On approximation of real, complex, and p-adic numbers by algebraic numbers of bounded degree. by K. I. Tsishchanka
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1 On approximation of real, complex, and p-adic numbers by algebraic numbers of bounded degree by K. I. Tsishchanka
2 I. On approximation by rational numbers Theorem 1 (Dirichlet, 1842). For any real irrational number ξ there exist infinitely many rational numbers p/q such that ξ p q < 1 q 2.
3 Example: Let ξ = e. Consider the continued fraction expansion: We have e = = = = and so on.
4 The first convergents are: δ 1 = 3 e 3 < 1 δ 2 = 8 3 e 8 3 < δ 3 = 11 4 e 11 4 < δ 4 = 19 7 e 19 7 < δ 5 = e < δ 6 = e <
5 We also note that δ 1 = 3 e 3 < δ 2 = 8 3 e 8 3 < δ 3 = 11 4 e 11 4 < δ 4 = 19 7 e 19 7 < δ 5 = e < δ 6 = e <
6 Theorem 2. For any real irrational number ξ there exist infinitely many rational numbers p/q such that ξ p q < 1 2q 2.
7 Theorem 2. For any real irrational number ξ there exist infinitely many rational numbers p/q such that ξ p q < 1 2q 2. Theorem 1. For any real irrational number ξ there exist infinitely many rational numbers p/q such that ξ p q < 1 q 2.
8 Finally, some convergents give even better approximation: δ 1 = 3 e 3 < δ 2 = 8 3 e 8 3 < δ 3 = 11 4 e 11 4 < δ 4 = 19 7 e 19 7 < δ 5 = e < δ 6 = e <
9 Theorem 3 (Hurwitz). For any real irrational number ξ there exist infinitely many rational numbers p/q such that ξ p q < 1. 5q 2
10 Theorem 3 (Hurwitz). For any real irrational number ξ there exist infinitely many rational numbers p/q such that ξ p q < 1. 5q 2 This result is the best possible.
11 Theorem 3 (Hurwitz). For any real irrational number ξ there exist infinitely many rational numbers p/q such that ξ p q < 1. 5q 2 This result is the best possible. Example: ξ = =
12 II. Polynomial Interpretation Theorem 1. For any real irrational number ξ there exist infinitely many rational numbers p/q such that ξ p q < 1 q 2.
13 II. Polynomial Interpretation Theorem 1. For any real irrational number ξ there exist infinitely many rational numbers p/q such that ξ p q < 1 = qξ p < q 1. q 2
14 II. Polynomial Interpretation Theorem 1. For any real irrational number ξ there exist infinitely many rational numbers p/q such that ξ p q < 1 = qξ p < q 1. q 2 Theorem 4. For any real irrational number ξ there exist infinitely many polynomials P Z[x] of the first degree such that P (ξ) P 1, where P denotes the height of P.
15 II. Polynomial Interpretation Theorem 1. For any real irrational number ξ there exist infinitely many rational numbers p/q such that ξ p q < 1 = qξ p < q 1. q 2 Theorem 5. For any real number ξ A n there exist infinitely many polynomials P Z[x] of degree n such that P (ξ) P n, where A n is the set of real algebraic numbers of degree n.
16 ξ p q < q 2
17 ξ p q < q 2 qξ p < q 1
18 ξ p q < q 2 qξ p < q 1 P (ξ) P n
19 ξ p q < q 2 qξ p < q 1 ξ α P (ξ) P n
20 ξ p q < q 2 qξ p < q 1 ξ α? P (ξ) P n
21 III. Conjecture of Wirsing Conjecture (Wirsing, 1961). For any real number ξ A n, there exist infinitely many algebraic numbers α A n with ξ α H(α) n 1+ɛ, ɛ > 0, where H(α) is the height of α. Further W. M. Schmidt conjectured that the exponent n 1 + ɛ can be replaced even by n 1.
22 ξ p q < q 2 qξ p < q 1 ξ α H(α) n 1 P (ξ) P n
23 ξ p q < q 2 qξ p < q 1 ξ α H(α) n 1 P (ξ) P n At the moment this Conjecture is proved only for n = 1 ξ α H(α) 2 (Dirichlet)
24 ξ p q < q 2 qξ p < q 1 ξ α H(α) n 1 P (ξ) P n At the moment this Conjecture is proved only for n = 1 ξ α H(α) 2 (Dirichlet) n = 2 ξ α H(α) 3 (Davenport Schmidt)
25 ξ p q < q 2 qξ p < q 1 ξ α H(α) n 1 P (ξ) P n At the moment this Conjecture is proved only for n = 1 ξ α H(α) 2 (Dirichlet) n = 2 ξ α H(α) 3 (Davenport Schmidt) n > 2???
26 Consider the polynomial P (x) = a n x n a 1 x + a 0 = a n (x α 1 )... (x α n ). Without loss of generality we can assume that α 1 is the root of P (x) closest to ξ.
27 It is known that ξ α 1 P (ξ) P (ξ). By Theorem 3 there are infinitely many polynomials P Z[x] of degree n such that Let n = 1. Then P (ξ) P n. P (ξ) = a 1 P ξ α 1 P 1 P = P 2
28 It is known that ξ α 1 P (ξ) P (ξ). By Theorem 3 there are infinitely many polynomials P Z[x] of degree n such that P (ξ) P n. Let n = 2. Then for some δ 1 we have P (ξ) = 2a 2 ξ + a 1 P δ ξ α 1 P 2 P δ = P 2 δ
29 Question: Can one prove that the following is impossible: All polynomials with P (ξ) P n have a small derivative P (ξ) P δ.
30 Question: Can one prove that the following is impossible: All polynomials with P (ξ) P n have a small derivative P (ξ) P δ. Answer: Yes, for n=1 (Dirichlet, 1842) n=2 (Davenport Schmidt, 1967)
31 IV. Theorem of Wirsing Theorem 6 (Wirsing, 1961). For any real number ξ A n there exist infinitely many algebraic numbers α A n with ξ α H(α) n 2 λ n, lim n λ n = 2.
32 By Dirichlet s Box principle there are infinitely many polynomials P (x) = a n (x α 1 )... (x α n ) such that P (ξ) P n, therefore ξ α 1... ξ α n P n a 1 n. Even if a n = P, we can only prove that ξ α 1... ξ α n P n 1 H(α 1 ) n 1, ξ α 1 H(α 1 ) n+1 n??? It is also clear, that the worth case for us is when ξ α 1 =... = ξ α n.
33 Question: Can one prove that for infinitely many polynomials P Z[x] with P (ξ) P n the situation ξ α 1 =... = ξ α n is impossible?
34 Question: Can one prove that for infinitely many polynomials P Z[x] with P (ξ) P n the situation ξ α 1 =... = ξ α n is impossible? Answer: For infinitely many polynomials P Z[x] with P (ξ) P n we have: ξ α 1 ξ α 2 1, ξ α 3,..., ξ α n are big.
35 Step 1: Construct infinitly many P, Q Z[x], deg P, Q n, such that P (ξ) P n Q(ξ) Q n P Q and P, Q have no common root
36 Step 2. Consider the resultant of P, Q : R(P, Q) = a l m bm l (α i β j ). 1 i,j n
37 Step 2. Consider the resultant of P, Q : R(P, Q) = a l m bm l (α i β j ). 1 i,j n On the one hand, R(P, Q) 0, since P, Q have no common root.
38 Step 2. Consider the resultant of P, Q : R(P, Q) = a l m bm l (α i β j ). 1 i,j n On the one hand, R(P, Q) 0, since P, Q have no common root. Moreover, R(P, Q) Z, since P, Q have integer coefficients.
39 Step 2. Consider the resultant of P, Q : R(P, Q) = a l m bm l (α i β j ). 1 i,j n On the one hand, R(P, Q) 0, since P, Q have no common root. Moreover, R(P, Q) Z, since P, Q have integer coefficients. Therefore we get R(P, Q) 1.
40 Step 3. On the other hand, R(P, Q) = a l m bm l α i β j. 1 i,j n
41 Step 3. On the other hand, R(P, Q) = a l m bm l α i β j 1 i,j n P 2n α i β j. 1 i,j n
42 Step 3. On the other hand, R(P, Q) = a l m bm l α i β j 1 i,j n P 2n α i β j P 2n 1 i,j n 1 i,j n max( ξ α i, ξ β j ).
43 Step 3. On the other hand, R(P, Q) = a l m bm l α i β j 1 i,j n P 2n α i β j P 2n 1 i,j n 1 i,j n max( ξ α i, ξ β j ). If ξ α 1 =... = ξ α n P 1 n, 1 ξ β 1 =... = ξ β n P 1 n 1
44 Step 3. On the other hand, R(P, Q) = a l m bm l α i β j 1 i,j n P 2n α i β j P 2n 1 i,j n 1 i,j n max( ξ α i, ξ β j ). If ξ α 1 =... = ξ α n P 1 n, 1 ξ β 1 =... = ξ β n P 1 n, 1 then R(P, Q) P 2n P ( 1 n)n 1 2 = P n n2 < 1
45 Step 3. On the other hand, R(P, Q) = a l m bm l α i β j 1 i,j n P 2n α i β j If then P 2n 1 i,j n 1 i,j n max( ξ α i, ξ β j ). ξ α 1 =... = ξ α n P 1 1 n, ξ β 1 =... = ξ β n P 1 1 n, R(P, Q) P 2n P ( 1 1 n)n 2 = P n n2 < 1, which gives a contradiction, since R(P, Q) 1 by Step 2.
46 Lemma (Wirsing, 1961): P (ξ) 1 2 Q(ξ) P n 3 2, ξ γ max P (ξ) Q(ξ) 1 2 P n 3 2, where γ is a root of P or Q closest to ξ. Since we get P (ξ) P n, Q(ξ) Q n, ξ γ P n 2 n+n 3 2 = P n
47 V. Big Derivative Method Theorem 7 (Bernik-Tsishchanka, 1993). For any real number ξ A n there exist infinitely many algebraic numbers α A n with ξ α H(α) n 2 λ n, lim n λ n = 3.
48 The following table contains the values of n 2 + λ n corresponding to Wirsing s Theorem (1961), the Theorem of Bernik-Tsishchanka (1993), and the Conjecture: n Conj
49 Fix some H > 0. By Dirichlet s Box Principle there exists an integer polynomial P such that a n H,..., a 2 H, a 1 H 1+ɛ, a 0 H 1+ɛ, (1) where ɛ > 0. P (ξ) H n ɛ,
50 Case A: Let max( a 1, a 0 ) H, that is max( a 1, a 0 ) = H 1+δ = P, 0 < δ ɛ. It is clear that in this case the derivative of P is big, that is P (ξ) H 1+δ. (2) We have the following well-known inequality ξ α P (ξ) P (ξ), (3) where α is the root of the polynomial P closest to ξ. Substituting (1) and (2) into (3), we get ξ α H n ɛ H 1+δ = H (1+δ) n+1+ɛ+δ 1+δ = P n+1+ɛ+δ 1+δ H(α
51 Case B: Let max( a 1, a 0 ) H, then P H. (4) Using Dirichlet s Box we construct an integer polynomial Q such that b n H,..., b 2 H, b 1 H 1+ɛ, b 0 H 1+ Q(ξ) H n ɛ, (5) If max( b 1, b 0 ) H, then ξ β H(β) n+1+2ɛ 1+ɛ. If max( b 1, b 0 ) H, then Q H. (6) Then we can apply Wirsing s Lemma: P (ξ) 1 2 Q(ξ) P n 3 2, ξ γ max P (ξ) Q(ξ) 1 2 P n 3 2, Substituting (4), (5), (6), and P (ξ) H n ɛ, we get: ξ γ H n ɛ H(γ) n ɛ.
52 Let us compare estimates in the Case A and Case B: Case A: Case B: ξ α H(α) n+1+2ɛ 1+ɛ ξ α H(α) n ɛ If we take ɛ = 0, then Case A: ξ α H(α) n 1 Case B: ξ α H(α) n On the other hand, if we take ɛ = 2, then Case A: ξ α H(α) n+5 3 Case B: ξ α H(α) n Finally, if we choose an optimal value of ɛ, namely we obtain ɛ = 1 6 n, ξ α H(α) n/2+λ n, lim n λ n = 3, in both cases.
53 VI. Improvement Let us consider an integer polynomial P such that a n H,..., a 3 H, a 2 H 1+ɛ, a 1 H 1+ɛ, a 0 H 1+ɛ, P (ξ) H n 2ɛ. We have Case A: ξ α H(α) n+1+3ɛ 1+ɛ Case B: ξ α H(α) n ɛ. Put then ɛ = 1 10 n, ξ α H(α) n/2+λ n, lim n λ n = 4.5, in both cases. However, the Case A does not work. In fact, max( a 2, a 1, a 0 ) H P (ξ) is big.
54 VII. Method of Polynomial Staircase In 1996 a new approach to this problem was introduced: Step 1. Let R (k) be a polynomial with k big coefficients. We construct the following n polynomials which are small at ξ. Q (3),..., Q (n+1), P (n+1), Step 2. We prove that they are linearly independent. Step 3. Using a linear combination of these polynomials, we construct the polynomial L (2) = d 1 Q (3) d n 1 Q (n+1) + d n P (n+1) with two big coefficients. The Case A does work for L. Moreover, it is possible to show that an influence of the numbers d 1,..., d n is very weak, so L(ξ) H n 2ɛ.
55 Theorem 8. For any real number ξ A n there exist infinitely many algebraic numbers α A n with ξ α H(α) n 2 λ n, lim n λ n = 4.
56 The following table contains the values of n 2 + λ n corresponding to Wirsing s Theorem (1961), the Theorem of Bernik-Tsishchanka (1993), Theorem 8 (2001), and the Conjecture: n Conj
57 VIII. Complex case Theorem 9 (Wirsing, 1961). For any complex number ξ A n there exist infinitely many algebraic numbers α A n with where ξ α H(α) A, A = n Method: Resultant
58 In 2000 this result was slightly improved: A = n 4 + λ n, where lim n λ n = 3 2. Method: Big Derivative. Method Polynomial Staircase :???
59 IX. P -adic case Theorem 10 (Morrison, 1978). Let ξ Qp. If ξ A n, then there are infinitely many algebraic numbers α A n with ξ α H(α) A, where A = when n = 2, n when n > 2.
60 Theorem 11 (Teulié, 2002). If ξ A 2, then there are infinitely many algebraic numbers α A 2 with ξ α H(α) 3. The second part of Morrison s theorem was also improved: A = n 2 + λ n, where lim n λ n = 3. Method: Big Derivative. Method Polynomial Staircase :???
61 X. Two Counter-Examples 1. Simultaneous case. Conjecture. For any two real numbers ξ 1, ξ 2 A n there exist infinitely many algebraic numbers α 1, α 2 such that { ξ1 α 1 P (n+1)/2, ξ 2 α 2 P (n+1)/2, where P (x) Z[x], P (α 1 ) = P (α 2 ) = 0, deg P n. The implicit constant in should depend on ξ 1, ξ 2, and n.
62 Counter-example (Roy-Waldschmidt, 2001). For any sufficiently large n there exist real numbers ξ 1 and ξ 2 such that max{ ξ 1 α 1, ξ 2 α 2 } > P 3 n. Theorem 12. For any real numbers ξ 1, ξ 2 A n at least one of the following assertions is true: (i) There exist infinitely many algebraic numbers α 1, α 2 of degree n such that ξ 1 α 1 P n 8 3 8, ξ 2 α 2 P n (ii) For some ξ {ξ 1, ξ 2 } there exist infinitely many algebraic numbers α of degree 2 k n+2 4 such that ξ α H(α) n
63 2. Approximation by algebraic integers. Theorem 13. (Davenport - Schmidt, 1968) Let n 3. Let ξ be real, but not algebraic of degree 2. Then there are infinitely many algebraic integers α of degree 3 which satisfy where 0 < ξ α H(α) η 3, η 3 = 1 2 (3 + 5) =
64 Conjecture. Let ξ be real, but is not algebraic of degree n. Suppose ɛ > 0. Then there are infinitely many real algebraic integers α of degree n with ξ α H(α) n+ɛ. Theorem 14 (Roy, 2001). There exist real numbers ξ such that for any algebraic integer α of degree 3, we have ξ α H(α) η 3.
65 XI. Most Recent Result Theorem 15. For any real number ξ A 3 there exist infinitely many algebraic numbers α A 3 such that ξ α H(α) A, where A = is the largest root of the equation 2x 3 11x x + 8 = 0.
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