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THEORY OF NUMBERS

TCHEBYCHEV'S THEOREM

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\begin{description}

\item[Prime numbers]

We denote $\prod(x)=\sum_{p\leq x}1.$ So $\prod(p_n)=n\
\prod(x)<x$.

\item[Theorem]

$\sum_p\frac{1}{p}$ diverges.

\item[Proof]

$\prod_{p\leq N}\left(1-\frac{1}{p}\right)^{-1}=\prod_{p\leq
N}\left(1+\frac{1}{p}+\frac{1}{p^2}+\ldots\right)=\sum_{n=1}^\infty\frac{\theta_n}{n}$

where $\theta_N=\left\{\begin{array}{cl}1&\textrm{if every prime
factor of $n$ is }\leq N\\
0&\textrm{otherwise}\end{array}\right.\\ \geq
\sum_{n=1}^N\frac{1}{n}\geq\log N\\ \log N\leq \prod_{p\leq
N}\left(1-\frac{1}{p}\right)^{-1}$

Therefore

\begin{eqnarray*}
\log\log N&\leq& \sum_{p\leq N}-\log\left(1-\frac{1}{p}\right)\\
&=&\sum_{p\leq N}\frac{1}{p}+\sum_{p\leq
N}\left(\frac{1}{p^2}+\frac{1}{p^3}+\ldots\right)\\ &=&\sum_{p\leq
N}\frac{1}{p}+C
\end{eqnarray*}

Therefore $\sum_{p\leq N}\frac{1}{p}>\log\log N-C$

\item[Tchebychev's Theorem]

$\exists$ positive constants $A,\ B$ such that $A\frac{x}{\log
x}<\prod(x)<B\frac{x}{\log x}$ for large $x$.

\item[Proof]\

\begin{description}

\item[Definition]

\begin{eqnarray*}
\Lambda(n)&=&\left\{\begin{array}{cl}\log p&\textrm{if
}n=p^l\\0&\textrm{otherwise}\end{array}\right.\\
\Psi(x)&=&\sum_{n\leq x}\Lambda(n)\\ \theta(x)&=&\sum_{p\leq
x}\log p\\ F(x)&=&\sum_{n\leq x}\log n
\end{eqnarray*}

\item[Lemma A]

$F(x)=x\log x-x+O(\log x)\ x\geq 2$

\item[Corollary]

$F(x)-2F\left(\frac{1}{2}x\right)=x\log 2+O(\log x)\ x\geq 2$

\item[Proof]

$F(x)=\sum_{2\leq n\leq x}\log n$

Now $\log t$ increases as $t$ increases and so

$$\int_{n-1}^n\log t\,dt\leq\log n\leq \int_n^{n+1}\log t\,dt
n\geq 2$$

So

$$\int_1^{[x]}\log t\,dt\leq F(x)\leq \int_1^{[x]+1}\log t\,dt$$

$$[t\log t-t]_1^{[x]}\leq F9x)\leq [t\log t-t]_1^{[x]+1}$$

Where $F(x)=x\log x-x+O(\log x)$

\item[Lemma B]

$\Psi(x)-\Psi\left(\frac{1}{2}x\right)\leq
F(x)-2F\left(\frac{1}{2}x\right)\leq \psi(x)\ x\geq2$

\item[Proof]

$\log n=\sum_{d|n}\Lambda(d)\ n=1,2,\ldots$

\begin{eqnarray*}
F(x)&=&\sum_{n\leq x}\log n=\sum_{n\leq x}\sum_{d|n}\Lambda(d)\\
&=&\sum_{d\leq x}\Lambda(d)\sum_{n\ d|n\leq x}1\\ &=&\sum_{d\leq
x}\lambda(d)\sum_{m\ m\leq\frac{x}{d}}1\\ &=&\sum_{d\leq
x}\Lambda(d)\left[\frac{x}{d}\right]
\end{eqnarray*}

\begin{eqnarray*}
F(x)-2F\left(\frac{1}{2}x\right)&=&\sum_{d\leq
x}\Lambda(d)\left[\frac{x}{a}\right]-2\sum_{d\leq\frac{1}{2}x}\Lambda(d)\left[\frac{\frac{1}{2}x}{d}\right]\\
&=&\sum_{\frac{1}{2}x<d\leq
x}\Lambda(d)\left[\frac{x}{d}\right]+\sum_{d\leq\frac{1}{2}x}\Lambda(d)\left\{\left[\frac{x}{d}\right]-2\left[\frac{\frac{1}{2}x}{d}\right]\right\}\\
&=&\sum_1+\sum_2
\end{eqnarray*}

$\sum_1=\sum\Lambda(d)=\Psi(x)-\Psi\left(\frac{1}{2}x\right)$

Now $f(\alpha)=[\alpha]-2\left[\frac{1}{2}\alpha\right]$ is
periodic with period 2.

$f(\alpha)=\left\{\begin{array}{cl}0&\textrm{if }0\leq
\alpha<1\\1&\textrm{if }1\leq\alpha<2\end{array}\right.$

Therefore $0\leq f(\alpha)\leq1$ for all real $\alpha$

therefore $0\leq
\sum_2\leq\sum{d\leq\frac{1}{2}x}\Lambda(d)=\Psi\left(\frac{1}{2}x\right)$

Hence the result.

\item[Lemma C]

\begin{description}

\item[(i)]
$\Psi(x)\geq x\log2+O(\log x)$

\item[(ii)]
$\Psi(x)\leq2x\log 2+O(\log^2x)$

\end{description}

\item[Proof]

\begin{description}

\item[(i)]

Immediate by $B$ and corollary $A$.

\item[(ii)]

Choose $l$ to satisfy $1\leq x2^{-l}<2$

Then
\begin{eqnarray*}
\Psi(x)&=&\Psi(x)-\Psi(x2^{-l})\\
&=&\sum_{n=0}^{l-1}\left\{\Psi(x2^{-n})-\Psi(x2^{-n-1})\right\}\\
&\leq&x\log2\sum_{n=0}^{l-1}+O(l\log x)\textrm{ by }B\\
&\leq&2x\log2+O(\log^2x)
\end{eqnarray*}

\end{description}

\item[Lemma D]

$\theta(x)=\Psi(x)+O\left\{(x^\frac{1}{2})\log^2x\right\}$

\item[Proof]

\begin{eqnarray*}
0&\leq&\Psi(x)-\theta(x)\\&=&\sum_{p,l (p^l\leq x,\ l\geq2)}\log
p\\ &\leq&\frac{\log x}{\log 2}.x^\frac{1}{2}\log x
\end{eqnarray*}

\item[Corollary]
$\theta(x)=O(x)$

\item[Lemma E]

\begin{eqnarray*}
\prod(x)&=&\frac{\theta(x)}{\log
x}+O\left(\frac{x}{\log^2x}\right)\\ &=&\frac{\Psi(x)}{\log
x}+O\left(\frac{x}{\log^2x}\right)
\end{eqnarray*}

\item[Proof]
Suppose $x\geq3$

\begin{eqnarray*}
\prod(x)&=&\sum_{p\leq x}1\\ &=&\sum_{2\leq n\leq
x}\frac{\theta(n)-\theta(n-1)}{\log n}\\
&=&\frac{\theta([x])}{\log([x])}+\sum_{2\leq n\leq
x-1}\theta(n)\left(\frac{1}{\log n}-\frac{1}{\log (n-1)}\right)\\
&=&\frac{\theta([x])}{\log([x])}+O\sum_{2\leq n\leq
x-1}\frac{1}{\log^2n}\\ &=&\frac{\theta(x)}{\log
x}+O\left(\frac{x}{\log^2 x}\right)
\end{eqnarray*}

The theorem follows from Lemmas E, D, C if we take $$A<\log2,\
B>2\log2.$$

It follows that $\exists K_1,k_2$ such that

$$k_1m\log n<p_n<k_2n\log n. $$

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