equation solving - Archimedes' Scheme to find $pi$

Given an Archimedes' scheme

$$a_{n+1}=\frac{2a_nb_n}{a_n+b_n}$$ $$b_{n+1}=\sqrt{a_{n+1}b_n}$$

with initial values $a_0=2\sqrt{3},b_0=3.$

How do I prove using Mathematica that this converges to $\pi$?

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)

Comments

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