一道极值问题的妙解

数学星空 发表于 2012-1-29 16:13:48

本帖最后由数学星空于 2012-1-30 22:55 编辑

数学星空 发表于 2012-1-29 16:18:47

本帖最后由数学星空于 2012-1-30 20:54 编辑

数学星空 发表于 2012-1-29 16:24:06

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 发表于 2012-1-30 08:58:54

https://bbs.emath.ac.cn/data/attachment/forum/month_1201/1201291613c152e15ba6342b1e.jpg
https://bbs.emath.ac.cn/data/attachment/forum/month_1201/12012916189d6353337688202a.jpg
https://bbs.emath.ac.cn/data/attachment/forum/month_1201/12012916183702c3cbcf966677.jpg

mathe 发表于 2012-1-30 09:03:42

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 发表于 2012-1-30 20:55:35

TO mathe:
4#的图片是如何处理的哟??

mathe 发表于 2012-1-30 22:14:51

查看了论坛帮助，使用\

gxqcn 发表于 2012-1-30 22:30:57

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 发表于 2012-1-30 22:43:00

http://static.tieba.baidu.com/tb/editor/images/jd/j_0006.gif

数学星空 发表于 2012-2-3 20:38:14

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 thevariables real quadricsf=X^T*A*X, whereX=(x_1,x_2,...,x_n)^T and x_{1}^2+x_2^2+...+x_n^2 =1.
Thenmin{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)/2ifi!=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.

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一道极值问题的妙解