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MathGroup Archive 2005

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Re: DirectedInfinity[1 + I], why does it get replaced by (1 + I)/(sqrt(2) ?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg55815] Re: [mg55802] DirectedInfinity[1 + I], why does it get replaced by (1 + I)/(sqrt(2) ?
  • From: Bob Hanlon <hanlonr at cox.net>
  • Date: Thu, 7 Apr 2005 05:09:54 -0400 (EDT)
  • Reply-to: hanlonr at cox.net
  • Sender: owner-wri-mathgroup at wolfram.com

To normalize the direction to a unit vector.

Abs[(1+I)/Sqrt[2]]

1


Bob Hanlon

> 
> From: "Matt" <anonmous69 at netscape.net>
To: mathgroup at smc.vnet.net
> Date: 2005/04/06 Wed AM 03:11:59 EDT
> To: mathgroup at smc.vnet.net
> Subject: [mg55815] [mg55802] DirectedInfinity[1 + I], why does it get replaced by (1 + 
I)/(sqrt(2) ?
> 
> 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
> 
> 


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