一道极值问题的妙解

数学星空

数学星空

数学星空

不知如何将图片最大化使其清晰可见,请行家帮忙处理一下....

mathe

mathe

其实凑系数结果看起来简单，得到过程还是挺困难的。这个题由于是二次型，应该可以转化为求矩阵的最大特征值问题

数学星空

TO mathe:
4#的图片是如何处理的哟??

mathe

查看了论坛帮助，使用\[img=width,height\]

gxqcn

发图时，确实要注意千万别被系统“自动缩小”了，所用方法就是楼上所述（参见：帮助）。

wayne

数学星空

Tourish 给出:

With the same condition, we can find the minimun value is    $(n+n^2)/4+sqrt((n^2*(n+1)*(2*n+1))/24)$ ,also a very interesting result

I created this problem when I found the following interesting result:
For the  variables real quadrics  $f=X^T*A*X$, where  $X=(x_1,x_2,...,x_n)^T$ and $x_{1}^2+x_2^2+...+x_n^2 =1$.
Then  $min{lambda_i}<=f<=max{lambda_i}$, where $lambda_i$ are the eigenvalue of matrix The proof for this conclusion is rather easy in linear algebra.
I won't write the detail here(also I don't know how to type out a matrix $A$ ).
For this problem ,
we have$ A={a_{ij}}$, where $a_{ij}=(i+j)/2$  if  $i!=j $and $a_{ij}=i$ if $ i=j$ .
Therefore
$||lambda*E-A||=lambda^(n-2)*((lambda-(n+n^2)/4)^2-(n^2*(n+1)*(2*n+1))/24)$
So from the conclusion, the maximum and minimum of the expression is exactly the result given in the above.