Tn=sin(n+1)π/(N+1))sin(nπ/(N+1))/{sin(π/(N+1))}{sin(2π/(N+1))

Tn=-1/2{cos(2n+1)π/(N+1))-cosπ/(N+1))｝/{sin(π/(N+1))}{sin(2π/(N+1))

ΣTnを求めるのは簡単ではない

パーセバルの公式を使う方法と

x=ΣTncos(n-1)θ

y=ΣTnsin(n-1)θ,n=1~N-1を求める方法があると思われる。

Tn=sin(n+1)π/(N+1))sin(nπ/(N+1))/{sin(π/(N+1))}{sin(2π/(N+1))

Tncos(n-1)θ=cos2(n-1)π/(N+1)sin(n+1)π/(N+1))sin(nπ/(N+1))/{sin(π/(N+1))}{sin(2π/(N+1))

Tnsin(n-1)θ=sin2(n-1)π/(N+1)sin(n+1)π/(N+1))sin(nπ/(N+1))/{sin(π/(N+1))}{sin(2π/(N+1))

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

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(2θ)=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+2)/2→訂正

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=5x^2+x-2

X^3=2X^2+X-1を代入すると

cos(3θ)=(X^3-3X^2+2)/2=(2X^2+X-1-3X^2+2)/2=(-X^2+X+1)/2→訂正

cos(4θ)=(X^4-4X^3+2X^2+4X-1)/2

cos(4θ)=(5x^2+x-2-8X^2-4X+4+2X^2+4X-1)/2=(-X^2+X+1)/2

1+cosθ+cos(2θ)+cos(3θ)cos(4θ)+cos(5θ)+cos(6θ)

=1+2cosθ+2cos(2θ)+2cos(3θ)cos(4θ)

=1+(X-1)+(X^2-2X-1)+(-X^2+X+1)=0

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

=x1+x2・(cosθ+cos3θ)+x3・cos2θ+x1・cos4θ

=x2・(2cos2θcosθ)+x3・cos2θ+x1・(1+cos3θ)

Y0=x2・sinθ+x3・sin2θ+x2・sin3θ+x1・sin4θ

=x2・(sinθ+sin3θ)+x3・sin2θ+x1・sin4θ

=x2・(2sin2θcosθ)+x3・sin2θ-x1・sin3θ

X0^2+Y0^2を求めることになる。

=(x2)^2・(2cos2θcosθ)^2+(x3)^2・(cos2θ)^2+(x1)^2・(1+cos3θ)^2

+(x2)^2・(2sin2θcosθ)^2+(x3)^2・(sin2θ)^2+(x1)^2・(sin3θ)^2

+2x2x3(2cos2θcosθcos2θ)+2x1x2(2cos2θcosθ)(1+cos3θ)+2x1x3cos2θ(1+cos3θ)

+2x2x3(2sin2θcosθsin2θ)-2x1x2(2sin2θcosθ)(sin3θ)-2x1x3sin2θ(sin3θ)

=4(x2)^2(cosθ)^2+(x3)^2+(x1)^2(2+2cos3θ)

+4x2x3cosθ+4x1x2cos5θcosθ+4x1x2(cos2θcosθ)+2x1x3cos5θ+2x1x3cos2θ→訂正

=4(x2)^2(cosθ)^2+(x3)^2+(x1)^2(2+2cos3θ)

+4x2x3cosθ+8x1x2cos2θcosθ+4x1x3cos2θ→訂正

=(2x2cosθ+x3)^2+(x1)^2(2+2cos3θ)+8x1x2cosθ(cos2θ)+4x1x3cos2θ→訂正

案外簡単な形になるかもしれない

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(X(X-1)/(X^2+X+2)+{X^2-X}/(X^2+X+2))^2+((2+(-X^2+X+1))/(X^2+X+2))^2

+2(X^2-2X-1)(X-1)(X)/(X^2+X+2)^2+2{X^2-X}(X^2-2X-1)/(X^2+X+2)^2

(2{X^2-X}/(X^2+X+2))^2+((2+(-X^2+X+1))/(X^2+X+2))^2

+2(X^2-2X-1)(X-1)(X)/(X^2+X+2)^2+2{X^2-X}(X^2-2X-1)/(X^2+X+2)^2

4(X~2-X)^2/(X^2+X+2))^2+(-X^2+X+3))/(X^2+X+2))^2

+4(X^2-2X-1)(X~2-X)/(X^2+X+2)^2

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

4(2X^4-5X^3+2X^2+X)/(X^2+X+2))^2+(-X^2+X+3))/(X^2+X+2))^2

4(5X^3-8X^2-7X+6)/(X^2+X+2))^2+(-X^2+X+3))/(X^2+X+2))^2

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

=(7X^2-7X+7)/(X^2+X+2)^2

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