How to evaluate (V / F) an expression of the form # = #
10-31-2014, 01:49 PM
Post: #4
 DrD Senior Member Posts: 1,133 Joined: Feb 2014
RE: How to evaluate (V / F) an expression of the form # = #
I also encountered this in working with a problem just the other day, where it was explained that "==" is a test for "representational" equivalence, but not mathematical equivalence. As long as the operands are numerical, or symbolically the same, the test is usable.

For me, there IS room for confusion on "==." The user guide definition: "Equality test. Returns 1 if the left side and right side are equal, and 0 otherwise."

Two expressions will fail the test if the expressions are equivalent, BUT represented differently:

(√(-2.*cos(t)+2.))== (2*ABS(sin((1/2)*t))) => 0

to pass the '==' test, make an assumption on the variable (t), in effect converting them to a number:

(((√((-2.)*cos(t)+2.))==(2*ABS(sin((1/2)*t))))|(t )= (π/4)) => 1

I think the "==" test would be more beneficial if it operated as the user guide described, for both cases. Especially, in a programmatic context. I don't know if this is worthy of a bug report, though.

Maybe others will have additional thinking to share.

-Dale-
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 Messages In This Thread How to evaluate (V / F) an expression of the form # = # - compsystems - 10-28-2014, 02:13 AM RE: How to evaluate (V / F) an expression of the form # = # - Gilles - 10-31-2014, 11:07 AM RE: How to evaluate (V / F) an expression of the form # = # - compsystems - 10-31-2014, 11:34 AM RE: How to evaluate (V / F) an expression of the form # = # - DrD - 10-31-2014 01:49 PM RE: How to evaluate (V / F) an expression of the form # = # - Mark Hardman - 10-31-2014, 05:16 PM RE: How to evaluate (V / F) an expression of the form # = # - DrD - 10-31-2014, 06:27 PM

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