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MathGroup Archive 2004

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AW: Infrequent Mathematica User

  • To: mathgroup at smc.vnet.net
  • Subject: [mg47059] AW: [mg47048] Infrequent Mathematica User
  • From: Matthias.Bode at oppenheim.de
  • Date: Mon, 22 Mar 2004 22:39:16 -0500 (EST)
  • Sender: owner-wri-mathgroup at wolfram.com

Hello Jim,

applying the KIS principle, try:

Clear[x, y, z]
func = x/(1 + x^2) + 
    y/(1 + x^2 + y^2) + 
    z/(1 + x^2 + y^2 + z^2); 
dfx = D[func, x]; 
dfy = D[func, y]; 
dfz = D[func, z]; 
Reduce[dfx == 0 && 
   dfy == 0 && dfz == 0, x]
N[%]

Best regards,
Matthias Bode
Sal. Oppenheim jr. & Cie. KGaA
Koenigsberger Strasse 29
D-60487 Frankfurt am Main
GERMANY
Tel.: +49(0)69 71 34 53 80
Mobile: +49(0)172 6 74 95 77
Fax: +49(0)69 71 34 95 380
E-mail: matthias.bode at oppenheim.de
Internet: http://www.oppenheim.de



-----Ursprüngliche Nachricht-----
Von: Jim Dars [mailto:jim-dars at comcast.net]
Gesendet: Montag, 22. März 2004 11:19
An: mathgroup at smc.vnet.net
Betreff: [mg47048] Infrequent Mathematica User


Hi All, (second post, first didn't display)

f is defined below as a function of x, y, and z.
I wish to take the partials set to zero and solve the 3 equations for x, y,
and z.
I've copied from Mathematica and had to clean up the paste, a bit.  I used
the partial symbol from the palette to define my partial derivatives.  The 3
lines on this page look nothing like what appeared when using Mathematica
to obtain the partials.  I've tried the "Solve equation" with just "a" and
a[x_,y_,z_] etc.
Mathematica replies "{{}}".
I sure would appreciate some advice.

Thanks, Best wishes, Jim
Jim-Dars at comcast.net

f[x_, y_, z_] =
      x/(1 + x^2) + y/(1 + x^2 + y^2) +
        z/(1 + x^2 + y^2 + z^2);
  a[x_, y_, z_] = \[PartialD]\_x f;\)\[IndentingNewLine]
  b[x_, y_, z_] = \[PartialD]\_y f;\)\[IndentingNewLine]
  c[x_, y_, z_] = \[PartialD]\_z\ f;\)\[IndentingNewLine]
  Solve[{a[x_, y_, z_] == 0, b[x_, y_, z_] == 0, c[x_, y_, z_] == 0}, {x, y,
      z}]


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