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# 数学代写|加性组合代写Additive Combinatorics代考|CSE291 Thin bases of higher order

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## 数学代写|加性组合代写Additive Combinatorics代考|Thin bases of higher order

We now return to the study of thin bases $B$ and their associated counting functions $r_{k, B}(n)$, initiated in Section 1.3. However, in this section we can use Theorem $1.37$ to present a proof of Theorem $1.15$, which asserted for each $k \geq 1$ the existence of a base $B$ of order $k$ with $r_{k, B}(n)=O_k(\log n)$ for all large $n$. This was proven in the $k=2$ case (see Theorem 1.13) using Chernoff’s inequality, but that method does not directly apply for higher $k$ because $r_{k, B}(n)$ cannot be easily expressed as the sum of independent random variables.

We begin with a simple lemma on boolean polynomials that shows that if $\mathbf{E}(X)$ is not too large, then at most points $\left(t_1, \ldots, t_n\right)$ of the sample space, the polynomial $X$ does not contain too many independent terms (cf. Exercise 1.3.12).

Lemma 1.40 Let $X=\sum_{A \in \mathcal{A}} \prod_{j \in A} t_j$ be a boolean polynomial of $n$ independent boolean variables $t_1, \ldots, t_n$, let $B \subseteq[1, n]$ be the random set $B:={j \in[1, n]$ : $\left.t_j=1\right}$, and let $D \in \mathbf{N}$ be the random variable, defined as the largest number of disjoint sets in $\mathcal{A}$ which are contained in $B$. Then for any integer $K \geq 1$ we have
$$\mathbf{P}(D \geq K) \leq \frac{\mathbf{E}(X)^K}{K !} .$$
Proof Observe that for $A_1, \ldots, A_k$ disjoint,
$$\mathbf{I}(D \geq K) \leq \frac{1}{K !} \sum_{A_1, \ldots, A_K \in \mathcal{A}, \text { disjoint }} \prod_{j \in A_1} t_j \ldots \prod_{j \in A_k} t_j .$$
Taking expectations of both sides and using linearity of expectation (1.3) followed by independence, we conclude
$$\mathbf{P}(D \geq K) \leq \frac{1}{K !} \sum_{A_1, \ldots, A_K \in \mathcal{A}} \mathbf{E}\left(\prod_{j \in A_1} t_j\right) \ldots \mathbf{E}\left(\prod_{j \in A_k} t_j\right) .$$
But by linearity of expectation again, the left-hand side is just $\mathbf{E}(X)^K / K !$, and the claim follows.

## 数学代写|加性组合代写ADDITIVE COMBINATORICS代考|Thin Waring bases

Recall that a thin basis of order $k$ is a set $B \subset \mathbf{N}$ such that $r_{k, B}(n)=O(\log n)$ for all large $n$. Theorem 1.15, proved above, asserts that $\mathbf{N}$ contains a thin basis of any order. Given the abundance of classical bases such as the squares and primes, it is then natural to pose the following question:

Question 1.46 Let A be any fixed basis of order $k$. Does A contain a thin subbasis $B$ ?

Note that Sidon’s original question can be viewed as the $k=2, A=\mathbf{N}$ case of this question. From (1.21) we know that a thin basis $B$ enjoys the bounds
$$|B \cap[0, N]|=\Omega_k\left(N^{1 / k}\right) ; \quad|B \cap[0, N]|=O_k\left(N^{1 / k} \log ^{1 / k} N\right)$$
for all large $N$. Thus we can consider the following weaker version of Question $1.46$ :

Question 1.47 Let $A$ be any fixed basis of order k. Does A contain a subbasis $B$ with $|B \cap[0, N]|=O_k\left(N^{1 / k} \log ^{1 / k} N\right)$ for all large $N$ ?

Question $1.47$ has been investigated intensively for the Waring bases $\mathbf{N}^{\wedge} r=$ $\left{0^r, 1^r, 2^r, \ldots\right}$, especially when $r=2$ [90, 56, 387, 388, 384, 331]. For these bases it is known that if $k$ is sufficiently large depending on $r$, then $\mathbf{N}^{\wedge} r$ is a basis of order $k$, and furthermore that
$$r_{k, \mathbf{N}^{\wedge} r}(n)=\Theta_{k, r}\left(n^{\frac{k}{r}-1}\right) ;$$
note that this is consistent with (1.21).

## 数学代写|加性组合代写ADDITIVE COMBINATORICS代考|THIN BASES OF HIGHER ORDER

$$\mathbf{P}(D \geq K) \leq \frac{\mathbf{E}(X)^K}{K !} .$$

$$\mathbf{I}(D \geq K) \leq \frac{1}{K !} \sum_{A_1, \ldots, A_K \in \mathcal{A}, \text { disjoint }} \prod_{j \in A_1} t_j \ldots \prod_{j \in A_k} t_j$$

$$\mathbf{P}(D \geq K) \leq \frac{1}{K !} \sum_{A_1, \ldots, A_K \in \mathcal{A}} \mathbf{E}\left(\prod_{j \in A_1} t_j\right) \ldots \mathbf{E}\left(\prod_{j \in A_k} t_j\right) .$$

## 数学代写|加性组合代写ADDITIVE COMBINATORICS代考|THIN WARING BASES

$$|B \cap[0, N]|=\Omega_k\left(N^{1 / k}\right) ; \quad|B \cap[0, N]|=O_k\left(N^{1 / k} \log ^{1 / k} N\right)$$

$90,56,387,388,384,331$

$$r_{k, \mathbf{N}^{\wedge} r}(n)=\Theta_{k, r}\left(n^{\frac{k}{r}-1}\right) ;$$

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