TITLE: Why is $G(k)$ “more fundamental” than the Hilbert-Waring function $g(k)$? 
QUESTION [8 upvotes]: In the Wikipedia entry for Waring's problem, the section on $G(k)$ starts as: “From the work of Hardy and Littlewood, more fundamental than $g(k)$ turned out to be $G(k)$, which is defined...” There is no real justification or citations. 
Do you believe this? Is there a specific, objective sense in which $G(k)$ is more fundamental? My answer is that the Hardy-Littlewood method itself works only for sufficiently large $x$ (let's say you are interested in density of subsets of $\mathbf N$ in the interval $[1;x]$), so it's better suited to handle $G(k)$ than $g(k)$. Is there a better answer?

REPLY [7 votes]: Most often in analytic number theory one is interested in asymptotic behavior.  The function $G(k)$ reflects that -- i.e., it ignores a finite number of exceptions.  If you actually try to compute the function $g(k)$, you find that it can be quite a bit larger than $G(k)$ but for reasons which feel somewhat "accidental": that is, you are focusing on the difficulty of representing very small numbers.  
[Incidentally, I surmise that both the wikipedia article and the paragraph above are influenced by the passage on this in Hardy and Wright's Introduction to the Theory of Numbers.  I think that Hardy and Wright said it better than both wikipedia and me, and I recommend you look to see what they said.]
Anyway, notice that whether one quantity is "more fundamental" than another is not a mathematical statement per se: it is a statement of opinion, aesthetic and experience.  One would be well within their rights to be more interested in the function $g(k)$ than $G(k)$: maybe the intricate, chaotic-looking behavior of small numbers appeals to you.