b***e
发帖数: 1419 | 1 (a_1*b_1 + ... + a_n*b_n)^2 <= (a_1^2 + ... + a_n^2)*(b_1^2 + ... + b_n^2)
书上有一个证明，但是属于天外飞仙型，无迹可寻。请问有没有好的几何解释？ |
N***m
发帖数: 4460 | 2 内积阿
【在 b***e 的大作中提到】
: (a_1*b_1 + ... + a_n*b_n)^2 <= (a_1^2 + ... + a_n^2)*(b_1^2 + ... + b_n^2)
: 书上有一个证明，但是属于天外飞仙型，无迹可寻。请问有没有好的几何解释？
|
b***e
发帖数: 1419 | 3 我知道是内积，不知道为什么内积和这个不等式有什么关系。我不是数学专业的，业余
的。请赐教。
【在 N***m 的大作中提到】
: 内积阿
|
N***m
发帖数: 4460 | 4 a内积b=|a||b|cos(夹角)。
Is this clear now?
【在 b***e 的大作中提到】
: 我知道是内积，不知道为什么内积和这个不等式有什么关系。我不是数学专业的，业余
: 的。请赐教。
|
n***p
发帖数: 7668 | 5 N维空间里夹角的概念是通过内积定义的，not conversely.
【在 N***m 的大作中提到】
: a内积b=|a||b|cos(夹角)。
: Is this clear now?
|
N***m
发帖数: 4460 | 6 he is trying to undestand this inequality, not trying to prove it.
【在 n***p 的大作中提到】
: N维空间里夹角的概念是通过内积定义的，not conversely.
|
n***p
发帖数: 7668 | 7 I agree.
【在 N***m 的大作中提到】
: he is trying to undestand this inequality, not trying to prove it.
|
S********n
发帖数: 71 | 8 Cauchy–Schwarz inequality
wiki之。
N维的概念是从2，3维generalize的。
楼主能明白在2D的情况就行了。内积就是点乘，楼上给出定义了。
多维（n>3）的情况就没有几何意义了，都不能构造这样的物理空间了。
发信人: nonpp (有错就改), 信区: Mathematics
标 题: Re: 不等式证明一问
发信站: BBS 未名空间站 (Thu Mar 17 11:44:22 2011, 美东)
N维空间里夹角的概念是通过内积定义的，not conversely. |
a****e
发帖数: 1247 | 9 Cauchy不等式证明再标准不过了。 考虑n=2的情况就够了。 两个向量本来就是平面
问题。
【在 b***e 的大作中提到】
: (a_1*b_1 + ... + a_n*b_n)^2 <= (a_1^2 + ... + a_n^2)*(b_1^2 + ... + b_n^2)
: 书上有一个证明，但是属于天外飞仙型，无迹可寻。请问有没有好的几何解释？
|
b***e
发帖数: 1419 | 10 In fact, I do want to prove it in some more intuitive ways. The standard
approach as referred from wiki:
http://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality
introduces a mysterious \sigma, which is from no-where. Well ok, I am not
as smart as Cauchy, so I cannot see through it. If I were to prove it
(without reading the wiki or text book), I don't see I can come up with
such a \sigma here. I guess I am more of asking a geometry mapping for
this \sigma.
【在 N***m 的大作中提到】
: he is trying to undestand this inequality, not trying to prove it.
|
|
|
N***m
发帖数: 4460 | 11 I didn't see the mysterious \sigma. where is it?
btw, the proof for the case of R^n on wiki is straightforward.
【在 b***e 的大作中提到】
: In fact, I do want to prove it in some more intuitive ways. The standard
: approach as referred from wiki:
: http://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality
: introduces a mysterious \sigma, which is from no-where. Well ok, I am not
: as smart as Cauchy, so I cannot see through it. If I were to prove it
: (without reading the wiki or text book), I don't see I can come up with
: such a \sigma here. I guess I am more of asking a geometry mapping for
: this \sigma.
|
S********n
发帖数: 71 | |
n*****1
发帖数: 172 | 13 I think s/he means the \delta
【在 S********n 的大作中提到】
: do u mean \delta?
|
z***c
发帖数: 102 | 14 Here is a geometrical explanation. Let u and v be vectors, |u| denote vector
norm, and the inner product. Cauchy's inequality
\le |u| |v|
is equivalent to
/|v| \le |u|.
The left hand side has a geometric meaning: it is the projection of u onto
the direction of v. Hence Cauchy's inequality simply means that the
projection of a vector has shorter or equal length than the vector itself.
This can be easily proved using a right triangle.
Now, this proof actually inspires the proof given in wikipedia. To simplify
notation I will use t instead of \delta used in wikipedia. I made a picture
for this, see attachment.
【在 b***e 的大作中提到】
: In fact, I do want to prove it in some more intuitive ways. The standard
: approach as referred from wiki:
: http://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality
: introduces a mysterious \sigma, which is from no-where. Well ok, I am not
: as smart as Cauchy, so I cannot see through it. If I were to prove it
: (without reading the wiki or text book), I don't see I can come up with
: such a \sigma here. I guess I am more of asking a geometry mapping for
: this \sigma.
|
b***e
发帖数: 1419 | 15 he, 纯爷们。
【在 n*****1 的大作中提到】
: I think s/he means the \delta
|
b***e
发帖数: 1419 | 16 My bad. I used to know all of those, but after 6 years...
【在 S********n 的大作中提到】
: do u mean \delta?
|
b***e
发帖数: 1419 | 17 多谢，让我研究一下。
vector
onto
itself.
simplify
【在 z***c 的大作中提到】
: Here is a geometrical explanation. Let u and v be vectors, |u| denote vector
: norm, and the inner product. Cauchy's inequality
: \le |u| |v|
: is equivalent to
: /|v| \le |u|.
: The left hand side has a geometric meaning: it is the projection of u onto
: the direction of v. Hence Cauchy's inequality simply means that the
: projection of a vector has shorter or equal length than the vector itself.
: This can be easily proved using a right triangle.
: Now, this proof actually inspires the proof given in wikipedia. To simplify
|
s*****s
发帖数: 1559 | 18 这个最好看看历史，wiki有。 把两边的差写出来，不等式就成了等式。 最初的两个变
量的就是
婆罗莫及多等式，差是平行四边形面积的平方。 三个变量的就是空间向量的内积和外
积之间的拉
格朗日等式。 Cauchy-Binet等式是高维的情形，用一点线性代数的知识，就能理解
Cauchy-Binet
公式里出现的很多平行四边形的几何意义了。
【在 b***e 的大作中提到】
: (a_1*b_1 + ... + a_n*b_n)^2 <= (a_1^2 + ... + a_n^2)*(b_1^2 + ... + b_n^2)
: 书上有一个证明，但是属于天外飞仙型，无迹可寻。请问有没有好的几何解释？
|
