■正単体の正多角形断面(その27)

4sinxsin2xTncos2(n-1)α={-cos3x+cos(2n-3)x+cos(2n-1)x-cos(4n-1)x}

4sinxsin2xTnsin2(n-1)α={sin3x+sin(2n-3)x+sin(2n-1)x-sin(4n-1)x}

4sinxsin2xΣTncos2(n-1)α={-Σcos3x+Σcos(2n-3)x+Σcos(2n-1)x-Σcos(4n-1)x}

4sinxsin2xΣTnsin2(n-1)α={Σsin3x+Σsin(2n-3)x+Σsin(2n-1)x-Σsin(4n-1)x}

{-Σcos3x+Σcos(2n-3)x+Σcos(2n-1)x-Σcos(4n-1)x}^2+{Σsin3x+Σsin(2n-3)x+Σsin(2n-1)x-Σsin(4n-1)x}^2

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合わせると

n^2

2{sin(nx)/sin(x)}^2

{sin(2nx)/sin(2x)}^2

-2ncos(n+1)xsin(nx)/sin(x)

-2ncos(n+3)xsin(nx)/sin(x)

2ncos(2n+4)xsin(2nx)/sin(2x)

2{sin(nx)/sin(x)}^2・cos2x

-2cos((n+3)x)・sin(nx)/sin(x)・sin(2nx)/sin(2x)

-2cos((n+1)x)・sin(nx)/sin(x)・sin(2nx)/sin(2x)

ここまでは数値的にも合致していることを確認した。

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収束判定

x=π/(N+1)

n=N-1→∞

sin(nx)→sinπ=0

sin(2nx)→sin2π=0

n^2

-n{cos(2n+4)x+cos2x}sin(nx)/sin(x)→?

2ncos(2n+4)xsin(2nx)/sin(2x)→?

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分子は

x=π/(N+1)

n=N-1→∞

sin(nx)→sinπ=0

sin(2nx)→sin2π=0

n^2

-n{cos(2n+4)x+cos2x}sin(nx)/sin(x)→?

2ncos(2n+4)xsin(2nx)/sin(2x)→?

n^2で割れば

n^2→1

-n{cos(2n+4)x+cos2x}sin(nx)/sin(x)→0

2ncos(2n+4)xsin(2nx)/sin(2x)→0

で1に収束する。

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一方、分母は

4sinαsin2αΣTn=Σsinrxsin(r+1)x={(N)sin2x-sin2Nx}/sinx

4sinαsin2αΣTn=Σsinrxsin(r+1)x={2(N)cosx}-{sin2Nx}/sinx

(4sinαsin2αΣTn)^2→4N^2(cosx)^2→4N^2

n^2で割れば→4に収束する

d^2→1/4 (QED)

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