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MathGroup Archive 1996

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Re: Bug??

  • To: mathgroup at smc.vnet.net
  • Subject: [mg3809] Re: Bug??
  • From: rhall2 at umbc.edu (hall robert)
  • Date: Sat, 27 Apr 1996 00:55:50 -0400
  • Organization: University of Maryland, Baltimore County
  • Sender: owner-wri-mathgroup at wolfram.com

In article <4l794q$1ov at ralph.vnet.net>,
Theo Pillay <pillay12 at shrike.und.ac.za> wrote:
>
>f[x,y] = a[x] y + b[x];
>
>F1 = f[x,y];
>
>F2 = y2 (D[f[x,y], y]  - D[r[x], x]) + D[f[x,y],x];
>
>F3 = r[x];
>
>eight = -y2 A[x,y] D[F2,y] - 2 y2 B[x,y,y2] D[F3,y] +
>        2 y2 D[F2,y,x] + y2^2 D[F2,{y,2}] - A[x,y] D[F2,x] +
>        B[x,y,y2] D[F2,y2] -
>        2 B[x,y,y2] D[F3,x] - D[B[x,y,y2], y] F1 - D[B[x,y,y2], y2] F2
>        - D[B[x,y,y2],x] F3 + D[F2, {x,2}];  (* 1 *)
>
>I get different answers for eight depending on the position of line (1). As
>shown I get the wrong answer, but it works if I move (1) up to the previous
>line (i.e next to F2).

The difference in output is the term 
	-2 B[x, y, y2] r'[x]
which comes from
	-2 B[x,y,y2] D[F3,x]
The problem is the derivative. If F3 isn't defined, then
	D[F3,x] = 0.
Placing line 1 before the definition
	F3 = r[x]
causes the entire term to evaluate to zero. This seems to
be the result you want.

If F3 is defined as the function r[x], then Mathematica
treats it as it would any other symbolic function, and 
stores the result as
	r'[x].

	F3 mates with x to survive. Call it marriageable.
	Differentiation kills the bachelor variable.

-- 
Bob Hall            | "Know thyself? Absurd direction!
rhall2 at gl.umbc.edu  |  Bubbles bear no introspection."  -Khushhal Khan Khatak

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