接近整数在趣味数学中,接近整数是指很接近整数的无理数。这类数字中,有些因为其数学上的特性使其接近整数,有些还找不到其特性,看起来似乎只是巧合。 Ed Pegg jr.先生发现上图中的线段d长度为 1 2 1 30 ( 61421 − 23 5831385 ) {\displaystyle {\frac {1}{2}}{\sqrt {{\frac {1}{30}}(61421-23{\sqrt {5831385}})}}} ,非常接近7(数值为7.0000000857)[1] 目录 1 有关黄金比例及其他皮索特-维贡伊拉卡文数 2 有关黑格纳数 3 有关π及e 4 其他例子 5 外部链接 6 注释 7 参考资料 有关黄金比例及其他皮索特-维贡伊拉卡文数 黄金比例 φ = 1 + 5 2 ≈ 1.61803398875 {\displaystyle \varphi ={\frac {1+{\sqrt {5}}}{2}}\approx 1.61803398875\,} 的高次方符合此特性。例如 φ 17 = 3571 + 1597 5 2 ≈ 3571.00028 ≈ F 16 + F 18 {\displaystyle \varphi ^{17}={\frac {3571+1597{\sqrt {5}}}{2}}\approx 3571.00028\approx F_{16}+F_{18}} φ 18 = 2889 + 1292 5 ≈ 5777.999827 ≈ F 17 + F 19 {\displaystyle \varphi ^{18}=2889+1292{\sqrt {5}}\approx 5777.999827\approx F_{17}+F_{19}} φ 19 = 9349 + 4181 5 2 ≈ 9349.000107 ≈ F 18 + F 20 {\displaystyle \varphi ^{19}={\frac {9349+4181{\sqrt {5}}}{2}}\approx 9349.000107\approx F_{18}+F_{20}} 其中 F n {\displaystyle F_{n}} 代表费波纳契数列的第 n {\displaystyle n} 项这是因为有恒等式 φ n = F n − 1 + F n × φ {\displaystyle \varphi ^{n}=F_{n-1}+F_{n}\times \varphi } [注 1],所以当 n {\displaystyle n} 为足够大的正整数时, φ n = F n − 1 + F n × φ ≈ F n − 1 + F n × ( F n + 1 F n ) = F n − 1 + F n + 1 {\displaystyle \varphi ^{n}=F_{n-1}+F_{n}\times \varphi \approx F_{n-1}+F_{n}\times \left({\frac {F_{n+1}}{F_{n}}}\right)=F_{n-1}+F_{n+1}} 这些数字接近整数的原因和黄金比例的特性有关,不是数学巧合。其原因是因为黄金比例为皮索特-维贡伊拉卡文数,而皮索特-维贡伊拉卡文数的高次方会是接近整数。 这些数字与费波纳契数有密切的关系,因为费波纳契数相邻两项的比值会趋近于黄金比例,而如果m整除n,则第m个费波纳契数也会整除第n个费波纳契数。 皮索特-维贡伊拉卡文数是指代数数本身大于1,而且其极小多项式中另一根的绝对值小于1。像黄金比例本身大于1, φ {\displaystyle \varphi } 的最小多项式为 x 2 − x − 1 = 0 {\displaystyle x^{2}-x-1=0} 另一根为 φ ¯ = 1 − 5 2 ≈ − 0.618 {\displaystyle {\overline {\varphi }}={\frac {1-{\sqrt {5}}}{2}}\approx -0.618\,} 绝对值小于1,因此黄金比例为皮索特-维贡伊拉卡文数,其高次方会是接近整数。 依照根和系数的关系,可得知 φ φ ¯ = − 1 {\displaystyle \varphi {\overline {\varphi }}=-1} φ + φ ¯ = 1 {\displaystyle \varphi +{\overline {\varphi }}=1} 而 φ n + φ ¯ n {\displaystyle \varphi ^{n}+{\overline {\varphi }}^{n}} 可以用 φ φ ¯ {\displaystyle \varphi {\overline {\varphi }}} 及 φ + φ ¯ {\displaystyle \varphi +{\overline {\varphi }}} 来表示,由于二根之和及二根之积均为整数,计算所得的结果也是一个正整数,假设为一正整数K,则 φ n {\displaystyle \varphi ^{n}} 可以用下式表示 φ n = K − φ ¯ n {\displaystyle \varphi ^{n}=K-{\overline {\varphi }}^{n}} 由于 φ ¯ {\displaystyle {\overline {\varphi }}} 的绝对值小于1,在n增大时,其高次方会趋于0,此时可得 φ n ≈ K {\displaystyle \varphi ^{n}\approx K} 除了黄金比例外,其他皮索特-维贡伊拉卡文数的无理数也符合此一条件,例如 1 + 2 {\displaystyle 1+{\sqrt {2}}} 。 有关黑格纳数 以下也是几个非巧合出现的接近整数,和最大三项的黑格纳数有关: e π 43 ≈ 884736743.999777466 {\displaystyle e^{\pi {\sqrt {43}}}\approx 884736743.999777466\,} e π 67 ≈ 147197952743.999998662454 {\displaystyle e^{\pi {\sqrt {67}}}\approx 147197952743.999998662454\,} e π 163 ≈ 262537412640768743.99999999999925007 {\displaystyle e^{\pi {\sqrt {163}}}\approx 262537412640768743.99999999999925007\,} 以上三式可以用以下的式子表示[2]: e π 43 = 12 3 ( 9 2 − 1 ) 3 + 744 − 2.225 ⋯ × 10 − 4 {\displaystyle e^{\pi {\sqrt {43}}}=12^{3}(9^{2}-1)^{3}+744-2.225\cdots \times 10^{-4}\,} e π 67 = 12 3 ( 21 2 − 1 ) 3 + 744 − 1.337 ⋯ × 10 − 6 {\displaystyle e^{\pi {\sqrt {67}}}=12^{3}(21^{2}-1)^{3}+744-1.337\cdots \times 10^{-6}\,} e π 163 = 12 3 ( 231 2 − 1 ) 3 + 744 − 7.499 ⋯ × 10 − 13 {\displaystyle e^{\pi {\sqrt {163}}}=12^{3}(231^{2}-1)^{3}+744-7.499\cdots \times 10^{-13}\,} 其中: 21 = 3 × 7 , 231 = 3 × 7 × 11 , 744 = 24 × 31 {\displaystyle 21=3\times 7,231=3\times 7\times 11,744=24\times 31\,} 由于艾森斯坦级数的关系,使得上式中出现平方项。常数 e π 163 {\displaystyle e^{\pi {\sqrt {163}}}\,} 有时会称为拉马努金常数。 有关π及e 许多有关π及e的常数也是接近整数,例如 e π − π = 19.999099979189 ⋯ {\displaystyle e^{\pi }-\pi =19.999099979189\cdots \,} 以及 e 5 π = 6635623.999341134233 ⋯ {\displaystyle e^{5\pi }=6635623.999341134233\cdots \,} 格尔丰德常数( e π {\displaystyle e^{\pi }\,} )接近 π + 20 {\displaystyle \pi +20\,} ,至2011年为止还没找到出现此特性的原因[1],因此只能视为一数学巧合。另一个有关格尔丰德常数的常数也是接近整数 e π − π − 1 6 π = 1.00793356 ⋯ {\displaystyle {\frac {e^{\pi }-\pi -1}{6\pi }}=1.00793356\cdots \,} 以下也是一些接近整数的例子 22 π 4 = 2143.0000027480 ⋯ {\displaystyle 22{\pi }^{4}=2143.0000027480\cdots \,} π 5 = 306.019684 ⋯ {\displaystyle {\pi }^{5}=306.019684\cdots \,} π 3 = 31.006276 ⋯ {\displaystyle {\pi }^{3}=31.006276\cdots \,} π 13 / 2 = 1704.017978 ⋯ {\displaystyle {\pi }^{13/2}=1704.017978\cdots \,} π 3 − π 500 = 30.999993494 ⋯ {\displaystyle {\pi }^{3}-{\frac {\pi }{500}}=30.999993494\cdots \,} π 2 + π 24 = 10.000504 ⋯ {\displaystyle {\pi }^{2}+{\frac {\pi }{24}}=10.000504\cdots \,} π 5 − 3 π 3 = 213.0008547 ⋯ {\displaystyle {\pi }^{5}-3{\pi }^{3}=213.0008547\cdots \,} e π − π + 0.0009 = 19.9999999791 ⋯ {\displaystyle e^{\pi }-\pi +0.0009=19.9999999791\cdots \,} e 3 = 20.0855369 ⋯ {\displaystyle e^{3}=20.0855369\cdots \,} e 9 / 2 = 90.0171313 ⋯ {\displaystyle e^{9/2}=90.0171313\cdots \,} e π 2 = 85.019695 ⋯ {\displaystyle e^{\pi {\sqrt {2}}}=85.019695\cdots \,} e 6 − π 5 − π 4 = 0.00001767 ⋯ {\displaystyle e^{6}-{\pi }^{5}-{\pi }^{4}=0.00001767\cdots \,} 9 π 5 − 2 e 3 = 2714.00608922 ⋯ {\displaystyle 9{\pi }^{5}-2e^{3}=2714.00608922\cdots \,} e 13 / 2 − π = 662.00004039 ⋯ {\displaystyle e^{13/2}-\pi =662.00004039\cdots \,} π 13 / 2 − e 9 / 2 = 1614.00084707 ⋯ {\displaystyle {\pi }^{13/2}-e^{9/2}=1614.00084707\cdots \,} 其他例子 cos { π cos [ π cos ln ( π + 20 ) ] } ≈ − 0.9999999999999999999999999999999999606783 {\displaystyle {}_{\cos \left\{\pi \cos \left[\pi \cos \ln \left(\pi +20\right)\right]\right\}\approx -0.9999999999999999999999999999999999606783}} sin 2017 2 5 ≈ − 0.9999999999999999785 {\displaystyle {}_{\sin 2017{\sqrt[{5}]{2}}\approx -0.9999999999999999785}} ∑ k = 1 ∞ ⌊ n tanh π ⌋ 10 n − 1 81 ≈ 1.11 × 10 − 269 {\displaystyle {}_{\sum _{k=1}^{\infty }{\frac {\lfloor n\tanh \pi \rfloor }{10^{n}}}-{\frac {1}{81}}\approx 1.11\times 10^{-269}}} 29 ( cos 2 π 59 − cos 24 π 59 ) − 19 5 ≈ 3.057684294154 × 10 − 6 {\displaystyle {}_{{\sqrt {29}}\left(\cos {\frac {2\pi }{59}}-\cos {\frac {24\pi }{59}}\right)-{\frac {19}{5}}\approx 3.057684294154\times 10^{-6}}} 1 + 103378831900730205293632 e 3 π 163 − 196884 e 2 π 163 − 262537412640768744 e π 163 ≈ 1.161367900476 × 10 − 59 {\displaystyle {}_{1+{\frac {103378831900730205293632}{e^{3\pi {\sqrt {163}}}}}-{\frac {196884}{e^{2\pi {\sqrt {163}}}}}-{\frac {262537412640768744}{e^{\pi {\sqrt {163}}}}}\approx 1.161367900476\times 10^{-59}}} ln 2 262537412640768744 π 2 − 163 ≈ 2.32167 × 10 − 29 {\displaystyle {}_{{\frac {\ln ^{2}262537412640768744}{\pi ^{2}}}-163\approx 2.32167\times 10^{-29}}} 10 tanh 28 15 π − π 9 e 8 ≈ 3.661398 × 10 − 8 {\displaystyle {}_{10\tanh {\frac {28}{15}}\pi -{\frac {\pi ^{9}}{e^{8}}}\approx 3.661398\times 10^{-8}}} 91 10 4 − 33 19 ≈ 3.661398 × 10 − 8 {\displaystyle {}_{{\sqrt[{4}]{\frac {91}{10}}}-{\frac {33}{19}}\approx 3.661398\times 10^{-8}}} γ − 10 81 ( 11 − 2 10 ) = ∫ 0 ∞ ( 1 e x − 1 − 1 x e x ) d x − 10 81 ( 11 − 2 10 ) ≈ 2.72 × 10 − 7 {\displaystyle {}_{\gamma -{10 \over 81}\left(11-2{\sqrt {10}}\right)=\int _{0}^{\infty }\left({\frac {1}{e^{x}-1}}-{\frac {1}{xe^{x}}}\right){\rm {d}}x-{10 \over 81}\left(11-2{\sqrt {10}}\right)\approx 2.72\times 10^{-7}}} ( 5 + 5 ) Γ ( 3 4 ) e 5 6 π ≈ 1.000000000000045422 {\displaystyle {}_{{\frac {\left(5+{\sqrt {5}}\right)\Gamma \left({3 \over 4}\right)}{e^{{\frac {5}{6}}\pi }}}\approx 1.000000000000045422}} 1 4 ( cos 1 10 + cosh 1 10 + 2 cos 2 20 cosh 2 20 ) ≈ 1.000000000000248 {\displaystyle {}_{{1 \over 4}\left(\cos {1 \over 10}+\cosh {1 \over 10}+2\cos {{\sqrt {2}} \over 20}\cosh {{\sqrt {2}} \over 20}\right)\approx 1.000000000000248}} e 6 − π 5 − π 4 ≈ 1.7673 × 10 − 5 {\displaystyle {}_{e^{6}-\pi ^{5}-\pi ^{4}\approx 1.7673\times 10^{-5}}} 29 ( cos 2 π 59 − cos 24 π 59 ) ≈ 3.0576842941540143382 × 10 − 6 {\displaystyle {}_{{\sqrt {29}}\left(\cos {\frac {2\pi }{59}}-\cos {\frac {24\pi }{59}}\right)\approx 3.0576842941540143382\times 10^{-6}}} ( 3 5 ) γ = ( 3 5 ) ∫ 0 ∞ ( 1 e x − 1 − 1 x e x ) d x ≈ 3.000060964 {\displaystyle {}_{\left(3{\sqrt {5}}\right)^{\gamma }=\left(3{\sqrt {5}}\right)^{\int _{0}^{\infty }\left({\frac {1}{e^{x}-1}}-{\frac {1}{xe^{x}}}\right){\rm {d}}x}\approx 3.000060964}} e ϕ 0 ( 2 + 3 4 ) = e ∫ 0 ∞ ( 1 t e t − e 2 − 3 4 t e t − 1 ) d t ≈ 1.99999969 {\displaystyle {}_{e^{\phi _{0}\left({\frac {2+{\sqrt {3}}}{4}}\right)}=e^{\int _{0}^{\infty }\left({\frac {1}{te^{t}}}-{\frac {e^{{\frac {2-{\sqrt {3}}}{4}}t}}{e^{t}-1}}\right){\rm {d}}t}\approx 1.99999969}} 9 3 3 ln 2 ≈ 1.00030887 {\displaystyle {}_{{\frac {\sqrt[{3}]{9}}{3\ln 2}}\approx 1.00030887}} ∑ k = − ∞ ∞ 10 − k 2 10000 − 100 π ln 10 = θ 3 ( 0 , 1 10 10000 ) − 100 π ln 10 ≈ 1.3809 × 10 − 18613 {\displaystyle {}_{\sum _{k=-\infty }^{\infty }10^{-{\frac {k^{2}}{10000}}}-100{\sqrt {\frac {\pi }{\ln 10}}}=\theta _{3}\left(0,{\frac {1}{\sqrt[{10000}]{10}}}\right)-100{\sqrt {\frac {\pi }{\ln 10}}}\approx 1.3809\times 10^{-18613}}} π 9 e 8 ≈ 9.998387 {\displaystyle {}_{{\pi ^{9} \over e^{8}}\approx 9.998387}} e π − π ≈ 19.999099979 {\displaystyle {}_{e^{\pi }-\pi \approx 19.999099979}} e π − ln 3 ln 2 − 4 5 ≈ 31.0000000033 {\displaystyle {}_{{\frac {e^{\pi }-\ln 3}{\ln 2}}-{\frac {4}{5}}\approx 31.0000000033}} π 11 e 3 − Γ [ Γ ( π + 1 ) + 1 ] = π 11 e 3 − ∫ 0 ∞ t ∫ 0 ∞ u π e u d u e t d t ≈ 7266.9999993632596 {\displaystyle {}_{{\frac {\pi ^{11}}{e^{3}}}-\Gamma \left[\Gamma \left(\pi +1\right)+1\right]={\frac {\pi ^{11}}{e^{3}}}-\int _{0}^{\infty }{\frac {t^{\int _{0}^{\infty }{\frac {u^{\pi }}{e^{u}}}{\rm {d}}u}}{e^{t}}}{\rm {d}}t\approx 7266.9999993632596}} 163 ( π − e ) ≈ 68.999664 {\displaystyle {}_{163\left(\pi -e\right)\approx 68.999664}} ( 23 9 ) 5 = 6436343 59049 ≈ 109.00003387 {\displaystyle {}_{\left({\frac {23}{9}}\right)^{5}={\frac {6436343}{59049}}\approx 109.00003387}} 88 ln 89 ≈ 395.00000053 {\displaystyle {}_{88\ln 89\approx 395.00000053}} 510 lg 7 ≈ 431.00000040727098 {\displaystyle {}_{510\lg 7\approx 431.00000040727098}} 272 log π 97 ≈ 1087.000000204 {\displaystyle {}_{272\log _{\pi }97\approx 1087.000000204}} 53453 ln 53453 ≈ 4910.00000122 {\displaystyle {}_{{\frac {53453}{\ln 53453}}\approx 4910.00000122}} 53453 ln 53453 + 163 ln 163 ≈ 4941.99999995925082 {\displaystyle {}_{{\frac {53453}{\ln 53453}}+{\frac {163}{\ln 163}}\approx 4941.99999995925082}} 2 4 ( π 17 − 4 e 2 π + 4 π e π ) 8 − 2 4 ( π 17 − 4 e 2 π − 4 π e π ) 8 ≈ 2.570287024592328869357 × 10 − 6 {\displaystyle {}_{{\sqrt[{8}]{{\frac {\sqrt {2}}{4}}\left(\pi ^{17}-4e^{2\pi }+4\pi e^{\pi }\right)}}-{\sqrt[{8}]{{\frac {\sqrt {2}}{4}}\left(\pi ^{17}-4e^{2\pi }-4\pi e^{\pi }\right)}}\approx 2.570287024592328869357\times 10^{-6}}} 10 − 2 4 ( π 17 − 4 e 2 π − 4 π e π ) 8 ≈ 2.57055302118 × 10 − 6 {\displaystyle {}_{10-{\sqrt[{8}]{{\frac {\sqrt {2}}{4}}\left(\pi ^{17}-4e^{2\pi }-4\pi e^{\pi }\right)}}\approx 2.57055302118\times 10^{-6}}} 10 − 2 4 ( π 17 − 4 e 2 π + 4 π e π ) 8 ≈ 2.65996596963 × 10 − 10 {\displaystyle {}_{10-{\sqrt[{8}]{{\frac {\sqrt {2}}{4}}\left(\pi ^{17}-4e^{2\pi }+4\pi e^{\pi }\right)}}\approx 2.65996596963\times 10^{-10}}} 163 ln 163 ≈ 31.9999987343 {\displaystyle {}_{{\frac {163}{\ln 163}}\approx 31.9999987343}} 2 2 2 / 3 ≈ 3.00507511272 {\displaystyle 2^{2^{2/3}}\approx 3.00507511272} 4 ln 2.117 ≈ 2.99999996861 {\displaystyle 4\ln 2.117\approx 2.99999996861} ln K 0 − ln ln K 0 ≈ 1.0000744 {\displaystyle {}_{\ln K_{0}-\ln \ln K_{0}\approx 1.0000744}} ,其中 K 0 {\displaystyle K_{0}} 是辛钦常数 10 81 − ∑ n = 1 ∞ ∑ k = 10 n − 1 10 n − 1 10 − n [ k − ( 10 n − 1 − 1 ) ] k 10 ∑ k = 0 n − 1 9 × 10 k − 1 k = 10 81 − ∑ n = 1 ∞ ∑ k = 10 n − 1 10 n − 1 k 10 k n − 9 ∑ k = 0 n − 1 10 k ( n − k ) ≈ 1.022344 × 10 − 9 {\displaystyle {}_{{\frac {10}{81}}-\sum _{n=1}^{\infty }{\frac {\sum _{k=10^{n-1}}^{10^{n}-1}10^{-n\left[k-(10^{n-1}-1)\right]}k}{10^{\sum _{k=0}^{n-1}9\times 10^{k-1}k}}}={\frac {10}{81}}-\sum _{n=1}^{\infty }\sum _{k=10^{n-1}}^{10^{n}-1}{\frac {k}{10^{kn-9\sum _{k=0}^{n-1}10^{k}(n-k)}}}\approx 1.022344\times 10^{-9}}} − 1 5 + e 6 5 4 F 3 ( − 1 5 , 1 20 , 3 10 , 11 20 ; 1 5 , 2 5 , 3 5 ; 256 3125 e 6 ) + 2 25 e 6 5 4 F 3 ( 1 5 , 9 20 , 7 10 , 19 20 ; 3 5 , 4 5 , 7 5 ; 256 3125 e 6 ) − 4 125 e 12 5 4 F 3 ( 2 5 , 13 20 , 9 10 , 23 20 ; 4 5 , 6 5 , 8 5 ; 256 3125 e 6 ) + 7 625 e 18 5 4 F 3 ( 3 5 , 17 20 , 11 10 , 27 20 ; 6 5 , 7 5 , 9 5 ; 256 3125 e 6 ) − π ≈ 2.89221114964408683 × 10 − 8 {\displaystyle {}_{-{\frac {1}{5}}+e^{\frac {6}{5}}{}_{4}F_{3}\left(-{\frac {1}{5}},{\frac {1}{20}},{\frac {3}{10}},{\frac {11}{20}};{\frac {1}{5}},{\frac {2}{5}},{\frac {3}{5}};{\frac {256}{3125e^{6}}}\right)+{\frac {2}{25e^{\frac {6}{5}}}}{}_{4}F_{3}\left({\frac {1}{5}},{\frac {9}{20}},{\frac {7}{10}},{\frac {19}{20}};{\frac {3}{5}},{\frac {4}{5}},{\frac {7}{5}};{\frac {256}{3125e^{6}}}\right)-{\frac {4}{125e^{\frac {12}{5}}}}{}_{4}F_{3}\left({\frac {2}{5}},{\frac {13}{20}},{\frac {9}{10}},{\frac {23}{20}};{\frac {4}{5}},{\frac {6}{5}},{\frac {8}{5}};{\frac {256}{3125e^{6}}}\right)+{\frac {7}{625e^{\frac {18}{5}}}}{}_{4}F_{3}\left({\frac {3}{5}},{\frac {17}{20}},{\frac {11}{10}},{\frac {27}{20}};{\frac {6}{5}},{\frac {7}{5}},{\frac {9}{5}};{\frac {256}{3125e^{6}}}\right)-\pi \approx 2.89221114964408683\times 10^{-8}}} Root of x 6 − 615 x 5 + 151290 x 4 − 18608670 x 3 + 1144433205 x 2 − 28153057165 x + 39605 = 0 {\displaystyle {}_{\qquad {\mbox{Root of }}x^{6}-615x^{5}+151290x^{4}-18608670x^{3}+1144433205x^{2}-28153057165x+39605=0}\,} 615 − 55 5 − 7451370 + 3332354 5 + 6 8890710030 + 3976046490 5 3 − 7451370 + 3332354 5 − 6 8890710030 + 3976046490 5 3 6 ≈ 1.40677447684 × 10 − 6 {\displaystyle {}_{{\frac {615-55{\sqrt {5}}-{\sqrt[{3}]{7451370+3332354{\sqrt {5}}+6{\sqrt {8890710030+3976046490{\sqrt {5}}}}}}-{\sqrt[{3}]{7451370+3332354{\sqrt {5}}-6{\sqrt {8890710030+3976046490{\sqrt {5}}}}}}}{6}}\approx 1.40677447684\times 10^{-6}}} Root of 312500000 x 5 − 6843750000 x 4 + 6826250000 x 3 + 10476025000 x 2 − 7886869750 x − 72099 = 0 {\displaystyle {}_{\qquad {\mbox{Root of }}312500000x^{5}-6843750000x^{4}+6826250000x^{3}+10476025000x^{2}-7886869750x-72099=0}\,} tan ( arctan 4 5 + 4 π 5 ) + 19 50 = 219 50 + − 1 − 5 + 10 − 2 5 i 4 884 + 799 i 5 + − 1 − 5 − 10 − 2 5 i 4 884 − 799 i 5 + − 1 + 5 − 10 + 2 5 i 4 1156 + 289 i 5 + − 1 + 5 + 10 + 2 5 i 4 1156 − 289 i 5 ≈ − 9.141538637378949398666277 × 10 − 6 {\displaystyle {}_{\tan \left({\frac {\arctan 4}{5}}+{\frac {4\pi }{5}}\right)+{\frac {19}{50}}={\frac {219}{50}}+{\frac {-1-{\sqrt {5}}+{\sqrt {10-2{\sqrt {5}}}}{\rm {i}}}{4}}{\sqrt[{5}]{884+799{\rm {i}}}}+{\frac {-1-{\sqrt {5}}-{\sqrt {10-2{\sqrt {5}}}}{\rm {i}}}{4}}{\sqrt[{5}]{884-799{\rm {i}}}}+{\frac {-1+{\sqrt {5}}-{\sqrt {10+2{\sqrt {5}}}}{\rm {i}}}{4}}{\sqrt[{5}]{1156+289{\rm {i}}}}+{\frac {-1+{\sqrt {5}}+{\sqrt {10+2{\sqrt {5}}}}{\rm {i}}}{4}}{\sqrt[{5}]{1156-289{\rm {i}}}}\approx -9.141538637378949398666277\times 10^{-6}}} e r f i ( e r f i 3 3 ) = 2 π ∫ 0 2 π ∫ 0 3 3 e t 2 d t e u 2 d u = 2 π e ( 2 e 3 π ∫ 0 ∞ sin ( 2 3 3 t ) e t 2 d t ) 2 ∫ 0 ∞ sin [ 4 u e 3 π ∫ 0 ∞ sin ( 2 3 3 t ) e t 2 d t ] e u 2 d u = 2 π ∫ 0 2 e 3 π ∫ 0 ∞ sin ( 2 3 3 t ) e t 2 d t e u 2 d u = 2 π e ( 2 π ∫ 0 3 3 e t 2 d t ) 2 ∫ 0 ∞ sin ( 4 u π ∫ 0 3 3 e t 2 d t ) e u 2 d u ≈ 1.00002087363809430195879 {\displaystyle {}_{\rm {{erfi}\left({\rm {{erfi}{\frac {\sqrt {3}}{3}}}}\right)={\frac {2}{\sqrt {\pi }}}\int _{0}^{{\frac {2}{\sqrt {\pi }}}\int _{0}^{\frac {\sqrt {3}}{3}}e^{t^{2}}{\rm {{d}t}}}e^{u^{2}}{\rm {{d}u={\frac {2}{\sqrt {\pi }}}e^{\left({\frac {2{\sqrt[{3}]{e}}}{\sqrt {\pi }}}\int _{0}^{\infty }{\frac {\sin \left({\frac {2}{3}}{\sqrt {3}}t\right)}{e^{t^{2}}}}{\rm {d}}t\right)^{2}}\int _{0}^{\infty }{\frac {\sin \left[{\frac {4u{\sqrt[{3}]{e}}}{\sqrt {\pi }}}\int _{0}^{\infty }{\frac {\sin \left({\frac {2}{3}}{\sqrt {3}}t\right)}{e^{t^{2}}}}{\rm {d}}t\right]}{e^{u^{2}}}}{\rm {d}}u={\frac {2}{\sqrt {\pi }}}\int _{0}^{{}_{{\frac {2{\sqrt[{3}]{e}}}{\sqrt {\pi }}}\int _{0}^{\infty }{\frac {\sin \left({\frac {2}{3}}{\sqrt {3}}t\right)}{e^{t^{2}}}}{\rm {d}}t}}e^{u^{2}}{\rm {d}}u={\frac {2}{\sqrt {\pi }}}e^{\left({\frac {2}{\sqrt {\pi }}}\int _{0}^{\frac {\sqrt {3}}{3}}e^{t^{2}}{\rm {{d}t}}\right)^{2}}\int _{0}^{\infty }{\frac {\sin \left({\frac {4u}{\sqrt {\pi }}}\int _{0}^{\frac {\sqrt {3}}{3}}e^{t^{2}}{\rm {{d}t}}\right)}{e^{u^{2}}}}{\rm {d}}u\approx 1.00002087363809430195879}}}}} sin 11 = − 0.999990207... {\displaystyle \sin 11=-0.999990207...} ,这是由于 3.5 π ≈ 10.9956 ≈ 11 {\displaystyle 3.5\pi \approx 10.9956\approx 11} 的缘故,另一个类似的例子为 sin 355 = − 0.00003014435335948844921433... {\displaystyle \sin 355=-0.00003014435335948844921433...} 1 2 + 2 2 + 3 2 + 4 2 + . . . . . . . + 552057 2 ≈ 236818619.0000004307 {\displaystyle {\sqrt {1^{2}+2^{2}+3^{2}+4^{2}+.......+552057^{2}}}\approx 236818619.0000004307} 外部链接 J.S. Markovitch Coincidence, data compression, and Mach's concept of economy of thought (页面存档备份,存于互联网档案馆)注释 ^ 此式可利用数学归纳法与性质 φ 2 = φ + 1 {\displaystyle \varphi ^{2}=\varphi +1} 证明。 参考资料 ^ 1.0 1.1 Eric Weisstein, "Almost Integer" (页面存档备份,存于互联网档案馆) at MathWorld ^ 存档副本. [2011-09-17]. (原始内容存档于2009-08-11).