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[组合] 两题征解(组合不等式与向量)

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2014-4-26 22:39

学下real,发点非常规的怪题,
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妙不可言,不明其妙,不着一字,各释其妙!

第2题,平移这些向量,使得起点都在同一点,取定一个向量,并记这个向量为(1,0),若有向量$v_k$的终点在x轴下方,用$v_k$相反向量替换,仍记为$v_k$,
如此使得这n个单位向量终点都不在x轴下放,重新标记下标,使得对应的复数$e^{iα_k},i是虚数单位,k=1,2,3,...,n$满足$0=α_1\le α_2 \le α_3\le....\le α_n\le \pi$.
那么猜想:$\abs{e^{iα_1}-e^{iα_2}+e^{iα_3}-e^{iα_4}+.....+e^{iα_n}}\le1$
只是证明了n=3.三个向量依次是$(1,0),(\cos α,\sin α ),(\cos β,\sin β),0\le α\le β$
计算长度$\abs{(1,0)-(\cos α,\sin α )+(\cos β,\sin β)}^2\le1$
等价于$\sin(\frac{β-α}{2})\sin \frac{α}{2} \sin \frac{β}{2} \ge 0 $

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回复 2# realnumber
到这一步还是不错了!

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本帖最后由 APPSYZY 于 2022-4-22 17:33 编辑

第一题
$n\le\max\{a,b\}$时容易验证. 直接归纳,假设$n$时命题已经成立. 在$n+1$的情形,只需证明
\[\frac{n}{a+b+1}\binom{n}{a}\binom{n}{b}+\binom{n}{a}\binom{n}{b}\le\frac{n+1}{a+b+1}\binom{n+1}{a}\binom{n+1}{b}.\]
利用$\binom{n+1}{a}=\frac{n+1}{n+1-a}\binom{n}{a}$得到
\[(n+1+a+b)(n+1-a)(n+1-b)\le(n+1)^3,\]

\begin{align*} \RHS-\LHS &= (n+1)^3-((n+1)^3-((a+b)^2-ab)(n+1)+ab(a+b)) \\ &= (a^2+ab+b^2)(n+1)-ab(a+b) \\ &\ge 0. \end{align*}

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本帖最后由 hbghlyj 于 2022-4-23 06:08 编辑

回复 2# realnumber
https://math.stackexchange.com/q ... um-i-1?noredirect=1
Note that the convex polygon generated by $\mathrm{C} = \text{conv}\{\pm z_j: j = 1, 2, \cdots, 2k+1\}$ lies in the closed unit disk $\overline{\mathbb{D}}$. We are trying to find a vector $\overline{\varepsilon} \in \{1,-1\}^{2k+1}$ s.t., $\displaystyle J(\overline{\varepsilon}) = \left|\sum\limits_{j=1}^{2k+1} \varepsilon_jz_j\right| \le 1$.

The problem remains unchanged if we were to perform rotations (multiplication by unimodular factors throughout) or flip of sign of $z_j$'s. Therefore, wlog (by a rotation followed by reindexing of points/flipping the sign if necessary) we may assume $z_1 = 1 \in \partial \mathbb{D}$ and $z_2 , \cdots, z_{2k+1}$ lie in the upper half circle and arranged in increasing order of their arguments in that order. If $z_{2k+1}$ coincides with $-z_{1}$ in the process, simply choose $\varepsilon_1 = \varepsilon_{2k+1} = 1$ so that the terms cancels off in $J(\overline\varepsilon)$ and we are left to work with $2k-1$ points. So, it is safe to assume $z_2, \cdots, z_{2k+1}$ are infact distinct and strictly in the upper half circle (arguments strictly between $(0,\pi)$). Let us further cyclically index the points $z_{2k+1 + j} = -z_j$ ($z_j$'s in the diagram) for $j=1 , 2, \cdots , 2k+1$ (so $z_{4k+3} = z_1$; $z_0 = z_{4k+2} = z_{2k+1}'$, etc.)



Now, let us denote the edge vectors by cyclic indexing $w_j := (z_{j+1} - z_j)$ for $j = 1, 2, \cdots, 4k+2$. We Claim: $$v := z_1 + \sum\limits_{j=1}^{k} w_{2j} = \sum\limits_{j=1}^{2k+1} (-1)^{j-1} z_j \in \overline{\mathbb{D}}.$$

Proof of claim: If $v = 0$ the we are through. So we may assume $v \neq 0$.

We note that $\displaystyle \sum\limits_{j=1}^{2k+1} (-1)^{j} w_j = 2\sum\limits_{j=1}^{2k+1} (-1)^{j-1} z_j = 2v$. More interestingly we have $$\sum\limits_{j=m+1}^{m+2k+1} (-1)^{j}w_j = 2v, \, \forall \, m \ge 0$$ i.e., the alternating sum of $2k+1$ consecutive edge vectors is always the fixed vector $2v$ and hence invariant under cyclic reindexing of the points. So at this point if we were to rotate the axis with $\hat{v} := e^{-i\arg{v}}$ so that $v$ coincides with the positive real line then we may cyclically shift the index s.t., $z_{1}, \cdots, z_{2k + 1}$ all lie is the closed upper half plane $\overline{\mathbb{H}}$.

So each of the edge vectors $w_{j}$'s for $j = 1, \cdots, 2k$ now lying in the upper half plane has disjoint projections on the real line (except at atmost one common point) all contained in the interval $[-1,1]$. Now there are two cases to consider.

First case, if the projection of $z_1' = -z_1$ on the real line lies to the left of projection of $z_{2k+1}$ i.e., if $\mathrm{Re}(z_{1}') < \mathrm{Re}(z_{2k+1})$, then projection of the edge $w_{2k+1}$ on the real line is disjoint from the rest and contained in $[-1,1]$ as well. Hence, $$2|v| = \left|\sum\limits_{j=1}^{2k+1} (-1)^{j}\langle \hat{v} , w_j \rangle\right| \le \mathrm{Re}(z_1) - \mathrm{Re}(z_1') \le 2.$$

The second case being, if projection of $z_{1}' = -z_1$ lies to the right of $z_{2k+1}$ i.e., $\mathrm{Re}(z_{2k+1}) < \mathrm{Re}(z_{1}')$ then in turn we must have the projection of $z_{2k+1}' = -z_{2k+1}$ lie to the right of $z_1$ i.e., $\mathrm{Re}(z_1) < \mathrm{Re}(z_{2k+1}')$. Therefore, we may consider the projection of the edge $w_{0} = w_{4k+2} = (z_{4k+3} - z_{4k+2}) = (z_1 - z_{2k+1}')$ on the real line instead which is disjoint from the rest. Hence, $$2|v| = \left|\sum\limits_{j=0}^{2k} (-1)^{j}\langle \hat{v} , w_j \rangle\right| \le \mathrm{Re}(z_{2k+1}') - \mathrm{Re}(z_{2k+1}) \le 2.$$

Eitherway, we have $2|v| \le 2$, hence, $|v| \le 1$ proving our claim.

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本帖最后由 hbghlyj 于 2022-4-23 06:26 编辑

第二题当n是偶数时还成立吗

不成立了,如$v_1=(1,0),v_2=(0,1)$时$|ε_1v_1+ε_2v_2|$只能是$\sqrt2$

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