■数直線上の集合(その89)
単純循環連分数
L=[a:b,b,b,b,・・・]
で表される数Lを求めてみることにしましょう.
L−a=R=[0:b,b,b,b,・・・]=1/(b+R)
R^2+bR−1=0 → R=(−b+(b^2+4)^(1/2))/2
L=a+R=a−b/2+(b^2/4+1)^(1/2)
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a=0,b=5の場合、
L=a+R=a−b/2+(b^2/4+1)^(1/2)
=−5/2+(25/4+1)^(1/2)
=(-5+√29)/2
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a=0,b=6の場合、
L=a+R=a−b/2+(b^2/4+1)^(1/2)
=−3+(9+1)^(1/2)
=(-3+√10)
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a=0,b=7の場合、
L=a+R=a−b/2+(b^2/4+1)^(1/2)
=−7/2+(49/4+1)^(1/2)
=(-7+√53)/2
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