A sequence of numbers X1 , X2 , X3 ,..., starts with X1 =2, and, if you know any term Xn , you can find the next term Xn+1 using the formula Xn=1 = 1 2 ( Xn + 3 Xn ).

Solution by Andaleeb Ahmed, Age 17, Woodhouse Sixthform College, London.

For the iteration
Xn=1 = 1 2 ( Xn + 3 Xn )

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X1 = 2,   X2 =1.75,   X2 =1.732142857,   X4 =1.73205081 X5 = 1.732050808,   X6 =1.732050808,   X7 =1.732050808 X8 = 1.732050808   and so on

We notice that when Xn =1.732050808, so is Xn+1 . Squaring these terms we get X1 2 =4, X2 2 =3.0625,..., X5 2 =3 and the rest of the other terms are the same!! This implies that when Xn ≈3 so is Xn+1 and the values of Xn tend to the limit 3. This special property can easily be proven. Assume that the limit exists, so Xn+1 = Xn =X, then solve the equation
X= 1 2 (X+ 3 N )

If we test it for N=3, we see that X29 =1.44224957, which is what the calculator gives for the cube root of 3. Testing it for N=8, we get X1 =2, which is the right answer. By experimentation you can soon discover for yourself that it is not safe to assume that the same method works finding fourth roots using the iteration formula.
Xn+1 = 1 2 ( Xn + N Xn 3 )

There is work to do to show that the iteration Xn+1 =F( Xn ) converges to a limit L if and only if -1 < F'(L) < 1.