在趣味數學中，接近整数是指很接近整數的無理數。這類數字中，有些因為其數學上的特性使其接近整数，有些還找不到其特性，看起來似乎只是巧合。


有關黃金比例及其他皮索特-维贾亚拉加文数 黃金比例
  
    
      
        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}}
  
代表費波納契數列的第
  
    
      
        n
      
    
    {\displaystyle n}
  
項
這是因為有恆等式
  
    
      
        
          φ
          
            n
          
        
        =
        
          F
          
            n
            −
            1
          
        
        +
        
          F
          
            n
          
        
        ×
        φ
      
    
    {\displaystyle \varphi ^{n}=F_{n-1}+F_{n}\times \varphi }
  
，所以當
  
    
      
        n
      
    
    {\displaystyle n}
  
為足夠大的正整數時，

  
    
      
        
          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}}
  

這些數字接近整數的原因和黃金比例的特性有關，不是數學巧合。其原因是因為黃金比例為皮索特-维贾亚拉加文数，而皮索特-维贾亚拉加文数的高次方會是接近整數。
這些數字與費波納契數有密切的關係，因為費波納契數相鄰兩項的比值會趨近於黃金比例，而如果m整除n，則第m個費波納契數也會整除第n個費波納契數。
皮索特-维贾亚拉加文数是指代數數本身大於1，而且其極小多項式中另一根的絕對值小於1。像黃金比例本身大於1，
  
    
      
        φ
      
    
    {\displaystyle \varphi }
  
的最小多項式為

  
    
      
        
          x
          
            2
          
        
        −
        x
        −
        1
        =
        0
      
    
    {\displaystyle x^{2}-x-1=0}
  

另一根為

  
    
      
        
          
            φ
            ¯
          
        
        =
        
          
            
              1
              −
              
                
                  5
                
              
            
            2
          
        
        ≈
        −
        0.618
        
      
    
    {\displaystyle {\overline {\varphi }}={\frac {1-{\sqrt {5}}}{2}}\approx -0.618\,}
  

絕對值小於1，因此黃金比例為皮索特-维贾亚拉加文数，其高次方會是接近整数。
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 }}}
  
來表示，由於二根之和及二根之積均為整數，計算所得的結果也是一個正整數，假設為一正整數K，則
  
    
      
        
          φ
          
            n
          
        
      
    
    {\displaystyle \varphi ^{n}}
  
可以用下式表示

  
    
      
        
          φ
          
            n
          
        
        =
        K
        −
        
          
            
              φ
              ¯
            
          
          
            n
          
        
      
    
    {\displaystyle \varphi ^{n}=K-{\overline {\varphi }}^{n}}
  

由於
  
    
      
        
          
            φ
            ¯
          
        
      
    
    {\displaystyle {\overline {\varphi }}}
  
的絕對值小於1，在n增大時，其高次方會趨於0，此時可得

  
    
      
        
          φ
          
            n
          
        
        ≈
        K
      
    
    {\displaystyle \varphi ^{n}\approx K}
  

除了黃金比例外，其他皮索特-维贾亚拉加文数的無理數也符合此一條件，例如
  
    
      
        1
        +
        
          
            2
          
        
      
    
    {\displaystyle 1+{\sqrt {2}}}
  
。


有關黑格納數 以下也是幾個非巧合出現的接近整數，和最大三項的黑格納數有關：

  
    
      
        
          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:

  
    
      
        
          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\,}
  

由於艾森斯坦級數的關係，使得上式中出現平方項。常數
  
    
      
        
          e
          
            π
            
              
                163
              
            
          
        
        
      
    
    {\displaystyle e^{\pi {\sqrt {163}}}\,}
  
有時會稱為拉馬努金常數。


有關π及e 許多有關π及e的常數也是接近整數，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年為止還沒找到出現此特性的原因，因此只能視為一數學巧合。另一個有關格尔丰德常数的常數也是接近整數

  
    
      
        
          
            
              
                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數的例子

  
    
      
        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 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...}
  
，這是由於
  
    
      
        3.5
        π
        ≈
        10.9956
        ≈
        11
      
    
    {\displaystyle 3.5\pi \approx 10.9956\approx 11}
  
的緣故，另一個類似的例子為
  
    
      
        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}
  

  
    
      
        
          
            
              5
              
                6
              
            
            
              6
              
                5
              
            
          
        
        ≈
        2
      
    
    {\displaystyle {\frac {5^{6}}{6^{5}}}\approx 2}
  


外部連結 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）


註釋 參考資料