■2次無理数の有理数近似(その72)
an+1+bn+1√2=(1+√2)^n(an+bn√2)
=(an+2bn)+(an+bn)√2
(1,1)
p^2-2q^2=+/-1のとき
(p+2q)^2-2(p+q)^2=-p^2+2q^2=-/+1
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an+1+√3bn+1=(2+√3)(an+√3bn)
=(2an+3bn)+√3(an+2bn)
(2,1)
p^2-3q^2=1のとき
(2p+3q)^2-3(p+2q)^2=p^2-3q^2=1
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an+1+bn+1√5=(2+√5)(an+bn√5)
=(2an+5bn)+√5(an+2bn)
(2,1)
p^2-5q^2=+/-1のとき
(2p+5q)^2-5(p+2q)^2=-p^2+5q^2=-/+1
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an+1+√6bn+1=(5+2√6)(an+√6bn)
=(5an+12bn)+√6(2an+5bn)
(5,2)
p^2-6q^2=1のとき
(5p+12q)^2-6(2p+5q)^2=p^2-6q^2=1
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an+1+√7bn+1=(8+3√7)(an+√7bn)
=(8an+21bn)+√7(3an+8bn)
(8,3)
p^2-7q^2=1のとき
(8p+21q)^2-7(3p+8q)^2=p^2-7q^2=1
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an+1+√8bn+1=(3+√8)(an+√8bn)
=(3an+8bn)+√8(an+3bn)
(3,1)
p^2-8q^2=1のとき
(3p+8q)^2-8(p+3q)^2=p^2-8q^2=1
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an+1+bn+1√10=(3+√10)(an+bn√10)
=(3an+10bn)+√10(an+3bn)
(3,1)
p^2-10q^2=+/-1のとき
(3p+10q)^2-10(p+3q)^2=-p^2+10q^2=-/+1
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