關於九宮數的一點研究
九宮旋轉1,8,3,4,9,2,7,6，針對這個週期序列計算它的序列公式，結果非常複雜：
1/8 cos(1/4 π (n - 8)) + 1/8 cos(1/2 π (n - 8)) + 1/8 cos(3/4 π (n - 8)) + 1/8 cos(π (n - 8)) + 1/8 cos(5/4 π (n - 8)) + 1/8 cos(3/2 π (n - 8)) + 1/8 cos(7/4 π (n - 8)) + 1/4 cos(1/4 π (n - 7)) + 1/4 cos(1/2 π (n - 7)) + 1/4 cos(3/4 π (n - 7)) + 1/4 cos(π (n - 7)) + 1/4 cos(5/4 π (n - 7)) + 1/4 cos(3/2 π (n - 7)) + 1/4 cos(7/4 π (n - 7)) - 3/8 cos(1/4 π (n - 6)) - 3/8 cos(1/2 π (n - 6)) - 3/8 cos(3/4 π (n - 6)) - 3/8 cos(π (n - 6)) - 3/8 cos(5/4 π (n - 6)) - 3/8 cos(3/2 π (n - 6)) - 3/8 cos(7/4 π (n - 6)) + 1/2 cos(1/4 π (n - 5)) + 1/2 cos(1/2 π (n - 5)) + 1/2 cos(3/4 π (n - 5)) + 1/2 cos(π (n - 5)) + 1/2 cos(5/4 π (n - 5)) + 1/2 cos(3/2 π (n - 5)) + 1/2 cos(7/4 π (n - 5)) - 1/8 cos(1/4 π (n - 4)) - 1/8 cos(1/2 π (n - 4)) - 1/8 cos(3/4 π (n - 4)) - 1/8 cos(π (n - 4)) - 1/8 cos(5/4 π (n - 4)) - 1/8 cos(3/2 π (n - 4)) - 1/8 cos(7/4 π (n - 4)) - 1/4 cos(1/4 π (n - 3)) - 1/4 cos(1/2 π (n - 3)) - 1/4 cos(3/4 π (n - 3)) - 1/4 cos(π (n - 3)) - 1/4 cos(5/4 π (n - 3)) - 1/4 cos(3/2 π (n - 3)) - 1/4 cos(7/4 π (n - 3)) + 3/8 cos(1/4 π (n - 2)) + 3/8 cos(1/2 π (n - 2)) + 3/8 cos(3/4 π (n - 2)) + 3/8 cos(π (n - 2)) + 3/8 cos(5/4 π (n - 2)) + 3/8 cos(3/2 π (n - 2)) + 3/8 cos(7/4 π (n - 2)) - 1/2 cos(1/4 π (n - 1)) - 1/2 cos(1/2 π (n - 1)) - 1/2 cos(3/4 π (n - 1)) - 1/2 cos(π (n - 1)) - 1/2 cos(5/4 π (n - 1)) - 1/2 cos(3/2 π (n - 1)) - 1/2 cos(7/4 π (n - 1)) 化簡下來，可得
1/2 Cos[n π] ((1+Sqrt[2]) Cos[(n π)/4]-(-1+Sqrt[2]) Cos[(3 n π)/4]+3 (1+Sqrt[2]) Sin[(n π)/4]+3 (-1+Sqrt[2]) Sin[(3 n π)/4]) 如果要在九宮上，試圖直接計算先天八卦序列也是很複雜的，會變成如下樣子：
(-3 Cos[((-8 + n) Pi)/4])/8 - (3 Cos[((-8 + n) Pi)/2])/8 - (3 Cos[(3 (-8 + n) Pi)/4])/8 - (3 Cos[(-8 + n) Pi])/8 - (3 Cos[(5 (-8 + n) Pi)/4])/8 - (3 Cos[(3 (-8 + n) Pi)/2])/8 - (3 Cos[(7 (-8 + n) Pi)/4])/8 + (3 Cos[((-7 + n) Pi)/4])/8 + (3 Cos[((-7 + n) Pi)/2])/8 + (3 Cos[(3 (-7 + n) Pi)/4])/8 + (3 Cos[(-7 + n) Pi])/8 + (3 Cos[(5 (-7 + n) Pi)/4])/8 + (3 Cos[(3 (-7 + n) Pi)/2])/8 + (3 Cos[(7 (-7 + n) Pi)/4])/8 - Cos[((-6 + n) Pi)/4]/2 - Cos[((-6 + n) Pi)/2]/2 - Cos[(3 (-6 + n) Pi)/4]/2 - Cos[(-6 + n) Pi]/2 - Cos[(5 (-6 + n) Pi)/4]/2 - Cos[(3 (-6 + n) Pi)/2]/2 - Cos[(7 (-6 + n) Pi)/4]/2 - Cos[((-5 + n) Pi)/4]/8 - Cos[((-5 + n) Pi)/2]/8 - Cos[(3 (-5 + n) Pi)/4]/8 - Cos[(-5 + n) Pi]/8 - Cos[(5 (-5 + n) Pi)/4]/8 - Cos[(3 (-5 + n) Pi)/2]/8 - Cos[(7 (-5 + n) Pi)/4]/8 - Cos[((-4 + n) Pi)/4]/4 - Cos[((-4 + n) Pi)/2]/4 - Cos[(3 (-4 + n) Pi)/4]/4 - Cos[(-4 + n) Pi]/4 - Cos[(5 (-4 + n) Pi)/4]/4 - Cos[(3 (-4 + n) Pi)/2]/4 - Cos[(7 (-4 + n) Pi)/4]/4 + Cos[((-3 + n) Pi)/4]/2 + Cos[((-3 + n) Pi)/2]/2 + Cos[(3 (-3 + n) Pi)/4]/2 + Cos[(-3 + n) Pi]/2 + Cos[(5 (-3 + n) Pi)/4]/2 + Cos[(3 (-3 + n) Pi)/2]/2 + Cos[(7 (-3 + n) Pi)/4]/2 + Cos[((-2 + n) Pi)/4]/4 + Cos[((-2 + n) Pi)/2]/4 + Cos[(3 (-2 + n) Pi)/4]/4 + Cos[(-2 + n) Pi]/4 + Cos[(5 (-2 + n) Pi)/4]/4 + Cos[(3 (-2 + n) Pi)/2]/4 + Cos[(7 (-2 + n) Pi)/4]/4 + Cos[((-1 + n) Pi)/4]/8 + Cos[((-1 + n) Pi)/2]/8 + Cos[(3 (-1 + n) Pi)/4]/8 + Cos[(-1 + n) Pi]/8 + Cos[(5 (-1 + n) Pi)/4]/8 + Cos[(3 (-1 + n) Pi)/2]/8 + Cos[(7 (-1 + n) Pi)/4]/8這實際在用三角函數來體現這個週期函數，進行了變換，總共有54項，如果僅考慮描述一半圖形的話，它們僅是180度顛倒，所以考慮以四組卦為序，並設：
A=Cos[1/2π(-4+x)]
B=Cos[π(-4+x)]
C=Cos[3/2π(-4+x)]
D=Cos[1/2π(-3+x)]
E=Cos[π(-3+x)]
F=Cos[3/2π(-3+x)]
G=Cos[1/2π(-2+x)]
H=Cos[π(-2+x)]
I=Cos[3/2π(-2+x)]
J=Cos[1/2π(-1+x)]
K=Cos[π(-1+x)]
L=Cos[3/2π(-1+x)]
顯然從A~L，初步可以認為這採用12個維度的描述，是從x-1、x-2、x-3、x-4，分別取乘1/2、1、3/2，然後得到的角度值，由於最終結果從-4至4，而3/2倍數的最大值為1.5，所以相當於被分解成了多個0~1.5間的小片斷進行相加，所以兩至三個為1.5的值便可以構成3~4.5之間的波動，然後再有兩個修正值應當足夠表達，所以化簡可以到四到五個項，然而化簡不利於觀察內部結構，此處不化簡。
坎艮震巽= -1 -1/4A-1/4B-1/4C-1/2D-1/2E-1/2F+3/4G+3/4H+3/4I-J-K-L 離坤兌乾= 1 +1/4A+1/4B+1/4C+1/2D+1/2E+1/2F-3/4G-3/4H-3/4I+J+K+L 震巽離坤=-(1/2) -3/4A-3/4B-3/4C+D+E+F-1/4G-1/4H-1/4I-1/2J-1/2K-1/2L 兌乾坎艮= 1/2 +3/4A+3/4B+3/4C-D-E-F+1/4G+1/4H+1/4I+1/2J+1/2K+1/2L 巽離坤兌=1/2 +1/2A+1/2B+1/2C-3/4D-3/4E-3/4F+G+H+I-1/4J-1/4K-1/4L 乾坎艮震=-(1/2) -1/2A-1/2B-1/2C+3/4D+3/4E+3/4F-G-H-I+1/4J+1/4K+1/4L 艮震巽離=1 +A+B+C-1/4D-1/4E-1/4F-1/2G-1/2H-1/2I+3/4J+3/4K+3/4L 坤兌乾坎=-1 -A-B-C+1/4D+1/4E+1/4F+1/2G+1/2H+1/2I-3/4J-3/4K-3/4L在格式上有意將常數項獨立了出來，這樣便可以看到後面的項是如何匹配的，同樣將先天往後天轉化也排出來：
離乾艮震= 3/2 -1/2A-1/2B-1/2C+3/4D+3/4E+3/4F+1/4G+1/4H+1/4I+J+K+L 艮震離乾= 3/2 +1/4A+1/4B+1/4C+D+E+F-1/2G-1/2H-1/2I+3/4J+3/4K+3/4L 震離乾艮= 3/2 +3/4A+3/4B+3/4C+1/4D+1/4E+1/4F+G+H+I-1/2J-1/2K-1/2L 乾艮震離= 3/2 +A+B+C-1/2D-1/2E-1/2F+3/4G+3/4H+3/4I+1/4J+1/4K+1/4L比較下來會得到有趣的結果：
坎艮震巽= -1 -1/4A-1/4B-1/4C-1/2D-1/2E-1/2F+3/4G+3/4H+3/4I-J-K-L 離坤兌乾= 1 +1/4A+1/4B+1/4C+1/2D+1/2E+1/2F-3/4G-3/4H-3/4I+J+K+L 艮震離乾= 3/2 +1/4A+1/4B+1/4C+D+E+F-1/2G-1/2H-1/2I+3/4J+3/4K+3/4L 震巽離坤=-(1/2) -3/4A-3/4B-3/4C+D+E+F-1/4G-1/4H-1/4I-1/2J-1/2K-1/2L 兌乾坎艮= 1/2 +3/4A+3/4B+3/4C-D-E-F+1/4G+1/4H+1/4I+1/2J+1/2K+1/2L 震離乾艮= 3/2 +3/4A+3/4B+3/4C+1/4D+1/4E+1/4F+G+H+I-1/2J-1/2K-1/2L 巽離坤兌=1/2 +1/2A+1/2B+1/2C-3/4D-3/4E-3/4F+G+H+I-1/4J-1/4K-1/4L 乾坎艮震=-(1/2) -1/2A-1/2B-1/2C+3/4D+3/4E+3/4F-G-H-I+1/4J+1/4K+1/4L 離乾艮震= 3/2 -1/2A-1/2B-1/2C+3/4D+3/4E+3/4F+1/4G+1/4H+1/4I+J+K+L 艮震巽離=1 +A+B+C-1/4D-1/4E-1/4F-1/2G-1/2H-1/2I+3/4J+3/4K+3/4L 坤兌乾坎=-1 -A-B-C+1/4D+1/4E+1/4F+1/2G+1/2H+1/2I-3/4J-3/4K-3/4L 乾艮震離=3/2 +A+B+C-1/2D-1/2E-1/2F+3/4G+3/4H+3/4I+1/4J+1/4K+1/4L其中的軌跡表現得便很明顯了。