# Stopping criterion for Newton-Raphson

I am currently doing a question on Newton-Raphson method and I am not sure what it means by 'explain your stopping criterion'.

Question

Using the Newton-Raphson method with initial guess $$x_0=1.5$$, solve the equation $$x^2=2$$ correct to four decimal places. Be sure to explain your stopping criterion

So my issue is not working out Newton-Raphson, you just follow the equation, to which I make it $$1.4142$$ after three iterations which is to 4 d.p but what dose it mean by 'stopping criterion'?

In an computer lab, we have done code for this and in a while loop we set the to |$$f(x_0)$$|>$$\epsilon$$

where epsilon was set by us, and the lower we set $$\epsilon$$ the more iterations were produced. But there was a limit on this, and from that I got the gist it was a convergence limit? But I am not sure if or how this relates to this question nor how one would workout the stopping criterion.

Possible things to take into consideration: the values $$f(x_n)$$, the derivatives $$f'(x_n)$$, the differences $$x_n - x_{n-1}$$. Whatever rule you come up with, it should be something that, in theory, you could put into a computer program as a condition to end that while loop. You don't have to write that program, of course - just explain your rule in natural language.
In reality, your objective is to produce a code which can compute $$\sqrt{\alpha}$$ with a small relative error for any $$\alpha > 0$$. It is easy to determine when $$\alpha - x_n^2$$ is small, but when is it small enough? The relative error is given by $$r_n = \frac{\sqrt{\alpha} - x_n}{\sqrt{\alpha}}.$$ We do not know the exact value of $$\sqrt{\alpha}$$, but we can nevertheless estimate the relative error as follows. We have $$r_n = \frac{(\sqrt{\alpha}-x_n)(\sqrt{\alpha}+x_n)}{\sqrt{\alpha} (\sqrt{\alpha}+x_n)} = \frac{\alpha-x_n^2}{\sqrt{\alpha} (\sqrt{\alpha}+x_n)} \approx \frac{\alpha-x_n^2}{2\alpha}.$$ The last approximation is good when $$\sqrt{\alpha} - x_n$$ is small.
Resist the temptation to use $$|f(x)| > \epsilon$$ to control a while loop. It is a recipe for disaster, because you depend on $$f$$ being implemented correctly. If $$f$$ returns NaN (not a number), then you exit the while loop believing $$|f(x)| \leq \epsilon$$. This is a consequence of IEEE standard for floating point arithmetic which specifies that all comparisons with NaN return false. The safe construction is to use a for loop with a user defined number of iterations. You should evaluate $$|f(x)|$$ inside the loop and exit the loop prematurely if $$|f(x)| \leq \epsilon$$ (all is well) or if $$f(x)$$ is not finite (serious problem).
To stop, you need to be sure you're right to four decimal places, so for example you might require consecutive estimates to differ by at most $$10^{-5}$$.
For a really advanced treatment you could estimate roughly how many steps that should take from first principles. In solving $$f(x)=0$$, in this case with $$f(x):=x^2-2$$, the error terms have an approximate iteration $$\epsilon_{n+1}\approx M\epsilon_n^2,\,M:=-\frac{f^{\prime\prime}(a)}{f^\prime(a)}$$ with $$a$$ the root. So $$M=-\frac{2}{2\sqrt{2}}=-\frac{1}{\sqrt{2}}$$ and the initial error approximates $$0.1$$. This suggests three iterations beyond $$1.5$$ is enough. That fits my data.