■平方根と連分数(その50)

α=[a0:a1,a2,・・・]

λ=[an+1:an+2,an+3,・・・]+[0:an,an-1,・・・,a1]

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α→λ=[2;2,1,1,2,2,1,1,,,]+[0;1,1,2,2,1,1,2,2,,,]

→(9+√221)/10+(-9+√221)/10=√221/5

α→λ=[2;1,1,1,1,,,]+[0;2,1,1,1,,,]

→φ+1+1/(φ+1)=3

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x=[2:1,2,2,1,2,・・・],y=1/x

x=2+1/(1+1/(2+1/x))

x=2+1/(1+x/(2x+1))

x=2+(2x+1)/(3x+1)=(8x+3)/(3x+1)

3x^2-7x-3=0

x={7+(85)^1/2}/6

y=6/{7+(85)^1/2}={(85)^1/2-7}/6

λ=(85)^1/2/3=(9・9+4)/3

λ=(9・Q^2+4)^1/2/Q

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x=[2:1,1,2,2,1,1,2,・・・],y=1/x

x=2+1/(1+1/(1+1/(2+1/x))

x=2+1/(1+1/(1+x/(2x+1))

x=2+1/(1+(2x+1)/(3x+1))

x=2+(3x+1)/(5x+2)

x=(13x+5)/(5x+2)

5x^2-11x-5=0

x={11+(221)^1/2}/10

y=10/{11+(221)^1/2}={(221)^1/2-11}/10

λ=(221)^1/2/5=(9・25+4)^1/2/5にならない

λ=(9・Q^2-4)^1/2/Qになる

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

λ=(9・25-4)^1/2/5=(9・Q^2-4)^1/2/Qになっている

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x=[2:1,1,1,2,2,1,1,1,2,・・・],y=1/x

x=2+1/(1+1/(1+1/(1+1/(2+1/x))

x=2+1/(1+1/(1+1/(1+x/(2x+1))

x=2+1/(1+1/(1+(2x+1)/(3x+1))

x=2+1/(1+(3x+1)/(5x+2)

x=2+(5x+2)/(8x+3)

x=(21x+8)/(8x+3)

8x^2-18x-8=0

x={9+(145)^1/2}/8

y=8/{9+(145)^1/2}={(145)^1/2-9}/8

λ=(145)^1/2/4=(9・16+4)^1/2/4

λ=(9・Q^2+4)^1/2/Q・・・になっていない

x={18+(580)^1/2}/16

y={-18+(580)^1/2}/16

λ=(580)^1/2/8=(9・64+4)^1/2/4

λ=(9・Q^2+4)^1/2/Q・・・になっている

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α→λ=[2;2,1,1,1,1,2,2,1,1,,,]+[0;1,1,1,1,2,2,1,1,1,1,,,]

x=2+1/(2+1/(1+1/(1+1/(1+1/(1+1/x))

x=2+1/(2+1/(1+1/(1+1/(1+x/(x+1))

x=2+1/(2+1/(1+1/(1+(x+1)/(2x+1))

x=2+1/(2+1/(1+(2x+1)/(3x+2))

x=2+1/(2+(3x+2)/(5x+3))

x=2+(5x+3)/(13x+8)=(31x+19)/(13x+8)

13x^2-23x-19=0

x={23+(1517)^1/2}/26

y=1/(1+1/(1+1/(1+1/(1+1/(2+1/(2+y))

y=1/(1+1/(1+1/(1+1/(1+(y+2)/(2y+5))

y=1/(1+1/(1+1/(1+(2y+5)/(3y+7))

y=1/(1+1/(1+(3y+7)/(5y+12))

y=1/(1+(5y+12)/(8y+19))

y=(8y+19)/(13y+31)

13y^2+23y-19=0

y={-23+(1517)^1/2}/26

λ=(1517)^1/2/13=(1517)^1/2/13

→(9・169-4)^1/2/13

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α→λ=[2;2,1,1,1,1,1,1,2,2,1,1,,,]+[0;1,1,1,1,1,1,2,2,1,1,1,1,,,]

x=2+1/(2+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/x))

x=2+1/(2+1/(1+1/(1+1/(1+1/(1+1/(1+x/(x+1))

x=2+1/(2+1/(1+1/(1+1/(1+1/(1+(x+1)/(2x+1))

x=2+1/(2+1/(1+1/(1+1/(1+(2x+1)/(3x+2))

x=2+1/(2+1/(1+1/(1+(3x+2)/(5x+3))

x=2+1/(2+1/(1+(5x+3)/(8x+5))

x=2+1/(2+(8x+5)/(13x+8))

x=2+(13x+8)/(34x+21)

x=(81x+50)/(34x+21)

34x^2-60x-50=0

x={60+(10400)^1/2}/68

y=1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(2+1/(2+y))

y=1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+(y+2)/(2y+5))

y=1/(1+1/(1+1/(1+1/(1+1/(1+(2y+5)/(3y+7))

y=1/(1+1/(1+1/(1+1/(1+(3y+7)/(5y+12))

y=1/(1+1/(1+1/(1+(5y+12)/(8y+19))

y=1/(1+1/(1+(8y+19)/(13y+31))

y=1/(1+(13y+31)/(21y+50))

y=(21y+50)/(34y+81)

34y^2+60y-50=0

y={-60+(10400)^1/2}/68

λ=(10400)^1/2/34=

→(9・1156-4)^1/2/34

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x=[2:1,1,1,1,1,2,2,1,1,1,1,1,2,・・・],y=1/x

x=2+1/(1+1/(1+1/(1+1/(1+1/(1+1/(2+1/x))

x=2+1/(1+1/(1+1/(1+1/(1+1/(1+x/(2x+1))

x=2+1/(1+1/(1+1/(1+1/(1+(2x+1)/(3x+1))

x=2+1/(1+1/(1+1/(1+(3x+1)/(5x+2))

x=2+1/(1+1/(1+(5x+2)/(8x+3))

x=2+1/(1+(8x+3)/(13x+5))

x=2+(13x+5)/(21x+8)

x=(55x+21)/(21x+8)

21x^2-47x-21=0

x={47+(3973)^1/2}/42

y=42/{-47+(3973)^1/2}={(3973)^1/2-47}/42

λ=(3973)^1/2/21=(9・441+4)^1/2/21

λ=(9・Q^2+4)^1/2/Q・・・なりたつ

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x=[2:1,1,1,1,1,1,1,2,2,1,1,1,1,1,1,1,2,・・・],y=1/x

x=2+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(2+1/x))

x=2+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+x/(2x+1))

x=2+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+(2x+1)/(3x+1))

x=2+1/(1+1/(1+1/(1+1/(1+1/(1+(3x+1)/(5x+2))

x=2+1/(1+1/(1+1/(1+1/(1+(5x+2)/(8x+3))

x=2+1/(1+1/(1+1/(1+(8x+3)/(13x+5))

x=2+1/(1+1/(1+(13x+5)/(21x+8))

x=2+1/(1+(21x+8)/(34x+13))

x=2+(34x+13)/(55x+21)

x=(144x+55)/(55x+21)

55x^2-123x-55=0

x={123+(27229)^1/2}/110

y=110/{-123+(27229)^1/2}={(27229)^1/2-123}/110

λ=(27229)^1/2/55=(9・3025+4)^1/2/55

λ=(9・Q^2+4)^1/2/Q・・・なりたつ

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これでフィボナッチ数になることははっきりした

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