■多面体的組み合わせ論(その11)
4次元の場合もやってみよう.
(y0−y1)=(y1−y2)/2=(y2−y3)/3=(y3−y4)/4
y0=1,y4=0
===================================
[1]形状ベクトル(1,0,0,0)の場合,
(y1−y2)/2=(y2−y3)/3=(y3−y4)/4=0
y1=y2=y3=y4=0
[2]形状ベクトル(0,1,0,0)の場合,
(y0−y1)=(y2−y3)/3=(y3−y4)/4=0
y0=y1=1,y2=y3=y4=0
[3]形状ベクトル(0,0,1,0)の場合
(y0−y1)=(y1−y2)/2=(y3−y4)/4=0
y0=y1=y2=1,y3=y4=0
[4]形状ベクトル(0,0,0,1)の場合
(y0−y1)=(y1−y2)/2=(y2−y3)/3=0
y0=y1=y2=y3=1,y3=y4=0
[5]形状ベクトル(1,1,0,0)の場合
(y0−y1)=(y1−y2)/2
(y2−y3)/3=(y3−y4)/4=0,y2=y3=y4=0
2(1−y1)=y1→y1=2/3
[6]形状ベクトル(1,0,1,0)の場合
(y0−y1)=(y2−y3)/3
(y1−y2)/2=(y3−y4)/4=0→y1=y2,y3=y4=0
3(1−y1)=y1→y1=y2=3/4
[7]形状ベクトル(1,0,0,1)の場合
(y0−y1)=(y3−y4)/4
(y1−y2)/2=(y2−y3)/3=0,y1=y2=y3
4(1−y1)=y1→y1=y2=y3=4/5
[8]形状ベクトル(0,1,1,0)の場合
(y0−y1)=(y3−y4)/4=0→y1=1,y3=y4=0
(y1−y2)/2=(y2−y3)/3
3(1−y2)=y2→y2=3/4
[9]形状ベクトル(0,1,0,1)の場合
(y0−y1)=(y2−y3)/3=0→y1=1,y2=y3
(y1−y2)/2=(y3−y4)/4
4(1−y2)=2y2→y2=y3=2/3
[10]形状ベクトル(0,0,1,1)の場合
(y0−y1)=(y1−y2)/2=0→y1=y2=1
(y2−y3)/3=(y3−y4)/4
4(1−y3)=3y3→y3=4/7
[11]形状ベクトル(1,1,1,0)の場合
(y0−y1)=(y1−y2)/2=(y2−y3)/3
(y3−y4)/4=0→y3=y4=0
2(1−y1)=(y1−y2)
3(y1−y2)=2y2→y1=5/6,y2=1/2
[12]形状ベクトル(1,1,0,1)の場合
(y0−y1)=(y1−y2)/2=(y3−y4)/4
(y2−y3)/3=0→y2=y3
2(1−y1)=y1−y2
4(y1−y2)=2y2→y1=6/7,y2=y3=4/7
[13]形状ベクトル(1,0,1,1)の場合
(y0−y1)=(y2−y3)/3=(y3−y4)/4
(y1−y2)/2=0→y1=y2
3(1−y2)=(y2−y3)
4(y2−y3)=3y3→y1=y2=3/4,y3=y4=0
[14]形状ベクトル(0,1,1,1)の場合
(y0−y1)=0→y1=1
(y1−y2)/2=(y2−y3)/3=(y3−y4)/4
3(1−y2)=2(y2−y3)
4(y2−y3)=3y3→y2=7/9,y3=4/9
[15]形状ベクトル(1,1,1,1)の場合
y1=1−1・2/n(n+1)
y2=1−2・3/n(n+1)
y3=1−3・4/n(n+1)
yn=1−n(n+1)/n(n+1)=0
→y1=9/10,y2=7/10,y3=4/10
===================================