With this question, I am coming back to a long-standing issue I have with Mathematica's concept of a "generically correct" result. The basic idea seems to be, vaguely, that if a result is true in the vast majority of cases, then the Wolfram Language decides that it is o.k. to just substitute such a result instead of alerting the user to the fact that this result may be wrong "sometimes". I note in passing that, in the fields of mathematics I am most familiar with, the concept of "generically true" is rigorously defined as "true with the exception of a set of measure zero", after appropriate definition of the measure(s) one is talking about, and that no such definition seems to exist for the Wolfram Language.

Example 1: Let's look at the expression e1, defined via

e1 = Sin[2 Pi j]/j

If we ask Mathematica to simplify the expression for the case of integer j, we get 0:

Assuming[j ∈ Integers, Simplify[e1]]

Given what I said above, I am not particularly happy about the concept of "generically correct" revealed by this choice. Yes, this result is incorrect only for one single value of the integer, j = 0, and correct for an infinite number of other cases. However, if we consider the important case of finite subsets of the integers, then the case of a set with just one element is not particularly small. Note also that, if we ask for, say

Assuming[j ∈ Integers ∧ 0 <= j <= 1, Simplify[e1]]

we will get the same answer of 0, which clearly makes no sense at all now. By the way, the mathematically equivalent

Assuming[j == 0 ∨ j == 1, Simplify[e1]]

happens to produce a correct result of $\sin(2 \pi j)/j$.

Now, one might argue that it would be asking too much to expect Mathematica to go through all possible cases of a finite set designated similarly to the above before producing an answer, but consider my

Example 2:

e2 = Sin[2 Pi (j - k)]/(j - k)

and ask for

Assuming[j ∈ Integers ∧ k ∈ Integers, Simplify[e2]]

Again, Mathematica claims the result is zero, even though in this case it is clear that there can be no mathematical justification for doing so. Notice that the set of pairs {i, j} for which this result is correct (the set that has i<>j) contains exactly as countably infinite many elements as the set where it is false.

So, here is my question: Given the difficulties revealed by the above examples, is there any safe way to use assumptions in Mathematica, or should they just be avoided, period? Note that, in the above trivial cases, it's easy to see when we receive a wrong answer. However, imagine us having Assumptions hidden in some piece of code that accepts arbitrary expressions as its input. The consequence will be that in general we cannot know if such code will produce a valid answer.

Alternatively, is there a way to know exactly what conclusions the various Mathematica functions that accept assumptions might draw from such assumptions? In other words, is there, at least in principle, any way to rationally test code that contains assumptions?

P.S.: I will say that I find the statement in Mathematica's documentation that "It is however often not practical to try to get formulas that are valid for absolutely every possible value of each variable" thoroughly unconvincing. I frankly don't see an issue with not returning a definite result when such a result does not exist. I would very much prefer Mathematica to force me to state assumptions such that a definite result can be obtained, rather than it silently and implicitly introducing assumptions that may turn out to be false. I guess ideally there would be a way to tell Mathematica to switch to a "rigorous mode", where no assumptions are introduced surreptitiously.

  • 1
    $\begingroup$ It's a Chrome extension: chrome.google.com/webstore/detail/mathematica-editor-button/… I'm not sure if Firefox has one too. $\endgroup$
    – Feyre
    Commented Aug 19, 2016 at 19:26
  • 2
    $\begingroup$ Do you know of any examples which don't include removable singularities? $\endgroup$
    – Feyre
    Commented Aug 19, 2016 at 19:27
  • 1
    $\begingroup$ Have you seen this? mathematica.stackexchange.com/questions/89990/… As far as I am aware, the problems only arise when removable singularites are involved. $\endgroup$
    – Feyre
    Commented Aug 19, 2016 at 19:49
  • 2
    $\begingroup$ (1) I would tend to avoid Assumptions->Element[xxx,Integers] as that is the case where notions of measure are most likely to be problematic. $\endgroup$ Commented Aug 19, 2016 at 20:28
  • 2
    $\begingroup$ (2) It is very not practical to get formulas correct if your definition of "correct" does not remove removable singularities. I'm not sure if that is a case under consideration here though. $\endgroup$ Commented Aug 19, 2016 at 20:30


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.