A formula to work out the square root of 2?

Answer 1

There is no simple analytical formula. One possible way to estimate the square root of a number is by iteration. This entails making a guess at the answer and then improving on it. Repeating the procedure should lead to a better estimate at each stage. One such is the Newton-Raphson method.
If you want to find the square root of 2, define f(x) = x^2 – 2. Then finding the square root of 2 is equivalent to solving f(x) = 0.

Let f’(x) = 2x. This is the derivative of f(x) but you do not need to know that to use the N-R method.

Start with x0 as the first guess. Then let xn+1 = xn - f(xn)/f’(xn) for n = 0, 1, 2, … Provided you made a reasonable choice for the starting point, the iteration will very quickly converge to the true answer. It works even if your first guess is not so good:
Suppose you start with x0 = 1 (a pretty poor choice since 1^2 is 1, not 2).
Even so, x3 = 1.4142156863, which is less than 0.0002% from the true value. Finally, remember that the negative value is also a square root.

Answer 2

There is no analytical formula to enable you to calculate it. One way to estimate the square root of a number is by iteration. This entails making a guess at the answer and then improving on it. Repeating the procedure should lead to a better estimate at each stage. One such is the Newton-Raphson method.

If you want to find the square root of 2, define f(x) = x^2 – 2. Then finding the square root of 2 is equivalent to solving f(x) = 0.

Let f’(x) = 2x. This is the derivative of f(x) but you do not need to know that to use the N-R method.

Start with x0 as the first guess. Then let xn+1 = xn - f(xn)/f’(xn) for n = 0, 1, 2, … Provided you made a reasonable choice for the starting point, the iteration will very quickly converge to the true answer. It works even if your first guess is not so good:
Suppose you start with x0 = 1 (a pretty poor choice since 1^2 is 1, not 2).
Even so, x3 = 1.4142156863, which is less than 0.0002% from the true value. Finally, remember that the negative value is also a square root.