Question

Newton Raphson - Error Analysis.

In the lecture in part 2B, Prof. Jerison makes the claim that the error of a subsequent guess using the Newton Raphson method should be approximately equal to the square of the error of the previous guess.
e.g/
E1 = 10^-1, E2 =10^-2, E3 = 10^-4, E4 = 10^-8

I'm asking what the theory behind this comes from. Also, a student asks what happens when the error of the initial guess is greater than 1 - and Prof. Jerison said Newton-Raphson would fail to converge. I've tested this and found this claim false.

Thanks for your help.

Answer 1

I think (think being the key word) there are two main points that explain why the error decreases exponentially in Newton's method.

1. The error from \[x_{0}\] is sorta the input used to calculate \[x_{1}\] and so forth.

2. The error of a linear approximation and a quadratic is quadratic in nature.

I'm not sure if that's the answer. I'm just putting it out there.

As for what Jerison said about the errors of the initial guesses, I think he explained that it's tricky, and you have to check to make sure that Newton's method is giving results that make sense. In the example from class \[f(x) = x^{2} - 5\] he pointed out that if we had guessed a negative x initially, we would have ended up with a very good approximation for the wrong x-intercept (we were looking for the positive x-intercept). He then further explained that any positive number guess would eventually give a good approximation (maybe after many iterations if we guess really far away). And finally he said that a guess of zero would have been undefined in our equation.

Sometimes a guess that has an error greater than 1, will still lead to a "good" approximation and sometimes not.