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Re: [Q] Implementing identities as rules

  • To: mathgroup at
  • Subject: [mg18714] Re: [mg18608] [Q] Implementing identities as rules
  • From: Carl Woll <carlw at>
  • Date: Sat, 17 Jul 1999 02:36:36 -0400
  • Organization: Physics Department, U of Washington
  • References: <>
  • Sender: owner-wri-mathgroup at


I've seen some other responses to your question, but I have a different
approach which I think is more general. The basic idea is to give rules
for the derivatives of your function, rather than defining the function
itself. For your example we have

Derivative[1,0][f][x_,y_]:=2x f[x,y]
Derivative[0,1][f][x_,y_]:=2y f[x,y]

Now, we can try things like


2 x f[x, y]

2 y f[x, y]

4 x y f[x, y]

2 f[x, y] + 4 x  f[x, y]

                   2                 3
24 y f[x, y] + 48 x  y f[x, y] + 16 y  f[x, y] +

      2  3
  32 x  y  f[x, y]

Now, admittedly, one could get the same answers using the methods
proposed by the other respondents in this case. However, as a trivial
example, suppose your function is something like E^(x^2 +y^2)+x? Now, the
application of rules like E^(x^2+y^2)+x -> f[x,y] will not work (although
you can get around this by using the rule E^(x^2+y^2) -> f[x,y]-x).

Did you have a more complicated situation in mind then the example you
provided? If so, and if none of the proposed methods are able to help
you, please feel free to share it.

Carl Woll
Dept of Physics
U of Washington

Kevin Jaffe wrote:

> 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?  I know that if I try the
> rule
> In[3]:= %2 /. Exp[a_^2 + b_^2] :> f[a, b]
>              2    2
>             x  + y
> Out[3]= 2 E        x
> I get the original expression, because when the pattern is replace,
> f[a, b] is immediately evaluated to reproduce the original expression.
> (I know that the replacement occurs because if instead I use a rule
> whose right hand side cannot be evaluated further
> In[4]:= %2 /. Exp[a_^2 + b_^2] :> g[a, b]
> Out[4]= 2 x g[x, y]
> I get the desired result.)
> Is there a way to instruct Mathematica not to evaluate the expression
> after the replacement has been made?
> Thanks,
> kj0 at
> Get your FREE Email at

Carl Woll
Dept of Physics
U of Washington

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