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 = (xn + 3/xn)/2 . Calculate the first six terms of
this sequence. What do you notice? Calculate a few more terms and
find the squares of the terms. Can you prove that the special
property you notice about this sequence will apply to all the later
terms of the sequence? Write down a formula to give an
approximation to the cube root of a number and test it for the cube
root of 3 and the cube root of 8. How many terms of the sequence do
you have to take before you get the cube root of 8 correct to as
many decimal places as your calculator will give? What happens when
you try this method for fourth roots or fifth roots etc.?

$2\wedge 3\wedge 4$ could be $(2^3)^4$ or $2^{(3^4)}$. Does it make any difference? For both definitions, which is bigger: $r\wedge r\wedge r\wedge r\dots$ where the powers of $r$ go on for ever, or $(r^r)^r$, where $r$ is $\sqrt{2}$?

Dalmatians

Stage: 4 and 5 Challenge Level:

Start with different numbers and write down the sequences. What
patterns do you notice? Can you explain what happens?