$∑a_i=1$,求$∑a_ia_{i+1}$的最大值

hbghlyj

根据弧中点多边形文中的计算, 当$a_i$都等于$1/n$时取等:

...we find the maximum value of $\sum_{i=0}^{n-1} a_{i} a_{i+1}$ subject to the constraint $\sum_{i=0}^{n-1} a_{i}=1$. Using Lagrange multipliers it can easily be determined that the maximum value of the above sum is attained when all the $a_{i}$ 's have the same value, that is, when they are all equal to $1 / n$.

然而在下面的帖子中, 结论是(n≥4时)当两个相等, 其它全为0时取最大值:
Let $a_1,a_2,a_3,....a_n\geq0$ and $a_1+a_2+a_3+....+a_n=2m$,then prove that the greatest possible value of $\sum_{i=1}^{n}a_i.a_{i+1}$.
The question is equivalent to proving that
$$4(a_1a_2+a_2a_3+...+a_na_1) \leq (a_1+a_2+...+a_n)^2.$$

Yes, we can prove it. WLOG, let $a_1=\max \{a_1,a_2,...,a_n\}$, then
\begin{eqnarray*}
4(a_1a_2+a_2a_3+...+a_na_1)
&\leq& 4a_1(a_2+a_4+a_5+...+a_n)+4a_2a_3 \\
&\leq& 4(a_1+a_3)(a_2+a_4+a_5+...+a_n) \\
&\leq& (a_1+a_2+...+a_n)^2. \text{(AM-GM)}
\end{eqnarray*}

The equality holds if and only if $a_1=a_2=m, a_3=a_4=...=a_n=0$.

hbghlyj

根据弧中点多边形文中的计算, 当$a_i$都等于$1/n$时取等:
...

原文只计算了区域内的驻点, 没有计算边界上的点, 导致它得出的最大值是错的?
注意$n=4$情况和$n>4$情况的不同:

1. n=4;
2. NullSpace[Array[RotateRight[PadRight[{1,-1,1,-1},n],#]&,n]]

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输出{{1, 0, 0, 1}, {-1, 0, 1, 0}, {1, 1, 0, 0}}
而当$n=5$时(任何$n>4$时):

1. n=5;
2. NullSpace[Array[RotateRight[PadRight[{1,-1,1,-1},n],#]&,n]]

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输出{{1, 1, 1, 1, 1}}

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当$n=4$时,$\left(\frac m2,\frac m2,\frac m2,\frac m2\right)$和$\left(m,m,0,0,0\right)$都是极大值点,而且取值相同:
$$f\left(\frac m2,\frac m2,\frac m2,\frac m2\right)=4\left(m\over2\right)^2=m^2=f\left(m,m,0,0,0\right)$$
这也是函数的全局最大值:

1. Maximize[{a1 a2 + a2 a3 + a3 a4 + a4 a1, a1 + a2 + a3 + a4 == 1}, {a1, a2, a3, a4}]

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输出{1/4,{a1->1/2,a2->0,a3->0,a4->1/2}}

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当$n=5$时,只有$\left(\frac{2m}5,\frac{2m}5,\frac{2m}5,\frac{2m}5,\frac{2m}5\right)$是极大值点,注意$(m,m,0,0,0)$是边界点但不是极大值点,但是:
$$f\left(\frac{2m}5,\frac{2m}5,\frac{2m}5,\frac{2m}5,\frac{2m}5\right)=5\left(2m\over5\right)^2=\frac45m^2<m^2=f(m,m,0,0,0)$$
这是因为, 函数在$\left(\frac{2m}5,\frac{2m}5,\frac{2m}5,\frac{2m}5,\frac{2m}5\right)$附近是小于$\frac45m^2$的, 但是在远处趋于∞, 所以没有全局最大值. 限定了$a_i>0$后, 就在边界上取最大值.

1. Maximize[{a1 a2 + a2 a3 + a3 a4 + a4 a5 + a5 a1, a1 + a2 + a3 + a4 + a5 == 1}, {a1, a2, a3, a4, a5}]

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输出是∞.