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数学代写|数论代写Number Theory代考|MATH2088 Mertens’ theorem

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数学代写|数论代写Number Theory代考|Mertens’ theorem

Our next goal is to prove the following theorem, which turns out to have a number of applications.
Theorem 5.10. We have
$$\sum_{p \leq x} \frac{1}{p}=\log \log x+O(1) .$$
The proof of this theorem, while not difficult, is a bit technical, and we proceed in several steps.
Theorem 5.11. We have
$$\sum_{p \leq x} \frac{\log p}{p}=\log x+O(1) .$$
Proof. Let $n:=\lfloor x\rfloor$. By Theorem $5.2$, we have
$$\log (n !)=\sum_{p \leq n} \sum_{k \geq 1}\left\lfloor n / p^k\right\rfloor \log p=\sum_{p \leq n}\lfloor n / p\rfloor \log p+\sum_{k \geq 2} \sum_{p \leq n}\left\lfloor n / p^k\right\rfloor \log p .$$
We next show that the last sum is $O(n)$. We have
\begin{aligned} \sum_{p \leq n} \log p \sum_{k \geq 2}\left\lfloor n / p^k\right\rfloor & \leq n \sum_{p \leq n} \log p \sum_{k \geq 2} p^{-k} \ & =n \sum_{p \leq n} \frac{\log p}{p^2} \cdot \frac{1}{1-1 / p}=n \sum_{p \leq n} \frac{\log p}{p(p-1)} \ & \leq n \sum_{k \geq 2} \frac{\log k}{k(k-1)}=O(n) . \end{aligned}
Thus, we have shown that
$$\log (n !)=\sum_{p \leq n}\lfloor n / p\rfloor \log p+O(n) .$$
Further, since $\lfloor n / p\rfloor=n / p+O(1)$, applying Theorem $5.5$, we have
$$\log (n !)=\sum_{p \leq n}(n / p) \log p+O\left(\sum_{p \leq n} \log p\right)+O(n)=n \sum_{p \leq n} \frac{\log p}{p}+O(n) .$$
We can also estimate $\log (n$ !) using a little calculus (see $\S \mathrm{A} 2)$. We have
$$\log (n !)=\sum_{k=1}^n \log k=\int_1^n \log t d t+O(\log n)=n \log n-n+O(\log n) .$$

数学代写|数论代写Number Theory代考|The sieve of Eratosthenes

As an application of Theorem $5.10$, consider the sieve of Eratosthenes. This is an algorithm for generating all the primes up to a given bound $k$. It uses an array $A[2 \ldots k]$, and runs as follows.
\begin{aligned} & \text { for } n \leftarrow 2 \text { to } k \text { do } A[n] \leftarrow 1 \ & \text { for } n \leftarrow 2 \text { to }\lfloor\sqrt{k}\rfloor \text { do } \ & \quad \text { if } A[n]=1 \text { then } \ & \qquad i \leftarrow 2 n \text {; while } i \leq k \text { do }{A[i] \leftarrow 0 ; i \leftarrow i+n} \end{aligned}
When the algorithm finishes, we have $A[n]=1$ if and only if $n$ is prime, for $n=2, \ldots, k$. This can easily be proven using the fact (see Exercise 1.1) that a composite number $n$ between 2 and $k$ must be divisible by a prime that is at most $\sqrt{k}$, and by proving by induction on $n$ that at the beginning of the $n$th iteration of the main loop, $A[i]=0$ iff $i$ is divisible by a prime less than $n$, for $i=n, \ldots, k$. We leave the details of this to the reader.
We are more interested in the running time of the algorithm. To analyze the running time, we assume that all arithmetic operations take constant time; this is reasonable, since all the quantities computed in the algorithm are bounded by $k$, and we need to at least be able to index all entries of the array $A$, which has size $k$.

数学代写|数论代写Number Theory代考|Mertens’ theorem

$$\sum_{p \leq x} \frac{1}{p}=\log \log x+O(1) .$$

$$\sum_{p \leq x} \frac{\log p}{p}=\log x+O(1) .$$

$$\log (n !)=\sum_{p \leq n} \sum_{k \geq 1}\left\lfloor n / p^k\right\rfloor \log p=\sum_{p \leq n}\lfloor n / p\rfloor \log p+\sum_{k \geq 2} \sum_{p \leq n}\left\lfloor n / p^k\right\rfloor \log p .$$

$$\sum_{p \leq n} \log p \sum_{k \geq 2}\left\lfloor n / p^k\right\rfloor \leq n \sum_{p \leq n} \log p \sum_{k \geq 2} p^{-k} \quad=n \sum_{p \leq n} \frac{\log p}{p^2} \cdot \frac{1}{1-1 / p}=n \sum_{p \leq n} \frac{\log p}{p(p-1)} \leq n \sum_{k \geq 2} \frac{\log k}{k(k-1)}=O(n) .$$

$$\log (n !)=\sum_{p \leq n}\lfloor n / p\rfloor \log p+O(n) .$$

数学代写|数论代写Number Theory代考|The sieve of Eratosthenes

$$\log (n !)=\sum_{p \leq n}(n / p) \log p+O\left(\sum_{p \leq n} \log p\right)+O(n)=n \sum_{p \leq n} \frac{\log p}{p}+O(n) .$$

$$\log (n !)=\sum_{k=1}^n \log k=\int_1^n \log t d t+O(\log n)=n \log n-n+O(\log n) .$$

for $n \leftarrow 2$ to $k$ do $A[n] \leftarrow 1 \quad$ for $n \leftarrow 2$ to $\lfloor\sqrt{k}\rfloor$ do $\quad$ if $A[n]=1$ then $\quad i \leftarrow 2 n$; while $i \leq k$ do $A[i] \leftarrow 0 ; i \leftarrow i+n$

1.1） $n$ 介于 2 和 $k$ 必须能被一个至多为的质数整除 $\sqrt{k}$ ，并通过归纳证明 $n$ 在开始的时候 $n$ 主循环的第 th 次迭代， $A[i]=0$ 当且仅当 $i$ 能被小于的责数整除 $n$ ，为了 $i=n, \ldots, k$. 我们将其细节留给续者。

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