定义 设 x = ( x 1 , x 2 , … , x n ) T ∈ R n x=(x_1,x_2,dots,x_n)^T in R^n x=(x1,x2,…,xn)T∈Rn,则称乘积 x j 1 x j 2 … x j d x_{j_1}x_{j_2}dots x_{j_d} xj1xj2…xjd为 x x x的一个 d d d阶多项式，其中 j 1 , j 2 , … , j d ∈ { 1 , 2 , … , n } j_1,j_2,dots,j_d in {1,2,dots,n} j1,j2,…,jd∈{1,2,…,n}
有序齐次多项式
考虑二维空间（ x ∈ R 2 x in R^2 x∈R2）的模式， x = ( x 1 , x 2 ) T x = (x_1,x_2)^T x=(x1,x2)T,其所有的二阶单项式为 x i 2 , x 2 2 , x 1 x 2 , x 2 x 1 x_i^2,x_2^2,x_1x_2,x_2x_1 xi2,x22,x1x2,x2x1 为有序单项式。
C
2
(
x
)
=
(
x
1
2
,
x
2
2
,
x
1
x
2
,
x
2
x
1
)
T
C_2(x)=(x_1^2,x_2^2,x_1x_2,x_2x_1)^T
C2(x)=(x12,x22,x1x2,x2x1)T
C
d
(
x
)
=
(
x
j
1
x
j
2
…
x
j
d
∣
j
1
,
j
2
,
…
,
j
d
∈
{
1
,
2
,
…
,
n
}
)
T
C_d(x)=(x_{j_1}x_{j_2}dots x_{j_d}|_{j_1,j_2,dots,j_d in {1,2,dots,n}})^T
Cd(x)=(xj1xj2…xjd∣j1,j2,…,jd∈{1,2,…,n})T
1、二次项推导
令
x
=
(
x
1
,
x
2
)
T
,
y
=
(
y
1
,
y
2
)
T
令 x=(x_1,x_2)^T,y=(y_1,y_2)^T
令x=(x1,x2)T,y=(y1,y2)T
C
2
(
x
)
⋅
C
2
(
y
)
=
(
x
1
2
,
x
2
2
,
x
1
x
2
,
x
2
x
1
)
T
⋅
(
y
1
2
,
y
2
2
,
y
1
y
2
,
y
2
y
1
)
T
=
x
1
2
y
1
2
+
x
2
2
y
2
2
+
x
1
x
2
y
1
y
2
+
x
2
x
1
y
2
y
1
=
(
x
⋅
y
)
2
begin{aligned} C_2(x)·C_2(y) &=(x_1^2,x_2^2,x_1x_2,x_2x_1)^T · (y_1^2,y_2^2,y_1y_2,y_2y_1)^T\ &=x_1^2y_1^2+x_2^2y_2^2+x_1x_2y_1y_2+x_2x_1y_2y_1\ &=(x·y)^2 end{aligned}
C2(x)⋅C2(y)=(x12,x22,x1x2,x2x1)T⋅(y12,y22,y1y2,y2y1)T=x12y12+x22y22+x1x2y1y2+x2x1y2y1=(x⋅y)2
2、多次幂推导
令
x
=
(
x
j
1
,
x
j
2
,
…
,
x
j
d
∣
j
1
,
j
2
,
…
,
j
d
∈
{
1
,
2
,
…
,
n
}
)
T
令 x=(x_{j_1},x_{j_2},dots,x_{j_d}|_{j_1,j_2,dots,j_d in {1,2,dots,n}})^T
令x=(xj1,xj2,…,xjd∣j1,j2,…,jd∈{1,2,…,n})T
y
=
(
y
j
1
,
y
j
2
,
…
,
y
j
d
∣
j
1
,
j
2
,
…
,
j
d
∈
{
1
,
2
,
…
,
n
}
)
T
y=(y_{j_1},y_{j_2},dots,y_{j_d}|_{j_1,j_2,dots,j_d in {1,2,dots,n}})^T
y=(yj1,yj2,…,yjd∣j1,j2,…,jd∈{1,2,…,n})T
C
d
(
x
)
⋅
C
d
(
y
)
=
(
x
j
1
,
x
j
2
,
…
,
x
j
d
∣
j
1
,
j
2
,
…
,
j
d
∈
{
1
,
2
,
…
,
n
}
)
T
⋅
(
y
j
1
,
y
j
2
,
…
,
y
j
d
∣
j
1
,
j
2
,
…
,
j
d
∈
{
1
,
2
,
…
,
n
}
)
T
=
∑
j
1
=
1
n
⋯
∑
j
d
=
1
n
x
j
1
…
x
j
d
⋅
y
j
1
…
y
j
d
=
∑
j
1
=
1
n
x
j
1
y
j
1
⋯
∑
j
d
=
1
n
x
j
d
y
j
d
=
(
∑
j
=
1
n
x
i
y
i
)
d
=
(
x
⋅
y
)
d
begin{aligned} C_d(x)·C_d(y) &=(x_{j_1},x_{j_2},dots,x_{j_d}|_{j_1,j_2,dots,j_d in {1,2,dots,n}})^T · (y_{j_1},y_{j_2},dots,y_{j_d}|_{j_1,j_2,dots,j_d in {1,2,dots,n}})^T\ &=sum_{j_1=1}^ndotssum_{j_d=1}^n x_{j_1}dots x_{j_d} · y_{j_1}dots y_{j_d}\ &=sum_{j_1=1}^n x_{j_1}y_{j_1} dots sum_{j_d=1}^n x_{j_d}y_{j_d}\ &=(sum_{j=1}^nx_iy_i)^d=(x·y)^d end{aligned}
Cd(x)⋅Cd(y)=(xj1,xj2,…,xjd∣j1,j2,…,jd∈{1,2,…,n})T⋅(yj1,yj2,…,yjd∣j1,j2,…,jd∈{1,2,…,n})T=j1=1∑n⋯jd=1∑nxj1…xjd⋅yj1…yjd=j1=1∑nxj1yj1⋯jd=1∑nxjdyjd=(j=1∑nxiyi)d=(x⋅y)d
最后
以上就是长情小鸭子最近收集整理的关于多项式核函数相关推导的全部内容，更多相关多项式核函数相关推导内容请搜索靠谱客的其他文章。
发表评论 取消回复