■平方根と連分数(その33)
α=[a0:a1,a2,・・・]
λ=[an+1:an+2,an+3,・・・]+[0:an,an-1,・・・,a1]
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α=Φのとき
λ>[1:1,1,1,・・・]+[0:1,1,1,・・・,1]→φ+φ-1=√5
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α=1+√2のとき
λ>[2:2,2,2,・・・]+[0:2,2,2,・・・,2]→1+√2+√2-1=√8
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α=(3+√13)/2のとき
λ>[3:3,3,3,・・・]+[0:3,3,3,・・・,3]→(3+√13)/2+(-3+√13)/2=√13
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α=(9+√221)/10=[2:2,1,1,2,2,1,1,・・・]のとき
λ>[2:2,1,1,2,2,1,1,・・・]+[0:1,1,2,2,1,1,2,2,・・・,2]→(9+√221)/10+(-9+√221)/10=√221/5
x=[2:2,1,1,2,2,1,1,・・・]
y=[0:1,1,2,2,1,1,2,2,・・・]=1/(1+1/(1+1/x))=1/(1+1/(x+1)/x)=1/(1+x/(x+1))=(x+1)/(2x+1)=(-9+√221)/10
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α=(√3)=[1:1,2,1,2,1,2,・・・]のとき
λ>[2:1,2,1,2,・・・]+[0:1,2,1,2,・・・,1]→1+(√3)-1+√3=√12
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α=(9√3+65)/22
(9√3+65)/22-3=(9√3-1)/22
22/(9√3-1)=(9√3+1)/11
(9√3+1)/11-1=(9√3-10)/11
11/(9√3-10)=(9√3+10)/13
(9√3+10)/13-1=(9√3-3)/13
13/(9√3-3)=(9√3+3)/18
(9√3+3)/18-1=(9√3-15)/18
18/(9√3-15)=(9√3+15)
(9√3+15)-30=(9√3-15)
1/(9√3-15)=(9√3+15)/18
(9√3+15)/18-1=(9√3-3)/18
18/(9√3-3)=(9√3+3)/13
(9√3+3)/13-1=(9√3-10)/13***
13/(9√3-10)=(9√3+10)/11
(9√3+10)/11-2=(9√3-12)/11
11/(9√3-12)=(9√3+12)/9
(9√3+12)/9-3=(9√3-15)/9
9/(9√3-15)=(9√3+15)/2
(9√3+15)/2-15=(9√3-15)/2
2/(9√3-15)=(9√3+15)/9
(9√3+15)/9-3=(9√3-12)/9
9/(9√3-12)=(9√3+12)/11
(9√3+12)/11-2=(9√3-10)/11
11/(9√3-10)=(9√3-10)/13***
α=(9√3+65)/22=[3:1,1,1,30,1,1,2,3,15,3,2,3,15,3,2,3,15,3,・・・]
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λ>[2:3,15,3,・・・]+[0:3,15,3,2,・・・]
あるいは[3:15,3,2,・・・]+[0:2,3,15,3,・・・]かもしれない
x=[3:15,3,2,・・・]= 3+1/(15+1/(3+1/(2+1/x))
= 3+1/(15+1/(3+x/(2x+1))
= 3+1/(15+(2x+1)/(7x+3))
= 3+(7x+3)/(107x+46)
=(328x+141)/(107x+43)
x(107x+43)=(328x+141)
107x^2-285-141=0
x={285+(285^2+4・107・141)^1/2}/214=3.09
y=1/(2+1/(3+1/(15+1/x))
y=1/(2+1/(3+x/(15x+1))
y=1/(2+(15x+1)/(46x+3))
y=(46x+3)/(107x+7)=0.44
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これはαではなく、λである.
λ=(9√3+65)/22=[3:1,1,1,30,1,1,2,3,15,3,2,3,15,3,2,3,15,3,・・・]
α=(p+√D)/q
α'=(p-√D)/q
α+1/α'=λとすると
(p+√D)/q+q/(p-√D)=(p+√D){1/q+q/(p^2-D)}=(p+√D){p^2-D+q^2}/q(P^2-D)=(9√3+65)/22
p=65,D=243
22{p^2-D+q^2}=q(P^2-D)
22{3982+q^2}=q(3982)
{3982+q^2}=181q
q={181+(181^2-4・3982)^1/2}/2
q={181+(32761-15928)^1/2}/2・・・整数にならない
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22x^2-130xy+cY^2を考える
65^2-22c=243、c=181
22x^2-130x+181Y^2を考える→λ=(9√3+65)/22
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