Then for $t$ a real number, $$ \begin{gathered} 0 \leq p(t) \equiv \sum_{j=1}^{P}\left(z_{j}+t \theta w_{j}\right)\left(\overline{z_{j}}+t \overline{\theta w_{j}}\right) \ =\overbrace{\sum_{j=1}^{p} z_{j} \overline{z_{j}}}^{p}+\sum_{j=1}^{P} z_{j} t \overline{\theta \omega_{j}}+\sum_{j=1}^{P} t \theta w_{j} \overline{z_{j}}+t^{2} \sum_{j=1}^{P} w_{j} \overline{w_{j}} \ \equiv a^{2}+2 t \operatorname{Re} \theta \sum_{j=1}^{p} w_{j} \overline{z_{j}}+t^{2} b^{2}=a^{2}+2 t\left|\sum_{j=1}^{P} w_{j} \overline{z_{j}}\right|+t^{2} b^{2} \end{gathered} $$ Since this is always nonnegative for all real $t$, it follows from the quadratic formula that $$ 4\left|\sum_{j=1}^{P} w_{j} \overline{z_{j}}\right|^{2}-4 a^{2} b^{2}=4\left|\sum_{j=1}^{P} w_{j} \overline{z_{j}}\right|^{2}-4\left(\sum_{j=1}^{P} z_{j} \overline{z_{j}}\right)\left(\sum_{j=1}^{P} w_{j} \overline{w_{j}}\right) \leq 0 $$ Indeed, $p(t)=0$ either has exactly one real root or no real roots. Thus the desired inequality follows. 2.16 Integer Multiples of Irrational Numbers This section will give a proof of a remarkable result. I think its proof, based on the pigeon hole principle, is even more interesting than the result obtained. Dirichlet proved it in the 1830 ‘s. Jacobi used similar ideas around the same time in studying elliptic functions. The theorem involves the sum of integer multiples of numbers whose ratio is irrational. If $a / b$ is irrational, then it is not possible that $m a+n b=0, m, m \in \mathbb{Z}$ because if this were so, you would have $\frac{-m}{n}=\frac{b}{a}$ and so the ratio of $a, b$ is rational after all. Even though you cannot get 0 (which you can get if the ratio of $a$ and $b$ is rational) you can get such an integer combination arbitrarily small. Dirichlet did this in the 1830 ‘s long before the real numbers were constructed from the rational numbers in 1858 and published in 1872 . Theorem $2.16 .1$ If $a, b$ are real numbers and $a / b$ is not rational, then for every $\varepsilon>0$ there exist integers $m, n$ such that $|m a+n b|<\varepsilon$. Proof: Let $P_{N}$ denote all combinations of the form $m a+n b$ where $m, n$ are integers and $|m|,|n| \leq N$. Thus there are $(2 N+1)^{2}$ of these integer combinations and all of them are contained in the interval $I \equiv[-N(|a|+|b|), N(|a|+|b|)]$. Now pick an integer $M$ such that $$ (2 N)^{2}2$. Now partition the interval $I$ into $M$ equal intervals. If $l$ is the length of one of these intervals, then $$ \begin{aligned} l M &=2 N(|a|+|b|), \ l &=\frac{2 N(|a|+|b|)}{M}<\frac{2 N(|a|+|b|)}{(2 N)^{2}}=\frac{2(|a|+|b|)}{4} \frac{1}{N} \equiv \frac{C}{N} \end{aligned} $$ Now as mentioned, all of the points of $P_{N}$ are contained in $I$ and there are more of these points than there are intervals in the partition of $I$ which has only $M$ intervals.

那么对于 $t$ 一个实数， $$ \开始{聚集} 0 \leq p(t) \equiv \sum_{j=1}^{P}\left(z_{j}+t \theta w_{j}\right)\left(\overline{z_{j}}+ t \overline{\theta w_{j}}\right) \\ =\overbrace{\sum_{j=1}^{p} z_{j} \overline{z_{j}}}^{p}+\sum_{j=1}^{P} z_{j} t \上划线{\theta \omega_{j}}+\sum_{j=1}^{P} t \theta w_{j} \overline{z_{j}}+t^{2} \sum_{j=1} ^{P} w_{j} \overline{w_{j}} \\ \equiv a^{2}+2 t \operatorname{Re} \theta \sum_{j=1}^{p} w_{j} \overline{z_{j}}+t^{2} b^{2 }=a^{2}+2 t\left|\sum_{j=1}^{P} w_{j} \overline{z_{j}}\right|+t^{2} b^{2} \结束{聚集} $$ 由于这对于所有实数 $t$ 始终为非负数，因此从二次公式得出 $$ 4\left|\sum_{j=1}^{P} w_{j} \overline{z_{j}}\right|^{2}-4 a^{2} b^{2}=4\left |\sum_{j=1}^{P} w_{j} \overline{z_{j}}\right|^{2}-4\left(\sum_{j=1}^{P} z_{j } \overline{z_{j}}\right)\left(\sum_{j=1}^{P} w_{j} \overline{w_{j}}\right) \leq 0 $$ 实际上，$p(t)=0$ 要么只有一个实根，要么没有实根。因此，期望的不等式如下。 2.16 无理数的整数倍 本节将证明一个显着的结果。我认为它的证明，基于鸽子洞原理，比得到的结果更有趣。 Dirichlet 在 1830 年代证明了这一点。 Jacobi 大约在同一时间使用了类似的想法来研究椭圆函数。该定理涉及比率为无理数的整数倍数之和。如果 $a / b$ 是无理数，那么 $m a+nb=0, m, m \in \mathbb{Z}$ 是不可能的，因为如果是这样，你将有 $\frac{-m} {n}=\frac{b}{a}$ 等 $a, b$ 的比例毕竟是合理的。即使你不能得到 0（如果 $a$ 和 $b$ 的比率是有理数，你可以得到），你可以得到这样一个任意小的整数组合。早在 1858 年从有理数构造实数并于 1872 年发表之前，狄利克雷在 1830 年代就这样做了。 定理 $2.16 .1$ 如果 $a, b$ 是实数并且 $a / b$ 不是有理数，那么对于每个 $\varepsilon>0$ 存在整数 $m, n$ 使得 $|m a+nb| 2$。现在将区间 $I$ 划分为 $M$ 相等的区间。如果 $l$ 是这些间隔之一的长度，则 $$ \开始{对齐} l M &=2 N(|a|+|b|), \\ l &=\frac{2 N(|a|+|b|)}{M}

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