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Approximations of pi 7/10 https://en.wikipedia.org/wiki/Approximations_of_pi reference science, encyclopedia 2026-05-05T16:19:48.727542+00:00 kb-cron

This is derived from Ramanujan's class invariant g100 = 25/8/(51/4 1). accurate to 30 decimal places:

          ln
          
          (
          
            640320
            
              3
            
          
          +
          744
          )
        
        
          163
        
      
    
    =
    3.14159
     
    26535
     
    89793
     
    23846
     
    26433
     
    
      83279
      
        +
      
    
  

{\displaystyle {\frac {\ln(640320^{3}+744)}{\sqrt {163}}}=3.14159\ 26535\ 89793\ 23846\ 26433\ 83279^{+}}

Derived from the closeness of Ramanujan constant to the integer 6403203+744. This does not admit obvious generalizations in the integers, because there are only finitely many Heegner numbers and negative discriminants d with class number h(d) = 1, and d = 163 is the largest one in absolute value. accurate to 52 decimal places:

          ln
          
          (
          
            5280
            
              3
            
          
          (
          236674
          +
          30303
          
            
              61
            
          
          
            )
            
              3
            
          
          +
          744
          )
        
        
          427
        
      
    
  

{\displaystyle {\frac {\ln(5280^{3}(236674+30303{\sqrt {61}})^{3}+744)}{\sqrt {427}}}}

Like the one above, a consequence of the j-invariant. Among negative discriminants with class number 2, this d the largest in absolute value. accurate to 52 decimal places:

          ln
          
          (
          
            2
            
              
              30
            
          
          (
          (
          3
          +
          
            
              5
            
          
          )
          (
          
            
              5
            
          
          +
          
            
              7
            
          
          )
          (
          
            
              7
            
          
          +
          
            
              11
            
          
          )
          (
          
            
              11
            
          
          +
          3
          )
          
            )
            
              12
            
          
          
          24
          )
        
        
          
            
              5
            
          
          
            
              7
            
          
          
            
              11
            
          
        
      
    
  

{\displaystyle {\frac {\ln(2^{-30}((3+{\sqrt {5}})({\sqrt {5}}+{\sqrt {7}})({\sqrt {7}}+{\sqrt {11}})({\sqrt {11}}+3))^{12}-24)}{{\sqrt {5}}{\sqrt {7}}{\sqrt {11}}}}}

This is derived from Ramanujan's class invariant G385. accurate to 161 decimal places:

          ln
          
          
            
              (
            
          
          (
          2
          u
          
            )
            
              6
            
          
          +
          24
          
            
              )
            
          
        
        
          3502
        
      
    
  

{\displaystyle {\frac {\ln {\big (}(2u)^{6}+24{\big )}}{\sqrt {3502}}}}

where u is a product of four simple quartic units,

    u
    =
    (
    a
    +
    
      
        
          a
          
            2
          
        
        
        1
      
    
    
      )
      
        2
      
    
    (
    b
    +
    
      
        
          b
          
            2
          
        
        
        1
      
    
    
      )
      
        2
      
    
    (
    c
    +
    
      
        
          c
          
            2
          
        
        
        1
      
    
    )
    (
    d
    +
    
      
        
          d
          
            2
          
        
        
        1
      
    
    )
  

{\displaystyle u=(a+{\sqrt {a^{2}-1}})^{2}(b+{\sqrt {b^{2}-1}})^{2}(c+{\sqrt {c^{2}-1}})(d+{\sqrt {d^{2}-1}})}

and,

            a
          
          
            
            =
            
              
                
                  1
                  2
                
              
            
            (
            23
            +
            4
            
              
                34
              
            
            )
          
        
        
          
            b
          
          
            
            =
            
              
                
                  1
                  2
                
              
            
            (
            19
            
              
                2
              
            
            +
            7
            
              
                17
              
            
            )
          
        
        
          
            c
          
          
            
            =
            (
            429
            +
            304
            
              
                2
              
            
            )
          
        
        
          
            d
          
          
            
            =
            
              
                
                  1
                  2
                
              
            
            (
            627
            +
            442
            
              
                2
              
            
            )
          
        
      
    
  

{\displaystyle {\begin{aligned}a&={\tfrac {1}{2}}(23+4{\sqrt {34}})\\b&={\tfrac {1}{2}}(19{\sqrt {2}}+7{\sqrt {17}})\\c&=(429+304{\sqrt {2}})\\d&={\tfrac {1}{2}}(627+442{\sqrt {2}})\end{aligned}}}

Based on one found by Daniel Shanks. Similar to the previous two, but this time is a quotient of a modular form, namely the Dedekind eta function, and where the argument involves

    τ
    =
    
      
        
        3502
      
    
  

{\displaystyle \tau ={\sqrt {-3502}}}

. The discriminant d = 3502 has h(d) = 16. accurate to 256 digits:

        15261343909396942111177730086852826352374060766771618308167575028500999
        48590509502030754798379641288876701245663220023884870402810360529259
      
    
    .
    .
    .
  

{\displaystyle {\frac {15261343909396942111177730086852826352374060766771618308167575028500999}{48590509502030754798379641288876701245663220023884870402810360529259}}...}




  
    .
    .
    .
    
      
        
          551152789881364457516133280872003443353677807669620554743
          
            
              10005
            
          
        
        3134188302895457201473978137944378665098227220269702217081111
      
    
  

{\displaystyle ...{\frac {551152789881364457516133280872003443353677807669620554743{\sqrt {10005}}}{3134188302895457201473978137944378665098227220269702217081111}}}
  • improved inverse of sum of the first nineteen terms of Chudnovsky series. The continued fraction representation of π can be used to generate successive best rational approximations. These approximations are the best possible rational approximations of π relative to the size of their denominators. Here is a list of the first thirteen of these:

         3
         1
    
    
     ,
    
    
         22
         7
    
    
     ,
    
    
         333
         106
    
    
     ,
    
    
         355
         113
    
    
     ,
    
    
         103993
         33102
    
    
     ,
    
    
         104348
         33215
    
    
     ,
    
    
         208341
         66317
    
    
     ,
    
    
         312689
         99532
    
    
     ,
    
    
         833719
         265381
    
    
     ,
    
    
         1146408
         364913
    
    
     ,
    
    
         4272943
         1360120
    
    
     ,
    
    
         5419351
         1725033
    

    {\displaystyle {\frac {3}{1}},{\frac {22}{7}},{\frac {333}{106}},{\frac {355}{113}},{\frac {103993}{33102}},{\frac {104348}{33215}},{\frac {208341}{66317}},{\frac {312689}{99532}},{\frac {833719}{265381}},{\frac {1146408}{364913}},{\frac {4272943}{1360120}},{\frac {5419351}{1725033}}}

Of these,

        355
        113
      
    
  

{\displaystyle {\frac {355}{113}}}

is the only fraction in this sequence that gives more exact digits of π (i.e. 7) than the number of digits needed to approximate it (i.e. 6). The accuracy can be improved by using other fractions with larger numerators and denominators, but, for most such fractions, more digits are required in the approximation than correct significant figures achieved in the result.

=== Summing a circle's area ===

Pi can be obtained from a circle if its radius and area are known using the relationship:

    A
    =
    π
    
      r
      
        2
      
    
    .
  

{\displaystyle A=\pi r^{2}.}

If a circle with radius r is drawn with its center at the point (0, 0), any point whose distance from the origin is less than r will fall inside the circle. The Pythagorean theorem gives the distance from any point (x, y) to the center:

    d
    =
    
      
        
          x
          
            2
          
        
        +
        
          y
          
            2
          
        
      
    
    .
  

{\displaystyle d={\sqrt {x^{2}+y^{2}}}.}

Mathematical "graph paper" is formed by imagining a 1×1 square centered around each cell (x, y), where x and y are integers between r and r. Squares whose center resides inside or exactly on the border of the circle can then be counted by testing whether, for each cell (x, y),

          x
          
            2
          
        
        +
        
          y
          
            2
          
        
      
    
    ≤
    r
    .
  

{\displaystyle {\sqrt {x^{2}+y^{2}}}\leq r.}

The total number of cells satisfying that condition thus approximates the area of the circle, which then can be used to calculate an approximation of π. Closer approximations can be produced by using larger values of r. Mathematically, this formula can be written:

    π
    =
    
      lim
      
        r
        →
        ∞
      
    
    
      
        1
        
          r
          
            2
          
        
      
    
    
      ∑
      
        x
        =
        
        r
      
      
        r
      
    
    
    
      ∑
      
        y
        =
        
        r
      
      
        r
      
    
    
      
        {
        
          
            
              1
            
            
              
                if 
              
              
                
                  
                    x
                    
                      2
                    
                  
                  +
                  
                    y
                    
                      2
                    
                  
                
              
              ≤
              r
            
          
          
            
              0
            
            
              
                if 
              
              
                
                  
                    x
                    
                      2
                    
                  
                  +
                  
                    y
                    
                      2
                    
                  
                
              
              >
              r
              .
            
          
        
        
      
    
  

{\displaystyle \pi =\lim _{r\to \infty }{\frac {1}{r^{2}}}\sum _{x=-r}^{r}\;\sum _{y=-r}^{r}{\begin{cases}1&{\text{if }}{\sqrt {x^{2}+y^{2}}}\leq r\\0&{\text{if }}{\sqrt {x^{2}+y^{2}}}>r.\end{cases}}}

In other words, begin by choosing a value for r. Consider all cells (x, y) in which both x and y are integers between r and r. Starting at 0, add 1 for each cell whose distance to the origin (0, 0) is less than or equal to r. When finished, divide the sum, representing the area of a circle of radius r, by r2 to find the approximation of π. For example, if r is 5, then the cells considered are:

The 12 cells (0, ±5), (±5, 0), (±3, ±4), (±4, ±3) are exactly on the circle, and 69 cells are completely inside, so the approximate area is 81, and π is calculated to be approximately 3.24 because 81/52 = 3.24. Results for some values of r are shown in the table below:

Similarly, the more complex approximations of π given below involve repeated calculations of some sort, yielding closer and closer approximations with increasing numbers of calculations.