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Saturday, December 14, 2013

1.2. Mann's Theorem - Canonical Extension

 Definition 1:  Base of the Canonical Extension : \beta_0

If a sequence of integers C does not possess the property of normality (see Definition 1, here), then there exists c,c' in the interval (0,n) such that c\notin C,c'\notin C, c + c' - n = a + b where C = A+B and a\in A and b\in B

Let \beta_0\in B be the least number b for which c + c' - n = a + b


Definition 2:  Canonical Extension of the set C : C_1

The equation c + c' - n = a + \beta_0 has a solution in the numbers c, c', a where c\notin C, c'\notin C; a\in A; c, c', a \in (0,n).

Let C^{*} be the set of all numbers c,c' included in the above solution so that C\cap C^{*} is the empty set.

We will call C_1 = C \cup C^{*} the "canonical extension" of the set C.


Definition 3:  Canonical Extension of the set B : B_1

Let B^{*} be the set of all values \beta_0 + n - c where c\in C^{*}

Each number b*\in B^{*} can be written in the form c' - a where c' \in C^{*} and a \in A [Since \beta_0+n- c = c' - a from the equation c + c' - n = a + \beta_0 above]

Since b* has the form \beta_0+n-c, it follows that b* \ge \beta_0 \ge 0

Since b* = c' - a, b* \le c' \le n

So, all numbers b*\in B^{*} lie in the interval (0,n).

For all numbers b*\in B^{*}, b*\notin B since otherwise, b* = c' - a would imply that since c' = b* + a, c' \in C which is false.

We will call B_1 = B\cup B^{*} the "canonical extension" of the set B.


Lemma 1:  A + B_1 = C_1

Proof:

(1)  Either b_1\in B or b_1\in B^{*}

(2)  If b_1 \in B, then a + b_1 \in C \subseteq C_1

(3)  If b_1 \in B^{*}, then b_1 = \beta_0 + n - c' where c' \notin C

(4)  Since c + c' - n = a + \beta_0, b_1 = c - a


QED







Wednesday, November 27, 2013

1.2 Mann's Theorem: Theorem 1.2.1. (H. B. Mann)

The details in these notes are taken from A. O. Gelfond & Yu. V. Linnik, elementary methods in analytic number theory, 1965 (translated by Amiel Fienstein/Revised and edited by L. J. Mordell).

Lemma 1.  If d(A) + d(B) > 1, we can assume that \exists A',B' such that:
  • d(A') + d(B') = 1 
  • A' \subseteq A
  • B' \subseteq B
Proof:

(1)  Assume that d(A) + d(B) > 1

(2)  There exists A',B' where A' \subseteq A and B' \subseteq B and d(A')+d(B') \le 1

(3)  In fact, we can assume d(A') + d(B') is as close to 1 as we want since:

     (a)  We can assume that d(B) \le d(A) < 1.

     Note:  If d(A)=1, then we set B' = B - \{1\} to get d(A)+d(B')=1
     (b)   Let \epsilon be any number such that 1 > \epsilon > 0

     (c)  We can assume that 1 - d(A) > \epsilon

     Note: If not, then we set B' = B - \{1\} to get \epsilon > 1 - d(A)-d(B')

     (d)  There exists integers x,y such that 1 - d(A) - \epsilon \le \frac{x}{y} < 1 - d(A)

     (e)  From the definition of density, we know that d(B) \le \frac{B(y)}{y}

     (f)  Using y, we can build B' by removing B(y)-x elements from B.

     Note 1:   \frac{x}{y} < d(B) \le \frac{B(y)}{y} since d(A) + d(B) > 1 but d(A) + \frac{x}{y} < 1.

     Note 2:  If for any reason the approach in step 3f doesn't work, we can choose a larger x' so that x' > x and \frac{x'}{y'} = \frac{x}{y}

QED

Lemma 2.  if for every n \ge 1, there exists m such that m \in [1,n] and C(n) - C(n-m) \ge [d(A) + d(B)]m, then d(C) \ge d(A) + d(B).

Proof:

(1)  Assume that for every n \ge 1, there exists m such that m \in [1,n] and C(n) - C(n-m) \ge [d(A) + d(B)]m.

(2)  Then, for any n, we have:
C(n) - C(n-m) \ge (d(A) + d(B))m
C(n-m)-C(n-m-m') \ge (d(A) + d(B))m'
C(n-m-m')-C(n-m-m-m'') \ge (d(A) + d(B))m''
\vdots
(3)  Adding all of these values up, gets us to:
C(n) \ge (d(A)+d(B))(m + m' + m'' + \cdots) = (d(A) + d(B))n
(4)  Divide both sides by n to get:
\frac{C(n)}{n} \ge d(A)+d(B)
(5)  Since this is true for all n, using the definition of infimum [see Definition 3, here], it follows that:
\inf_n{\frac{C(n)}{n}} \ge d(A)+d(B)
(6)  This gets us to:  d(C) \ge d(A) + d(B).  [see Definition 4, here]

QED

Lemma 3.  If A(n) + B(n) > n-1, then n occurs in A+B

Proof:

(1)  Assume A(n) + B(n) > n-1

(2)  Assume n \notin C = A+B  [If it is, we are already done.]

(3)  Then: A(n-1) = A(n) and B(n-1) = B(n) so that:
A(n-1)+B(n-1)>n-1
(4)  Let a_1,a_2,\cdots,a_r and b_1,b_2\cdots,b_s be the numbers of [1,n-1] which appear in A and B respectively.

(5)  So that, r = A(n-1) and s = B(n-1)

(6)  Then, the numbers a_1, a_2, \cdots, a_r and n - b_1, n_b2, \cdots, n - b_s are all distinct and belong to [1,n-1]

(7)  But by assumption, there are more than n-1 of them which is impossible.

(8)  Therefore, we reject our assumption in step #2.

QED

Definition 1:  normal

A system of numbers H \subset [0,n] is normal if for all f \in [1,n]; f' \in [1,n]; f \notin H; f'\notin H:
f + f' - n \notin H
Lemma 4.  If C is normal, then for every n \ge 1 one can find m \in [1,n] such that:
C(n) - C(n-m) \ge (\alpha + \beta)m
where:
  • C=A+B
  • \alpha = d(A)
  • \beta = d(B)
Proof:

(1)  We can assume that n \notin C

Note: if n \in C, then there is a trivial solution for m=1 since:
C(n)-C(n-m) = C(n)-C(n-1) = 1 \ge (d(A)+d(B))1

(2)  Let m be the least positive integer not belonging to C

(3)   It is clear that m \le n

(4)  We can assume that m < n since:

    (a)  Assume that m = n

    (b)  C(n) - C(n-m) = C(n) - C(0) = n-1

   (c)  From Lemma 3 above, A(n) + B(n) \le n-1

   (d)  So, it follows that C(n) - C(0) \ge A(n) + B(n) \ge (d(A) + d(B))n

(5)  Let s \in [n-m,n] such that 0 < s+m-n < m

(6)  Now s \in C.  If not, then we would have s \notin C; m \notin C and s+m-n \notin C but s+m-n < m which goes against m being the least positive integer not belonging to C.

(7)  So, from step #5 and step #6, it follows that C(n) - C(n-m) = m-1

(8)  Using Lemma 3 above, since m \notin C, it follows that A(m) + B(m) \le m-1.

(9)  So, that, we have C(n) - C(n-m) \ge A(m) + B(m) \ge (d(A) + d(B))m

QED

Theorem 1.2.1 (H. B. Mann)
d(A+B) \ge \min(d(A) + d(B),1)






1.1 Additive Properties of Sequences - Theorem 1.1.2 (L. G. Shnirelman)

The details in these notes are taken from A. O. Gelfond & Yu. V. Linnik, elementary methods in analytic number theory, 1965 (translated by Amiel Fienstein/Revised and edited by L. J. Mordell).

Definition 5: Additive Basis

A subset A \subseteq \mathbb{N} with the property that A + A + \cdots + A = \mathbb{N} for a finite sum.

Theorem 1.1.2 (L. G. Shnirelman)  Every sequence with a positive density is a basis for the natural numbers.


Proof:

(1)  Let A^{(k)} = \underbrace{A + A + \dots + A}_{k\,\text{times}}

(2)  Using the Corollary from Theorem 1.1.1, we have:
d(A^{(k)}) \ge 1 - (1 - d(A))^k
(3)  For sufficiently large k, we have:
d(A^{(k)})>\frac{1}{2}
since 0 \le\frac{A(n)}{n}\le 1 and d(A)=\inf\limits_{n}\frac{A(N)}{n}.

(4)  For this k, the number of terms of A^{(k)} in the interval [1,n] will be greater than \frac{n}{2}.

(5)  If a_i ranges over the terms of A^{(k)} which do not exceed n, then there exists i_1,i_2 such that:  a_{i_1} = n - a_{i_2}.

Note:  If i_1,i_2 don't exist, then there are more than 2*(\frac{n}{2}+1)=n+2 values in [0,n] which is impossible.

(6)  So, there exists k such that for any n, there exists a_{i_1}, a_{i_2} \in A^{(k)} such that n = a_{i_1} + a_{i_2}.

QED








Friday, November 22, 2013

1.1 Additive Properties of Sequences - Theorem 1.1.1 (L. G. Shnirelman)

The details in these notes are taken from A. O. Gelfond & Yu. V. Linnik, elementary methods in analytic number theory, 1965 (translated by Amiel Fienstein/Revised and edited by L. J. Mordell).


Definition 1:  infinite sequence of integers

Let any capital Latin letter (e.g. A, B, C, ...) denote an infinite sequence where each term is distinct and in numerical order with the first term 0.

For example, A would represent:
0,a_1,a_2,a_3, \dots
where
0 < a_1 < a_2 < a_3 < \dots

Definition 2:  number of terms not exceeding n

Let A(n) be numerical function on sequences defined as the number of terms not exceeding n so that:
A(n) = \sum_{0 < a_i \le n} 1

For example:
0\le\frac{A(n)}{n}\le1

Definition 3:  infinitum (inf)

For a given set S, the greatest integer that is less than or equal to all elements in S.

Examples:

\inf\, \{1,2,3\}=1
\inf\, \{ x \in \mathbb{Z} : 0 < x < n \} = 1

Definition 4:  density of a sequence

The density d(A) of a sequence A is defined as:
d(A) = \inf_n \frac{A(n)}{n}

Examples:

  • if 1 \notin A, then d(A)=0
  • d(A)=1 if and only if A contains all the positive integers.
  • The densities of squares, cubes, and prime numbers equal 0.
  • \forall{n}\, A(n) \ge n*d(A) 


Theorem 1.1.1 (L. G. Shnirelman)

d(A+B) \ge d(A) + d(B) - d(A)d(B)

Proof:
(1)  Let C = A+B such that c \in C implies c = a + b where a \in A, b \in B

(2)  Let \alpha=d(A), \beta=d(B), \gamma=d(C)

(3)  There are A(n) numbers in A in the interval [1,n] each of which appears in C since 0 \in B.

(4)  Let a_k, a_{k+1} be any two adjacent numbers in A.

(5)  Let l_k = a_{k+1}-a_k-1, so that if l_k > 0, we have the following numbers not in A:

a_k+1, a_k+2, \dots, a_k+l_k = a_{k+1}-1
(6)  For a given a_k,a_{k+1}, B(l_k) is equal to count of the numbers in between a_k and a_{k+1} that are in C since 1 \le r \le l_k and r \in B implies that a_k + r \in C.

(7)  From step(6), we have:
C(n) \ge A(n) + \sum_{k \le A(n)}{B(l_k)}
where each l_k is taken from the intervals between numbers of the form a_k and a_{k+1}

(8)  For each k, B(l_k) \ge \beta{l_k}. [From Definition 4 above]

(9)  So, from step(7), we have:
C(n) \ge A(N) + \beta\sum_{k \le A(n)}{l_k} = A(n) + \beta(n - A(n))
=A(n)(1 - \beta) + \beta{n} \ge \alpha(1 - \beta)n + \beta{n}

(10)  For all n \ge 1, we have:
\gamma=d(A+B)=\inf_n\frac{C(n)}{n} \ge \alpha + \beta - \alpha\beta

QED

Corollary:

d(A_1 + \dots + A_k) \ge 1 - \prod_{i=1}^k{(1 - d(A_i))}

Proof:

(1)  From Theorem 1.1.1 above:
1 - d(A+B) \le 1 - d(A) - d(B) + d(A)d(B) = (1 - d(A))(1 - d(B))
(2)  So that:
d(A+B)-1 \ge -(1-d(A))(1-d(B))
(3)  Which shows the case for k=2 where:
d(A+B) \ge 1 - (1 - d(A))(1 - d(B))
(4)  Assume that it is true up to some k \ge 2 so that:
d(A_1 + \dots + A_k) \ge 1 - \prod_{i=1}^k{(1 - d(A_i))}
(5)  Let B = A_1 + \dots + A_k

(6)  From step(3) above, we have:
d(B + A_{k+1}) \ge 1 - (1 - d(B))(1 - d(A_{k+1}))
(7) From step(4), we have:
-d(B) \le -1 +\prod_{i=1}^{k}(1 - d(A_i))
(8)  Which gives us:
1-d(B) \le  \prod_{i=1}^{k}(1 - d(A_i))
(9)  Combining step(8) with step(6) gives us:
d(A_1 + \dots + A_{k+1}) \ge 1 - \prod_{i=1}^k(1 - d(A_i))(1 - d(A_{k+1}))
QED