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Two problems

  • Subject: Two problems
  • From: Richard Petti <petti at ALLEGHENY.SCRC.Symbolics.COM>
  • Date: Thu, 22 Jun 89 13:51:01 -0500
  • Apparently-to: mathgroup-out at

    Date: Wed, 21 Jun 89 00:35:34 -0500
    From: Dallas_Burtraw at

    Greetings.  I am in the Economics Dept at Michigan and have been using
    Mathematica for a little while.  Two questions have surfaced that we
    have not been able to resolve.  Can you help?
	    First, I want to attach a penalty in a function within a program, so
    for example, I always want x > xo. The number x is a real.  I want to write
    ((x - xo)^2)^(.5) - (x - xo) which is 0 for x nonnegative and greater than
    0 for x negative.  The problem is Mathematica automatically simplifies this
    algebra.  Is there a way to prevent that?

I can only suggest how these problems are solved using MACSYMA. Perhaps
Mathematica has similar features. MACSYMA is quite careful about assumptions
like SQRT(X^2) -> ABS(X), and has a data base for storing facts about
symbolic variables. It helps in this case:

(C1) F1:SQRT((X-X0)^2)-(X-X0);

(D1) 							 ABS(X0 - X) + X0 - X

(C2) ASSUME(X-X0 > 0);

(D2) 							       [X > X0]

(C3) F2:SQRT((X-X0)^2)-(X-X0);

(D3) 								   0

(C4) FORGET(X-X0 > 0);

(D4) 							       [X > X0]

(C5) F3:SQRT((X-X0)^2)-(X-X0);

(D5) 							 ABS(X0 - X) + X0 - X

(C6) ASSUME(X-X0 < 0);

(D6) 							       [X0 > X]

(C7) F4:SQRT((X-X0)^2)-(X-X0);

(D7) 							      2 X0 - 2 X

(C8) FORGET(X-X0 < 0);

(D8) 							       [X0 > X]

The MACSYMA data base can store many other types of properties and use them
in calculuations. For example consider the greatest integer function ENTIER:

(C16) ENTIER(3/2);

(D16) 								   1

(C17) ENTIER(Y);

(D17) 							       ENTIER(Y)


(D18) 								 DONE

(C19) ENTIER(Y);

(D19) 								   Y


    The second, more substantial question is the reason the first came up.
    It seems Mathematica can not use its problem solving techniques, either
    Solve or FindRoot, over conditional statements.  For example,
    f(x) = x if ....
    f(x) = x^2 otherwise.
    Writing such a conditional with If or Which, then FindRoot can not
    solve because it can not compute the derivatives.  Is there a way around
    this other than calculating maximization conditions separately and then
    rewriting a new conditional?  It seems like an easy fix within the 
    program if it truly is a bug.

    The real issue on this second question is the construction of a constrained
    maximization or Lagrangian problem.  It seems clumsy, but I have had to
    solve each f(x) = x, etc. and then test the "if" statement.  This wastes
    time and I lose any numerical value for the Lagrangian weight on the 
    constraint.  Do you have suggestions?

	    Regards, Dallas Burtraw
	    Dallas_Burtraw at

MACSYMA's ROOT_BY_BISECTION command handles functions, including compiled 
user-defined functions. The functions can include logical operations like
conditional branching (if-then) statements. Example:

(C1) F(X):=IF X > 0 THEN 2*(X-3) ELSE 3*(X-3);

(D1) 			    F(X) := IF X > 0 THEN 2 (X - 3) ELSE 3 (X - 3)


(D2) 						  [F]

(C3) ROOT_BY_BISECTION(F,0.0,10.0);

(D3) 						  3.0

Dick Petti
Computer Aided Mathematics Group
Symbolics, Inc.
email: petti at

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