割圆术 (赵友钦)维基百科,自由的 encyclopedia 赵友钦割圆术是元代数学家赵友钦在所著的《革象新书》卷五《乾象周髀》篇研究的割圆术。与刘徽从内接正六角形开始不同,赵氏割圆术从分割内接正方形开始[1]。 赵友钦割圆术 赵友钦《革象新书》卷五《乾象周髀》篇割圆术书影 如图,圆的半径为r; 内接正方形的边长为 ℓ {\displaystyle \ell } ,由圆心到正方形一边倒垂直距离为 d d = r 2 − ( ℓ 2 ) 2 {\displaystyle d={\sqrt {r^{2}-({\frac {\ell }{2}})^{2}}}} e = r − d = r − r 2 − ( ℓ 2 ) 2 {\displaystyle e=r-d=r-{\sqrt {r^{2}-({\frac {\ell }{2}})^{2}}}} d 的延长线与圆周相交点将圆周等分为正八边形。 令正八边形的边长为 ℓ 2 {\displaystyle \ell _{2}} ℓ 2 = ( ℓ / 2 ) 2 + e 2 {\displaystyle \ell _{2}={\sqrt {(\ell /2)^{2}+e^{2}}}} ℓ 2 = 1 2 ∗ ℓ 2 + 4 ∗ ( r − 1 2 ∗ 4 ∗ r 2 − ℓ 2 ) 2 {\displaystyle \ell _{2}={\frac {1}{2}}*{\sqrt {\ell ^{2}+4*(r-{\frac {1}{2}}*{\sqrt {4*r^{2}-\ell ^{2}}})^{2}}}} 设 ℓ 3 {\displaystyle \ell _{3}} 为分割圆成正16边形之边长,赵友钦正确地推断 ℓ 3 {\displaystyle \ell _{3}} 与 ℓ 2 {\displaystyle \ell _{2}} 的迭代关系: ℓ 3 = 1 2 ∗ ( ℓ 2 ) 2 + 4 ∗ ( r − 1 2 ∗ 4 ∗ r 2 − ( ℓ 2 ) 2 ) 2 {\displaystyle \ell _{3}={\frac {1}{2}}*{\sqrt {(\ell _{2})^{2}+4*(r-{\frac {1}{2}}*{\sqrt {4*r^{2}-(\ell _{2})^{2}}})^{2}}}} 推而广之: ℓ n + 1 = 1 2 ∗ ( ℓ n ) 2 + 4 ∗ ( r − 1 2 ∗ 4 ∗ r 2 − ( ℓ n ) 2 ) 2 {\displaystyle \ell _{n+1}={\frac {1}{2}}*{\sqrt {(\ell _{n})^{2}+4*(r-{\frac {1}{2}}*{\sqrt {4*r^{2}-(\ell _{n})^{2}}})^{2}}}} 令 r=1; ℓ 1 = ( 2 ) {\displaystyle \ell _{1}={\sqrt {(}}2)} ℓ 2 = 2 − ( 2 ) {\displaystyle \ell _{2}={\sqrt {2-{\sqrt {(}}2)}}} ℓ 3 = 2 − 2 + ( 2 ) {\displaystyle \ell _{3}={\sqrt {2-{\sqrt {2+{\sqrt {(}}2)}}}}} ℓ 4 = 2 − 2 + 2 + ( 2 ) {\displaystyle \ell _{4}={\sqrt {2-{\sqrt {2+{\sqrt {2+{\sqrt {(}}2)}}}}}}} ℓ 5 = 2 − 2 + 2 + 2 + ( 2 ) {\displaystyle \ell _{5}={\sqrt {2-{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {(}}2)}}}}}}}}} ……
赵友钦割圆术是元代数学家赵友钦在所著的《革象新书》卷五《乾象周髀》篇研究的割圆术。与刘徽从内接正六角形开始不同,赵氏割圆术从分割内接正方形开始[1]。 赵友钦割圆术 赵友钦《革象新书》卷五《乾象周髀》篇割圆术书影 如图,圆的半径为r; 内接正方形的边长为 ℓ {\displaystyle \ell } ,由圆心到正方形一边倒垂直距离为 d d = r 2 − ( ℓ 2 ) 2 {\displaystyle d={\sqrt {r^{2}-({\frac {\ell }{2}})^{2}}}} e = r − d = r − r 2 − ( ℓ 2 ) 2 {\displaystyle e=r-d=r-{\sqrt {r^{2}-({\frac {\ell }{2}})^{2}}}} d 的延长线与圆周相交点将圆周等分为正八边形。 令正八边形的边长为 ℓ 2 {\displaystyle \ell _{2}} ℓ 2 = ( ℓ / 2 ) 2 + e 2 {\displaystyle \ell _{2}={\sqrt {(\ell /2)^{2}+e^{2}}}} ℓ 2 = 1 2 ∗ ℓ 2 + 4 ∗ ( r − 1 2 ∗ 4 ∗ r 2 − ℓ 2 ) 2 {\displaystyle \ell _{2}={\frac {1}{2}}*{\sqrt {\ell ^{2}+4*(r-{\frac {1}{2}}*{\sqrt {4*r^{2}-\ell ^{2}}})^{2}}}} 设 ℓ 3 {\displaystyle \ell _{3}} 为分割圆成正16边形之边长,赵友钦正确地推断 ℓ 3 {\displaystyle \ell _{3}} 与 ℓ 2 {\displaystyle \ell _{2}} 的迭代关系: ℓ 3 = 1 2 ∗ ( ℓ 2 ) 2 + 4 ∗ ( r − 1 2 ∗ 4 ∗ r 2 − ( ℓ 2 ) 2 ) 2 {\displaystyle \ell _{3}={\frac {1}{2}}*{\sqrt {(\ell _{2})^{2}+4*(r-{\frac {1}{2}}*{\sqrt {4*r^{2}-(\ell _{2})^{2}}})^{2}}}} 推而广之: ℓ n + 1 = 1 2 ∗ ( ℓ n ) 2 + 4 ∗ ( r − 1 2 ∗ 4 ∗ r 2 − ( ℓ n ) 2 ) 2 {\displaystyle \ell _{n+1}={\frac {1}{2}}*{\sqrt {(\ell _{n})^{2}+4*(r-{\frac {1}{2}}*{\sqrt {4*r^{2}-(\ell _{n})^{2}}})^{2}}}} 令 r=1; ℓ 1 = ( 2 ) {\displaystyle \ell _{1}={\sqrt {(}}2)} ℓ 2 = 2 − ( 2 ) {\displaystyle \ell _{2}={\sqrt {2-{\sqrt {(}}2)}}} ℓ 3 = 2 − 2 + ( 2 ) {\displaystyle \ell _{3}={\sqrt {2-{\sqrt {2+{\sqrt {(}}2)}}}}} ℓ 4 = 2 − 2 + 2 + ( 2 ) {\displaystyle \ell _{4}={\sqrt {2-{\sqrt {2+{\sqrt {2+{\sqrt {(}}2)}}}}}}} ℓ 5 = 2 − 2 + 2 + 2 + ( 2 ) {\displaystyle \ell _{5}={\sqrt {2-{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {(}}2)}}}}}}}}} ……