Path: newsspool2.news.pas.earthlink.net!stamper.news.pas.earthlink.net!elnk-nf2-pas!newsfeed.earthlink.net!newshub.sdsu.edu!headwall.stanford.edu!newsfeed.stanford.edu!postnews1.google.com!not-for-mail Message-ID: <3c65f87.0308310458.3bb6b21d@posting.google.com> From: jstevh@msn.com (James Harris) Newsgroups: sci.physics,sci.math,sci.math.num-analysis,alt.writing Subject: Re: My prime research, focus on a feature REVISED Date: 31 Aug 2003 05:58:30 -0700 References: <3c65f87.0308270833.66eedc8e@posting.google.com> <3c65f87.0308271550.325219f5@posting.google.com> <585ab5d8.0308281046.b3ebad5@posting.google.com> <3c65f87.0308300705.403324df@posting.google.com> <3c65f87.0308301524.4b58602@posting.google.com> <3F513626.771F2246@ix.netcom.com> Lines: 55 Organization: http://groups.google.com/ NNTP-Posting-Host: 67.192.35.4 Content-Type: text/plain; charset=ISO-8859-1 Content-Transfer-Encoding: 8bit X-Trace: posting.google.com 1062334710 29385 127.0.0.1 (31 Aug 2003 12:58:30 GMT) X-Complaints-To: groups-abuse@google.com NNTP-Posting-Date: 31 Aug 2003 12:58:30 GMT X-Received-Date: Sun, 31 Aug 2003 05:58:31 PDT (newsspool2.news.pas.earthlink.net) Xref: lexi2.athghost7038suus.net sci.physics:76016 sci.math:34819 OOPS! Forgot to include the program in my last post. Notice the "dx" in it as it is setup so that a small change, which is to take away the (int) cast will shift it to the continuous field. Then its just a matter of letting your dx approach 0. Thing is the number of calculations for small dx get HUGE rather quickly. I'm VERY interested in some experts on numerical methods for evaluating partial differential equations stepping forward, but my attempts so far to get an answer in that area from the math world have failed. My guess is that it would be a slamdunk in my favor, so they refuse to talk about it. James Harris ------------------------------------ #include #include double S(double x, double yin); double pi(double xin, double yin) { return ((int)xin-S(xin,yin)-1); } double min(double x, double y) { return x>y?y:x; } double S(double x, double yin) { double sum=0, i, sum1, sum2, dx=1; for(int i=2; i<=yin; i++) { sum1 = ( pi(x/i,min(i-dx,sqrt(x/i))) - pi(i-dx,sqrt(i-dx))) ; sum2 = ( pi(i,sqrt(i)) - pi(i-dx,sqrt(i-dx))); sum+=(sum1 * sum2); } return sum; } int main() { int input; cout <<"Input positive integer"<>input; cout << pi(input,sqrt(input)) << endl; return 0; }