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

For the iteration

Xn=1 = 1 2 æ ç è Xn + 3 Xn ö ÷ ø

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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 X_n = 1.732050808, so is X_n+1. Squaring these terms we get X₁² = 4, X₂² = 3.0625, ... , X₅² = 3 and the rest of the other terms are the same!! This implies that when X_n » Ö3 so is X_n+1 and the values of X_n tend to the limit Ö3. This special property can easily be proven. Assume that the limit exists, so X_n+1 = X_n = X, then solve the equation

X = 1 2 æ ç è X + 3 N ö ÷ ø

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 æ ç è Xn + N Xn3 ö ÷ ø

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