Re: Have I found a bug?
- To: mathgroup at smc.vnet.net
- Subject: [mg84232] Re: Have I found a bug?
- From: Louise Hoffman <louise.hoffman at gmail.com>
- Date: Fri, 14 Dec 2007 14:59:14 -0500 (EST)
- References: <200712140855.DAA07725@smc.vnet.net> <fjtrgn$cv8$1@smc.vnet.net>
> Integrating k[x] by hand, you obtain : (1/2)*Log[Abs[1-x^2]] , which is = > (1/2)Log[1-x^2] if -1<x<1. > So, if -1<x<1, you obtain as final result : > E^(-(1/2)Log[1-x^2]]=1/Sqrt[1-x^2] > Is it so? Yes, exactly =) > Well, you have not to forget that if Abs[x]>1, then > 1/2)*Log[Abs[1-x^2]]=(1/2)*Log[x^2-1] and you would so obtain the same = > result as Mathematica. How did you figure out that Abs[1-x^2]=x^2-1 ? > The reason is very simple : "Integrate" gives a primitive which is = > defined only modulo an arbitrary constant. Mathematica works in the = > complex space, so you can for example choose the integrating constant = > I*Pi/2. This is very interesting. Is it possible to tell Mathematica to only work in R^2? > Integrating k[x], you would, with -1<x<1, have obtained : > (1/2)Log[Abs[1-x^2]]+I*Pi/2 = (1/2)Log[1-x^2]+I*Pi/2 > And so, your final result would be : > E^(-((1/2)Log[1-x^2]+I*Pi/2))=1/Sqrt[1-x^2]*E^(-I*Pi/2)=1/Sqrt[1-x^2]= > *(-I) > But as -1<x<1, Sqrt[1-x^2]=Sqrt[-1*(x^2-1)]=-I*Sqrt[x^2-1] > (because if a and b are negative, Sqrt[a*b]=-Sqrt[a]*Sqrt[b] ) > So you finally obtain : 1/(-I*Sqrt[x^2-1])*(-I)=1/Sqrt[x^2-1] > Which is also the same result as Mathematica. > > So I definitively think Mathematica is doing very well! I can sure see that now =) The strange thing though is, that 1/Sqrt[1-x^2] is the "right" answer, as I need it to bring a Chebyshev's ODE on SL form. Thank you very much for your detailed answer =) Hugs, Louise
- References:
- Have I found a bug?
- From: Louise Hoffman <louise.hoffman@gmail.com>
- Have I found a bug?