In the tutorial questions for solving the number of iterations until an error less than some number (e.g., 10e-5) we solved the inequality of current error^(convergence rate ^ n minus number of iterations to get to current error) < desired error like 0.0365^2^n-2 < 10e-5 from question 6. But this wouldn’t work for large errors, right, such as an error > 1? Would there be an analogous way of guessing the number of iterations or would that not be something we have to deal with?
The easy answer is that this is something you do not have to deal with.
A more refined answer is that, when studying such (high-order) methods, you need to be reasonably close to the root to get convergence.
See the notes pg IV-195-196.
Someone may (resonably) suggest to apply a few bisections to lessen the error, then apply Newton’s.
In the particular tutorial, though there is convergence starting from error 1 (fairly large), I explicitly left out 1-2 iterations, and, when the error was ~1e-2, applied the formula.
For an arbitrary case, where you do not know the exact solution (root), you need to be aware of these issues, try a few iterations, see how the iterates behave, then decide whether to go with low order but guaranteed convergence, until the point where the error is small enough to apply the formula and the high-order method, or go directly to high-order.
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