I am having trouble understanding the concept of finding the rate of convergence. My homework question is a more beefed-up version of the following: Find the rate of convergence for $$\lim_{h\rightarrow 0} \dfrac{\sin(h)}{h}=1$$ as $h$ approaches $0.$
The only reference I can get from the book is to use the Taylor series expansion of $\sin(h)$ around $x_0=0$ so that $\frac{\sin(h)}{h}$ simplifies to $$1 - \dfrac{h^2}{6} + \dfrac{h^4}{120} - \dfrac{h^6}{5040} + \cdots.$$
Every other example of this sort that I see (ie when the problem involves trig functions) uses the series expansion. But then the series is arbitrarily cut off at a certain point! I'm not understanding 1) why the cut off point is chosen where it is and 2) how this affects the R.O.C.
So, to illustrate, the book might cut off the series and write it as $$\dfrac{\sin(h)}{h} \approx 1 - \dfrac{h^2}{6} + \mathcal{O}(h^4).$$ Well, when written this way then sure: the R.O.C. is $\mathcal{O}(h^4).$ Why did they stop the series there? One could just as arbitrarily write out the series as $$\dfrac{\sin(h)}{h} \approx 1 - \dfrac{h^2}{6} + \dfrac{h^4}{120} - \dfrac{h^6}{5040} + \mathcal{O}(h^8)$$ so that the R.O.C. is $\mathcal{O}(h^8).$
So what am I missing?
Additionally, here is the definition I am using to understand this: Suppose that $\lim_{h \rightarrow 0} G(h)=0$ and $\lim_{h \rightarrow 0} F(h)= L.$ If a positive constant $K$ exists with $$|F(h)-L| \leq K|G(h)|, \text{ for sufficiently small }h,$$ then we write $F(h)= L + \mathcal{O}(G(h)).$
I guess I could ask at this point how the $G(h)$ is chosen but I think I intuitively see how it's chosen (at least I hope). How do we choose $K$ and verify this?
Any assistance with understanding the process behind all this would help. I've looked up many examples online and not of it is sticking with me. I'm hoping here that I can use the back-and-forth to understand it better.
Also, if you need any more info or context from me, please let me know. Thanks!
Go the other way: the canonical way to express the most significant part of the difference is as
$$ -\frac{h^2}{6} + \mathcal{o}(h^2)$$
or even just $\mathcal{O}(h^2)$ depending on whether or not you care about the constant of proportionality.
(aside: I've not heard the phrase "rate of convergence" used in this context: I usually see it used in regards to sequences. It's probably worth checking your text to find exactly what they're asking for)
For practical applications, you take however many terms you need for your application: if you can solve your problem using just the fact that the difference is $\mathcal{O}(h^2)$, then you do. If you need $-h^2/6 + \mathcal{o}(h^2)$, then use that. Or $-h^2/6 + \mathcal{O}(h^3)$ or $-h^2/6 + \mathcal{O}(h^4)$ or $-h^2/6 + h^4/120 + \mathcal{o}(h^4)$ or whatever.