Proof Hypercube 1 - ประวัติรุ่นแก้ไข

Jittat: Reverted edit of 82.111.20.202, changed back to last version by Parinya

2007-10-28T08:39:35Z

Reverted edit of 82.111.20.202, changed back to last version by Parinya

82.111.20.202 เมื่อ 22:58, 27 ตุลาคม 2550

2007-10-27T22:58:48Z

Parinya: /* Theorem 1 */

2007-04-10T04:33:49Z

Theorem 1

Parinya เมื่อ 04:32, 10 เมษายน 2550

2007-04-10T04:32:23Z

Parinya เมื่อ 04:30, 10 เมษายน 2550

2007-04-10T04:30:47Z

หน้าใหม่

'''Theorem''': Let <math>v \in \mathbb{S}</math>.
: <math>N(v) \subseteq \mathbb{S} \Rightarrow vol(P_v) = 1/2^n</math>

''Proof:'' Let <math>e_i = 0\ldots 1 \ldots 0</math> where 1 appears at ith position.
:<math> N(v) = \{ v \oplus e_i\}_{i=1}^n</math>

There will be some technicalities in the proof. One way to get rid of them is to consider <math>v= 0^n</math>.

Consider a hyperplane <math>H_i</math> halving the middle point between v and <math>v \oplus e_i</math>
The hyperplane is defined by <math>(x-(v+ v \oplus e_i)/2)) \cdot e_i = 0</math>. Working out the calculation,

: <math>x_i - 1/2 = 0</math>

So, P_v is defined by the intersection of halfspaces <math>x_i \le 1/2</math> for <math>i= 1, \ldots, n</math>. It is obvious that the volume of this intersection is <math>1/2^n</math> (If you don't believe, you can do Reimann integration of this set).

Since the volume can't be any smaller, we conclude that, in this case, <math>Vol(P_v) = 1/2^n</math>

←รุ่นแก้ไขก่อนหน้า		รุ่นแก้ไขเมื่อ 08:39, 28 ตุลาคม 2550
แถว 1:		แถว 1:
−	~~acellicalet~~
	== Theorem 1 ==		== Theorem 1 ==
	'''Theorem''': Let <math>v \in \mathbb{S}</math>.		'''Theorem''': Let <math>v \in \mathbb{S}</math>.

←รุ่นแก้ไขก่อนหน้า		รุ่นแก้ไขเมื่อ 22:58, 27 ตุลาคม 2550
แถว 1:		แถว 1:
		+	acellicalet
	== Theorem 1 ==		== Theorem 1 ==
	'''Theorem''': Let <math>v \in \mathbb{S}</math>.		'''Theorem''': Let <math>v \in \mathbb{S}</math>.

@@ แถว 13: / แถว 13: @@
 : <math>x_i - 1/2 = 0</math>
-So, P_v is defined by the intersection of halfspaces <math>x_i \le 1/2</math> for <math>i= 1, \ldots, n</math>. It is obvious that the volume of this intersection is <math>1/2^n</math> (If you don't believe, you can do Reimann integration of this set).
+So, P_v is defined by the intersection of halfspaces <math>x_i \le 1/2</math> for <math>i= 1, \ldots, n</math>. It is obvious that the volume of this intersection is <math>1/2^n</math> (If you don't believe, you can do Reimann integration of this set :P ).
 Since the volume can't be any smaller, we conclude that, in this case, <math>Vol(P_v) = 1/2^n</math>. A more rigorous proof can also be achieved by considering other halfspaces and argue that it contains at least one of <math>h_i</math>.

@@ แถว 1: / แถว 1: @@
 '''Theorem''':  Let <math>v \in \mathbb{S}</math>.
 : <math>N(v) \subseteq \mathbb{S} \Rightarrow vol(P_v) = 1/2^n</math>
@@ แถว 14: / แถว 15: @@
 So, P_v is defined by the intersection of halfspaces <math>x_i \le 1/2</math> for <math>i= 1, \ldots, n</math>. It is obvious that the volume of this intersection is <math>1/2^n</math> (If you don't believe, you can do Reimann integration of this set).
-Since the volume can't be any smaller, we conclude that, in this case, <math>Vol(P_v) = 1/2^n</math>
+Since the volume can't be any smaller, we conclude that, in this case, <math>Vol(P_v) = 1/2^n</math>. A more rigorous proof can also be achieved by considering other halfspaces and argue that it contains at least one of <math>h_i</math>.