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(1.1)Definition: Metric Space: By a metric space M, we mean a set M = {x,y,z,...}, whose elements are called “points”, with a distance function ρ(x,y):M^2→ℝ , with the property that ∀ x,y ∈ M
(a.) ρ(x, y)>0 if x≠y, and also ρ(x, x) = 0 ∀x∈M
(b.) ρ(x, y) = ρ(y, x) ∀x, y∈M
(c.) ρ(x, z) ≤ ρ(x, y) + ρ(y, z) ∀x, y, z∈M

ρ(x,y) is called the metric (distance) of the metric space M

Now consider the n-dim real space ℝn (if you don't know it, see Linear Algebra) given 2 vectors from it, x = (ξ1,ξ2,...,ξn), y =(η1,η2,...,ηn), define the (1.2) scalar product of x and y as a function (x·y):ℝn^2↦ℝ having the property that:
(1.) (x·x)≥0
(2.) (x·y) = (y·x)
(3.) (αx·y) = α(x·y), ∀α∈ℝ
(4.) (x + y·z) = (x·z)+(y·z)


and define:
the length (norm) of x∈ℝn as |x|≡√(x·x) ≥0,
the distance of x and y  ∈ℝn as ρ(x,y)≡|x-y|=√(x-y·x-y) ≥0,
and if a vector x∈ℝn and |x|≡1, we call it unit vector or normalized vector.

Now we can construct our most familiar Euclidean space.

Define: the n-dim Euclidean space as the n-dim real space ℝn equipped with a scalar product (x·y)≡Σ(ξkηk), k=1 to n (Obviously it satisfies(1.2) 1.-4.) so the length of x in n-Euclidean space is |x|≡√(x·x) =√Σ(ξkξk) ≥0 and the distance of x and y in Euclidean space is ρ(x,y)≡|x-y|=√(x-y·x-y)=√Σ(ξk-ηk)*(ξk-ηk) ≥0

Note that there are other ways to define the scalar product as long as it satisfies out (1.~4.) requirements. we do not need to define scalar product as Σ(ξkηk), k=1 to n (a stronger condition) to prove the following theorems. Σ(ξkηk) is just a special case for Euclidean space.

Now we claim every n-dim real space ℝn with a valid scalar product is a metric space.(So Euclidean space is also a metric space) we check the distance function ρ(x,y)≡|x-y|=√(x-y·x-y) ≥0 against the axioms of a metric space. (1.1) Obviously, (1.1a),(1.1b) are true. For (1.1c), we first prove by the following 
lema1.2: ρ(x,y) ≤ ρ(x,z) + ρ(y,z) ⇔ |x+y| ≤ |x|+|y|

proof:


replace y by -y, z by 0, we have

ρ(x,-y) ≤ ρ(x,0) + ρ(-y,0)

⇒|x-(-y)| ≤ |x-0|+|-y-0|

⇒|x+y| ≤ |x|+|-y| = |x|+|y|

// Note |-y|=|y| becasue ρ(x,y)≡|x-y| and ρ(x,y)=ρ(y,x)≡|y-x|, let x=0



replace x by x-z, y by z-y,

|x-z + z-y| ≤ |x-z| + |z-y|

⇒ρ(x,y) ≤ ρ(x,z) + ρ(y,z)


so to prove (1.1c) is to prove |x+y| ≤ |x|+|y|, now we state a theorem:

Thm1.3: Let Rn equipped with a valid scalar product satisfying properties(1.~4.), then |x+y|≤|x|+|y| holds ∀x,y,z∈Rn , where |x|≡√(x·x) (i.e. √(x+y·x+y) ≤ √(x·x) + √(y·y)),

proof:

consider the expression φ(λ)≡(x-λy·x-λy), whereλ∈ℝ,   by 1.2(1.)

we have φ(λ)≥0, by 1.2(2.)(3.)(4.), we have
φ(λ)≡(x-λy·x-λy) 
= (x·x-λy) + (-λy·x-λy) 
= (x·x) + (x·-λy) + (-λy·x)  + (-λy·-λy) 
= (x·x)-2λ(x·y)+λ2(y·y) 
= A-2λB+λ2C, where A=(x·x) B =(x·y) C=(y·y)

Here, the polynomial can’t have distinct roots cuz φ(λ)≥0, so we have
(-2B)2-4AC ≤0 ⇒ (x·y)*(x·y) - (x·x)*(y·y)≤ 0

⇒(x·y)2 ≤ (x·x)*(y·y) ≡ |x|2|y|2

⇒(x·y)2≤(|x||y|)2  and (x·y)≥0 and |x||y|≥0, we have

⇒√(x·y)2 ≤ |x||y|

⇒(x·y)≤ |x||y| .........This is called the Schwarz Inequality(1.4)

Now, consider |x+y|≡√((x+y)·(x+y))

|x+y|2=((x+y)·(x+y)) = (x·x) + 2(x·y) + (y·y)

≤(by Schwarz) |x|2+2|x||y| + |y|2 = (|x|+|y|)2

⇒|x+y| ≤ |x| + |y| 

Q.E.D

Thus we've proved that any ℝn with a (valid) scalar product is a metric space!

Now, let's re-write Schwarz Inequality in terms of components of the vector in n-Euclidean space.

(x·y)≤ |x||y| (remember now (x·y)≡Σ(ξkηk), k=1 to n )

⇒Σ(ξkηk) ≤ √(∑ξk2)*√(∑ηk2) .... This is called Cauchy’s inequality(1.41)

Sometimes, we just call (x·y)≤ |x||y| as Cauchy–Schwarz inequality

共 8 則回應

1
半夜貼數學
.....
1
同學你好優秀
0
認真就是不一樣[▓▓]ε:)
1
B1 sleeping pill for you 0w0
B2 B3 Thanks;)
0
看到一堆符號就暈了[▓▓]ε:)
1
打了那麼多
重點應該是最後一段
0
嘿哥 這邊寫錯了喔
Post images
底下還有一些括號沒加,導致加減和乘的順序錯亂🤣
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