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Approximations of pi 6/10 https://en.wikipedia.org/wiki/Approximations_of_pi reference science, encyclopedia 2026-05-05T16:19:48.727542+00:00 kb-cron

where

    f
    (
    y
    )
    =
    (
    1
    
    
      y
      
        4
      
    
    
      )
      
        1
        
          /
        
        4
      
    
  

{\displaystyle f(y)=(1-y^{4})^{1/4}}

, the sequence

    1
    
      /
    
    
      a
      
        k
      
    
  

{\displaystyle 1/a_{k}}

converges quartically to π, giving about 100 digits in three steps and over a trillion digits after 20 steps. Even though the Chudnovsky series is only linearly convergent, the Chudnovsky algorithm might be faster than the iterative algorithms in practice; that depends on technological factors such as memory sizes and access times. For breaking world records, the iterative algorithms are used less commonly than the Chudnovsky algorithm since they are memory-intensive. The first one million digits of π and 1π are available from Project Gutenberg. A former calculation record (December 2002) by Yasumasa Kanada of Tokyo University stood at 1.24 trillion digits, which were computed in September 2002 on a 64-node Hitachi supercomputer with 1 terabyte of main memory, which carries out 2 trillion operations per second, nearly twice as many as the computer used for the previous record (206 billion digits). The following Machin-like formulae were used for this:

        π
        4
      
    
    =
    12
    arctan
    
    
      
        1
        49
      
    
    +
    32
    arctan
    
    
      
        1
        57
      
    
    
    5
    arctan
    
    
      
        1
        239
      
    
    +
    12
    arctan
    
    
      
        1
        110443
      
    
  

{\displaystyle {\frac {\pi }{4}}=12\arctan {\frac {1}{49}}+32\arctan {\frac {1}{57}}-5\arctan {\frac {1}{239}}+12\arctan {\frac {1}{110443}}}

(Kikuo Takano (1982))



  
    
      
        π
        4
      
    
    =
    44
    arctan
    
    
      
        1
        57
      
    
    +
    7
    arctan
    
    
      
        1
        239
      
    
    
    12
    arctan
    
    
      
        1
        682
      
    
    +
    24
    arctan
    
    
      
        1
        12943
      
    
  

{\displaystyle {\frac {\pi }{4}}=44\arctan {\frac {1}{57}}+7\arctan {\frac {1}{239}}-12\arctan {\frac {1}{682}}+24\arctan {\frac {1}{12943}}}

(F. C. M. Størmer (1896)).

These approximations have so many digits that they are no longer of any practical use, except for testing new supercomputers. Properties like the potential normality of π will always depend on the infinite string of digits on the end, not on any finite computation.

=== Miscellaneous approximations === As well as the formulas and approximations such as

          22
          7
        
      
    
  

{\displaystyle {\tfrac {22}{7}}}

and

          355
          113
        
      
    
  

{\displaystyle {\tfrac {355}{113}}}

discussed elsewhere in this article, The following expressions have been used to estimate π:

Accurate to three digits:

        2
      
    
    +
    
      
        3
      
    
    =
    
      3.146
      
        +
      
    
    .
  

{\displaystyle {\sqrt {2}}+{\sqrt {3}}=3.146^{+}.}

Karl Popper conjectured that Plato knew this expression, that he believed it to be exactly π, and that this is responsible for some of Plato's confidence in the universal power of geometry and for Plato's repeated discussion of special right triangles that are either isosceles or halves of equilateral triangles. Accurate to four digits:

    1
    +
    e
    
    γ
    =
    
      3.1410
      
        +
      
    
    ,
  

{\displaystyle 1+e-\gamma =3.1410^{+},}

where

    e
  

{\displaystyle e}

is the natural logarithmic base and

    γ
  

{\displaystyle \gamma }

is Euler's constant, and

        31
        
          3
        
      
    
    =
    
      3.1413
      
        +
      
    
    .
  

{\displaystyle {\sqrt[{3}]{31}}=3.1413^{+}.}

Accurate to four digits (or five significant figures):

        7
        +
        
          
            6
            +
            
              
                5
              
            
          
        
      
    
    =
    
      3.1416
      
        +
      
    
    .
  

{\displaystyle {\sqrt {7+{\sqrt {6+{\sqrt {5}}}}}}=3.1416^{+}.}

An approximation by Ramanujan, accurate to 4 digits (or five significant figures):

        9
        5
      
    
    +
    
      
        
          9
          5
        
      
    
    =
    
      3.1416
      
        +
      
    
    .
  

{\displaystyle {\frac {9}{5}}+{\sqrt {\frac {9}{5}}}=3.1416^{+}.}

Accurate to five digits:

          7
          
            7
          
        
        
          4
          
            9
          
        
      
    
    =
    
      3.14156
      
        +
      
    
    ,
  

{\displaystyle {\frac {7^{7}}{4^{9}}}=3.14156^{+},}




  
    
      
        306
        
          5
        
      
    
    =
    
      3.14155
      
        +
      
    
    ,
  

{\displaystyle {\sqrt[{5}]{306}}=3.14155^{+},}

and (by Kochański)

            40
            3
          
        
        
        2
        
          
            3
          
        
         
      
    
    =
    
      3.14153
      
        +
      
    
    .
  

{\displaystyle {\sqrt {{40 \over 3}-2{\sqrt {3}}\ }}=3.14153^{+}.}

accurate to six digits:

        (
        
          2
          
          
            
              
                2
                
                  
                    2
                  
                
                
                2
              
              
                2
                
                  2
                
              
            
          
        
        )
      
      
        2
      
    
    =
    3.14159
     
    
      6
      
        +
      
    
    .
  

{\displaystyle \left(2-{\frac {\sqrt {2{\sqrt {2}}-2}}{2^{2}}}\right)^{2}=3.14159\ 6^{+}.}

accurate to eight digits:

        (
        
          
            
              
                58
              
              4
            
          
          
          
            
              
                37
                
                  
                    2
                  
                
              
              33
            
          
        
        )
      
      
        
        1
      
    
    =
    
      
        
          66
          
            
              2
            
          
        
        
          33
          
            
              29
            
          
          
          148
        
      
    
    =
    3.14159
     
    
      263
      
        +
      
    
  

{\displaystyle \left({\frac {\sqrt {58}}{4}}-{\frac {37{\sqrt {2}}}{33}}\right)^{-1}={\frac {66{\sqrt {2}}}{33{\sqrt {29}}-148}}=3.14159\ 263^{+}}

This is the case that cannot be obtained from Ramanujan's approximation (22). accurate to nine digits:

            3
            
              4
            
          
          +
          
            2
            
              4
            
          
          +
          
            
              1
              
                2
                +
                (
                
                  
                    2
                    3
                  
                
                
                  )
                  
                    2
                  
                
              
            
          
        
        
          4
        
      
    
    =
    
      
        
          2143
          22
        
        
          4
        
      
    
    =
    3.14159
     
    
      2652
      
        +
      
    
  

{\displaystyle {\sqrt[{4}]{3^{4}+2^{4}+{\frac {1}{2+({\frac {2}{3}})^{2}}}}}={\sqrt[{4}]{\frac {2143}{22}}}=3.14159\ 2652^{+}}

This is from Ramanujan, who allegedly claimed the Goddess of Namagiri appeared to him in a dream and told him the true value of π. On the other hand, he also describes a method for obtaining this approximation through a clever geometric construction. accurate to ten digits (or eleven significant figures):

            10
            
              100
            
          
          11222.11122
        
        
          193
        
      
    
    =
    3.14159
     
    
      26536
      
        +
      
    
  

{\displaystyle {\sqrt[{193}]{\frac {10^{100}}{11222.11122}}}=3.14159\ 26536^{+}}

This approximation follows the observation that the 193rd power of 1/π yields the sequence 1122211125... Replacing 5 by 2 completes the symmetry without reducing the correct digits of π, while inserting a central decimal point remarkably fixes the accompanying magnitude at 10100. accurate to 12 decimal places:

        (
        
          
            
              
                163
              
              6
            
          
          
          
            
              181
              
                10005
              
            
          
        
        )
      
      
        
        1
      
    
    =
    3.14159
     
    26535
     
    
      89
      
        +
      
    
  

{\displaystyle \left({\frac {\sqrt {163}}{6}}-{\frac {181}{\sqrt {10005}}}\right)^{-1}=3.14159\ 26535\ 89^{+}}

This is obtained from the Chudnovsky series (truncate the series (1.4) at the first term and let E6(τ163)2/E4(τ163)3 = 151931373056001/151931373056000 ≈ 1). accurate to 16 digits:

          2510613731736
          
            
              2
            
          
        
        1130173253125
      
    
    =
    3.14159
     
    26535
     
    89793
     
    
      9
      
        +
      
    
  

{\displaystyle {\frac {2510613731736{\sqrt {2}}}{1130173253125}}=3.14159\ 26535\ 89793\ 9^{+}}
  • inverse of sum of first two terms of Ramanujan series.

         165707065
         52746197
    
    
     =
     3.14159
    
     26535
    
     89793
    
    
       4
    
         +
    

    {\displaystyle {\frac {165707065}{52746197}}=3.14159\ 26535\ 89793\ 4^{+}}

accurate to 18 digits:

        (
        
          
            
              
                253
              
              4
            
          
          
          
            
              
                643
                
                  
                    11
                  
                
              
              903
            
          
          
          
            
              223
              172
            
          
        
        )
      
      
        
        1
      
    
    =
    3.14159
     
    26535
     
    89793
     
    
      2387
      
        +
      
    
  

{\displaystyle \left({\frac {\sqrt {253}}{4}}-{\frac {643{\sqrt {11}}}{903}}-{\frac {223}{172}}\right)^{-1}=3.14159\ 26535\ 89793\ 2387^{+}}

This is the approximation (22) in Ramanujan's paper with n = 253. accurate to 19 digits:

          3949122332
          
            
              2
            
          
        
        1777729635
      
    
    =
    3.14159
     
    26535
     
    89793
     
    
      2382
      
        +
      
    
  

{\displaystyle {\frac {3949122332{\sqrt {2}}}{1777729635}}=3.14159\ 26535\ 89793\ 2382^{+}}
  • improved inverse of sum of first two terms of Ramanujan series. accurate to 24 digits:

           2286635172367940241408
    
    
               2
    
    
    
         1029347477390786609545
    
    
     =
     3.14159
    
     26535
    
     89793
    
     23846
    
    
       2649
    
         +
    

    {\displaystyle {\frac {2286635172367940241408{\sqrt {2}}}{1029347477390786609545}}=3.14159\ 26535\ 89793\ 23846\ 2649^{+}}

  • inverse of sum of first three terms of Ramanujan series. accurate to 25 decimal places:

         1
         10
    
    
     ln
     
    
       (
    
    
    
    
               2
    
                 21
    
    
    
               (
    
    
                   5
    
                     4
    
    
    
               
               1
    
                 )
    
                   24
    
    
    
    
    
         +
         24
    
       )
    
     =
     3.14159
    
     26535
    
     89793
    
     23846
    
     26433
    
    
       9
    
         +
    

    {\displaystyle {\frac {1}{10}}\ln \left({\frac {2^{21}}{({\sqrt[{4}]{5}}-1)^{24}}}+24\right)=3.14159\ 26535\ 89793\ 23846\ 26433\ 9^{+}}