ΣTn

ΣTncos2(n-1)α

ΣTnsin2(n-1)α

と合致するか、確認しておきたい

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

α=π/(N+1)

Tn=sin(n+1)αsinnα/sinαsin2α

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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X=1+2cos2π/(N+1)

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)α-cosα｝/sinαsin2α

次の課題は

Tncos2(n-1)α

Tnsin2(n-1)α

を求めることである。

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Tncos2(n-1)α=sin(n+1)αsinnαcos(2n-2)α/sinαsin2α

=1/4{-cos3x+cos(2n-3)x+cos(2n-1)x-cos(4n-1)x}/sinxsin2x

=1/4{-cos3x+cos2nxcos3x+sin2nxsin3x+cos2nxcosx+sin2nxsinx-cos4nxcosx-sin4nxsinx}/sinxsin2x

Tnsin2(n-1)α=sin(n+1)αsinnαsin(2n-2)α/sinαsin2α

=1/4{sin3x+sin(2n-3)x+sin(2n-1)x-sin(4n-1)x}/sinxsin2x

=1/4{sin3x+sin2nxcos3x-cos2nxsin3x+sin2nxcosx-cos2nxsinx-sin4nxcosx+cos4nxsinx}/sinxsin2x

Σsinrx=sinx+・・・+sinnx=sin((n+1)x/2)sin(nx/2)/sin(x/2)

Σcosrx=cosx+・・・+cosnx=cos((n+1)x/2)sin(nx/2)/sin(x/2)

Σsin2rx=sin((n+1)x)sin(nx)/sin(x)

Σcos2rx=cos((n+1)x)sin(nx)/sin(x)

Σsin4rx=sin(2(n+1)x)sin(2nx)/sin(2x)

Σcos4rx=cos(2(n+1)x)sin(2nx)/sin(2x)

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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

=

(Σcos3x)^2+(Σsin3x)^2

(Σcos(2n-3)x)^2+(Σsin(2n-3)x)^2

(Σcos(2n-1)x)^2+(Σsin(2n-1)x)^2

(Σcos(4n-1)x)^2+(Σsin(4n-1)x)^2

-2(Σcos3x)(Σcos(2n-3)x)+2(Σsin3x)(Σsin(2n-3)x)

-2(Σcos3x)(Σcos(2n-1)x)+2(Σsin3x)(Σsin(2n-1)x)

+2(Σcos3x)(Σcos(2n-1)x)-2(Σsin3x)(Σsin(2n-1)x)

+2(Σcos(2n-3)x)(Σcos(2n-1)x)+2(Σsin(2n-3)x)(Σsin(2n-1)x)

-2(Σcos(2n-3)x)(Σcos(4n-1)x)-2(Σsin(2n-3)x)(Σsin(2n-1)x)

-2(Σcos(2n-1)x)(Σcos(4n-1)x)-2(Σsin(2n-1)x)(Σsin(2n-1)x)

(Σcos(2n-3)x)=(Σcos2nxcos3x+Σsin2nxsin3x)=cos((n+1)x)sin(nx)cos3x/sin(x)+sin((n+1)x)sin(nx)sin3x/sin(x)

(Σsin(2n-3)x)=(Σsin2nxcos3x-Σcos2nxsin3x)=sin((n+1)x)sin(nx)cos3x/sin(x)-cos((n+1)x)sin(nx)sin3x/sin(x)

(Σcos(2n-1)x)=(Σcos2nxcosx+Σsin2nxsinx)=cos((n+1)x)sin(nx)cosx/sin(x)+sin((n+1)x)sin(nx)sinx/sin(x)

(Σsin(2n-1)x)=(Σsin2nxcosx-Σcos2nxsinx)=sin((n+1)x)sin(nx)cosx/sin(x)-cos((n+1)x)sin(nx)sinx/sin(x)

(Σcos(4n-1)x)=(Σcos4nxcosx+Σsin4nxsinx)=cos(2(n+1)x)sin(2nx)cosx/sin(2x)+sin(2(n+1)x)sin(2nx)sinx/sin(2x)

(Σsin(4n-1)x)=(Σsin4nxcosx-Σcos4nxsinx)=sin(2(n+1)x)sin(2nx)cosx/sin(2x)-cos(2(n+1)x)sin(2nx)sinx/sin(2x)

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n項まで

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

(Σcos(2n-3)x)=(Σcos2nxcos3x+Σsin2nxsin3x)=cos((n+1)x)sin(nx)cos3x/sin(x)+sin((n+1)x)sin(nx)sin3x/sin(x)

(Σcos(2n-1)x)=(Σcos2nxcosx+Σsin2nxsinx)=cos((n+1)x)sin(nx)cosx/sin(x)+sin((n+1)x)sin(nx)sinx/sin(x)

(Σcos(4n-1)x)=(Σcos4nxcosx+Σsin4nxsinx)=cos(2(n+1)x)sin(2nx)cosx/sin(2x)+sin(2(n+1)x)sin(2nx)sinx/sin(2x)

4sinxsin2xΣTncos2(n-1)α=-ncos3x+cos((n+1)x)sin(nx)(cos3x+cosx)/sin(x)+sin((n+1)x)sin(nx)(sin3x+sinx)/sin(x)

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

4sinxsin2xΣTncos2(n-1)α=-ncos3x+cos((n+1)x)sin(nx)2(cos2xcosx)/sin(x)+sin((n+1)x)sin(nx)(2sin2xcosx)/sin(x)

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

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

(Σsin(2n-3)x)=(Σsin2nxcos3x-Σcos2nxsin3x)=sin((n+1)x)sin(nx)cos3x/sin(x)-cos((n+1)x)sin(nx)sin3x/sin(x)

(Σsin(2n-1)x)=(Σsin2nxcosx-Σcos2nxsinx)=sin((n+1)x)sin(nx)cosx/sin(x)-cos((n+1)x)sin(nx)sinx/sin(x)

(Σsin(4n-1)x)=(Σsin4nxcosx-Σcos4nxsinx)=sin(2(n+1)x)sin(2nx)cosx/sin(2x)-cos(2(n+1)x)sin(2nx)sinx/sin(2x)

4sinxsin2xΣTnsin2(n-1)α=nsin3x+sin((n+1)x)sin(nx)(cos3x+cosx)/sin(x)-cos((n+1)x)sin(nx)(sin3x+sinx)/sin(x)

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

4sinxsin2xΣTnsin2(n-1)α=nsin3x+sin((n+1)x)sin(nx)2(cos2xcosx)/sin(x)-cos((n+1)x)sin(nx)2(sin2xcosx)/sin(x)

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

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α=π/(N+1)とおくと

求めたいのはr=N-1とおいて、

4sinxsin2xΣTn=Σsinrxsin(r+1)x={(N)sin2x-sin2Nx}/sinx

ここで、x=π/(N+1)であるから

sin2Nx=sin(2(N+1)-2)x=-sin2x

{(N)sin2x-sin2Nx}=(n+1)sin2x

ΣTn=Σsinrxsin(r+1)x={(N)sin2x-sin2Nx}/sinx4sinxsin2x

ΣTn=Σsinrxsin(r+1)x=(n+1)sin2x/sinx4sinxsin2x

ΣTn=Σsinrxsin(r+1)x=(n+1)/(2sinx)^2

Ti=sin(i+1)αsiniα/sinαsin2α

Ti/ΣTn=4sinαsin(i+1)αsiniα/sin2α

Ti/ΣTn=2sin(i+1)αsiniα/(n+1)cosα

Ti/ΣTn=-{cos(2i+1)α-cosα}/(n+1)cosα

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検してみる

n=4,i=1

Ti/ΣTn=-{cos3α-cosα}/5cosα

Ti/ΣTn=-{4(cosα)^2-4}/5

Ti/ΣTn=-{τ^2-4}/5

Ti/ΣTn=-{τ-3}/5

Ti/ΣTn=τ^-1/√5・・・OK

n=4,i=2

Ti/ΣTn=-{cos5α-cosα}/5cosα

Ti/ΣTn=-{16(cosα)^4-20(cosα)^2+4}/5

Ti/ΣTn=-{τ^4-5τ^2+4}/5

Ti/ΣTn=-{3τ+2-5τ-5+4}/5

Ti/ΣTn=-{-2τ+1}/5

Ti/ΣTn=1/√5・・・OK

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　　φ^-4＝−３φ＋５、 √５φ^-4＝７φ−１１

　　φ^-3＝２φ−３、 √５φ^-3＝-４φ＋７

　　φ^-2＝−φ＋２、 √５φ^-2＝３φ−４

　　φ^-1＝φ−１、 √５φ^-1＝−φ＋３

　　φ^0＝１、 √５φ^0＝２φ−１

　　φ^1＝φ、 √５φ^1＝φ＋２

　　φ^2＝φ＋１、 √５φ^2＝３φ＋１

　　φ^3＝２φ＋１、 √５φ^3＝４φ＋３

　　φ^4＝３φ＋２、 √５φ^4＝７φ＋４

　　φ^5＝５φ＋３、 √５φ^5＝11φ＋７

　　φ^6＝８φ＋５、 √５φ^6＝18φ＋11

　右辺ｍφ＋ｎの係数ｍ，ｎはフィボナッチ数列をなす．

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