そろそろｄ→1/2証明しておきたい。

x=Σancosnθ

y=Σansinnθ

と有限フーリエ級数で表されるので、n→∞のとき、パーセバルの公式を使えば極限値を求めることができると思われる。

したがって、{an}を求めなければならない。

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Tn=-{(β-γ)α^n+1+(γ-α)β^n+1+(α-β)γ^n+1}/(α-β)(β-γ)(γ-α)

α=1

β=ω

γ=ω^-1

ω=cos(2π/(N+1))+isin(2π/(N+1))

ω^-1=cos(2π/(N+1))-isin(2π/(N+1))

(β-γ)=2isin(2π/(N+1))

(γ-α)β^n+1+(α-β)γ^n+1=(β^n-γ^n)-(β^n+1-γ^n+1)=2i{sin(2nπ/(N+1))-sin(2(n+1)π/(N+1))}

=-4icos((2n+1)π/(N+1))sin(π/(N+1))

(α-β)(γ-α)=(1-ω)(ω^-1-1) =ω^-1-1-1+ω＝2cos(2π/(N+1))-2=-4{sin(π/(N+1))}^2

分子=2isin(2π/(N+1))-4icos((2n+1)π/(N+1))sin(π/(N+1))

=4isin(π/(N+1))cos(π/(N+1))-4icos((2n+1)π/(N+1))sin(π/(N+1))

分母=-4{sin(π/(N+1))}^2・2isin(2π/(N+1))

=-16i{sin(π/(N+1))}^2・sin(π/(N+1))cos(π/(N+1))

Tn=1/4・{cos(π/(N+1))-cos((2n+1)π/(N+1))}/{sin(π/(N+1))}^2cos(π/(N+1))

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

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

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

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n=0のときTn=0

n=1のとき、Tn=sin(2π/(N+1))sin(π/(N+1))/{sin(π/(N+1))}{sin(2π/(N+1))＝1

n=2のとき、Tn=sin(3π/(N+1))sin(2π/(N+1))/{sin(π/(N+1))}{sin(2π/(N+1))＝-4{sin(π/(N+1))}^2+3

X=1+2cos(2π/(n+1))＝1+2（1-2{sin(π/(N+1))}^2）より

n=2のとき、Tn=sin(3π/(N+1))sin(2π/(N+1))/{sin(π/(N+1))}{sin(2π/(N+1))＝-4{sin(π/(N+1))}^2+3=X

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