2

I am solving the following problem:

Use Newton's method to find solutions accurate to within $10^{-5}$ for the following problem.

$e^x + 2^{-x} + 2 \cos x - 6 = 0$ for $1 < x < 2$

I am familiar with claiming a $p_0$ and using the newtons formula, but I am completely confused on how the solution should be accurate to within $10^{-5}$. could someone explain that this means, and how I am supposed to do it.

I am thinking that I just have to ensure $|x_{n+1}-x_n|<10^{-5}$ is this correct?

user1729
  • 31,015
lo leafs
  • 19

1 Answers1

1

A rule of thumb is that if a region of quadratic convergence is reached, then the root is contained in an interval/disk around $x_{n+1}$ with radius equal to the step $|x_{n+1}-x_n|$. So a valid test would be that the step size reduced by a factor more than two over the last three steps and that the last step size is less than the error bound.

Lutz Lehmann
  • 126,666
  • You are correct, no need to get the diameter that small. – Lutz Lehmann Sep 07 '21 at 11:19
  • so the last step size just needs to be within $10^-5$ correct? – lo leafs Sep 08 '21 at 23:27
  • Standing alone it is not a good certificate, thus the additional test to reasonably exclude linear convergence like it happens at a multiple root or close to a cluster of roots. Consider the harmonic series to see that just the step size is not sufficiently informative. Or a geometric series with a factor below but close to $1$. Newton's method can also move through regions far from a solution where the step size becomes very small. However this is more typical for multi-dimensional problems. – Lutz Lehmann Sep 09 '21 at 07:10