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From: Bill Rowe on 7 Jan 2010 02:32 On 1/6/10 at 5:59 AM, fateman(a)cs.berkeley.edu (Richard Fateman) wrote: >For example, Exp[I x] -Exp[- I x] /. Exp[I x] -> s should >probably result in s-1/s. In Mathematica, one gets s-E^(-Ix). >can either (1) Make this come out s-1/s >or >(2) Argue that Mathematica already does the right thing, blame the >user, blame the documentation, blame the nature of mathematics, >claim that it is impossible to "read the user's mind" etc. The argument isn't that Mathematica does the "right" thing. "Right" is quite subjective here. The result Mathematic does return for this example is clearly mathematically correct even if it is not the desired result. The real choice is either (1) understand how Mathematica is designed and make use of that to get the result you want or (2) find another software package that works more like what you want. Arguing Mathematic does either the "wrong thing" or the "right thing" here is pointless. >To me, the question is simply, by what programming technique >can we make Mathematica do the truly expected thing. In this >case, and I >believe in every other case, a transformation of the rules will >help. In particular, using the rule >x-> -I Log[s] instead of Exp[x I] -> s. >Is it possible that Mathematica could make this change? How could >it possibly make such a transformation? (hint. Solve for variable s) =46or me, I never want Mathematica designed in a way where it tries to guess or otherwise divine my intention and do something other than what I specifically told it to do via my input. I totally detest any software that does that. Which is one of the main reasons a greatly dislike Microsoft Word. >For another example, x/5 /. 1/5->Fifth results in Fifth x but >3/5 /. 1/5 -> Fifth is unchanged. >Is it possible that Mathematica could do this consistently? What is inconsistent here? Mathematica internally evaluates 1/5 as Rational[1,5], x/5 as Times[x, Rational[1,5]] and 3/5 as Rational[3,5]. In every case Rational[1,5] is being replaced with Fifth. Mathematica behaves in a consistent manner even though this is clearly not immediately apparent to a new user. Again, the choice is either understand this behavior and live with it or find different software. There isn't any other productive choice. Any software package that comes close to approximating the capability Mathematica offers has to make some set of design decisions. It simply is not possible to make those decisions in a manner that will please all potential users or not cause some level of confusion to a new user. This level of capability will always require significant learning on the part of any user to master. So, the real choice is learn and understand the way Mathematica works and live with it or find another software package more to your likely. There is no other productive choice.
From: Noqsi on 7 Jan 2010 02:32 On Jan 6, 3:59 am, Richard Fateman <fate...(a)cs.berkeley.edu> wrote: > > For example, Exp[I x] -Exp[- I x] /. Exp[I x] -> s should p= robably > result in s-1/s. > In Mathematica, one gets s-E^(-Ix). You're confusing two different kinds of "substitution". It is extremely important to preserve the distinction. > > can either > (1) Make this come out s-1/s > or > (2) Argue that Mathematica already does the right thing, blame the user, > blame the documentation, blame the nature of mathematics, claim that it > is impossible to "read the user's mind" etc. Or blame those who don't understand critical distinctions and want to erase them. > > To me, the question is simply, by what programming technique can we make > Mathematica do the truly expected thing. Well, in this case it does what anybody who understands how Replace works and what it's good for expects. If that's not the "truly expected thing", I don't know what is. > In this case, and I believe in every other case, a transformation of the > rules will help. In particular, using the rule > x-> -I Log[s] instead of Exp[x I] -> s. Mathematica has a tool that can do what you want here: Reduce[a == Exp[I x] - Exp[-I x] && Exp[I x] == s, {a}, {x}] s != 0 && a == (-1 + s^2)/s It's a bit fussier than perhaps you'd like, but that's mathematics for you ;-)
From: Vince Virgilio on 7 Jan 2010 02:33 On Jan 6, 5:59 am, Richard Fateman <fate...(a)cs.berkeley.edu> wrote: SNIP > > What I would say is > > Because of the disparity in internal representations (see below) , > for replacements on parts of complex expressions, don't use /. ... > use /. BetterRules[ ...] > This expects too much from ReplaceAll. > or use transformation or selection programs such as Conjugate, Im, Re, .. . > Yes. Not unlike something Andrzej said elsewhere about PolynomialReduce (others said similar as well). SNIP Vince Virgilio ( Incidentally, my FullForm of x+4*I differs from yours, if I read yours correctly: In[1]:= x+4I//FullForm Out[1]//FullForm= Plus[Complex[0,4],x] )
From: David Bailey on 7 Jan 2010 02:34 Richard Fateman wrote: > I think it is interesting that the same issue came up in the design of > another computer algebra system, years ago. > That is, which objects are "atomic" and which are decomposable for > purposes of substitution. And further, > of those which are decomposable, how much cleverness should be applied > during substitution > > For example, Exp[I x] -Exp[- I x] /. Exp[I x] -> s should probably > result in s-1/s. > In Mathematica, one gets s-E^(-Ix). > > can either > (1) Make this come out s-1/s > or > (2) Argue that Mathematica already does the right thing, blame the user, > blame the documentation, blame the nature of mathematics, claim that it > is impossible to "read the user's mind" etc. > > To me, the question is simply, by what programming technique can we make > Mathematica do the truly expected thing. Notice that using the transformation rule Exp[I x] -> s (or f[x]->s in general) in the way you require, involves inverting it to produce x->g[s] for some g. In general g may not be unique, which is why the following code generates a warning, but essentially does what you want to do: Solve[Eliminate[{ans == Exp[I x] - Exp[-I x], Exp[I x] == s}, {x}], ans] Reduce (rather than Eliminate) yields a more mathematically precise answer, but the result is considerably more clumsy. Note also that ReplaceAll *can* be used to do mathematical operations without complications provided the LHS of each rule is a variable (but not a constant such as I, Pi, etc). David Bailey http://www.dbaileyconsultancy.co.uk
From: Richard Fateman on 8 Jan 2010 04:15
Noqsi wrote: > > Mathematica has a tool that can do what you want here: > > Reduce[a == Exp[I x] - Exp[-I x] && Exp[I x] == s, {a}, {x}] > s != 0 && a == (-1 + s^2)/s > > It's a bit fussier than perhaps you'd like, but that's mathematics for > you ;-) > Reduce is a really neat program in Mathematica, one that I especially admire since it was improved to work on more than polynomials. Unfortunately, it won't work for I->-I, maybe because that is based on a decision involving constants represented differently from expressions that construct them. (David Bailey already pointed this out. Do people get instantaneous unfiltered feed from this newsgroup??) but also.. Bill Rowe said ... "Again, the choice is either understand this behavior and live with it or find different software. There isn't any other productive choice." Well, reporting something as a bug and hoping it will be fixed is another choice. And writing a version of the facility that does the right thing is another choice. (Any takers?) Either of these could be "productive". Are Mathematica design decisions sacred or something? RJF |