Mathematica Precision

*To*: mathgroup at smc.vnet.net*Subject*: [mg8489] Mathematica Precision*From*: David Djajaputra <dd4b at virginia.edu>*Date*: Tue, 2 Sep 1997 16:15:31 -0400*Organization*: University of Virginia*Sender*: owner-wri-mathgroup at wolfram.com

--------------E7F4C761F09A86D88F419661 To all Mathematica lovers out there, I would really appreciate it if anyone can give me helpful comment on this problem. I need to work with the following equation: Solve[a(x + y) + c (Sqrt[1 - x^2] + Sqrt[1 - y^2])== e, y] It gives two roots. Fine. I then define one root as a new function: Arguemin[x_, a_, c_, e_] := (a e -a^2 x - a c Sqrt[1 - x^2] -Sqrt[-c^2 e^2 + 2 a c^2 e x + c^4 x^2 + 2 c^3 e Sqrt[1-x^2] - 2 a c^3 Sqrt[1-x^2] + a^2 c^2 (1 - x^2)])/(a^2 + c^2) and plug it back into the original equation: Checking[x_, a_, c_, e_]:= a( x + Arguemin[x,a,c,e]) + c (Sqrt[1 - x^2] + Sqrt[1 -(Arguemin[x,a,c,e])^2]) I set the precision to infinity: $MaxExtraPrecision = Infinity If I run Checking[x,a,c,e], it should give me result=e, right? (In my problem, a=1-Cos(s) and c=Sin(s), and I choose s=0.1 here.) Checking[Cos[0.4], 1-Cos[0.1], Sin[0.1], 0.1] 0.100252 It doesn't. For other values of s, it is even worse. Couldn't Mathematica do better than this? Much thanks in advance, David --------------E7F4C761F09A86D88F419661 <HTML> To all Mathematica lovers out there, <P>I would really appreciate it if anyone can give me helpful comment on this problem. <BR>I need to work with the following equation: <PRE>Solve[a(x + y) + c (Sqrt[1 - x^2] + Sqrt[1 - y^2])== e, y]</PRE> It gives two roots. Fine. I then define one root as a new function: <PRE>Arguemin[x_, a_, c_, e_] := (a e -a^2 x - a c Sqrt[1 - x^2] -Sqrt[-c^2 e^2 + 2 a c^2 e x + c^4 x^2 + 2 c^3 e Sqrt[1-x^2] - 2 a c^3 Sqrt[1-x^2] + a^2 c^2 (1 - x^2)])/(a^2 + c^2)</PRE> and plug it back into the original equation: <PRE>Checking[x_, a_, c_, e_]:= a( x + Arguemin[x,a,c,e]) + c (Sqrt[1 - x^2] + Sqrt[1 -(Arguemin[x,a,c,e])^2])</PRE> I set the precision to infinity: <PRE>$MaxExtraPrecision = Infinity</PRE> If I run Checking[x,a,c,e], it should give me result=e, right? (In my problem, a=1-Cos(s) and c=Sin(s), and I choose s=0.1 here.) <PRE>Checking[Cos[0.4], 1-Cos[0.1], Sin[0.1], 0.1]</PRE> <PRE>0.100252</PRE> It doesn't. For other values of s, it is even worse. Couldn't Mathematica do better than <BR>this? <BR> <P>Much thanks in advance, <P>David</HTML> --------------E7F4C761F09A86D88F419661--