投影上の距離については解決したが、実際の距離Dも1/2に収束するのだろうか？

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

x=π/(N+1)とおくと

Tn=sin(n+1)x)sin(nx)/sin(x)sin(2x)

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n=1〜n-1

D^2=Σ(Tn-1/(n+1))^2+2/(n+1)^2

→訂正

Σ(Tn)=Sとして

D^2=Σ(Tn/S-1/(n+1))^2+2/(n+1)^2

D^2=Σ(Tn)^2/S^2-2/(n+1)/SΣTn+(n-1)/(n+1))^2+2/(n+1)^2

D^2=Σ(Tn)^2/S^2-2/(n+1)+1/(n+1)

D^2=Σ(Tn)^2/S^2-1/(n+1)

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Σ(Tn)は計算済みで、n=1〜n-1

Σsin(n+1)xsin(nx)＝{nsin2x-sin(2nx)}/4sinx・・・ｎ-1項まで

S=Σ(Tn)={nsin2x-sin(2nx)}/4sinxsinxsin2x

sin(n+1)xsin(nx)=-1/2・{cos(2n+1)x-cosx}

Σcos(2n+1)x=sin(2nx)/2sinx-cos(x)={sin(2nx)-2sinxcosx}/2sinx・・・ｎ-1項まで

Σcosx=(n-1)cosx

Σsin(n+1)xsin(nx)=-1/2・{sin(2nx)-2sinxcosx-2sinx(n-1)cosx}/2sinx

Σsin(n+1)xsin(nx)＝{nsin2x-sin(2nx)}/4sinx

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(sin(x)sin(2x))^2Σ(Tn)^2=1/4Σ{cos(2n+1)x-cosx}^2=1/4Σ{cos(2n+1)x}^2-cosx/2Σcos(2n+1)x+(n-1)(cosx)^2/4

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Σ{cos(2n+1)x}^2=Σ(1+cos(4n+2)x)/2

Σcos(4n+2)xが求められればよい。

ここで、

Σcos(2n+1)x=sin(2nx)/2sinx-cos(x)={sin(2nx)-2sinxcosx}/2sinx・・・ｎ-1項まで,を使うと

Σcos(4n+2)x=Σcos(2n+1)(2x)=sin(4nx)/2sin2x-cos(2x)={sin(4nx)-2sin2xcos2x}/2sin2x

Σ{cos(2n+1)x}^2=Σ(1+cos(4n+2)x)/2=(n-1)/2+sin(4nx)/4sin2x-cos(2x)/2

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(sin(x)sin(2x))^2Σ(Tn)^2=1/4Σ{cos(2n+1)x-cosx}^2=1/4Σ{cos(2n+1)x}^2-cosx/2Σcos(2n+1)x+(n-1)(cosx)^2/4

=(n-1)/8+sin(4nx)/16sin2x-cos(2x)/8

-cosx/2・sin(2nx)/2sinx+(cos(x))^2/2+(n-1)(cosx)^2/4

=(n)/8+sin(4nx)/16sin2x-cosx/2・sin(2nx)/2sinx+(n)(cosx)^2/4

S^2={nsin2x-sin(2nx)}^2/(4sinxsinxsin2x)^2

D^2=Σ(Tn)^2/S^2-1/(n+1)

n=3のとき,ｘ=π/4

(sin(x)sin(2x))^2Σ(Tn)^2=3/8+1/4+3/8=1

Σ(Tn)^2=2

S^2=(4)^2/(4/2)^2=4

D^2=Σ(Tn)^2/S^2-1/(n+1)=2/4-1/4=1/4・・・一致

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(sin(x)sin(2x))^2S^2={nsin2x-sin(2nx)}^2/(4sinx)^2

(sin(x)sin(2x))^2Σ(Tn)^2=(n)/8+sin(4nx)/16sin2x-cosx/2・sin(2nx)/2sinx+(n)(cosx)^2/4

Σ(Tn)^2/S^2={2n(sinx)^2+sin(4nx)(sinx)^2/sin2x-4(sinx)^2cosx・sin(2nx)/sinx+4(sinx)^2(n)(cosx)^2}/{nsin2x-sin(2nx)}^2

Σ(Tn)^2/S^2={2n(sinx)^2+sin(4nx)(sinx)^2/sin2x-2sin2x・sin(2nx)+n(sin2x)^2}/{nsin2x-sin(2nx)}^2

D^2=Σ(Tn)^2/S^2-1/(n+1)

n→∞とするとsin(4nx)→0，sin(2nx)→0

Σ(Tn)^2/S^2→{2n(sinx)^2+n(sin2x)^2}/{nsin2x}^2→ {2n(sin2x)^2/4(cosx)^2+n(sin2x)^2}/n^2(sin2x)^2

→ {n/2(cosx)^2+n}/n^2

D^2=Σ(Tn)^2/S^2-1/(n+1)→0・・・一致

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