k***s
发帖数: 206 | 1 I have two vectors:
X=
Y=
There's a function F(X,Y), and if you expand that function, you get
1+x1*y1 + x2*y2 + x1*x2*y1*y2
Find F(X,Y).
(Hint: dot product of X * Y is (x1*y1+x2*y2) )
这个我想 F(X,Y) 至少因该是 X*Y + 1, 但是这个 x1*x2*y1*y2 项怎么处理？ | a****9
发帖数: 418 | 2 设I=(1,1)
(I*X)^2 = x1^2+x2^2+2*x1*x2
X^2 = x1^2+x2^2
=> x1*x2 = ((I*X)^2 - X^2)/2
同理可以得到 y1*y2 = ((I*Y)^2 - Y^2)/2
【在 k***s 的大作中提到】
: I have two vectors:
: X=
: Y=
: There's a function F(X,Y), and if you expand that function, you get
: 1+x1*y1 + x2*y2 + x1*x2*y1*y2
: Find F(X,Y).
: (Hint: dot product of X * Y is (x1*y1+x2*y2) )
: 这个我想 F(X,Y) 至少因该是 X*Y + 1, 但是这个 x1*x2*y1*y2 项怎么处理？
| c********e
发帖数: 141 | 3 How about this:
F(X,Y) = 1+x1*y1 + x2*y2 + x1*x2*y1*y2
= (1+x1y1)(1+x2y2)
= (1+X'[1 0; 0 0]Y)(1+X'[0 0; 0 1]Y) | p*****k
发帖数: 318 | 4 you need to be more specific than that. are constant vectors allowed in the
expression F(x,y) (then it's kinda trivial as already pointed out), or only
X, Y and dot operation can be used? | k***s
发帖数: 206 | 5 say only X,Y and dot, +,-,power | m********0
发帖数: 2717 | 6 as far as I understand, no constant vector is allowed.
maybe 'expend' means taylor expension. and that's the
expension to the second order.
the
only
【在 p*****k 的大作中提到】
: you need to be more specific than that. are constant vectors allowed in the
: expression F(x,y) (then it's kinda trivial as already pointed out), or only
: X, Y and dot operation can be used?
| J****g
发帖数: 103 | | o*****n
发帖数: 2098 | 8 I feel it is not a correct answer, but it is so far I can get.
1+X*Y+[(X*([0 1;1 0]Y))^2+(X*Y)^2-X^2*Y^2]/4 |
|
