■入れ子の平方根の無限和(その11)

 √(1−√(1−1/2√(1−1/4√(1−1/8√1−・・・))))=1/2であるが,

[Q]√(1+√(1+1/2√(1+1/4√(1+1/8√1−・・・))))=?

を求めてみたい.

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 n→∞のとき

{1+1/2^n-1(√1+・・・)}^1/2=1+1/2^n

√1+・・・=33/32=1+1/2^5

√(1+1/8√1+・・・)=17/16=1+1/2^4

√(1+1/4√(1+1/8√1+・・・))=9/8=1+1/2^3

√(1+1/2√(1+1/4√(1+1/8√1+・・・)))=5/4=1+1/2^2

√(1+√(1+1/2√(1+1/4√(1+1/8√1−・・・))))=1+1/2=3/2

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 3√(a+3√(a+3√(a+3√(a+・・・))))=b

 a+3√(a+3√(a+3√(a+・・・)))=b^3

 3√(−a+3√(−a+3√(−a+・・・)))=b^3−a

b=b^3−a→a=b^3−bであればよいことになる.

 b=1とおくとa=0であるから,b=2とおくと

 3√(6+3√(6+3√(6+3√(6+・・・))))=2

b=3とおくと

 3√(24+3√(24+3√(24+3√(24+・・・))))=3

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