■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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