artificial intelligence courses
min?xf(x)\min_{x} f(x)xmin?f(x)
gi(x)g_i(x)gi?(x)
min?x∣∑i=1m(x1a1+x2?bi)2∣\min_{x}|\sum_{i=1}^{m} (x_1a_1+x_2-b_i)^2|xmin?∣i=1∑m?(x1?a1?+x2??bi?)2∣
x1slopex_1 slopex1?slope
x2interceptx_2 interceptx2?intercept
h(a)=x1a+x2h(a)=x_1a+x_2h(a)=x1?a+x2?
C=x∈Rn:Ax≤b,A∈Rm?n,b∈RnC = {x \in \R^n:Ax\le b, A \in \R^{m*n},b \in \R^n}C=x∈Rn:Ax≤b,A∈Rm?n,b∈Rn
x,y∈Rn∈C?Ax≤b,Ay≤bx,y \in \R^n \in C \Rightarrow Ax \le b, Ay \le b x,y∈Rn∈C?Ax≤b,Ay≤b
?θAx≤θb,(1?θ)Ay≤(1?θ)b\Rightarrow \theta Ax \le \theta b, (1-\theta) Ay \le (1-\theta)b?θAx≤θb,(1?θ)Ay≤(1?θ)b
?θAx\Rightarrow \theta Ax ?θAx
Lp(x,y)=(∑k=1n∣xi?yi∣p)1pL_p(x,y)=(\sum_{k=1}^{n}|x_i-y_i|^p)^{\frac{1}{p}}Lp?(x,y)=(k=1∑n?∣xi??yi?∣p)p1?
L1normL_1 normL1?norm
L2normL_2 normL2?norm
DM=(x?μ)TΣ?1(x?μ)???D_M = (x-\mu)^T \Sigma^{-1}(x-\mu) ???DM?=(x?μ)TΣ?1(x?μ)???
注意這個老哥
這個sigma是一個矩陣
如果這個矩陣是一個III, 一個單位矩陣,實際上乘起來是一個圓形
之后能夠拓展出來橢圓
這樣橢圓上面的點到圓心的距離相同
神奇
(x1?u1)2+(x2?u2)2=1(x_1-u_1)^2+(x_2-u_2)^2=1(x1??u1?)2+(x2??u2?)2=1
(x1?u1)2λ1+(x2?u2)2λ2=1\frac{(x_1-u_1)^2}{\lambda_1}+\frac{(x_2-u_2)^2}{\lambda_2}=1λ1?(x1??u1?)2?+λ2?(x2??u2?)2?=1
KL(p∣∣q)=∑ipilogpiqiKL(p||q)=\sum_i p_i log \frac{p_i}{q_i}KL(p∣∣q)=i∑?pi?logqi?pi??
KL(p∣∣q)=∫p(x)logp(x)q(x)dxKL(p||q)=\int p(x) log\frac{p(x)}{q(x)}dxKL(p∣∣q)=∫p(x)logq(x)p(x)?dx
p=qp=qp=q
KL=0KL=0KL=0
d(x,y)<d(x,z)+d(z,y)d(x,y)<d(x,z)+d(z,y)d(x,y)<d(x,z)+d(z,y)
123abc\begin{matrix} 1 & 2 & 3\\ a & b & c \end{matrix}1a?2b?3c?
123abc\begin{matrix} 1 & 2 & 3\\ a & b & c \end{matrix} 1a?2b?3c?
A=[a1,a2,...an]A=[a_1,a_2,...a_n]A=[a1?,a2?,...an?]
歐氏距離是一個特殊的馬氏距離
[α1α2?αm]\begin{bmatrix} \alpha_{1} \\ \alpha_{2} \\ \vdots \\ \alpha_{m} \\ \end{bmatrix}??????α1?α2??αm????????
[a1a2?anT]\begin{bmatrix} a_1 \\ a_2 \\ \vdots a_n^T \end{bmatrix}????a1?a2??anT??????
f:Rn→Rmf: R^n \rightarrow R^mf:Rn→Rm
vectored function
[abc123]\begin{bmatrix} a & b & c \\ 1 & 2 & 3 \end{bmatrix}[a1?b2?c3?]
[qweasd]\begin{bmatrix} q & w& e \\ a & s & d \end{bmatrix}[qa?ws?ed?]
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