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equation solving - Archimedes' Scheme to find pi



Given an Archimedes' scheme


an+1=2anbnan+bn bn+1=√an+1bn


with initial values a0=2√3,b0=3.


How do I prove using Mathematica that this converges to π?


My attemp is to use RSolve or RSolveValue. But my when I input


RSolve[{a[n + 1] == 2 a[n] b[n]/(a[n] + b[n]), 
  b[n + 1] == Sqrt[a[n + 1] b[n]], a[0] == 2 Sqrt[3], 
  b[0] == 3}, {a[n], b[n]}, n]

It gives an output the same as input



RSolve[{a[1 + n] == (2 a[n] b[n])/(a[n] + b[n]), 
  b[1 + n] == Sqrt[a[1 + n] b[n]], a[0] == 2 Sqrt[3], 
  b[0] == 3}, {a[n], b[n]}, n]

What is wrong here?



Answer



You should use RecurrenceTable:


N@RecurrenceTable[{a[n + 1] == 2 a[n] b[n]/(a[n] + b[n]), 
   b[n + 1] == Sqrt[a[n + 1] b[n]], a[0] == 2 Sqrt[3], 
   b[0] == 3}, {a[n], b[n]}, {n, 1, 10}]

N@Pi

with result:



{{3.21539, 3.10583}, {3.15966, 3.13263}, {3.14609, 3.13935}, {3.14271, 3.14103}, {3.14187, 3.14145}, {3.14166, 3.14156}, {3.14161, 3.14158}, {3.1416, 3.14159}, {3.14159, 3.14159}, {3.14159, 3.14159}}


3.14159



You can also change the number of digit, with N[expr,n] (n is the desired number of digits). See documentation.


EDIT Following @RunnyKine suggestion:


RecurrenceTable[{a[n + 1] == 2. a[n] b[n]/(a[n] + b[n]), 

  b[n + 1] == Sqrt[a[n + 1] b[n]], a[0] == 2. Sqrt[3.], 
  b[0] == 3.}, {a[n], b[n]}, {n, 1, 20}]

is much faster (on my laptop, AbsoluteTiming gives 0.5 sec for the first version and 0.01 for the second)


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