How to select convergence criterion in numerical analysis?

Question

When doing old exams in basic numerical analysis, I encountered this problem:

Solution proposal from lecturer:

My idea was to select $|f(x)| \le 0.5 \times 10^{-5}$ as the convergence criterion. With this criterion, $x=0.652910$ would have been accepted as a root. When looking at the solution, it seems like the convergence criterion is way smaller.

What is the correct way of selecting convergence criterion using the Bisection, Newton-Raphson or Fixed-Point Iterative method for the given problem?

Answer 1

You are asked to find $r$ to 6 decimals, but your proposal only ensures that $f(r)$ is correct to 6 decimals. You are calculating successive approximations to $r$ ($x_0,x_1,\cdots$). Plot $x_i$ vs. $i$: you should be able to see it converge, and identify a point beyond which you are confident that you have approximated the limit value to 6 decimal points.

Besides, your answer is as good as correct.

Answer 2

I don't know why you select function-value as a criterion, it's quite obvious the output is root. So, we iterate with criterion for roots, x, because the output values have fluctuating roots. Therefore, we need to use absolute error or relative error for root. $$\text{Abs. Error : }|x_{i+1}-x_i|$$ $$\text{Rel. Error : }\left|\frac{x_{i+1}-x_i}{x_i}\right|$$ Moreover, the problem requires 10^-6 tolerance, it means Abs. Error has the value less than 10^-6.

In my code, my answer is 0.652918640419205 in the long format of Matlab.