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MathGroup Archive 1999

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Re: [Q] Implementing identities as rules

  • To: mathgroup at smc.vnet.net
  • Subject: [mg18678] Re: [mg18608] [Q] Implementing identities as rules
  • From: BobHanlon at aol.com
  • Date: Thu, 15 Jul 1999 01:45:58 -0400
  • Sender: owner-wri-mathgroup at wolfram.com

A definition is always applied.  Consequently, avoid the definition.  One 
approach for symbolic manipulation is as follows:

Notebook[{
Cell[BoxData[
    \(\(toEqn[a_\  -> \ b_]\  := \ \((a\  == \ b)\);\)\)], "Input"],

Cell[BoxData[
    \(\(eqn1\  = \ f[x, \ y]\  == \ Exp[x^2\  + \ y^2];\)\)], "Input"],

Cell[CellGroupData[{

Cell[BoxData[
    \(eqn2\  = \ D[eqn1, \ x]\)], "Input"],

Cell[BoxData[
    RowBox[{
      RowBox[{
        SuperscriptBox["f", 
          TagBox[\((1, 0)\),
            Derivative],
          MultilineFunction->None], "[", \(x, y\), "]"}], 
      "==", \(2\ \[ExponentialE]\^\(x\^2 + y\^2\)\ x\)}]], "Output"]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
    \(eqn3\  = \ 
      eqn2\  /. \ \((subst1\  = \ 
            eqn1[\([2]\)]\  -> \ eqn1[\([1]\)])\)\)], "Input"],

Cell[BoxData[
    RowBox[{
      RowBox[{
        SuperscriptBox["f", 
          TagBox[\((1, 0)\),
            Derivative],
          MultilineFunction->None], "[", \(x, y\), "]"}], 
      "==", \(2\ x\ f[x, y]\)}]], "Output"]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
    \(eqn4\  = \ D[eqn1, \ y]\)], "Input"],

Cell[BoxData[
    RowBox[{
      RowBox[{
        SuperscriptBox["f", 
          TagBox[\((0, 1)\),
            Derivative],
          MultilineFunction->None], "[", \(x, y\), "]"}], 
      "==", \(2\ \[ExponentialE]\^\(x\^2 + y\^2\)\ y\)}]], "Output"]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
    \(eqn5\  = \ eqn4\  /. \ subst1\)], "Input"],

Cell[BoxData[
    RowBox[{
      RowBox[{
        SuperscriptBox["f", 
          TagBox[\((0, 1)\),
            Derivative],
          MultilineFunction->None], "[", \(x, y\), "]"}], 
      "==", \(2\ y\ f[x, y]\)}]], "Output"]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
    \(\(Solve[{eqn3, \ eqn5}, \ eqn2[\([1]\)], \ eqn1[\([1]\)]]\)[\([1, \ 
          1]\)]\  // \ toEqn\)], "Input"],

Cell[BoxData[
    RowBox[{
      RowBox[{
        SuperscriptBox["f", 
          TagBox[\((1, 0)\),
            Derivative],
          MultilineFunction->None], "[", \(x, y\), "]"}], "==", 
      FractionBox[
        RowBox[{"x", " ", 
          RowBox[{
            SuperscriptBox["f", 
              TagBox[\((0, 1)\),
                Derivative],
              MultilineFunction->None], "[", \(x, y\), "]"}]}], 
        "y"]}]], "Output"]
}, Open  ]]
},
FrontEndVersion->"4.0 for Macintosh",
ScreenRectangle->{{0, 1024}, {0, 748}},
WindowSize->{520, 626},
WindowMargins->{{Automatic, 118}, {Automatic, 4}},
ShowCellLabel->False,
MacintoshSystemPageSetup->"\<\
00<0001804P000000]P2:?oQon82n at 960dL5:0?l0080001804P000000]P2:001
0000I00000400`<300000BL?00400 at 0000000000000006P801T1T00000000000
00000000000000000000000000000000\>"
]


Bob Hanlon

In a message dated 7/13/99 7:26:26 AM, kj0 at mailcity.com writes:

>During symbolic manipulations it often important to be able to use
>both sides of an identity or definition.  For example, let f be a
>function defined by
>
>In[1]:= f = Function[{x, y}, Exp[x^2 + y^2]];
>
>Its partial derivative with respect to x is:
>
>In[2]:= D[f[x, y], x]
>
>            2    2
>           x  + y
>Out[2]= 2 E        x
>
>Now, I want to recast this result in the form 2 x f[x, y], i.e. I want
>to revert to the "left-hand side" of the original definition of f[x,
>y].  How does one do this in Mathematica?
>


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