tag:blogger.com,1999:blog-8231784566931768362.post7399956758422750446..comments2023-09-09T08:21:55.454-04:00Comments on MathNotations: In-Depth Investigation of Patterns: Lattice Points, Algebra,...Dave Marainhttp://www.blogger.com/profile/13321770881353644307noreply@blogger.comBlogger8125tag:blogger.com,1999:blog-8231784566931768362.post-16966016548464635002008-07-06T18:51:00.000-04:002008-07-06T18:51:00.000-04:001111111111111111<A HREF="http://cb-2008.blogspot.com/" REL="nofollow">1</A><A HREF="http://buy2008.org/map.asp" REL="nofollow">1</A><A HREF="http://buy2008.org/200807/map.asp" REL="nofollow">1</A><A HREF="http://buy2008.org/200806/map.asp" REL="nofollow">1</A><A HREF="http://cb-2008.blogspot.com/" REL="nofollow">1</A><A HREF="http://buy2008.org/map.asp" REL="nofollow">1</A><A HREF="http://buy2008.org/200807/map.asp" REL="nofollow">1</A><A HREF="http://buy2008.org/200806/map.asp" REL="nofollow">1</A><A HREF="http://cb-2008.blogspot.com/" REL="nofollow">1</A><A HREF="http://buy2008.org/map.asp" REL="nofollow">1</A><A HREF="http://buy2008.org/200807/map.asp" REL="nofollow">1</A><A HREF="http://buy2008.org/200806/map.asp" REL="nofollow">1</A><A HREF="http://cb-2008.blogspot.com/" REL="nofollow">1</A><A HREF="http://buy2008.org/map.asp" REL="nofollow">1</A><A HREF="http://buy2008.org/200807/map.asp" REL="nofollow">1</A><A HREF="http://buy2008.org/200806/map.asp" REL="nofollow">1</A>Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-11283506357977993392007-03-28T20:12:00.000-04:002007-03-28T20:12:00.000-04:00tim--thank you for your awesome insight and extens...tim--<BR/>thank you for your awesome insight and extensions!Practical problems with spreadsheets? I suspect you have a passion for math that goes beyond the practical! I will consider developing some examples around your results if that's ok with you.<BR/><BR/>I considered generalizing like you did but eric jablow got me 'going in circles!'<BR/><BR/>The circle problem is really a magnificent piece of math analyzed fairly extensively by Gauss. I decided to bring Pick's Thm into it because it just doesn't get the attention in schools that it so richly deserves...Dave Marainhttps://www.blogger.com/profile/13321770881353644307noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-50862575326867635042007-03-28T07:15:00.000-04:002007-03-28T07:15:00.000-04:00If you do a change of basis, you count diagonally;...If you do a change of basis, you count diagonally; you get two sets of grids and quickly see that the answer is (N+1)^2 + N^2.<BR/><BR/>I tricked myself this way though, since I overlooked the offsetting grid of 6x6 inside the 7x7; my first thought was 49. I had to graph it all to see...<BR/><BR/>Going to a circle would be obnoxious. I say this mostly for reasons of "squaring the circle" - the answer ends up being scale dependent, as certain points just make it in, or not.<BR/><BR/>To increase the difficulty somewhat, but not the inherent difficulty, perhaps going to:<BR/> |2x| + |y| <= 8.<BR/><BR/>Using the graphical approach and diagonal counting, this yields 5^2 + 3*4^2.<BR/><BR/>The next step is:<BR/> |2x| + |y| <= 9.<BR/>Here, max(x) is still 4, which complicates things a little. Still, the diagonal approach gives 3*5^2 + 4^2 = 91.<BR/><BR/>This suggests that the form of the general solution will be:<BR/>Points = a*(N+1)^2 + b*N^2<BR/>, where N = Min(max(x), max(y)). In other words, a linear combination of "big" (N+1) and "small" (N) diagonal squares.<BR/><BR/>Now that we have the general solution, all we need is the general problem:<BR/> |Ax| + |By| <= C.<BR/>For convenience, take A>B.<BR/><BR/>Then, N= Int(C/A)= max(x)<BR/>and M= Int(C/B) = max(y).<BR/><BR/>Now, the range, R, of y is (-M,+M) for a total (lattice points on y axis) of<BR/> R= 2*M +1<BR/><BR/>A hasty conjecture is that a,b satisfy:<BR/><BR/>R = a*(N+1) + b*N<BR/><BR/>Can we prove this? If the "big" <BR/>diagonal square is counted "a" times, then the diagonals of each will cross the y-axis (N+1) times. The remainder of the y-axis values must be associated with the "small" diagonal squares. So, yes, we can prove this and a,b <B>must</B> satisfy the equation.<BR/><BR/>At this point, if we have a, we have b. The value of a can be found by examining the range of y at x = N. Denoting this lower case r, and the maximum of y at x=N as lower case m, we see:<BR/><BR/>m = C - A*N<BR/>and<BR/><BR/>r = 2m + 1<BR/><BR/>However, r is simply how many times the maximum value of x is repeated, and thus, how many times the "big" diagonal square is repeated. In other words, this is "a" that we are looking for:<BR/><BR/>a = 2m + 1<BR/><BR/><BR/>To summarize:<BR/><BR/>For:<BR/><BR/> |Ax| + |By| <= C<BR/>x,y,A,B,C integers, A>B>0,<BR/>(A+B)<=C<BR/>has lattice points that number<BR/><BR/>P = a(N+1)^2 + bN^2<BR/><BR/>where<BR/><BR/>N= Int(C/A)<BR/>M= Int(C/B)<BR/>m = C - A*N<BR/>R = 2M + 1<BR/>a = 2m + 1<BR/>b = (R - a(N+1))/N<BR/><BR/>I think this works. FYI, this solution is influenced by years of solving various practical problems with spreadsheets.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-2489008525857998092007-03-26T13:57:00.000-04:002007-03-26T13:57:00.000-04:00Here's a different insight into the lattice point ...Here's a different insight into the lattice point problem. I will post the general formulas later on.<BR/>The area of the square formed by |x| + |y| = 6 is easily shown to be 72. <A HREF="http://mathworld.wolfram.com/PicksTheorem.html" REL="nofollow">Pick's Theorem</A> relates the area bounded by this square with the number of lattice points on the boundary, B, and the number of lattice points in the interior, I, as follows:<BR/>Area = I + B/2 - 1.<BR/>It is easy to show that B = 24 for this graph, so 72 = I + 24/2 - 1 which leads to I = 61. Adding we obtain that the total number of lattice points is 61+24 = 85. Pick's Theorem is beautiful although not easy to prove. Worth showing this to middle school students though (particularly if they have experimented with geoboards).Dave Marainhttps://www.blogger.com/profile/13321770881353644307noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-15124722686281584952007-03-26T09:49:00.000-04:002007-03-26T09:49:00.000-04:00great ideas, eric!I was thinking of going to the d...great ideas, eric!<BR/>I was thinking of going to the disk next. I do think the 9th graders can graph most of the solutions for the interior and on the disk but would need more machinery to verify their conjectures (Pythagorean, distance formula, etc). I will work on it. Now as a real challenge, how about solving the problem in n-dimensional space!Dave Marainhttps://www.blogger.com/profile/13321770881353644307noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-83572357094994382942007-03-26T07:27:00.000-04:002007-03-26T07:27:00.000-04:00When all this is done, you can talk about the numb...When all this is done, you can talk about the number of lattice points contained within the closed <I>disk</I> of radius <I>r</I>. This is <I>Gauß' circle problem</I>. MathWorld has an entertaining discussion; you can use it as a starting point for estimation, the O-notation, and probably a few more things. Of course, this isn't something 9th graders could do too much work on.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-10630680711091390802007-03-25T23:22:00.000-04:002007-03-25T23:22:00.000-04:00jonathan--I would expect them to add one of those ...jonathan--<BR/>I would expect them to add one of those 2 ways. Actually, the original SAT problem I saw asked only for the points on the boundary which students have a better chance of formulating in general since it's just '4N'. <BR/><BR/>For whatever reason, I added horizontally, but, from symmetry, it doesn't matter. Personally, I was more interested in the relationship between the points inside and those outside. For N = 6, the outer square has 13^2 = 169 points. This leads to 85 points inside or on the 'diamond' and 84 points outside. This could be explained visually by noting there are 6+5+4+3+2+1 = 21 'outside' points in each quadrant. Thus the number of 'outside' points is 4 times T_6 where I'm using the notation to stand for the triangular number of rank 6. <B>Triangular numbers</B> seem to play a huge role in many of these geometric combinatorial problems. <BR/><BR/>These beautiful connections are so important for the mathematical development of our students, IMO.Dave Marainhttps://www.blogger.com/profile/13321770881353644307noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-80604407808506223362007-03-25T22:21:00.000-04:002007-03-25T22:21:00.000-04:00You don't want them to count horizontally or verti...You don't want them to count horizontally or vertically? There is a neat x^2 + (2x + 1) +x^2 that drops straight out.Anonymousnoreply@blogger.com