http://www.mapleprimes.com/posts/39857-16-Years-Later en-us 2026 Maplesoft, A Division of Waterloo Maple Inc. Maplesoft Document System Thu, 11 Jun 2026 05:30:38 GMT Thu, 11 Jun 2026 05:30:38 GMT The latest comments added to the Post, 16 years later http://www.mapleprimes.com/images/mapleprimeswhite.jpg http://www.mapleprimes.com/posts/39857-16-Years-Later http://www.mapleprimes.com/posts/39857-16-Years-Later?ref=Feed:MaplePrimes:16 years later:Comments#comment73033 <p>You had a UUCP link to a server that bridged to internet?</p> The latest comments added to the Post, 16 years later 73033 Thu, 06 Mar 2008 05:12:10 Z jakubi jakubi http://www.mapleprimes.com/posts/39857-16-Years-Later?ref=Feed:MaplePrimes:16 years later:Comments#comment84002 <p>That is a correct answer to question a.&nbsp; Of course watmath.uwaterloo.edu doesn't exist anymore, it is now math.waterloo.ca.&nbsp; The IBM AIX machine that as called wmsical is also long long gone - but it served as Maplesoft's main mail server for many years.</p> The latest comments added to the Post, 16 years later 84002 Thu, 06 Mar 2008 05:19:35 Z JacquesC JacquesC http://www.mapleprimes.com/posts/39857-16-Years-Later?ref=Feed:MaplePrimes:16 years later:Comments#comment83999 I do not know if improvement, but I would have made it otherwise, calculating by residues: <pre> with(IntegrationTools): with(plots): J := Int(1/(y^4-2*y^3+2*I*r*y^2-2*I*r*y-r^2),y=-1..1); f:=GetIntegrand(J); s:=singular(f,y); </pre> <maple> s := {y = (-I*r)^(1/2)}, {y = -(-I*r)^(1/2)}, {y = 1-(1-I*r)^(1/2)}, {y = 1+(1-I*r)^(1/2)}; </maple> Numerically: <pre> L:=map(x->op([1,2],x),[s]); L1:=eval(L,r=0.1); L1 := [.2236067978-.2236067978*I, -.2236067978+.2236067978*I, -.1246114e-2+.4993777184e-1*I, 2.001246114-.4993777184e-1*I] </pre> And ploted: <pre> complexplot(L1,view=[-3..3,-0.3..0.3],style=point,symbol=solidbox,color=red,axes=normal); </pre> I choose as path on the complex plane the segment on the real axis (-1,1), the straight lines from its end points down to +/1-I*infinity, and closed at infinity. It encloses the first pole above, and its residue times (2*I*Pi) can be calculated as: <pre> f1:=factor(f); (2*I*Pi)*residue(f1,op(s[1])); </pre> <maple> 1/2*Pi/r </maple> With a minus sign gives the value of the integral along this path made in clockwise sense. Now, the contribution at infinity and the half infinity stright lines is finite and can be expanded in powers of r (easy to do). Hence the leading contribution to J as r->0 is <maple> -1/2*Pi/r </maple> The latest comments added to the Post, 16 years later 83999 Thu, 06 Mar 2008 08:22:39 Z jakubi jakubi http://www.mapleprimes.com/posts/39857-16-Years-Later?ref=Feed:MaplePrimes:16 years later:Comments#comment92938 <p>All you need to do nowadays is</p> <p>&gt; int(1/(y^4-2*y^3+2*I*r*y^2-2*I*r*y-r^2),y=-1..1) assuming r &gt; 0;</p> <p><maple>-1/8*(ln(r^2+9)*(-1+r*I)^(1/2)*I+2*arctan(3/r)*(-1+r*I)^(1/2)+4*I*arctan(2/(-1+r*I)^(1/2))-I*ln(r^2+1)*(-1+r*I)^(1/2)+2*arctan(1/r)*(-1+r*I)^(1/2))/r/(-1+r*I)^(1/2)</maple></p> <p>&gt; map(normal, series(%,r)) assuming r &gt; 0;</p> <p><maple>series((-1/2*Pi)*r^(-1)+(-1/8*ln(3)-1/8*I*Pi+1/2)+(3/32*Pi-3/32*I*ln(3)+5/24*I)*r+(-25/144+5/64*I*Pi+5/64*ln(3))*r^2(-1415/10368*I-35/512*Pi+35/512*I*ln(3))*r^3+(-63/1024*ln(3)-63/1024*I*Pi+12223/103680)*r^4+O(r^5),r,5)</maple></p> <p>&nbsp;</p> The latest comments added to the Post, 16 years later 92938 Thu, 06 Mar 2008 09:10:43 Z Robert Israel Robert Israel http://www.mapleprimes.com/posts/39857-16-Years-Later?ref=Feed:MaplePrimes:16 years later:Comments#comment73018 <p>Was the corporate name Maplesoft coined by a woman?</p> <p>Dave Linder<br /> Mathematical Software, Maplesoft</p> The latest comments added to the Post, 16 years later 73018 Thu, 06 Mar 2008 10:02:27 Z Dave L Dave L http://www.mapleprimes.com/posts/39857-16-Years-Later?ref=Feed:MaplePrimes:16 years later:Comments#comment83995 <p>That probably gives it away...</p> The latest comments added to the Post, 16 years later 83995 Thu, 06 Mar 2008 18:09:41 Z JacquesC JacquesC