2 q for all integers p, p 2, q with q > 0 where > c q λ λ = + log2 + 24) log27 2 )/2) c = Mathematics Subject Classification J68, J82 Key words phrases Simultaneous approximation, Padé approximation, Contour integrals

2 2 SARADHA AD DIVYUM SHARMA This result leads to bounds for solutions to certain simultaneous Pelltype equations as soon as λ < 2 See [] for details By taking = 49 in the above result, he derives that ) max { 2 p q }, p 2 q > c 0 q λ 0 with c 0, λ 0 ) = 0 7, 9) for all integers p, p 2, q with q > 0 In [4], Bennett improved the value of λ in 2) asymptotically, ie for q q 0 where q 0 is effectively computable but not explicitly given In fact, this is a special case of a more general result [4, Theorem ]) proved by elaborating on the ideas of Chudnovsky [5] which included estimating more precisely the common denominators of the coefficients of the approximants As a result, he showed that ) holds with c 0, λ 0 ) =, 7955) for all integers p, p 2, q with q q 0, where q 0 is a large inexplicit constant In [], he made this result explicit with q 0 = c 0, λ 0 ) = 0 0, 86) This is a particular case of the following result [, Theorem 4] Let θ, θ 2 be given by ) Then max { θ p q, θ 2 p } 2 q > for all integers p, p 2, q with q > 0 where q λ λ = λ) = + log8 + 6) log8 2 )/64) Let a, be positive integers let 0 < ν < be a rational number Write ν = s/n with gcds, n) = Our aim in this paper is to extend the above result of Bennett for numbers θ, θ 2 ) where θ, θ 2 ) { a To state our results, we put ) ν, + a e 0 = p n ) ν ), + a p max{ordp n 2a )+ p,0} ) ν, + 2a ) ν )} 2, 4/, 68) if n = 2 4) e, e, e ) = 4,, 679) if n = 4 e 0, e 0 /2, ) otherwise

3 SIMULTAEOUS RATIOAL APPROXIMATIO VIA RICKERT S ITEGRALS Theorem Let θ = a ) ν θ2 = + a ) ν with 6794e 2 a Assume that at least one of θ θ 2 is irrational Then 2) is valid with c = 0005 ae e λ = + log6 e + 2ae ) log27 2 a 2 )/6a e )) a ) ν + 2a ) ν ) 9ae ν ν 2 λ+ ) 2 a) Remark 2 Taking a = ν =, in Theorem, we get 2 4 Theorems 4 4 of [] ote that λ < 2 whenever if ν = /2 62 if ν = /4 Example By taking = 8, a = ν = /, we obtain λ 7, 58) 8264 Theorem Let θ = + a ) ν θ2 = + 2a ) ν with 867e 2 a Assume that at least one of θ θ 2 is irrational Then 2) is valid with λ = + log6 e + 6ae 2 + )) log27 a) 2a)/6a e )) c = ) ν ) )a 9ae ae e ν ν 2 λ+ ) 2 2a) Remark 4 Let θ θ 2 be algebraic numbers such that, θ, θ 2 are linearly independent over Q Then it is known by a celebrated theorem of Schmidt [2] that given ɛ > 0 there exists c = cθ, θ 2, ɛ) > 0 such that max { θ p q, θ 2 p } 2 q > c q λ with λ = 5 + ɛ, for all integers p, p 2, q with q > 0 By the classical theorem of Dirichlet on Diophantine approximation, it follows that such a result cannot hold for λ = 5 Thus Schmidt s result is optimal, but it is ineffective in the sense that c cannot be computed In Theorems, c is explicitly determined for the specified numbers θ, θ 2 ) Further, the exponent λ of q is found to be less than 2

4 4 SARADHA AD DIVYUM SHARMA Let θ θ 2 be either as in Theorem or Theorem, with ν = /2 Take = 0 2 write αi θ i = s i 0 with 0, s i positive integers α i square free for i =, 2 By the irrationality assumption on θ θ 2, at least one of the α i s exceeds We apply Theorems with q replaced by 0 q p i replaced by s i p i to get { s α max s p 0 0 q, s 2 α2 s } 2p 2 c 0 0 q > 0 q) λ This shows that max { α p q, α2 p } 2 q > c q λ with c = c/ λ 0 maxs, s 2 )) We give few values of a, 0, the corresponding α, α 2 λ in Tables 2 below for Theorems, respectively In all these cases c may be taken as 0 7 The values of a that we have chosen are taken from Table of [4] in which an asymptotic value of λ is given The values of λ given in the tables below are valid for all q Table Examples for Theorem a 0 α α 2 λ Table 2 Examples for Theorem a 0 α α 2 λ

5 SIMULTAEOUS RATIOAL APPROXIMATIO VIA RICKERT S ITEGRALS 5 Remark 5 The main idea in the proofs of Theorems is to dilate with or without translating) the contour considered by Rickert for representing the Padé approximants as contour integrals See Lemma 42 Further, when ν =,, precise estimates for the coefficients of the approximants are used as in [] See Lemma To present our next result, we introduce some notation Let a 0 < a < a 2 be integers with one of them equal to zero Suppose that is a positive integer exceeding M = max a 0, a, a 2 ) For 0 j 2, put ξ j = + a ) ν j Denote by θ θ 2 the two values from ξ 0, ξ ξ 2, which are not equal to Define e = lcma a 0 )a 2 a 0 ), a a 0 )a 2 a ), a 2 a 0 )a 2 a )), e 2 = lcma a 0, a 2 a 0, a 2 a ), e = p max{ordpn/e2)+/p ),0}, p n e 4 = e e 2 e In the following theorem we show that, under certain conditions, one can obtain positive numbers c λ such that 2) holds for all integers p, p 2, q with q > 0 Although the method leads to explicit values of c, these values are very small we do not present them here We will only give the values of λ explicitly Theorem 6 Put G = G 0 = M + a a 0 ) a a 0 ) 2 2a 2 a 0 a ), 42+2M+a a 0 ) a a 0 ) 2 2a 2 +a 0 a if a ) a 0 a 2 a 42+2M+a 2 a ) a 2 a ) 2 a 2a 0 a 2 if a ) a 0 > a 2 a, G 2 = M + a 2 a ) a 2 a ) 2 a 2 2a 0 + a ), P = maxg 0, G, G 2 ), e 4 if ν = /2 D = 9e 4 /8 if ν = /4 e 4 /2 otherwise

6 6 SARADHA AD DIVYUM SHARMA Then 2) holds with λ = + provided the denominator is positive Examples logp D) log668 M) /D) ) By taking = 7 2, ν = /2 a 0, a, a 2 ) = 6, 0, 2), we obtain λ, 2425) ) By taking = , ν = /2 a 0, a, a 2 ) = 4, 0, 6), we obtain λ, 64) 8705 Remark 7 The proof of Theorem 6 depends on expressing the contour integral appearing in the Padé approximation as an infinite integral We connect this integral to the values of the classical Beta function, which are then estimated using Stirling s formula In Sections 2-6 we present several lemmas for the proofs of the theorems These lemmas deal with simultaneous rational approximation to algebraic numbers ) ν, aj Z, 0 < ν < + a j for 0 j m with m 2 one of the a j s equals zero theorems are obtained in Section 7 by specializing to m = 2 2 Approximating Forms The The lemma presented in this section is used to find effective lower bounds for simultaneous rational approximations to real numbers θ,, θ m by the construction of suitable independent approximation forms This is a mild variant of Lemma 2 from [] Lemma 2 Let θ,, θ m R Put θ 0 = Suppose that there are positive real numbers l, p, d, L, P, D, r 0 having the following property For each integer r r 0 0, there exist p ijr Q 0 i, j m), with non-zero determinant C r Z such that for all i, j = 0,, m, we have 5) 6) C r dd r, C r p ijr Z, p ijr pp r

7 7) SIMULTAEOUS RATIOAL APPROXIMATIO VIA RICKERT S ITEGRALS 7 m p ijr θ j l L r j=0 If L D >, then for all non-zero integer tuples p,, p m, q), q L 8) max where θ p q 9) χ =,, θ m p ) m q logp D) logl/d) > c q +χ, c = /2m + )pdp Dmax, 2ld)) χ ) Further for all q, we can take χ as in 9) where c = /2m + )pdp Dmax, 2ld)) χ W +χ ), W = [ ) r0 ] L + D D) r0, we have Remark 2 Our aim is to find χ as small as possible For this, we see from 9) that P D should be small L should be large The auxiliary parameters p, d l are introduced to facilitate this Proof of Lemma 2 Let p,, p m, q be integers with q L r0 D) Put p 0 = q δ = max θ p q,, θ m p ) m q Take where Then r = + [ ] logcq), logl/d) C = max, 2ld) L/D) r > Cq q L/D) r 0

8 8 SARADHA AD DIVYUM SHARMA Since L > D, this implies that r r 0 Therefore, by hypothesis, there exist numbers p ijr Q, 0 i, j m, with detp ijr ) 0, C r Z satisfying the conditions 5) 7) Then m m p ijr p j = q m p ijr θ j q p ijr θ j p j /q) j=0 j=0 j=0 0) qll r + m + )qpp r δ m Since detp ijr ) 0, there exists i such that p ijr p j j=0 is a nonzero rational number further, its denominator divides C r, which is dd r Therefore we get m ) p ijr p j dd r ) j=0 ote that L/D) r Cq 2ldq Comparing 0) ) we get implying that δ = m + )qpp r δ dd r ql L r 2dD r 2m + )pdqp D) r 2m + )pdqp DCq) χ c q +χ This proves the first part of the lemma If q < L r0 D) we apply 8) with q replaced by W q p i replaced by W p i, for i =,, m This gives the second part of the lemma Rickert s Contour Integral Let a 0 < a < < a m be integers, with one of them equal to zero Suppose that x > max a i Let r be a non-negative integer For 0 i m i = 0,, m, consider the contour integral 2) I i x, r, γ) = 2π γ + zx) r+ν z a i )Az)) r dz, where Az) = z a 0 ) z a m ) γ is a closed, counter-clockwise contour containing a 0,, a m, but not passing through any of them Then I i x) = I i x, r, γ) defines a function of x which is analytic near the origin The following lemma summarizes the results in Lemmas, 4 of []

9 SIMULTAEOUS RATIOAL APPROXIMATIO VIA RICKERT S ITEGRALS 9 Lemma Let 0 i m Then ) I i x) has a zero of order at least m + )r at the origin 2) There are polynomials q ijr x) 0 j m) of degree at most r such that m I i x) = q ijr x) + a j x) ν j=0 Here, q ijr x) + a j x) ν is the residue of the integr in 2) at z = a j for j = 0,, m Also, det q ijrx)) 0 0 i,j m Thus for every i with 0 i m r 0, the polynomials q ijr x) are the Padé approximants to the function + a j x) ν, 0 j m The polynomials q ijr x) were given explicitly in [] as ) q ijr x) = h 0 ++h m= r ij, h l 0 r + ν h j ) x h j +a j x) r h j 0 l m, l j ril h l ) a j a l ) r il+h l, where r il = r + δ il where δ il is the Kronecker delta function We will now take x = / where is a natural number exceeding max a i 0 i m For 0 i, j m put ξ j = + a ) ν j q ijr = q ijr /) ote that one of the ξ i s, say ξ u is equal to For 0 i, j m, put if j = 0 θ j = ξ j if j u if u + j m ξ j q iur if j = 0 p ijr = q ij )r if j u q ijr if u + j m We approximate θ i s simultaneously by using Lemma 2 For this purpose, we need to determine the values of d, p, l, D, P, L occurring in Lemma 2 Thus we need to obtain upper bounds of the shape m given in Lemma 2 for C r, p ijr p ijr θ j This is equivalent to j=0

10 0 SARADHA AD DIVYUM SHARMA m finding such bounds for C r, q ijr q ijr ξ j as p i0r,, p imr ) is a j=0 permutation of q i0r,, q imr ) for all i, r 4 Determination of p P Let M = max a t Put R 0 = a a 0, R 0 t m 2 m = am a m for j = 2,, m, define aj a j R j = min, a ) j+ a j 2 2 Let 0 i, j m If j = m or R j = a j a j, take 2 a j a j if i = j T ji = a j 2a i + a j if i < j 2a i a j + a j if i > j If j = 0 or R j = a j+ a j, take 2 a j+ a j if i = j T ji = a j 2a i a j+ if i < j 2a i a j a j+ if i > j Further we define g ji = 2R j + M + R j ) ν ξ j T ji ν G j = 2m+ + M + R j ) m T ji Lemma 4 Condition 6) in Lemma 2 is satisfied by taking i=0 p = max g ji) P = max G j i,j j Proof Fix j with 0 j m Let γ j denote the circle of radius R j centered at a j Then 4) q ijr x) + a j x) ν = 2π + zx) r+ν z a i )Az)) dz r otice that, for z γ j, 5) + z + a j + R j + M + R j γ j

11 SIMULTAEOUS RATIOAL APPROXIMATIO VIA RICKERT S ITEGRALS j z a i Az) r = z a j r+δ ji z a h r+δ hi Case : R r+δ ji j h=0 j a j a h R j ) r+δ hi h=0 Let R j = a j a j 2 In this case, z a i Az) r a j a j ) r+δ ji 2 m+)r+ By 4) 5), we get m h=j+ m h=j+ m h=j+ z a h r+δ hi a h a j R j ) r+δ hi j a j 2a h + a j ) r+δ hi h=0 2a h a j + a j ) r+δ hi q ijr g ji G r j Case 2: Let R j = a j+ a j 2 In this case z a i Az) r a j+ a j ) r+δ ji 2 m+)r+ m h=j+ Again, using 4) 5), we get This proves the lemma j a j 2a h a j+ ) r+δ hi h=0 2a h a j a j+ ) r+δ hi q ijr g ji G r j If the integers a j s are equally spaced, we deduce the following result Corollary 4 Let a 0,, a m satisfy a j+ a j = a j a j for all j =,, m Then condition 6) in Lemma 2 is satisfied by taking p = min θ + M + R ) ν m + 2) 2 + M + R) P = j 2 m+2 R m+ [m/2]!m [m/2])! 0 j m Here R = a a 0 2

12 2 SARADHA AD DIVYUM SHARMA Proof Since T ji 2R for all i, we get that 6) Further G j = g ji min 0 j m ξ j + M + R ) ν + M + R j ) ) m R m+ 2j 2h) 2h 2j) h=0 h=j+ The denominator in the above expression equals R m+ 2j)!2m 2j)! 2 m j!m j)! From the inequality ) 2t 4t t t + for the central binomial coefficient, it follows that ) ) 2j 2m 2j 4 m > j m j j + )m j + ) 4m+ m + 2) 2 Also min j!m j)! = [m/2]!m [m/2])! Therefore the denominator 0 j m of G j is at least This implies that 7) G j 2 m+2 R m+ [m/2]!m [m/2])! m + 2) 2 m + 2) 2 + M + R) 2 m+2 R m+ [m/2]!m [m/2])! The corollary follows from 6), 7) Lemma 4 ow we consider the case m = 2 in Corollary 4 In this case, we take the lemniscate used by Rickert as the contour get better values for p P Lemma 42 Suppose i) a 0, a, a 2 ) = a, 0, a), then condition 6) in Lemma 2 is satisfied by taking p = a ) ν a ) ν P = + 2a ) 2a

13 SIMULTAEOUS RATIOAL APPROXIMATIO VIA RICKERT S ITEGRALS ii) a 0, a, a 2 ) = 0, a, 2a), then condition 6) in Lemma 2 is satisfied by taking p = ) ν )a P = ) )a 2a Proof In the first case, Az) = zz 2 a 2 ) Let γ be the lemniscate defined by Az) = 2a By taking z = at, this can be transformed to γ : tt 2 ) = 2 around, 0, This splits into three contours γ 0, γ γ 2 such that γ i is a contour around i not including the other two points Let γ i z) = γ iz/a) for 0 i 2 From 4) with x =, we get q ijr + a j ) ν = 2π + z ) r+ν z a i )zz 2 a 2 )) dz r γ j + z r+ν 2π z a i z z 2 a 2 ) dz r γ j Thus q ijr + a j ) ν 2πa r + at r+ν t a i a tt 2 ) r dt γ j ow we use some numerical calculations from the proof of [, Lemma 4] On γ 0, we have t a i a > 044 t / Also, γ 0 has length < 2775 Therefore 8) q i0r + a 0 ) ν a ) ν ) r + a ) r 2a ) r Similarly, on γ γ 2 we have t a i a > 054 t 2/

14 4 SARADHA AD DIVYUM SHARMA Further, the lengths of both γ γ 2 can be bounded by 496 Therefore, for j =, 2 we obtain q ijr + a j ) ν a ) ν ) r + 2a ) r 2a ) r which is also valid for j = 0 by 8) Thus we may take p = a ) ν a ) ν P = + 2a ) 2a Similarly, when a 0, a, a 2 ) = 0, a, 2a), we have Az) = zz a)z 2a) Taking γ j = γ j z a ) following the above argument, we get a p = ) ν )a P = ) )a 2a 5 Determination of l L Lemma 5 Condition 7) in Lemma 2 is satisfied by taking l = e + ) /2 L = 099m M) m+ + ) m+ M) sinπν) m m if r 00 Proof Following [4, Lemma, p 72], it is possible to write I i ) = )mr e πν e 2πν ) 2π m+)r Observe that x + + a i = x + ) + a Therefore 9) I i ) = π m+)r 0 0 x r+ν x + + a i ) m dx x + + a l ))r l=0 i x+) x + ) M ) x r+ν x + ) m+)r+ M Br + ν +, mr ν) π M) m+)r+ )m+)r+ dx

15 SIMULTAEOUS RATIOAL APPROXIMATIO VIA RICKERT S ITEGRALS 5 where Bm, n) is the classical Beta function, see [6, Exercise 4, p 228] for the above assertion It is well known that 20) Br + ν +, mr ν) = πr + ν) νmr ν) ν) sinπν)m + )r)! πr + )!mr)! sinπν)m + )r)! since 0 < ν < If x 2 is an integer, then by Stirling s formula [, Formula 68], we have x + 05) log x x + 9 < log x! < x + 05) log x x + Using this, we obtain ) r + )!mr)! log m + )r)! r + 5) logr + ) + mr + 05) logmr) mr + r + 05) logmr + r) + 0 = r log + /r) + 5 logr + ) r log m mr + r + 05) log + /m) + 0 Hence, 2) r + )!mr)! m + )r)! e0 + /r) r r + ) 5 e + ) /2 ) r + ) 5/r r m r + /m) mr+r+05 m m + /m) m+ Combining 9) 2) using r + ) 5/r 099 for r 00, we get the assertion of the lemma For some special numbers a i the values of l L can be improved as shown in the lemma below Lemma 52 Suppose that the set S = {a 0,, a m } satisfies one of the following two properties: i) All the elements of S are non-negative ii) a i S if only if a i S Then condition 7) in Lemma 2 is satisfied by taking ) l = + ν m + ) m+ L = m + )m+ a 0 ) a m ) m + m m a m ) m m Proof We follow [, Lemma ] Let γ be a closed contour enclosing all the a i s, traced counter clockwise Then ) r + ν I i x, r, γ) = x h J ih, h h=m+)r

16 6 SARADHA AD DIVYUM SHARMA where J ih = 2π γ z h z a i )Az)) r dz Therefore ) I i, r, γ b h J ih h, h=m+)r where b h = ) r+ν h For h r, bh decreases as h increases Hence we have ) 22) I i, r, γ b m+)r J ih h h=m+)r Further, by induction on r it follows that ) ) b m+)r + ν m m r 2) m + m + ) m+ Let J i x) = x h J ih h=0 Suppose that x < /M that γ is a circle centered at the origin, with radius r satisfying M < r < x Arguing as in the proof of [, Lemma 2], we get that J i x) = 2π γ xz)z a i )Az)) r dz = Res z=/x xz)z a i )Az)) r = a i x)a/x) r Thus when property i) holds for S, the coefficients J ih of J i x) are all positive Taking x = / in the above expression, we obtain that 24) J i /) = a i )A) = J r ih h This is also true when property ii) holds for S a i is non-negative Further, when property ii) holds, we have A) = A ) ow suppose that a i is negative Then we get 25) J ih h = J i /) = a i )A) r h=0 h=0

17 SIMULTAEOUS RATIOAL APPROXIMATIO VIA RICKERT S ITEGRALS 7 Hence from 22) 25), we find in both cases ) ) ) I i, r, γ + ν m m r m + m + ) m+ a i A) r which gives the values of l L as claimed Define 6 Determination of d D c = lcm{ 0 h m h i a h a i : 0 i m}, c 2 = lcm{ a l a t : 0 l < t m}, c = p n p max{ordpn/c 2)+/p ),0}, c 4 = c c 2 c Further, for k with k < n, let Sk) be the set of all primes p > nr + s, gcdp, nr) =, { r } > max nm k, k ) pk s mod n) p nm n Then it follows from [4, Lemma 42, p 728] that a denominator for p ijr can be taken as c 4 ) r c 2 rhn, r) where Hn, r) = k n, p Sk) k,n)= p Further, in [, Lemma ], Bennett used estimates from Prime umber Theory to show that H2, r) < 682/) r H4, r) < 679/4) r Using these results, we prove the following lemma Lemma 6 Let e e be given by 4) i) Condition 5) in Lemma 2 is satisfied by taking d = c 2 e c 4 if ν = /2 D = 9c 4 /8 if ν = /4 c 4 /2 otherwise ii) If m = 2 a 0, a, a 2 ) = a, 0, a) or 0, a, 2a), then condition 5) in Lemma 2 is satisfied by taking d = 2ae D = 4a e

18 8 SARADHA AD DIVYUM SHARMA Proof i) ote that e 2/) r if ν = /2 Hn, r) e /4) r if ν = /4 e otherwise Since a denominator for q ijr can be taken as c 4 ) r c 2 rhn, r) which is at most c 4 ) r c 2 5 r Hn, r), we can take d D as in the statement of the lemma ii) Observe that c = 2a 2, c 2 = 2a c = e 0 When ν,, we 2 4 bound the denominator of q ijr by c 4 ) r c 2 5 r e Let ν =, Hence 2 4 n = 2 or 4 Any summ in ) with m = 2 x = / is of the form j + a j ) r h j n h j hj! r a r iα+r iβ +h α+h β2 r iα +h α ) ) riα riβ where j = nr + ) nr + nh j )), 0 α, β 2, α, β not equal to j h α + h β + h j = r ij ote that 2 does not divide j ow, as in [], we get that divides riα j h α not exceed ) riβ h β h α h j! 2 ord 2h j!) Hn, r) ) Hence a denominator for the summ does h β Hn, r)2 2r+δ iα+δ ij n h j r a r+δ iα+δ iβ +δ ij h j 2Hn, r)4a ) r 2 r maxord 2n/a),0) Thus a denominator for q ijr can be taken as c 4 ) r c 2 Hn, r) Further, since n = 2 or 4, we have e 0 n Using the above bounds for Hn, r) we conclude that we may take d = 2ae D = 4a e 7 Proof of the Theorems For the proofs of the theorems we take m = 2 Then c i = e i for i 4 Proof of Theorem Let a 0, a, a 2 ) = a, 0, a), a 6794e 2 a Then by Lemmas 42i), 52 6ii), we can take p = a ) ν a ) ν, P = + 2a ) 2a,

19 SIMULTAEOUS RATIOAL APPROXIMATIO VIA RICKERT S ITEGRALS 9 l = 9ν ν2 ), L = 27 8 a) 4 2 a 2 ), 26) d = 2ae D = 4a e Thus L = 27 2 a 2 ) L/D > by the assumption on The values D 6a e of c χ are calculated from the expressions given by Lemma 2 In order that χ <, we require Thus χ < whenever P D 2 < L 9 2 a 2 ) > 2 a e 2 + 2a ) Since e 4/ a, the above quadratic inequality holds if 6794e 2 a This proves the theorem Proof of Theorem Let a 0, a, a 2 ) = 0, a, 2a), a Then by Lemmas 42ii), 52 6ii), we can take p = ) ν )a, P = ) )a 2a, l = 9ν ν2 ) 8 2a), L = a + 2a 2 ) d D as in 26) The values of c χ are calculated from the expressions given by Lemma 2 In this case, χ < whenever 9 2 a + 2a 2 ) > 2 a e ) )a which holds for 867e 2 a This completes the proof of the theorem Proof of Theorem 6 We apply Lemmas 4 6i) with m = 2 to get the values of P D as in the statement of the theorem From Lemma 5, for r 00 we get L = M) /2) 668 M) Thus by Lemma 2, the value of χ can be taken as logp D) log668 M) /D)

20 20 SARADHA AD DIVYUM SHARMA Acknowledgement We thank Professor R Tijdeman for his many useful suggestions References [] M Abramowitz I A Stegun, Hbook of Mathematical Functions: with Formulas, Graphs, Mathematical Tables, Dover Publications 964) [2] A Baker, Simultaneous rational approximations to certain algebraic numbers, Proc Camb Phil Soc 6 967), [] M A Bennett, Simultaneous approximation to pairs of algebraic numbers, Canad Math Soc Proc 5 995), [4] M A Bennett, Simultaneous rational approximation to binomial functions, Trans Amer Math Soc ), [5] G V Chudnovsky, On the method of Thue-Siegel, Ann Math II) 7 98), [6] E T Copson, An introduction to the theory of functions of a complex variable, Oxford University Press, London 95) [7] I Fel dman, An estimate of an incomplete linear form in several algebraic variables, Math otes 7 970), 4-49 [8] I Fel dman, Effective bounds for the number of solutions of certain Diophantine equations, Math otes 8 970), [9] K Mahler, Ein Beweis des Thue-Siegelschen Satzes über die Approximation algebraischen Zahlen für binomische Gleichungen, Math Ann 05 9), [0] C F Osgood, The simultaneous Diophantine approximation of certain kth roots, Proc Camb Phil Soc ), [] J H Rickert, Simultaneous rational approximations related Diophantine equations, Math Proc Camb Phil Soc 99), [2] W M Schmidt, Simultaneous approximation to algebraic numbers by rationals, Acta Math ), Saradha School of Mathematics Tata Institute of Fundamental Research Homi Bhabha Road, Mumbai India address: saradha@mathtifrresin Divyum Sharma School of Mathematics Tata Institute of Fundamental Research Homi Bhabha Road, Mumbai India address: divyum@mathtifrresin