math_review
topics
mathmatics
- supreme and optimum
- Norm and Linear product
- topology of R*
- Continuious Function
supreme and optimum
Def 1: 非空有界集合必有上确界
common norm
(1) x ∈ \in ∈ Rn, ||x||2= x 1 2 + x 2 2 + . . . + x n 2 \sqrt {x_1^2+x_2^2+...+x_n^2} x12+x22+...+xn2
(2) x ∈ \in ∈ R
Norm of vector
Def 2: a function f : Rn → \rightarrow → R is called norm,if
(1) f(x)>=0 & f(x)=0 if x= 0 ⃗ \vec 0 0
(2) homogenous: f(tx)=|t| f(x)
(3) triangle inquality: f(x+y)<=f(x)+f(y)
f(x)=||x||k
common norm
(1) x ∈ \in ∈ Rn, ||x||2= x 1 2 + x 2 2 + . . . + x n 2 \sqrt {x_1^2+x_2^2+...+x_n^2} x12+x22+...+xn2
(2) x ∈ R n \in\ R^n ∈ Rn, ∣ ∣ x ∣ ∣ p = ( x 1 p + . . . + x n p ) 1 p ||x||_p=(x_1^p+...+x_n^p)^{\frac 1 p} ∣∣x∣∣p=(x1p+...+xnp)p1
(2) xKaTeX parse error: Got function '\inf' with no arguments as subscript at position 15: \in R^n,||x||_\̲i̲n̲f̲=max\{|x_1|,...…
Def 3: Unit Ball
B = x ∈ R n , ∣ ∣ x ∣ ∣ ≤ 1 B={x\in R^n,||x||\leq 1} B=x∈Rn,∣∣x∣∣≤1
norm of matrix
P ∈ S + + n , ∣ ∣ x ∣ ∣ p = ( < x , P x > ) 1 2 P\in S_{++}^n,||x||_p=(<x,P_x>)^\frac 1 2 P∈S++n,∣∣x∣∣p=(<x,Px>)21
- KaTeX parse error: Got function '\sum' with no arguments as argument to '\sqrt' at position 15: ||A||_F=\sqrt \̲s̲u̲m̲_{i=1}^m\sum_{j…
- ∣ ∣ A ∣ ∣ p = m a x ∣ ∣ x ∣ ∣ p = 1 ∣ ∣ A x ∣ ∣ p ||A||_p=\underset {||x||_p=1}{max}||Ax||_p ∣∣A∣∣p=∣∣x∣∣p=1max∣∣Ax∣∣p
- ∣ ∣ A ∣ ∣ = m a x j ∑ i ∣ a i j ∣ ||A||=\underset j {max} \sum_i |aij| ∣∣A∣∣=jmax∑i∣aij∣
- ∣ ∣ A ∣ ∣ 2 = σ ( A ) ||A||_2=\sigma(A) ∣∣A∣∣2=σ(A)(奇异值,特征值绝对值最大)
- KaTeX parse error: Got function '\inf' with no arguments as subscript at position 7: ||A||_\̲i̲n̲f̲=\underset i{ma…
Inner product
f: R n ⊙ R n → R R^n \odot R^n \rightarrow R Rn⊙Rn→R
- Nonnegative: f ( x , x ) ≥ 0 , f ( x , x ) = 0 i f f x = 0 f(x,x)\geq 0,f(x,x)=0 iff x=0 f(x,x)≥0,f(x,x)=0iffx=0
- Symmetric: f ( x , y ) = f ( y , x ) f(x,y)=f(y,x) f(x,y)=f(y,x)
- Bilinear: f ( a x + b y , z ) = a f ( x , z ) + b f ( y , z ) f(ax+by,z)=af(x,z)+bf(y,z) f(ax+by,z)=af(x,z)+bf(y,z)
f ( z , a x + b y ) = a f ( z , x ) + b f ( z , y ) f(z,ax+by)=af(z,x)+bf(z,y) f(z,ax+by)=af(z,x)+bf(z,y)
f(x,y)=<x,y>
定义了内积of vector,we can find corresponding norm;法
< x , x > = ∣ ∣ x ∣ ∣ 2 2 <x,x>=||x||_2^2 <x,x>=∣∣x∣∣22, < x , x > p = ∣ ∣ x ∣ ∣ p 2 <x,x>_p=||x||_p^2 <x,x>p=∣∣x∣∣p2
given norm ,we can’t find corresponding inner product always(like KaTeX parse error: Got function '\inf' with no arguments as subscript at position 7: ||x||_\̲i̲n̲f̲)
example of inner product
- comply with above 3 properties
- A , B ∈ R m + n , < A , B > = t r ( A T B ) = ∑ i j a i j b i j A,B\in R^{m+n},<A,B>=tr(A^TB)=\sum_{ij}a_{ij} b_{ij} A,B∈Rm+n,<A,B>=tr(ATB)=∑ijaijbij
- l 2 ( R ) = x = ( x 1 , . . . ) , x 1 ∈ R , ∑ i = 1 n ∣ x ∣ 2 < inf l^2(R)={x=(x_1,...),x_1\in R,\sum_{i=1}^n |x|^2<\inf} l2(R)=x=(x1,...),x1∈R,∑i=1n∣x∣2<inf
KaTeX parse error: Got function '\inf' with no arguments as superscript at position 18: …,y>=\sum_{i=1}^\̲i̲n̲f̲ ̲x_iy_i < \inf - L 2 ( R ) = f : R → R , ∫ R ∣ f ( x ) ∣ 2 d x < inf L^2(R)={f:R\rightarrow R,\int_R|f(x)|^2 dx<\inf} L2(R)=f:R→R,∫R∣f(x)∣2dx<inf
< f , g > L 2 = ∫ R f ( x ) g ( x ) d x <f,g>_{L^2}=\int_Rf(x)g(x)dx <f,g>L2=∫Rf(x)g(x)dx
Def 5: x , y ∈ R n x,y \in R^n x,y∈Rn
included angle, a n g ( x , y ) = a r c c o s < x , y > ∣ ∣ x ∣ ∣ ∣ . ∣ ∣ y ∣ ∣ ang(x,y)=arccos \frac {<x,y>}{||x|||.||y||} ang(x,y)=arccos∣∣x∣∣∣.∣∣y∣∣<x,y>
Topology of IRn
logistic chain: N o r m → U n i t B a l l → N e i g h b o u r h o o d → L i m i t Norm\rightarrow Unit Ball\rightarrow Neighbourhood\rightarrow Limit Norm→UnitBall→Neighbourhood→Limit
Def 6: given ϵ \epsilon ϵ,the ϵ \epsilon ϵ-neighbourhood of a point x
N ϵ ( x ) = y , y ∈ I R n , ∣ ∣ y − x ∣ ∣ < ϵ N_{\epsilon}(x)={y,y\in IR^n,||y-x||<\epsilon} Nϵ(x)=y,y∈IRn,∣∣y−x∣∣<ϵ
Def 7: interor point of S, ∃ ϵ > 0 , N ϵ ( x ) ∈ S \exists \epsilon>0,N_{\epsilon}(x)\in S ∃ϵ>0,Nϵ(x)∈S
Def 8: interior of S,the set of all interior points of S
Def 9: Open set O=int O
Def 10: Closed Set IR\F is open
Def 11: ( x k ) (x_k) (xk) converges to x,if ∀ ϵ > 0 , ∀ p , ∣ ∣ x k − x ∣ ∣ p < ϵ , i f k > N \forall \epsilon>0,\forall p,||x_k-x||_p<\epsilon,if\ k>N ∀ϵ>0,∀p,∣∣xk−x∣∣p<ϵ,if k>N
Def 12:(limited point) x ∈ I R n , S ∈ I R n , ∃ ( x k ) , x k ≠ x , ( x k ) ∈ S , s . t . x k → x x\in IR^n,S\in IR^n,\exists (x_k),x_k\neq x,(x_k)\in S,s.t. x_k\rightarrow x x∈IRn,S∈IRn,∃(xk),xk=x,(xk)∈S,s.t.xk→x
Def 13: F is closed if it c ontains all its limit points