Re: DirectedInfinity[1 + I], why does it get replaced by (1 + I)/(sqrt(2) ?
- To: mathgroup at smc.vnet.net
- Subject: [mg55823] Re: [mg55802] DirectedInfinity[1 + I], why does it get replaced by (1 + I)/(sqrt(2) ?
- From: DrBob <drbob at bigfoot.com>
- Date: Thu, 7 Apr 2005 05:10:02 -0400 (EDT)
- References: <200504060711.DAA13652@smc.vnet.net>
- Reply-to: drbob at bigfoot.com
- Sender: owner-wri-mathgroup at wolfram.com
> I realize that (1 + I)/sqrt(2) is in the same direction as (1 + I), > but why did Mathematica change it to the more 'strange' form? Because directions have unit length. Norm[(1 + I)/Sqrt[2]] 1 DirectedInfinity[2 + I*3] First[%] Norm[%] DirectedInfinity[(2 + 3*I)/Sqrt[13]] (2 + 3*I)/Sqrt[13] 1 Bobby On Wed, 6 Apr 2005 03:11:59 -0400 (EDT), Matt <anonmous69 at netscape.net> wrote: > Hello, > This isn't particularly important probably, however, I am trying to > learn as much about Mathematica as possible, and I thought this might > shed some light on a 'Why Mathematica does this or that' principle. > > I'm working my way through 'The Mathematica Guidebook for > Programming' and on page 177, he gives an example as follows: > > In[39]:= DirectedInfinity[1 + I] DirectedInfinity[I] > Out[39]:= DirectedInfinity[-(1 - I)/sqrt(2)] > > That puzzled me a bit, so I decided to see what Mathematica would do > with just the first part: > > In[40]:= DirectedInfinity[1 + I] > Out[40]:= DirectedInfinity[(1 + I)/sqrt(2)] > > I realize that (1 + I)/sqrt(2) is in the same direction as (1 + I), but > why did Mathematica change it to the more 'strange' form? > > Thanks, > > Matt > > > > -- DrBob at bigfoot.com
- References:
- DirectedInfinity[1 + I], why does it get replaced by (1 + I)/(sqrt(2) ?
- From: "Matt" <anonmous69@netscape.net>
- DirectedInfinity[1 + I], why does it get replaced by (1 + I)/(sqrt(2) ?