{3}}+{\frac {1}{3\times 4}}-{\frac {1}{3\times 4\times 34}}={\frac {577}{408}}=1.41421{\overline {56862745098039}}.}
{60}}+{\frac {51}{60^{2}}}+{\frac {10}{60^{3}}}=1.41421{\overline {296}}.}
It makes use of compass and straightedge construction, proving the theorem by a method similar to that employed by ancient Greek geometers.
Konsep ini menggunakan konstruksi kompas dan sisi lurus klasik, membuktikan teorema ini dengan metode yang sama yang digunakan anggota geometri Yunani kuno.
Suppose m and n are integers.
Katakan m dan n adalah bilangan bulat.
Let m:n be a ratio given in its lowest terms.
Biarkan m:n menjadi rasio yang diberikan melalui sebutan terendah.
Therefore, the triangles ABC and ADE are congruent by SAS.
Maka segitiga ABC dan ADE adalah kongruen melalui SAS.
}{2}},{\frac {\sqrt {2}}{2}}\right)\!.} Each coordinate satisfies 2 21 21 2sin ⁡ 45 ∘cos ⁡ 45 ∘ .
}{2}},{\frac {\sqrt {2}}{2}}\right).} Nomor ini bertepatan dengan 2 21 21 2cos ⁡ ( 45 ∘ )sin ⁡ ( 45 ∘ ) .
⋯ {\sqrt {2}}}=\prod _{k=0}^{\infty }\left(1-{\frac {1}{(4k+2)^{2}}}\right)=\left(1-{\frac {1}{4}}\right)\!\left(1-{\frac {1}{36}}\right)\!\left(1-{\frac {1}{100}}\right)\cdots } and 2∏ k0 ∞ ( 4 k + 2 ) 2 ( 4 k + 1 ) ( 4 k + 3 )
}=\prod _{k=0}^{\infty }\left(1+{\frac {1}{4k+1}}\right)\left(1-{\frac {1}{4k+3}}\right)=\left(1+{\frac {1}{1}}\right)\left(1-{\frac {1}{3}}\right)\left(1+{\frac {1}{5}}\right)\left(1-{\frac {1}{7}}\right)\cdots .} Nomor tersebut dapat dinyatakan dengan mengambil deret Taylor untuk fungsi trigonometri.
( 2 ⋅ 2 1 ⋅ 3 ) ( 6 ⋅ 6 5 ⋅ 7 ) ( 10 ⋅ 10 9 ⋅ 11 ) ( 14 ⋅ 14 13 ⋅ 15 ) ⋯ }=\prod _{k=0}^{\infty }{\frac {(4k+2)^{2}}{(4k+1)(4k+3)}}=\left({\frac {2\cdot 2}{1\cdot 3}}\right)\!\left({\frac {6\cdot 6}{5\cdot 7}}\right)\!\left({\frac {10\cdot 10}{9\cdot 11}}\right)\!\left({\frac {14\cdot 14}{13\cdot 15}}\right)\cdots } or equivalently, 2∏ k0 ∞ ( 1 + 1 4 k + 1 ) ( 1 − 1 4 k + 3 )( 1 + 1 1 ) ( 1 − 1 3 ) ( 1 + 1 5 ) ( 1 − 1 7 ) ⋯ .
}=\sum _{k=0}^{\infty }{\frac {(2k+1)!}{(k!)^{2}2^{3k+1}}}={\frac {1}{2}}+{\frac {3}{8}}+{\frac {15}{64}}+{\frac {35}{256}}+{\frac {315}{4096}}+{\frac {693}{16384}}+\cdots .} Tidak diketahui apakah √2 dapat diwakilikan dengan rumus BBP-type.
}=\prod _{k=0}^{\infty }\left(1+{\frac {1}{4k+1}}\right)\left(1-{\frac {1}{4k+3}}\right)=\left(1+{\frac {1}{1}}\right)\!\left(1-{\frac {1}{3}}\right)\!\left(1+{\frac {1}{5}}\right)\!\left(1-{\frac {1}{7}}\right)\cdots .} The number can also be expressed by taking the Taylor series of a trigonometric function.
}=\prod _{k=0}^{\infty }\left(1+{\frac {1}{4k+1}}\right)\left(1-{\frac {1}{4k+3}}\right)=\left(1+{\frac {1}{1}}\right)\left(1-{\frac {1}{3}}\right)\left(1+{\frac {1}{5}}\right)\left(1-{\frac {1}{7}}\right)\cdots .} Nomor tersebut dapat dinyatakan dengan mengambil deret Taylor untuk fungsi trigonometri.
For example, the series for cos ⁠π/4⁠ gives 1 2∑ k0 ∞ ( − 1 ) k ( π 4 ) 2 k ( 2 k ) ! .
Misalnya, seri untuk cos(p/4) adalah: 1 2∑ k0 ∞ ( − 1 )
1 + 1 2 − 1 2 ⋅ 4 + 1 ⋅ 3 2 ⋅ 4 ⋅ 6 − 1 ⋅ 3 ⋅ 5 2 ⋅ 4 ⋅ 6 ⋅ 8 + ⋯1 + 1 2 − 1 8 + 1 16 − 5 128 + 7 256 + ⋯ .
1 + 1 2 − 1 2 ⋅ 4 + 1 ⋅ 3 2 ⋅ 4 ⋅ 6 − 1 ⋅ 3 ⋅ 5 2 ⋅ 4 ⋅ 6 ⋅ 8 + ⋯ .
}=\sum _{k=0}^{\infty }(-1)^{k+1}{\frac {(2k-3)!!}{(2k)!!}}=1+{\frac {1}{2}}-{\frac {1}{2\cdot 4}}+{\frac {1\cdot 3}{2\cdot 4\cdot 6}}-{\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 8}}+\cdots1+{\frac {1}{2}}-{\frac {1}{8}}+{\frac {1}{16}}-{\frac {5}{128}}+{\frac {7}{256}}+\cdots .} The convergence of this series can be accelerated with an Euler transform, producing 2∑ k0 ∞ ( 2 k + 1 ) ! 2 3 k + 1 ( k ! ) 21 2 + 3 8 + 15 64 + 35 256 + 315 4096 + 693 16384 + ⋯ .
}=\sum _{k=0}^{\infty }(-1)^{k+1}{\frac {(2k-3)!!}{(2k)!!}}=1+{\frac {1}{2}}-{\frac {1}{2\cdot 4}}+{\frac {1\cdot 3}{2\cdot 4\cdot 6}}-{\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 8}}+\cdots .} Konvergensi seri ini bisa dicepatkan dengan konversi Euler, menghasilkan 2∑ k0 ∞ ( 2 k + 1 ) ! ( k ! ) 2 2 3 k + 11 2 + 3 8 + 15 64 + 35 256 + 315 4096 + 693 16384 + ⋯ .
}=\sum _{k=0}^{\infty }{\frac {(2k+1)!}{2^{3k+1}{(k!)}^{2}}}={\frac {1}{2}}+{\frac {3}{8}}+{\frac {15}{64}}+{\frac {35}{256}}+{\frac {315}{4096}}+{\frac {693}{16384}}+\cdots .} It is not known whether 2 }} can be represented with a BBP-type formula.
}=\sum _{k=0}^{\infty }{\frac {(2k+1)!}{(k!)^{2}2^{3k+1}}}={\frac {1}{2}}+{\frac {3}{8}}+{\frac {15}{64}}+{\frac {35}{256}}+{\frac {315}{4096}}+{\frac {693}{16384}}+\cdots .} Tidak diketahui apakah √2 dapat diwakilikan dengan rumus BBP-type.
3 2 − 1 2 ( 1 6 + 1 204 + 1 235416 + … ) .
1 + 1 2 − 1 2 ⋅ 4 + 1 ⋅ 3 2 ⋅ 4 ⋅ 6 − 1 ⋅ 3 ⋅ 5 2 ⋅ 4 ⋅ 6 ⋅ 8 + ⋯ .
}={\frac {3}{2}}-{\frac {1}{2}}\sum _{n=0}^{\infty }{\frac {1}{a(2^{n})}}={\frac {3}{2}}-{\frac {1}{2}}\left({\frac {1}{6}}+{\frac {1}{204}}+{\frac {1}{235416}}+\dots \right).} The square root of two has the following continued fraction representation: 21 + 1 2 + 1 2 + 1 2 + 1 ⋱ .
}=\sum _{k=0}^{\infty }{\frac {(2k+1)!}{(k!)^{2}2^{3k+1}}}={\frac {1}{2}}+{\frac {3}{8}}+{\frac {15}{64}}+{\frac {35}{256}}+{\frac {315}{4096}}+{\frac {693}{16384}}+\cdots .} Tidak diketahui apakah √2 dapat diwakilikan dengan rumus BBP-type.
}=1+{\cfrac {1}{2+{\cfrac {1}{2+{\cfrac {1}{2+{\cfrac {1}{\ddots }}}}}}}}.}
}{2}}={\sqrt {\frac {1}{2}}}={\frac {1}{\sqrt {2}}}=\cos(45^{\circ })
{\sqrt {2}}&={\tfrac {3}{2}}-2\left({\tfrac {1}{4}}-\left({\tfrac {1}{4}}-{\bigl (}{\tfrac {1}{4}}-\cdots {\bigr )}^{2}\right)^{2}\right)^{2}\\[10mu]&={\tfrac {3}{2}}-4\left({\tfrac {1}{8}}+\left({\tfrac {1}{8}}+{\bigl (}{\tfrac {1}{8}}+\cdots {\bigr )}^{2}\right)^{2}\right)^{2}.\end{aligned}}}
}=\prod _{k=0}^{\infty }\left(1+{\frac {1}{4k+1}}\right)\left(1-{\frac {1}{4k+3}}\right)=\left(1+{\frac {1}{1}}\right)\left(1-{\frac {1}{3}}\right)\left(1+{\frac {1}{5}}\right)\left(1-{\frac {1}{7}}\right)\cdots .} Nomor tersebut dapat dinyatakan dengan mengambil deret Taylor untuk fungsi trigonometri.
"Square Root Approximations in Old Babylonian Mathematics: YBC 7289 in Context".
"Square Root Approximations in Old Babylonian Mathematics: YBC 7289 in Context" (PDF).
Historia Mathematica. 25 (4): 366–378. doi:10.1006/hmat.1998.2209.
Historia Mathematica. 25 (4): 366–378.