数分定理

Posted 2021-11-30 eufisky

接下来介绍高等概率论课程强大数定律中用到的一些重要结论.

引理1.假设$A=(a_{ij})$为一无穷实数矩阵,对每个固定的$j$, $\lim_{n\to\infty}a_{nj}=0$,存在非负常数$c$,使得$\sum_{j=1}^{\infty}|a_{nj}|\leq c$, $\forall n$.如果$\{x_n\}$为一有界实数列,定义
\[y_{n}=\sum_{j=1}^{\infty} a_{n j} x_{j},\quad n=1,2, \cdots\]
则
(1)如果$x_n\to 0$,则$y_n\to 0$;

(2)如果$\sum_{j=1}^{\infty}a_{nj}\to 1,x_n\to x\in\mathbb{R}$,则$y_n\to x$.

证. (1)由$y_n$的定义,有
\[\left|y_{n}\right| \leq \sum_{j=1}^{N}\left|a_{n j}\right|\left|x_{j}\right|+\sum_{j=N+1}^{\infty}\left|a_{n j} \| x_{j}\right|.\]
对任意$\varepsilon>0$,存在$N$使得$|x_j|\leq\frac{\varepsilon}{c},\forall j>N$.于是上式右边第二项不超过$c\cdot \frac{\varepsilon}{c}=\varepsilon$.由条件$\lim_{n\to\infty}a_{nj}=0$可知,对任意给定的$N$,上式右边第一项趋于$0$.因此$y_n\to 0$.

(2)由(1)知$\sum_{j=1}^{\infty}a_{nj}(x_{j}-x)\to x$.由此得$y_n\to x$.

引理2. (Toeplitz)假定$\{a_n\}$为一非负实数序列,令$b_n=\sum_{j=1}^{n}a_j$.假设$b_n>0,\lim_{n\to\infty}b_n=\infty$.如果$\{x_n\}$是一实数序列,收敛到实数$x$,则
\[\frac{1}{b_{n}} \sum_{j=1}^{n} a_{j} x_{j} \rightarrow x.\]

证.定义无穷矩阵$A$,它的第$n$行为
\[\left( {\frac{a_{1}}{b_{n}}}, {\frac{a_{2}}{b_{n}}}, {\cdots}, {\frac{a_{n}}{b_{n}}}, {0}, {0}, {\cdots}\right),\]
再利用引理1(1)即可.

(Kronecker)假设$\{b_n\}$是一列单增的正实数,满足$b_n\to\infty$, $\{x_n\}$为一列实数,满足$\sum_{n=1}^{\infty}x_n=x\in\mathbb{R}$,则
\[\lim _{n \rightarrow \infty} \frac{1}{b_{n}} \sum_{j=1}^{n} b_{j} x_{j}=0.\]

证.令$s_n=\sum_{j=1}^{n}x_j(n\geq 1),s_0=0,b_0=0$,由阿贝尔求和公式可知
\begin{align*} \sum_{j=1}^{n} b_{j} x_{j} &=\sum_{j=1}^{n} b_{j}\left(s_{j}-s_{j-1}\right) \\ &=b_{n} s_{n}-b_{0} s_{0}-\sum_{j=1}^{n} s_{j-1}\left(b_{j}-b_{j-1}\right) \\ &=b_{n} s_{n}-\sum_{j=1}^{n} s_{j-1}\left(b_{j}-b_{j-1}\right). \end{align*}
于是有
\[\frac{1}{b_{n}} \sum_{j=1}^{n} b_{j} x_{j}=s_{n}-\frac{1}{b_{n}} \sum_{j=1}^{n} a_{j} s_{j-1},\]
其中$a_j=b_j-b_{j-1}\geq 0$.利用$\lim_{n\to\infty}s_n=\lim_{j\to\infty}s_{j-1}=x$及引理2可得结论.

Wolstenholme嵌入不等式.对$\triangle ABC$和任意的实数$x,y,z$,均有
\[x^2+y^2+z^2\geq 2 y z \cos A+2 z x \cos B+2 x y \cos C,\]
当且仅当$x:y:z=\sin A:\sin B:\sin C$时取等成立.

证:进行配方法证明.
\begin{align*} & x^{2}+y^{2}+z^{2}-(2 x y \cos C+2 y z \cos A+2 z x \cos B)\\
=& x^{2}-2 x(y \cos C+z \cos B)+y^{2}+z^{2}-2 y z \cos A\\
=&(x-y \cos C-z \cos B)^{2}-(y \cos C+z \cos B)^{2}+y^{2}+z^{2}+2 y z \cos (B+C) \\
&=(x-y \cos C-z \cos B)^{2}+(y \sin C-z \sin B)^{2} \geq 0.
\end{align*}
当且仅当$\left\{\begin{array}{l}{x-y \cos C-z \cos B=0} \\ {y \sin C-z \sin B=0}\end{array}\right.$,即$x:y:z=\sin A:\sin B:\sin C$时取等成立.

考虑与$\overrightarrow{A B}, \overrightarrow{B C}, \overrightarrow{C A}$同向的单位向量$\vec{e}_{1}, \vec{e}_{2}, \vec{e}_{3}$,则对任意实数$x,y,z$,均有
\[\left(z \cdot \overrightarrow{e_{1}}+x \cdot \overrightarrow{e_{2}}+y \cdot \overrightarrow{e_{3}}\right)^{2} \geq 0,\]
展开便得原不等式.

令$x=y=z=1$,可得
\[\cos A+\cos B+\cos C \leq \frac{3}{2}.\]
令$p=2 y z, q=2 z x, r=2 x y$,解得$(x, y, z)=\left(\sqrt{\frac{q r}{2 p}}, \sqrt{\frac{r p}{2 q}}, \sqrt{\frac{p q}{2 r}}\right)$,代入嵌入不等式可得
\[p \cos A+q \cos B+r \cos C \leq \frac{1}{2}\left(\frac{q r}{p}+\frac{r p}{q}+\frac{p q}{r}\right).\]

(1998韩国)已知正实数$a,b,c$满足$a+b+c=abc$,求证:
\[\frac{1}{\sqrt{1+a^{2}}}+\frac{1}{\sqrt{1+b^{2}}}+\frac{1}{\sqrt{1+c^{2}}} \leq \frac{3}{2}.\]
令$a=\tan A, b=\tan B, c=\tan C$,原不等式成为
\[\cos A+\cos B+\cos C \leq \frac{3}{2}.\]

已知正实数$a,b,c$满足$a+b+c=abc$,实数$u,v,w$满足$uvw>0$,求证:
\[\frac{u}{\sqrt{1+a^{2}}}+\frac{v}{\sqrt{1+b^{2}}}+\frac{w}{\sqrt{1+c^{2}}} \leq \frac{1}{2}\left(\frac{v w}{u}+\frac{w u}{v}+\frac{u v}{w}\right).\]

(1998印度)已知$x,y,z$为正实数, $x y+y z+z x+x y z=4$,求证:
\[x+y+z \geq x y+y z+z x.\]

证.联想恒等式
\[\cos ^{2} A+\cos ^{2} B+\cos ^{2} C+2 \cos A \cos B \cos C=1.\]
将条件$x y+y z+z x+x y z=4$写成
\[\left(\frac{\sqrt{x y}}{2}\right)^{2}+\left(\frac{\sqrt{y z}}{2}\right)^{2}+\left(\frac{\sqrt{z x}}{2}\right)^{2}+2 \frac{\sqrt{x y}}{2} \cdot \frac{\sqrt{y z}}{2} \cdot \frac{\sqrt{z x}}{2}=1.\]
于是可设$\left\{\begin{array}{l}{\frac{\sqrt{y z}}{2}=\cos A} \\ {\frac{\sqrt{z x}}{2}=\cos B} \\ {\frac{\sqrt{x y}}{2}=\cos C}\end{array}\right.$,解得
\[(x, y, z)=\left(\frac{2 \cos B \cos C}{\cos A}, \frac{2 \cos C \cos A}{\cos B}, \frac{2 \cos A \cos B}{\cos C}\right),\]
原不等式成为
\[\frac{\cos B \cos C}{\cos A}+\frac{\cos C \cos A}{\cos B}+\frac{\cos A \cos B}{\cos C} \geq 2\left(\cos ^{2} A+\cos ^{2} B+\cos ^{2} C\right),\]
在嵌入不等式中令$x=\sqrt{\frac{\cos B \cos C}{\cos A}}, y=\sqrt{\frac{\cos C \cos A}{\cos B}}, z=\sqrt{\frac{\cos A \cos B}{\cos C}}$即可.

(2007年IMO中国国家集训队测试题)设正数$u,v,w$满足$u+v+w+\sqrt{u v w}=4$,求证:
\[\sqrt{\frac{v w}{u}}+\sqrt{\frac{w u}{v}}+\sqrt{\frac{u v}{w}} \geq u+v+w.\]

证.把条件改写为
\[\left(\frac{\sqrt{u}}{2}\right)^{2}+\left(\frac{\sqrt{v}}{2}\right)^{2}+\left(\frac{\sqrt{w}}{2}\right)^{2}+2 \frac{\sqrt{u}}{2} \cdot \frac{\sqrt{v}}{2} \cdot \frac{\sqrt{w}}{2}=1,\]

可设$\left\{\begin{array}{l}{\frac{\sqrt{u}}{2}=\cos A} \\ {\frac{\sqrt{v}}{2}=\cos B} \\ {\frac{\sqrt{w}}{2}=\cos C}\end{array}\right.$,只需证
\[\frac{\cos B \cos C}{\cos A}+\frac{\cos C \cos A}{\cos B}+\frac{\cos A \cos B}{\cos C} \geq 2\left(\cos ^{2} A+\cos ^{2} B+\cos ^{2} C\right).\]

设$\triangle ABC$为锐角三角形,求证:
\[\left(\frac{\cos A}{\cos B}\right)^{2}+\left(\frac{\cos B}{\cos C}\right)^{2}+\left(\frac{\cos C}{\cos A}\right)^{2}+8 \cos A \cos B \cos C \geq 4.\]

证.由恒等式$\cos ^{2} A+\cos ^{2} B+\cos ^{2} C+2 \cos A \cos B \cos C=1$得
\[4-8 \cos A \cos B \cos C=4\left(\cos ^{2} A+\cos ^{2} B+\cos ^{2} C\right),\]
只需证
\[\left(\frac{\cos A}{\cos B}\right)^{2}+\left(\frac{\cos B}{\cos C}\right)^{2}+\left(\frac{\cos C}{\cos A}\right)^{2} \geq 4\left(\cos ^{2} A+\cos ^{2} B+\cos ^{2} C\right),\]
下面证明
\[\left(\frac{\cos A}{\cos B}\right)^{2}+\left(\frac{\cos B}{\cos C}\right)^{2}+\left(\frac{\cos C}{\cos A}\right)^{2} \geq 2\left(\frac{\cos B \cos C}{\cos A}+\frac{\cos C \cos A}{\cos B}+\frac{\cos A \cos B}{\cos C}\right).\]
在嵌入不等式中,令$x=\frac{\cos B}{\cos C}, y=\frac{\cos C}{\cos A}, z=\frac{\cos A}{\cos B}$,则
\begin{align*}
&\left(\frac{\cos A}{\cos B}\right)^{2}+\left(\frac{\cos B}{\cos C}\right)^{2}+\left(\frac{\cos C}{\cos A}\right)^{2} =x^{2}+y^{2}+z^{2} \\
& \geq 2(x y \cos C+y z \cos A+z x \cos B)\\ &=2\left(\frac{\cos B \cos C}{\cos A}+\frac{\cos C \cos A}{\cos B}+\frac{\cos A \cos B}{\cos C}\right). \end{align*}

设$p,q$是两个互质的奇正整数,证明:
\[\sum_{0<x<\frac{q}{2}}\left[\frac{p}{q} x\right]+\sum_{0<y<\frac{p}{2}}\left[\frac{q}{p} y\right]=\frac{p-1}{2} \cdot \frac{q-1}{2}.\]

二次互反律

设$a,b$是两个非零整数,我们定义雅克比符号$\left(\frac{a}{b}\right)$:若存在整数$x$,使得$x^2\equiv a(\bmod b)$,那么就记$\left(\frac{a} {b}\right)=1$;否则就记$\left(\frac{a} {b}\right)=-1$.在$b$是素数时这个符号也叫做勒让德符号.

高斯二次互反律:设$p$和$q$为不同的奇素数,则
\[\left(\frac{q}{p}\right) \cdot\left(\frac{p}{q}\right)=(-1)^{(p-1) / 2 \cdot(q-1) / 2}.\]

三角形中三角恒等式

\begin{enumerate}
\item $\sin A+\sin B+\sin C=4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$.


\item $\sin 2 A+\sin 2 B+\sin 2 C=4 \sin A \sin B \sin C$.

\item $\sin 3 A+\sin 3 B+\sin 3 C=-4 \sin \frac{3}{2} A \sin \frac{3}{2} B \sin \frac{3}{2} C$

\item $\sin 4 A+\sin 4 B+\sin 4 C=-4 \sin 2 A \sin 2 B \sin 2 C$.

\item $\cos A+\cos B+\cos C=1+4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$.

\item $\cos 2 A+\cos 2 B+\cos 2 C=-1-4 \cos A \cos B \cos C$.

\item $\cos 3 A+\cos 3 B+\cos 3 C=1-4 \cos \frac{3}{2} A \cos \frac{3}{2} B \cos \frac{3}{2} C$.

\item $\cos 4 A+\cos 4 B+\cos 4 C=-1-4 \cos 2 A \cos 2 B \cos 2 C$.

\item $\tan A+\tan B+\tan C=\tan A\tan B\tan C$.

\item $\cot A \cot B+\cot B \cot C+\cot C \cot A=1$.

\item $\tan \frac{A}{2} \tan \frac{B}{2}+\tan \frac{B}{2} \tan \frac{C}{2}+\tan \frac{C}{2} \tan \frac{A}{2}=1$.

\item $\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}=\cot \frac{A}{2} \cot \frac{B}{2} \cot \frac{C}{2}$.

\item $\sin ^{2} A+\sin ^{2} B+\sin ^{2} C=2+2 \cos A \cos B \cos C$.

\item $\cos ^{2} A+\cos ^{2} B+\cos ^{2} C=1-2 \cos A \cos B \cos C$.

\item $\sin A \cos B \cos C+\cos A \sin B \cos C+\cos A \cos B \sin C=\sin A\sin B\sin C$.

\item $\cos A \sin B \sin C+\sin A \cos B \sin C+\sin A \sin B \cos C=-1+\cos A \cos B \cos C$.

\item $a \cos A+b \cos B+c \cos C=4 R \sin A \sin B \sin C$.

\item $p=4 R \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$.

\item $\tan \frac{A}{2} \tan \frac{B}{2} \tan \frac{C}{2}=\frac{r}{p}$.

\item $\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}=\frac{r}{4 R}$.

\item $\tan \frac{A}{2} \tan \frac{B}{2}=\frac{p-c}{p}$.

\item $\frac{a\cos A+b \cos B+\cos C}{a+b+c}=\frac{r}{R}$.

\item $\sin A+\sin B+\sin C=\frac{p}{R}$.

\item $\cos A+\cos B+\cos C=1+\frac{r}{R}$.
\end{enumerate}