By elementary compute, we can easily get an asymptotic formula for the number of solutions for the equation n_1+n_2+n_3=N. Here N is a fixed positive integer, sufficiently large. My question is: can we apply circle method to this equation and get a same asymptocic formula for the number of solutions?
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$\begingroup$ What is the "circle method"? $\endgroup$– Lee MosherMar 15, 2012 at 15:35
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7$\begingroup$ @Lee The Hardy-Littlewood circle method is a well-known technique in analytic number theory. en.wikipedia.org/wiki/Circle_method $\endgroup$– Felipe VolochMar 15, 2012 at 15:50
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I don't think the circle method works well this case. The method is based on the idea that the generating function has local peaks at rational numbers with sufficiently rapid local decay there (in particular it is small at points which are not close to a rational number with small denominator). In our case the generating function is the cube of the geometric series $\sum_{n=0}^N e(n\alpha)$ whose only peak is at $\alpha=0$, namely it is rather small when $\|\alpha\|$ is considerably larger than $1/N$. In other words, in our case only one rational number (namely $\alpha=0$) has a real impact at the integral, but as a by-product the decay is not as definite as in higher degree cases: the transitional range $\|\alpha\|\asymp 1/N$ is rather subtle. To put differently, in this case the mass in the integral is very unevenly distributed, hence it seems as if one would need to dissect the circle in a completely trivial and useless way: there would be one major arc (the whole circle) and no minor arc at all.
EDIT. It is instructive to look at the treatment of Waring's problem in Vaughan's book "The Hardy-Littlewood method", and see "what goes wrong" when $k=1$. Well, nothing goes wrong but it turns out that the method itself furnishes no information in this case. Indeed, the singular integral given by (2.9) and (2.15) is approximated by the full integral (2.20) which is really the number of solutions to $n_1+\cdots+n_s=N$ when $k=1$. This is then evaluated (for any $k$) in an elementary fashion in (2.22) of the book.
answered Mar 15, 2012 at 17:07
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2$\begingroup$ So I guess you could say it's a 'circular' argument...? $\endgroup$– samian86Mar 15, 2012 at 17:52
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$\begingroup$ @samian86: It is not a circular argument. The circle method reduces a count over $k$-th powers (or primes etc.) to a count over integers (with simple analytic weights), which then can be handled by more elementary methods. In the case of the OP's question, the reduction furnished by the circle method is trivial: it reduces the problem to itself. $\endgroup$ Mar 15, 2012 at 18:28
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1$\begingroup$ @GH: Something tells me samian86 was making a joke. $\endgroup$ Mar 17, 2012 at 1:44
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1$\begingroup$ There is a way to break the circular argument: Choose $[-1/\sqrt{N},1/\sqrt{N}]$ as the major arc, and the rest of $[-1/2,1/2]$ as the minor arcs. The treatment of the minor arcs is easy. When $\alpha$ is in the major arc note that $\sum_{n\le N}e(n\alpha)=(1+O(1/\sqrt{N}))(1-e(N\alpha))/(2\pi i\alpha)$. The integral of $\int_{|\alpha|\le1/\sqrt{N}}((1-e(\alpha N)/(2\pi i\alpha))^3e(-N\alpha)d\alpha$ can be computed asymptotically easily (it is $cN^2$ for some constant $c$). Of course, all this is not necessary, since it is much easier to count the solutions directly (combinatorially). $\endgroup$ Mar 19, 2012 at 18:10
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1$\begingroup$ @GH: Set $x=N\alpha$, so that the integral becomes $$N^2\int_{|x|\le\sqrt{N}}((e(x)-1)/(2\pi ix))^3e(-x)dx=(N^2+O(N))\int_{\mathbb{R}}((e(x)-1)/(2\pi ix))^3e(-x)dx$$ (there is a sign mistake above). This complete integral is the constant $c$ I was referring to. Setting $z=2\pi i x$, $c=\frac1{2\pi i}\int_{\Re(z)=0}((e^z-1)/z)^3e^{-z}dz$. Then $c$ can be computed by Cauchy's theorem (e.g. we can set $w=1/(z-1)$ to transform the line to a circle and then calculate the residue of the resulting function). $\endgroup$ Mar 20, 2012 at 22:19