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MathGroup Archive 2006

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Re: Possible Bug in ArcTan ?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg64832] Re: Possible Bug in ArcTan ?
  • From: "Jens-Peer Kuska" <kuska at informatik.uni-leipzig.de>
  • Date: Sun, 5 Mar 2006 03:18:50 -0500 (EST)
  • Organization: Uni Leipzig
  • References: <du6o44$5rg$1@smc.vnet.net> <du83m5$sv3$1@smc.vnet.net> <du8are$fp7$1@smc.vnet.net> <dubgv0$fm7$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Paul,

why you want avoid the two-argument form. The two-argument form help
a lot during programing, because one has not to type

If[x=!=0,ArcTan[y/x]]

and a division by zero is in the most programing languages
a very hard and evil error.

Regards
  Jens

"Paul Abbott" <paul at physics.uwa.edu.au> schrieb im 
Newsbeitrag news:dubgv0$fm7$1 at smc.vnet.net...
| In article <du8are$fp7$1 at smc.vnet.net>,
| "David W. Cantrell" <DWCantrell at sigmaxi.org> 
wrote:
|
| > "Jens-Peer Kuska" 
<kuska at informatik.uni-leipzig.de> wrote:
| > > Hi,
| > >
| > > why can ArcTan[] have two arguments 
ArcTan[x,y]
| >
| > As the Help Browser says,
| > "taking into account which quadrant the point 
(x,y) is in."
| >
| > For example, suppose you want to convert -2 + 
I to polar form,
| > r*E^(I*theta). One can't simply say theta = 
ArcTan[1/-2]. The range of the
| > single-argument ArcTan is [-Pi/2, Pi/2], i.e., 
fourth and first quadrants,
| > while our point (-2, 1) is in the third 
quadrant. But we can conveniently
| > say theta = ArcTan[-2, 1].
|
| Or we can say theta = 2 ArcTan[1/(Sqrt[5]-2)] 
(see below).
|
| > Note: Some other languages implement the 
two-argument form under the
| > name ATAN2. Furthermore, the order of the two 
arguments is often backwards
| > compared to Mathematica's, that is, 
ATAN2(y,x).
|
| There is another way that avoids the 
two-argument form altogether. Using
| the half-angle formula for tan,
|
|  Simplify[Tan[t/2] == Sin[t]/(Cos[t] + 1)]
|
|  True
|
| then in polar coordinates, x=r Cos[t], y=r 
Sin[t], r=Sqrt[x^2+y^2],
|
|  Tan[t/2] == y/(x+r) ==> t == 2 
ArcTan[y/(x+Sqrt[x^2+y^2])]
|
| This formula is also valid when x == 0, whereas 
ArcTan[y/x] is
| problematic there (ArcTan[0,y] is ok, of 
course).
|
| Cheers,
| Paul
|
| 
_______________________________________________________________________
| Paul Abbott 
Phone:  61 8 6488 2734
| School of Physics, M013 
Fax: +61 8 6488 1014
| The University of Western Australia 
(CRICOS Provider No 00126G)
| AUSTRALIA 
http://physics.uwa.edu.au/~paul
| 



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