【 加法展開を繰り返す 】
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加法展開を逐次繰り返すことで、無数の展開式が得られる。下記にその例を示す。
項数が増えるに従って、次の展開式の数は幾何級数的に増大する。
π/4 = arctan(1/1)
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└→ arctan(1/2)+arctan(1/3)
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├→ 2arctan(1/2)-arctan(1/7)
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├→ arctan(1/2)+arctan(1/3)
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├→ 2arctan(1/2)-arctan(1/5)+arctan(1/18)
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├→ 2arctan(1/2)-arctan(1/6)+arctan(1/43)
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├→ 2arctan(1/2)-arctan(1/8)-arctan(1/57)
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├→ 2arctan(1/2)-arctan(1/9)-arctan(1/32)
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├→ 2arctan(1/2)-arctan(1/12)-arctan(1/17)
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└→ 2arctan(1/3)+arctan(1/7)
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├→ arctan(1/2)+arctan(1/4)+arctan(1/13)
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├→ arctan(1/2)+arctan(1/3)
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├→ arctan(1/2)+arctan(1/4)+arctan(1/8)-arctan(1/21)
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├→ arctan(1/2)+arctan(1/4)+arctan(1/11)-arctan(1/72)
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├→ arctan(1/2)+arctan(1/4)+arctan(1/12)-arctan(1/157)
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├→ arctan(1/2)+arctan(1/4)+arctan(1/14)+arctan(1/183)
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├→ arctan(1/2)+arctan(1/4)+arctan(1/15)+arctan(1/98)
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├→ arctan(1/2)+arctan(1/4)+arctan(1/18)+arctan(1/47)
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├→ arctan(1/2)+arctan(1/4)+arctan(1/23)+arctan(1/30)
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├→ arctan(1/2)+arctan(1/5)+arctan(1/13)+arctan(1/21)
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└→ arctan(1/3)+arctan(1/4)+arctan(1/7)+arctan(1/13)
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├→ arctan(1/2)+arctan(1/5)+arctan(1/8)
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├→ arctan(1/2)+arctan(1/3)
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├→ arctan(1/2)+arctan(1/4)+arctan(1/8)-arctan(1/21)
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├→ arctan(1/2)+arctan(1/5)+arctan(1/7)-arctan(1/57)
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├→ arctan(1/2)+arctan(1/5)+arctan(1/9)+arctan(1/73)
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├→ arctan(1/2)+arctan(1/5)+arctan(1/13)+arctan(1/21)
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├→ arctan(1/2)+arctan(1/6)+arctan(1/8)+arctan(1/31)
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├→ arctan(1/2)+arctan(1/7)+arctan(1/8)+arctan(1/18)
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└→ arctan(1/3)+arctan(1/5)+arctan(1/7)+arctan(1/8)
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└→ 2arctan(1/3)+arctan(1/7)
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├→ arctan(1/2)+arctan(1/3)
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├→ 2arctan(1/2)-arctan(1/7)
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├→ 2arctan(1/3)+arctan(1/5)-arctan(1/18)
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├→ 2arctan(1/3)+arctan(1/6)-arctan(1/43)
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├→ 2arctan(1/3)+arctan(1/8)+arctan(1/57)
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├→ 2arctan(1/3)+arctan(1/9)+arctan(1/32)
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├→ 2arctan(1/3)+arctan(1/12)+arctan(1/17)
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├→ 2arctan(1/4)+arctan(1/7)+2arctan(1/13)
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└→ 2arctan(1/5)+arctan(1/7)+2arctan(1/8)
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