Show that Newton's method applied to x^n - c = 0 produces the iterative scheme...

Show that Newton's method applied to$\displaystyle x^{n} - c = 0$ (where n and c are positive constants) produces the iterative scheme $\displaystyle x_{n+1} =\frac{1}{n}[(n-1)x_n + cx_n^{1-n}]$ for approximating $\displaystyle c^{\frac{1}{n}}$. We learned Newton's method, but we haven't applied it to "schemes" and my teacher told us to look at this problem tonight. Can someone explain, please! Thanks.

Re: Show that Newton's method applied to x^n - c = 0 produces the iterative scheme...

I'm inept with Latex, so I apologize for the mess this may be.

Newton's method, you might remember, takes some guess say "a" as a zero for a function and makes a better approximation so long as the method doesn't break down (which it can). So the formula for this better approximation say "b", b=a-(f(a)/f'(a)).

So what your teacher gave you is another form of this approximation "b" being x(n+1) in your formula for a certain case "x^n-c=0". So b=a-(a^(n)-c/na^(n-1)). So at this point a is x(n) in your formula and a little algebra will give your formula. Start with a common denominator on the RHS, then divide each term out by the denominator. Then factor out a (1/n) and voila, there you have it. Once again sorry about not using Latex, I ought to learn how to use it.