The title pretty much says it.
Ummmm,
Yes by definition I would think. If you know that A is equivalent to B, then B can be “simplified” to A simply by replacing B with A. Now there are some cases where attempts to solve a particular expression, such as the solution to certain differential equations, have failed, so we don’t know what certain expressions are equivalent to and thus can’t simplify them.
You’re going to have to define the problem more carefully. What sort of mathematical expressions do you have in mind, and what sort of simplifications are allowed? For example, sin(2x) and 2sin(x)cos(x) are equivalent expressions, but you can “simplify” one to the other only after recognizing their equality as one of the trigonometric identities [or as a consequence of the usual identity for sin(x+y)].
In various contexts, the existence of standard simplified forms is an interesting and often nontrivial result. Polynomial expressions (involving addition and multiplication but not division), for example, can be simplified to standard polynomial form, and then it is a theorem that two polynomials in standard form are equivalent if and only if their coefficients are equal. Logical expressions can be written in standard form as disjunctions of conjunctions, and then compared (this is useful in designing logic circuits).
Lists of identities, like the trig identities, are attempts to give a complete list of substitutions that will allow you to convert an expression of a given type into any equivalent expression. The attempt may or may not be successful.
So, again, what context do you have in mind?
I’m trying and failing to think of any circumstances where it would NOT be the case. For instance, since you know 1+1=2, then anywhere in an equation where you can put (2), you can put (1+1), or vice versa.
IANA mathmetician, though.
The specific problem I have in mind is this:
(2(3sqrt(3)-5))^(1/3)-(2(3sqrt(3)+5))^(1/3)
Can this expression be simplified to -2?
Usually, homework cannot be simplified any further around here.
That big equation looks like the kind of thing that comes out when you solv a cubic equation.
For the specific question, Maple tells me that your expression is nearly equal to -2, and I could see the difference being rounding error. But it can’t evaluate the expression symbolically–maybe you could get somewhere with the binomial theorem.
Forgot the second part…
For the general question, you need to specify what you’re talking about. I think you can phrase the question this way: given two expressions A and B with A = B, is there a definable sequence of definable functions <f[sub]n[/sub]> such that for some k, f[sub]k/sub = B?
I suspect the answer is no, but I don’t know for sure.
Mathematica simplified it to -2. Just FYI.
I was unable to reproduce this feat by hand though.
PS I didn’t try that hard.
OK, now I tried and did not find it that hard to reproduce it by hand. Just tedious.
Do you mean you reproduced the numerical calculations to get -2, or that you used algebraic rules to simplify it?
FWIW, I don’t think it’d be that hard to simplify algebraically, simply, as you said, tedious.
Yes, that’s where I got it. I made a program for my calculator to solve cubic equations, and the formula I gave is based on my calculator’s output for the real root of x^3+x^2+x+1. The entire expression was over three, and included a -1 at the end, so I got rid of them to get an expression equal to -2. By factoring, it can be shown that the real root is exactly equal to -1, and so my modified fomula is exactly equal to -2. What I’m looking for is what ultrafilter described. More generally, I would like to know if it is always possible to find such a sequence of functions, for any pair of equivalent expressions.
How’d you do it?
a = 3*sqrt(3) - 5
b = (2*a)[sup]1/3[/sup]
c = 3*sqrt(3) + 5
d = (2*c)[sup]1/3[/sup]
b - d = -2
I cubed the equation, simplified and took the cube root to get -2. Algebraic manipulation. Though it is much easier to simplify if you note that
(sqrt(3)-1)^3 = 6 sqrt(3) - 10
and
(sqrt(3)+1)^3 = 6 sqrt(3) + 10
Using this it is trivial.
Really depends what you mean by “equivalent” and “reduce”. If you’re dealing with a formal system with a notion of equivalence generated as the symmetric closure of the transitive closure of a notion of one-step reduction, then that notion of reduction is called “Church-Rosser” if whenever
A->B and A->C
there exists a D such that
B->D and C->D
In this case, yes, given any two “equivalent” expressions (connected by a finite sequence of expressions, each of which either one-step reduces to or is a one-step reduction of the next) there is another expression such that each of the two equivalent expressions reduces to it. In general, though, no.
As others have said, 1. the question may be ambiguous, and 2. on some interpretations, the question answers itself in the affirmative: if two expressions are equivalent, one can be “simplified” to the other. Perhaps you mean to get at questions like “if two expressions are equal, does there always exist a proof that they are equal?” I think that the answer to this depends on what kind of expressions you allow, and may be negative due to the incompleteness theorem (e.g., it may be that Chaitin’s constant is equal to a certain expression, but there exists no proof of this). I presume that if we limit ourselves to, say, expressions involving +, -, *, /, powers, and integers, which covers anything you’ll get from the cubic equations for a polynomial with rational coefficients, the answer is yes. I can’t offer a proof off the top of my head, but I would be very interested in seeing a counterexample.
As for the Church-Rosser condition, isn’t that guaranteed to hold if all “reductions” work both ways, i.e., if whenever A->B, B->A? Isn’t that always the case with the sort of transformation of expressions we’re talking about?
No. Consider Church’s ?-calculus with his notion of one-step ?-reduction:
(?x.M)N ->[sub]?[/sub] [N/x]M
(where the right side is metalanguage for substituting N for each free occurrence of x in M). For example:
(?x.xx)(?y.zy) ->[sub]?[/sub] (?y.zy)(?y.zy) ->[sub]?[/sub] z(?y.zy)
or, by the transitive closure:
(?x.xx)(?y.zy) ->>[sub]?[/sub] z(?y.zy)
but it is not the case that
z(?y.zy) ->>[sub]?[/sub] (?x.xx)(?y.zy)
?-equivalence is the symmetric closure of this transitive closure of one-step ?-reduction.