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I saw recent thread on whether incorrect answers should be deleted. What about answers that are "not even wrong", because they use notation in ways that simply make no sense? I'm thinking about two of the answers here.

They really are nonsense. I mean, when people write $\int e^{itx}\,dt=\delta(x)$ that's bad enough, since the integral doesn't exist. But that formula can be "interpreted" as saying the Fourier transform of the function $1$ is $\delta$, which is true for the tempered-distribution Fourier transform. But here the answerers are talking about $\delta(x-i)$, which simply has no reasonable "interpretation". (Or none that I can see - the answerers have so far declined to explain what the notation is supposed to mean.)

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    $\begingroup$ I suppose there's no harm in deleting in such cases, but I'm not sure it's any better than downvoting and leaving explanatory comments. $\endgroup$
    – Gerry Myerson
    Commented Nov 26, 2017 at 23:31
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    $\begingroup$ In my understanding this is covered by the other discussion; it might make sense to make it explicit there though. $\endgroup$
    – quid Mod
    Commented Nov 27, 2017 at 0:10
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    $\begingroup$ For the specific case we could be in a case where the answer(s) might be relevant to preserve to highlight a common misunderstanding. It appears two answers have the same issue; in that sense it might be a somewhat common misconception. $\endgroup$
    – quid Mod
    Commented Nov 27, 2017 at 1:34
  • $\begingroup$ I will add that if the question is supposed to be specifically about this particular case, then the tag (specific-answer) should be added. (But I suppose the link is used only as an example, in which case this tag should not be used here.) $\endgroup$
    – Martin Sleziak
    Commented Nov 27, 2017 at 2:27
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    $\begingroup$ @quid It can't be a common misconception because there's no concept there, just blind manipulation of formulas. Regarding the duplication, my conjecture is one of them is a sock puppet. "Two days ago" is too coarse to know, but as far as I can see the second guy didn't exist until after my comments on the first guy's solution... $\endgroup$
    – David C. Ullrich
    Commented Nov 27, 2017 at 3:37
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    $\begingroup$ @DavidC.Ullrich I continue to believe that this type of approach, "blind manipulation of formulas", is somewhat common in such contexts. I do not insist on calling it a misconception. If you suspect sock-puppetry please raise a moderators flag. // By now one of the answers is self-deleted. $\endgroup$
    – quid Mod
    Commented Nov 27, 2017 at 12:14
  • $\begingroup$ @quid And now they're both gone. How does this work? In particular, yesterday I didn't see any "delete" link under the answer. This morning it said "delete(2)", I clicked on that and it was deleted. What makes that "delete" appear? $\endgroup$
    – David C. Ullrich
    Commented Nov 27, 2017 at 16:48
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    $\begingroup$ It would have been really nice if you had included the answer in a block quote since it has now been deleted $\endgroup$
    – gen-ℤ ready to perish
    Commented Nov 27, 2017 at 16:58
  • $\begingroup$ I don't think it is a good idea, because: 1- An answer is not bad because X persons think it is, I'm not trying to advertise anything but following this answer it's already voted to delete 3 times before i start to defend it in chat, and detail more explanations to save it, not all people are present in chat, nor even all are ready to defend their approaches. 2- I think marking a thread as protected suffices to prevent it from overloaded with junk. $\endgroup$
    – Abr001am
    Commented Nov 27, 2017 at 17:40
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    $\begingroup$ Users with 20k+ points can vote to delete answer-posts with score -1 or lower. As long as the score is non-negative there is no delete link for 20k+ users. Something similar is true for questions put on hold recently: at score -3 or lower they can be deleted, but not at a better score (once they are on hold/close for at least 48 hours one can vote to delete, regardless the score, and already at 10k+ points). $\endgroup$
    – quid Mod
    Commented Nov 27, 2017 at 18:02
  • $\begingroup$ @mathreadler If I don't get what? $\endgroup$
    – David C. Ullrich
    Commented Nov 27, 2017 at 22:27
  • $\begingroup$ @DavidC.Ullrich If you judge something to be meaningless. Maybe it makes sense to someone who has studied more / different things than you have even if you are a high ranking moderator. $\endgroup$
    – mathreadler
    Commented Nov 27, 2017 at 22:29
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    $\begingroup$ @mathreadler Guffaw. Maybe "jhk gorlbe" is a proof of Fermat's Last Theorem. $\endgroup$
    – David C. Ullrich
    Commented Nov 27, 2017 at 22:47
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    $\begingroup$ @mathreadler Someone who actively answers questions for a tag is likely to at least be aware of all relatively common approaches. If someone is using a non-standard approach/notation, they should be aware of this and clearly state it (as well as provide explanation). Even if they aren't aware of it, they should have plenty of opportunity and prompting to add explanation before their question is deleted. I'm definitely sympathetic as I have a very non-traditional background, but if someone is not aware that what they wrote would be perceived as nonsense, then they probably made a mistake. $\endgroup$
    – Derek Elkins left SE
    Commented Nov 28, 2017 at 1:39
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    $\begingroup$ @DerekElkins They both had plenty of prompting and explanation; neither one ever acknowledged that there was a problem. And it wasn't a non-standard approach - they insist that, omitting irrelvancies, the Fourier transform of $e^t$ is $\delta(x-i)$. The function $e^t$ simply does not have a Fourier transform in any of the standard senses, and they never defined a different version. And in fact $\delta(x-i)$" simply makes no sense - if it means anything it means that $\int \phi(t)\delta(t-i)=\phi(i)$ for Schwarz functions $\phi$. But there's simply no such thing as $\phi(i)$ here. $\endgroup$
    – David C. Ullrich
    Commented Nov 28, 2017 at 2:38

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Personally, I am not a complex analysis pro. It took me a minute to figure out why $\delta(x-i)$ was no good. If I encountered a problem like this, I may have made the same misstep. It's the type of mistake a beginner would make, so your comments on the incorrect answers do have expository value.

If you don't like the whole "downvote and comment" approach, you could preserve the mistake by adding a "One might naively expect..." paragraph to your answer. It wouldn't take long to simply point out that there is a discrepency here between blind formalism and real math. That way, the mistake would still be highlighted, whether the other answers were deleted or not. You would also have better control over the presentation of your correction.

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    $\begingroup$ Amusingly, it's not clear that it's actually a misstep. If the domain of $x$ is the reals, then clearly $\delta(x - i)$ ought to be the zero distribution. And WA informs me that the CPV of the integral involved is, in fact, zero. $\endgroup$
    – user14972
    Commented Nov 27, 2017 at 4:09
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    $\begingroup$ @Hurkyl Which integral? If $$\int_{-\infty}^{+\infty} e^{-(1+ ix)t}\cos t\,dt$$ then WA is talking out of its behind, that integral doesn't have a Cauchy principal value. If $$\int_{-\infty}^{+\infty} \hat{g}(t)e^{-t}\cos t\,dt,$$ then there's no reason to expect that integral to be $0$ in general (and, depending on the assumptions on $g$, no reason to expect that the integral makes sense). $\endgroup$
    – Daniel Fischer Mod
    Commented Nov 27, 2017 at 11:46
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    $\begingroup$ @Hurkyl If we say that $\delta(x-i)$ is the zero distribution then we're claiming that the Fourier transform of $e^t$ is $0$. That seems unfortunate. Also it raises the question of exactly what we mean by "Fourier transform" - the function $e^t$ does not have a Fourier transform in any standard sense - it's not a tempered distribution. $\endgroup$
    – David C. Ullrich
    Commented Nov 27, 2017 at 16:28
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    $\begingroup$ The idea of including in my answer an explanation of what was wrong with the other answers hadn't occurred to me. If it had, it might have seemed a little tacky. But now that the other answers have disappeared I may well do just that when I have a minute. $\endgroup$
    – David C. Ullrich
    Commented Nov 27, 2017 at 16:52
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I do not see that strict a difference between wrong answers and not-even-wrong answers. If my answer uses the inverse function $f^{-1}$ of a function that is not invertible, my answer is not-even-wrong. If I extend the answer by introducing a lemma saying that $f$ is invertible, my answer becomes merely wrong.

There is of course also the case of entirely meaningless salad of words and symbols, but the case under discussion is IMHO subsumed by the discussion on how to deal with wrong answers.

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I've caught myself in a similar situation before. Explicitly, one answer was fundamentally wrong via manipulations which were not valid.

I flagged the answer for moderation in the grounds that, based on the "Why and how are some answers deleted?" page, it should be deleted, since it did

"not fundamentally answer the question (...)".

The flag got declined. The reasoning: "declined - flags should not be used to indicate technical inaccuracies, or an altogether wrong answer".

At the time, I got a little upset with this (luckily, I was able to engage in a time-consuming, tiresome but ultimately fruitful conversation with the answerer, and he ended up kindly fixing his answer after a while). But I've come to understand that the interpretation that the community has about deletion of answers is not what is on the page I linked above, but rather:

Answers that do not fundamentally TRY to answer the question may be removed.

Whether this is fair or not is beyond me ("correctness" can be elusive sometimes, thus justificating such approach), but it seems that this is the policy about deletion of answers. I will stand corrected if this is not the case.

That said, under this "rule", there is no reason to delete (forcibly) this particular meaningless answer you link. If you think it is not useful, then by all means downvote. It is clear, however, that who answered attempted to give an answer.

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    $\begingroup$ You may be conflating flagging and voting. Moderators are not meant to make judgements on mathematical technicalities, but the rest of us certainly are, and surely can vote to delete wrong answers. $\endgroup$
    – Gerry Myerson
    Commented Nov 28, 2017 at 22:08
  • $\begingroup$ @GerryMyerson I think you are right, I am indeed conflating both. But I don't think this undermines the main purpose of the answer (the first part is just contextualization), which is to state the status quo of the community as a whole: it is not very eager to remove wrong answers. It is quite eager to remove gibberish answers (spam, nonsense, "comment-answers"), but not wrong ones. As I said, as of today, I don't know if this is good or bad. But accepting this made me live better in the platform. $\endgroup$
    – Aloizio Macedo Mod
    Commented Nov 28, 2017 at 22:31
  • $\begingroup$ Related to the point about (not) flagging incorrect answers: Is it in the Moderator's Job Description to patrol for correctness? and other posts linked there. $\endgroup$
    – Martin Sleziak
    Commented Dec 3, 2017 at 15:45
  • $\begingroup$ Re: it is not very eager to remove wrong answers. My impression is exactly the opposite. I have seen some deletions of answers where I am not entirely sure they are wrong. (But I should probably believe the judgement of the delete voters.) Recently I have posted this question: Should we vote to delete wrong answers? The top-voted answer - currently at score 24 - is in favor of deleting incorrect answers. $\endgroup$
    – Martin Sleziak
    Commented Dec 5, 2017 at 1:48
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There is a matter of convention here. Integral signs don't always mean Riemann (or Lebesgue) integral of functions.

In particular, in my observations there is a convention where an integral such as $\int_{-\infty}^{\infty} f(x,y) \, \mathrm{d}x $ is never meant to be read as an ordinary integral; since there is a free variable $y$, this should be interpreted as an operator on distributions. Specifically, the distribution defined by

$$ g(y) \mapsto \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x, y) g(y) \, \mathrm{d}y \, \mathrm{d}x $$

I don't like the convention any more than you do, but it is rather widespread, even if the practitioners often don't realize they mean something different from what they learned in introductory calculus.

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    $\begingroup$ Are the answers there make sense if they use the definition here? $\endgroup$
    – user99914
    Commented Nov 27, 2017 at 4:36
  • $\begingroup$ @JohnMa If one restricts $g$ to the space of Fourier transforms of functions with compact support, then they do make sense in a way. With less artificial assumptions on the space $g$ is taken from, not so much. $\endgroup$
    – Daniel Fischer Mod
    Commented Nov 27, 2017 at 11:50
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    $\begingroup$ I acknowledged that convention in my post. If the only crime was saying $\delta(x)=\int e^{itx}\,dt$ I wouldn't be so upset, because yes that is a common malapropism, with a reasonable interpretation. The problem here is much worse, asserting that $\delta(x-i)=\int e^te^{-itx}\,dt$. That's not a standard abuse of \int. If it means anything it means that $\int\hat\phi(t)e^t\,dt=\phi(i)$ for $\phi\in C^\infty_c(\mathbb R)$, and that's simply nonsense, because there's no such thing as $\phi(i)$. The problem is not the non-existent integral, the problem is $\delta(x-i)$. $\endgroup$
    – David C. Ullrich
    Commented Nov 27, 2017 at 15:59
  • $\begingroup$ @DanielFischer Yes, it occurred to me that if we restrict our "test functions" to some space of entire functions of exponential type then we could make sense of the assertion that the Fourier transform of $e^t$ is $\delta(x-i)$. But if we do that we're not talking about "the" Fourier transform in any of its standard incarnations; if we do that we're setting up a whole new framework, which needs to be explained. Imagine manabu giving an exposition of all that... $\endgroup$
    – David C. Ullrich
    Commented Nov 27, 2017 at 16:21

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