I know how to do square roots out by hand. Is there a formula/method, beside using logarithms, to get cube roots by hand.

One of the most straightforward methods is known as the method of bisection:

For this method we need f(x) to be a continuous function. Find two points a < b such that f(a) and f(b) have opposite signs. This implies that f(x) has (at least) one root in the interval (a, b). Let c=(a+b)/2. If f(c) has the same sign as f(b), there is a root in (a, c), otherwise a root lies in (c, b). We have therefore halved the size of the interval containing the root, so by repeating the above procedure we can get closer and closer to the root. If we want to guarantee being within, say e, of the actual root, (i.e. if the actual root is r, we want our approximation to lie in the interval (r-e, r+e)) how many times must we repeat the procedure?

Another method is to use what is known as a point iteration. Here the idea is to express f(x)=0 in the form x=g(x) for some function g(x). One way of doing this is letting g(x)=f(x)+x although this isn't the only way. Then pick some starting point x₀ and let x_n = g(x_n-1) for n = 1, 2, 3, the sequence x₀, x₁, x₂, to say r, then r will be a solution of f(x)=0. The main problem with this method is that the sequence x₀, x₁, not converge (e.g. it may keep getting bigger and bigger). Draw some graphs showing y=g(x) and y=x and try and work out the behaviour of the sequence x₀, x₁, different shapes of g(x). Can you find some conditions on g(x) that will guarantee convergence? Try and find different functions g(x) for f(x)=x³ -a. Which ones give the fastest convergence? (A good measure of how close you are to the actual solution is to look at how big f(x_n) is).

If you know some calculus, a special case of the above is known as the Newton-Raphson Method. This converges extremely quickly and is probably the method your calculator uses to work out things like square roots! In this case we let g(x)=x-f(x)/f ' (x). (Then f(x)=0 if and only if g(x)=x). So we iterate using x_n+1=x_n-f(x_n)/f ' (x_n). (In case you don't know any calculus, for f(x)=x³ -a, g(x)=x-(x³ -a)/(3x²). Try it and compare how quickly it converges in comparison with the other methods.) How does this method work? (Hint: look at tangents on the graph of y=f(x))

Try and see why all of the above methods work (or don't, in the case of some functions in the second method). if you don't understand anything I have written or have any further questions, please ask me.

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