e********0
发帖数: 259 | 1 请见图：欲证明claim，但是证到问号处就卡住了。
最好这个claim本身是一个定理，这样我就可以直接用了。
注释：claim中的middle hand side: 对x-y的绝对值分之一求m阶导（注意x,y都是3元
的向量，所以j1,...jm是可以相等的。），再对导函数取y=0.
第一个把这个问题证出来或者告知定理出处的兄弟给20个包子！ |
e********0
发帖数: 259 | 2 不好意思，您的意思是f对xi的导数与f对yi的导数是“对称”的？我不太能理解您
的argument. 如果您能帮我把证明的过程写出来，将非常有帮助。我会给您20包子。
谢谢！
do you mean it suffice to prove:
the partial derivative of
y\ |
B********e
发帖数: 10014 | 3 haha, sorry, 大嘴巴了
准确说两者只差个符号(for all x\neq y) |
e********0
发帖数: 259 | 4 can you prove it or not? This is a 3-D problem and m
can be arbirary integer, so you have to do some induction...
【在 B********e 的大作中提到】
: haha, sorry, 大嘴巴了
: 准确说两者只差个符号(for all x\neq y)
|
B********e
发帖数: 10014 | 5 forget all the |y=0 in your calculation first
work on your last second equation
move \partial y's out and move \partial x in
use what i said
【在 e********0 的大作中提到】
: can you prove it or not? This is a 3-D problem and m
: can be arbirary integer, so you have to do some induction...
|
e********0
发帖数: 259 | 6 sorry.. I don't follow what you are saying
Can you write it down and upload it? I'll give you 20 baozi!
【在 B********e 的大作中提到】
: forget all the |y=0 in your calculation first
: work on your last second equation
: move \partial y's out and move \partial x in
: use what i said
|
n***p
发帖数: 7668 | 7 让我尝试着解释一下。 It is just about symmetry.
Let r =|y-x|. Then
\frac{\partial r}{\partial y_i} = (y_i-x_i)/r
and
\frac{\partial r}{\partial x_i} = (x_i- y_i)/r.
That is
\frac{\partial r}{\partial y_i}
= - \frac{\partial r}{\partial x_i}.
For any function f(r), if you take partial derivative in
y, it is just chain rule. Say
\frac{\partial}{\partial y_i} f(r)
= f'(r) \frac{\partial r}{\partial y_i}
= - f'(r)\frac{\partial r}{\partial x_i}
= - \frac{\partial}{\partial x_i} f(r).
When you take partial derivatives m times, you apply
similar arguments m times and ultimately you have a constant
C = (-1)^m. |
e********0
发帖数: 259 | 8 thanks for your reply. This looks very promising.
What do you exactly mean by "\frac{\partial}{\partial y_i} f(r)"?
the derivative of f(r) with respect to y_i? then what is the "\frac"
for?
And why are we concerned about "any function f(r)"? We are only
interested in f(r)=1/r here, aren't we?
【在 n***p 的大作中提到】
: 让我尝试着解释一下。 It is just about symmetry.
: Let r =|y-x|. Then
: \frac{\partial r}{\partial y_i} = (y_i-x_i)/r
: and
: \frac{\partial r}{\partial x_i} = (x_i- y_i)/r.
: That is
: \frac{\partial r}{\partial y_i}
: = - \frac{\partial r}{\partial x_i}.
: For any function f(r), if you take partial derivative in
: y, it is just chain rule. Say
|
n***p
发帖数: 7668 | 9 I wrote the expressions in latex commands. \frac is a latex command
indication a fraction.
\frac{\partial}{\partial y_i} f(r) means
the partial derivative of f(r) with respect to y_i.
As to "any function f(r)", if you get the idea for any function,
immediately you get the idea for the specific case when f(r) = 1/r, right?
【在 e********0 的大作中提到】
: thanks for your reply. This looks very promising.
: What do you exactly mean by "\frac{\partial}{\partial y_i} f(r)"?
: the derivative of f(r) with respect to y_i? then what is the "\frac"
: for?
: And why are we concerned about "any function f(r)"? We are only
: interested in f(r)=1/r here, aren't we?
|
e********0
发帖数: 259 | 10 10 baozi sent, because you did not show why "take derivative to y, then let y=0" and "let y=0, then take derivative to x" are the same, i.e, why "differentiation" commutes with "taking value", which is the critical step.
【在 n***p 的大作中提到】
: I wrote the expressions in latex commands. \frac is a latex command
: indication a fraction.
: \frac{\partial}{\partial y_i} f(r) means
: the partial derivative of f(r) with respect to y_i.
: As to "any function f(r)", if you get the idea for any function,
: immediately you get the idea for the specific case when f(r) = 1/r, right?
:
|
n***p
发帖数: 7668 | 11 首先呢，一个包子是10伪币， 这个我不太关心。
本着把问题讲清楚的原则，let me try again.
Because of the symmetry,
taking derivative in y = - taking derivative in x.
Hence
"take derivative to y, then let y=0"
= - "take derivative to x, then let y=0"
= - "let y=0, then take derivative to x".
This is why when you take partial derivatives m times, there must be
a constant C = (-1)^m.
let y=0" and "let y=0, then take derivative to x" are the same, i.e, why
"differentiation" commutes with "taking value", which is the critical
step.
【在 e********0 的大作中提到】
: 10 baozi sent, because you did not show why "take derivative to y, then let y=0" and "let y=0, then take derivative to x" are the same, i.e, why "differentiation" commutes with "taking value", which is the critical step.
|
l******r
发帖数: 18699 | 12 看不清
【在 e********0 的大作中提到】
: 请见图：欲证明claim，但是证到问号处就卡住了。
: 最好这个claim本身是一个定理，这样我就可以直接用了。
: 注释：claim中的middle hand side: 对x-y的绝对值分之一求m阶导（注意x,y都是3元
: 的向量，所以j1,...jm是可以相等的。），再对导函数取y=0.
: 第一个把这个问题证出来或者告知定理出处的兄弟给20个包子！
|
e********0
发帖数: 259 | 13 100 weibi sent.
【在 n***p 的大作中提到】
: 让我尝试着解释一下。 It is just about symmetry.
: Let r =|y-x|. Then
: \frac{\partial r}{\partial y_i} = (y_i-x_i)/r
: and
: \frac{\partial r}{\partial x_i} = (x_i- y_i)/r.
: That is
: \frac{\partial r}{\partial y_i}
: = - \frac{\partial r}{\partial x_i}.
: For any function f(r), if you take partial derivative in
: y, it is just chain rule. Say
|
