Saturday, April 2, 2016

How can one calculate this Limit?



Let f(n) denote the number of integer solutions of the equation 3x2+2xy+3y2=n



How can one evaluate the limit limnf(1)+...f(n)n




Thanks


Answer



Let us elaborate the solution by Thomas Andrews. Though the idea is simple and repeatedly used in other answers, here we want to give a meticulously detailed solution.



As we have observed, the number f(1)++f(n) is equal to the integer solution (x,y), or equivalently the number of integer points (x,y)Z2, of the inequality



3x2+2xy+3y2n.



Now let




Sn={(x,y)Z2:(x,y) satisfies (1)}



be the set of integer solutions (x,y) of (1), and let



En={(x,y)R2:(x,y) satisfies (1)}



be the set of points (x,y)R2 satisfying (1). Plainly, Sn is the set of integer points contained in En and in this notation we have



#(Sn)=f(1)++f(n),




where # denotes the cardinality of a set. We also define



An={(x,y)Zn:C(x,y)En},



and



Bn={(x,y)Zn:C(x,y)En},



Now for each integer point p=(x,y)Z2, we associate a unit square




C(p)=C(x,y)=[x12,x+12]×[y12,y+12]



centered at p=(x,y). Then it is plain to observe that



AnSnBn



Also, by definition it is clear that we have



#(An)=Area(pAnC(p))Area(En)Area(pBnC(p))=#(Bn),



Now let us digress to a rudimentary real analysis and prove some basic facts. Let (M,d) be a metric space. For pM and AM, consider the minimum distance between p and A



dist(p,A)=inf{d(p,q):qA}



and correspondingly the ϵ-neighborhood Aϵ of K defined by




Nϵ(A)={qM:dist(q,A)<ϵ}.



Then we have the following proposition.




Proposition. If F is closed, ϵ>0Nϵ(F)=F.



Proof. Let F=ϵ>0Nϵ(F). Since FNϵ(F) for each ϵ>0, we have FF. To show the converse, let pF. Then for each n we have pN1/n(F) and hence there exists qnF such that d(p,qn)<2n. This shows that qnp. Since F is closed, the limit must lie in F. Therefore pF and hence F=F. ////





Now let us return to the original problem. For each pBnAn, we have C(p)En. This implies that



C(p)N2(En).



But by the scaling and the homogeneity of the polynomial 3x2+2xy+3y2, clearly we have



En=nE1={(nx,ny):(x,y)E1}



and accordingly




N2(En)=nN2/n(E1)



This shows that



#(BnAn)=Area(pBnAnC(p))Area(N2(En))=Area(nN2/n(E1))=nArea(N2/n(E1))



Now we are ready.




  • From (3), we have
    0#(Bn)#(An)nArea(N2/n(E1)),


    which goes to zero as n by the monotonicity of the Lebesgue measure together with the fact that Area(E1)=0. Here we used the proposition above.


  • Thus from (2), we have
    Area(E1)=Area(En)n#(Bn)n#(An)n+#(Bn)#(An)nArea(En)n+o(1)=Area(E1)+o(1)


    and therefore
    Area(E1)=limn#(An)n=limn#(Sn)n=limn#(Bn)n=Area(E1).




Now it remains to evaluate the area of the ellipse E1. Let



A=(3113)



so that x,Ax=3x2+2xy+3y2 for x=(x,y). Then by spectral theory for symmetric matrices, there exists a rotation matrix R such that Rx,DRx for D=diag(λ1,λ2). Then with a new coordinate (X,Y)=Rx, we have λ1X2+λ2Y21. Therefore the area of the ellipse E1 is given by




Area(E1)=πλ1λ2=πdetA=π22.


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