■正多面体の正多角形断面(その33)

 7点

  (1,0,0,0,0,0,0)

  (0,1,0,0,0,0,0)

  (0,0,1,0,0,0,0)

  (0,0,0,1,0,0,0)

  (0,0,0,0,1,0,0)

  (0,0,0,0,0,1,0)

  (0,0,0,0,0,0,1)

が,xy平面上の6点

  (cos0π/7,sin0π/7)

  (cos2π/7,sin2π/7)

  (cos4π/7,sin4π/7)

  (cos6π/7,sin6π/7)

  (cos8π/7,sin8π/7)

  (cos10π/7,sin10π/7)

  (cos12π/7,sin12π/7)

に投影されるためには,2×7行列

M=[cos0π/7,cos2π/7,cos4π/7,cos6π/7,cos8π/7,cos10π/7,cos12π/7]

  [sin0π/7,sin2π/7,sin4π/7,sin6π/7,sin8π/7,sin10π/7,sin12π/7]

が必要になる.

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x5=1/(X^4-2X^3+2X+2)

x4=X/(X^4-2X^3+2X+2)

x3={X^2-X}/(X^4-2X^3+2X+2)

x2={X^3-2X^2+1}/(X^4-2X^3+2X+2)

x1={X^4-3X^3+X^2+2X}/(X^4-2X^3+2X+2)

この中のいくつかは等しいと思われるが・・・

(X-1)/2=cos(360/7)=cosθ

cos(3θ)=cos(4θ)より

4{(X-1)/2}^3-3{(X-1)/2}=8{(X-1)/2}^4-8{(X-1)/2}^2+1

8{(X-1)}^3-24{(X-1)}=8{(X-1)}^4-32{(X-1)}^2+16

{(X-1)}^3-3{(X-1)}={(X-1)}^4-4{(X-1)}^2+2

x^3-3x^2+3x-1-3x+3=x^4-4x^3+6x^2-4x+1-4x^2+8x-4+2

x^3-3x^2+2=x^4-4x^3+2x^2+4x-1

x^4-5x^3+5x^2+4x-3=0→X=3を解に持つはずである。(x-3)(x^3-2x^2-x+1)=0

x^4-3x^3+x^2+2x=2x^3-4x^2-2x+3

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{(X-1)}^3-3{(X-1)}={(X-1)}^4-4{(X-1)}^2+2

Y=(X-1)

Y^4-Y^3-4Y^2+3Y+2=0

(Y-2)(Y^3+Y^2-2Y-1)=0

Yは2ではないので(Y^3+Y^2-2Y-1)=0

{(X-1)}^3+{(X-1)}^2-2{(X-1)}-1=0

X^3-3X^2+3X-1+X^2-2X+1-2X+2-1=0

X^3-2X^2-X-1=0, X^3=2X^2+X+1→ここに誤り

X^3-2X^2-X+1=0, X^3=2X^2+X-1→訂正

x^4-3x^3+x^2+2x=2x^3-4x^2-2x+3=4X^2+2X-2-4X^2-2X+3=1

X^3-2X^2+1=2X^2+X-1-2X^2+1=X

x1=x5,x2=x4

(X^4-2X^3+2X+2)=5x^3-5x^2-4x+3-2X^3+2X+2=3x^3-5X^2-2x+5

=3(2X^2+X-1)-5X^2-2x+5=x^2+x+2

x5=1/(X^2+X+2)

x4=X/(X^2+X+2)

x3={X^2-X}/(X^2+X+2)

x2=X/(X^2+X+2)

x1=1/(X^2+X+2)

x1+x2+x3+x4+x5=1

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(X-1)/2=cos(360/7)=cosθ

cos(3θ)=2{(X-1)/2}^2-1=(X^2-2X+1)/2-1=(X^2-2X-1)/2

cos(3θ)=4{(X-1)/2}^3-3{(X-1)/2}=(X-1)^3/2-(3X-3)/2=(X^3-3X^2-3X+1-3X+3)/2=(X^3-3X^2-6X+4)/2

cos(4θ)=8{(X-1)/2}^4-8{(X-1)/2}^2+1=(X-1)^4/2-4(X-1)^2/2+1=(X^4-4X^3+6X^2-4X+1)/2-4(X^2-2X+1)/2+2/2=(X^4-4X^3+2X^2+4X-1)/2

x^4-5x^3+5x^2+4x-3=0

X^3=2X^2+X-1を代入すると

cos(3θ)=(X^3-3X^2-6X+4)/2=(2X^2+X-1-3X^2-6X+4)/2=(-X^2-5X+3)/2

cos(4θ)=(5x^3-5x^2-4x+3-4X^3+2X^2+4X-1)/2=(x^3-3x^2+2)/2=(2X^2+X-1-3x^2+2)/2=(-X^2+X+1)/2

x5=1/(X^2+X+2)

x4=X/(X^2+X+2)

x3={X^2-X}/(X^2+X+2)

x2=X/(X^2+X+2)

x1=1/(X^2+X+2)

X0=x1+x2・cosθ+x3・cos2θ+x4・cos3θ+x5・cos4θ

=x1+x2・cosθ+x3・cos2θ+x2・cos3θ+x1・cos4θ

Y0=x2・sinθ+x3・sin2θ+x2・sin3θ+x1・sin4θ

X0^2+Y0^2を求めることになる。

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