Timeline for Is it always possible to calculate the limit of an elementary function?
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| Apr 21, 2020 at 17:00 | history | edited | Alexandre Eremenko | CC BY-SA 4.0 |
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| Mar 28, 2018 at 16:43 | comment | added | Alexandre Eremenko | @oliver: as I said many times, this depends on the EXACT definition of "elementary function". The definition in your question is ambiguous. | |
| Mar 28, 2018 at 16:07 | comment | added | Olivier Esser | Is it? It is elementary for ${\rm dom}(g)=\mathbb{R}\setminus\{0\}$ but if we say $g(0)=0$, it does not seems to be elementary and that is the function we get. I think that we can (easily) show that if an elementary function $g$ is defined on an open interval then it is analytic on that interval, because that's the case for the "primitive" elementary functions. | |
| Mar 27, 2018 at 2:11 | comment | added | Alexandre Eremenko | @oliver: It is not analytic at zero but it is an elementary function. | |
| Mar 25, 2018 at 17:20 | comment | added | Olivier Esser | @Alexandre Eremenko I am perplex when you say that the answer is ``yes'' without trigonometric function. We can define $f(x,a)=\exp({ -1 \over x^2+a^2})$. If we take $g(a)=\lim\limits_{x\rightarrow0}f(x,a)$, we get $g(a)=\exp({-1\over a^2})$, with $g(0)=0$. This function is known to be non analytic (at 0). | |
| Mar 24, 2018 at 0:21 | history | edited | Alexandre Eremenko | CC BY-SA 3.0 |
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| Mar 23, 2018 at 16:59 | history | edited | Alexandre Eremenko | CC BY-SA 3.0 |
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| Mar 23, 2018 at 16:52 | history | edited | Alexandre Eremenko | CC BY-SA 3.0 |
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| Mar 23, 2018 at 13:27 | comment | added | Alexandre Eremenko | @Matt F.: OK, this discussion becomes too long. I will start a new question. | |
| Mar 23, 2018 at 11:26 | comment | added | user44143 | Your update -- about functions rather than numbers, with a different set of functions, and motivation that requires some background -- is interesting and different enough to be moved to a new question. I'd be happy to comment on it in that context! | |
| Mar 23, 2018 at 2:03 | history | edited | Alexandre Eremenko | CC BY-SA 3.0 |
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| Mar 22, 2018 at 22:41 | history | edited | Alexandre Eremenko | CC BY-SA 3.0 |
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| Mar 22, 2018 at 19:37 | comment | added | Alexandre Eremenko | @Matt F.: Two and three incommensurable frequencies lead to elementary integrals. You have to start with $n=4$ to obtain a complicated constant. | |
| Mar 22, 2018 at 18:57 | history | edited | Alexandre Eremenko | CC BY-SA 3.0 |
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| Mar 22, 2018 at 13:21 | comment | added | user44143 | What is $m$ for $a_1=\lambda_2=1$, $a_2=\lambda_1=\sqrt{2}$? This looks like a good test case. I can calculate $$\phi(t)=\arctan\left(\frac{\sin(\sqrt{2}t)+\sqrt{2}\sin(t)}{ \cos(\sqrt{2}t)+\sqrt{2}\cos(t)}\right) + \pi\#(\{\cos(\sqrt{2}u)+\sqrt{2}\cos(u)=0, 0<u<t\})$$. Then I see $\phi(t)/t$ staying close to 1, but I haven't been able to calculate the integrals. | |
| Mar 21, 2018 at 20:09 | comment | added | Olivier Esser | @Alexandre Eremeno. I have notice it and I agree with it. However I have suppose that the function is well defined on an open interval around the elementary real $a$. So $f(x)$ is defined in an interval $]a \ b[$ for some $b$. This is the reason I expect my definition to be not entirely different, but I agree this is not completely clear. | |
| Mar 21, 2018 at 12:28 | comment | added | Alexandre Eremenko | @oliver: if you read Laczkovich and Ruzsa attentively, you see that there is a great difference between arcsin on [-1,1] and arcsin on (-1,1). Then, the domain of a function (with your definition) can be a complicated set, much more complicated than an interval. | |
| Mar 21, 2018 at 10:31 | comment | added | Olivier Esser | The definition of elementary function of "Laczkovich and I. Ruzsa" (see the proposed answer of Alexandre Eremenko) is very close to my definition. The elementary constant would be the same but there can be variation of the domain of the functions considered. I would be very surprised if it make a real difference in the end. | |
| Mar 20, 2018 at 0:28 | history | edited | Alexandre Eremenko | CC BY-SA 3.0 |
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| Mar 20, 2018 at 0:10 | history | edited | Alexandre Eremenko | CC BY-SA 3.0 |
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| Mar 19, 2018 at 23:10 | history | edited | Alexandre Eremenko | CC BY-SA 3.0 |
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| Mar 19, 2018 at 16:03 | history | edited | Alexandre Eremenko | CC BY-SA 3.0 |
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| Mar 19, 2018 at 12:33 | history | edited | Alexandre Eremenko | CC BY-SA 3.0 |
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| Mar 19, 2018 at 6:34 | comment | added | Olivier Esser | But this is the usual domain as described in elementary calculus books; see for example: en.wikipedia.org/wiki/Inverse_trigonometric_functions. $\arctan: ]-\infty\ \infty[ \rightarrow ]-\pi\ \pi[$; $\ln \mathbb{R^+} \rightarrow \mathbb{R^+}$; $\tan: \{x\in\mathbb{R}\ |\ \forall k\in \mathbb{Z} \ x\neq k\pi\}$. You can define $\sin$, $\cos$, $\arcsin$, $\arccos$ by means of these with common identities. I don't think it will matter as long as we use common real functions. | |
| Mar 18, 2018 at 23:30 | comment | added | Alexandre Eremenko | @oliver: your definition is still incomplete: you have to specify the domains of log, arcsin, arctan, etc. Or explain what you mean by them. | |
| Mar 18, 2018 at 20:08 | comment | added | Olivier Esser | It seems indeed that the "closed form" real are exactly what I call "elementary reals". Also the general idea as given in the introduction. I want to calculate the limit of a function in a closed-form. I initially thought that it was always possible; but although not yet completely clear; the answer of Alexandre Eremenko make me doubt it. Proving that a given real is not "elementary" or in "closed-form" might indeed be difficult. | |
| Mar 18, 2018 at 19:28 | comment | added | Timothy Chow | Regarding the definition of "elementary real," I gave one possible definition in this paper: timothychow.net/closedform.pdf I think that this definition is close to, if not exactly the same as, what olivier wants. The definition has the advantage of sidestepping some of the issues about domains of definition of elementary functions. On the other hand, difficulties remain; e.g., it remains unclear whether $\gamma$ is an elementary real. | |
| Mar 18, 2018 at 18:02 | comment | added | Olivier Esser | I hope the definition is now clear (see the updated question) | |
| Mar 18, 2018 at 17:21 | history | edited | Alexandre Eremenko | CC BY-SA 3.0 |
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| Mar 18, 2018 at 17:14 | history | edited | Alexandre Eremenko | CC BY-SA 3.0 |
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| Mar 18, 2018 at 16:50 | comment | added | Alexandre Eremenko | @oliver: once you give an exact definition of "pure real elementary function", I will address your question again, with this new definition. | |
| Mar 18, 2018 at 16:47 | comment | added | Alexandre Eremenko | @oliver: "multivalued function" is not a function, therefore you cannot integrate it. What you can integrate is a branch. There are infinitely many branches of arctan on the real line, they are different functions and each of them you can integrate. "Elementary functions" (as defined by Liouville and Ritt) are multivalued. If you propose some other approach, please give a precise definition of an "elementary function". Explain exactly what you mean by $x^{1/3}$ to begin with, and then think of $\arctan(1/x)$. | |
| Mar 18, 2018 at 16:23 | comment | added | Olivier Esser | How would you define (for example): $\int_0^1 \arctan(x) dx$ if $arctan$ is multi-valued? We use these expression all the times, it is supposed to be well defined. If you look in text over real analysis, $arctan$ is defined carefully (the image is $]-\pi\ \pi[$). | |
| Mar 18, 2018 at 16:13 | comment | added | Olivier Esser | I do not claim that complex or multi-valued functions are not interesting. But you seem to completely disregard pure real functions in a strict sense. Multi valued functions is a concept that is not so natural either. They are usually not accepted as "functions" outside complex analysis (just read the definition of function as stated here: en.wikipedia.org/wiki/Function_(mathematics)). How would you define a multi-valued function in general? What would it mean for these general multi-valued functions to be continuous? integrable? etc. | |
| Mar 18, 2018 at 14:32 | history | edited | Alexandre Eremenko | CC BY-SA 3.0 |
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| Mar 18, 2018 at 13:52 | comment | added | Alexandre Eremenko | @olivier: Just think on this example carefully: $\arctan(1/x)$ What is the domain? It surely must contain $R\backslash\{0\}$. Now, how exactly you want to define it? As having a jump at $0$? How much jump? It seems that the only reasonable definition is such that it has no jump, but is infinitely-valued. | |
| Mar 18, 2018 at 13:37 | comment | added | Alexandre Eremenko | @oliver: My function $\phi$ is real. But to write its expression in terms of exponential and log I have to use analytic continuation. Otherwise, if you wish to restrict strictly to the real domain, you have huge difficulties with domains of definition when you try to define your class. These difficulties will invalidate most theorems about elementary functions. For example $\arctan(1/x)$, is it defined and continuous in a neighborhood of $0$? I adhere to the point of view that $0$ is a removable singularity of this function, where the branches are smoothly connected. | |
| Mar 18, 2018 at 7:28 | comment | added | Olivier Esser | For the elementary function, I had in mind real functions only. My initial idea was over the reals only; with the definition given by the point 1-9 mentioned here: en.wikipedia.org/wiki/Elementary_function except that I initially only recognize constant rational as elementary (and not arbitrary constant). Note that I am increasingly convinced that you are right. | |
| Mar 18, 2018 at 0:02 | history | edited | Alexandre Eremenko | CC BY-SA 3.0 |
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| Mar 17, 2018 at 23:56 | comment | added | Alexandre Eremenko | @oliver: You do not give a definition of elementary function, referring to Wikipedia. Wikipedia does not have a precise definition, but gives complex logarithm as an example. It refers to Ritt's papers, and I follow the Ritt definition. | |
| Mar 17, 2018 at 23:34 | comment | added | Alexandre Eremenko | There will be no harm if you just replace $\phi(t)=\log f(t)$, as it differs from the old $\phi$ by a bounded summand. | |
| Mar 17, 2018 at 23:30 | comment | added | Alexandre Eremenko | $\phi(t)={\mathrm{Im}}\log f(t)$, this permits an analytic continuation along the real axis as long as $\phi(t)\neq 0$, and this is certainly an elementary function. | |
| Mar 17, 2018 at 19:44 | comment | added | Olivier Esser | But how do you ensure that the functions are elementary? OK for $r(t)$ but how do you get $\phi(t)$? You can't make piecewise definition. You use what is commonly known as $atan2(x,y) = \pm arctan(y/x)$ plus the special treatment for x=0. That's not elementary. | |
| Mar 17, 2018 at 19:27 | comment | added | Eric Towers | @olivier : Really? You don't recognize representing $f$ in polar coordinates? | |
| Mar 17, 2018 at 19:12 | comment | added | Olivier Esser | I may miss the obvious, but how do you get $f(t)=r(t)\exp(i\phi(t))$, could you maybe give a simple explanatory example. | |
| Mar 17, 2018 at 18:37 | history | edited | Alexandre Eremenko | CC BY-SA 3.0 |
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| Mar 17, 2018 at 18:04 | history | edited | Alexandre Eremenko | CC BY-SA 3.0 |
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| Mar 17, 2018 at 17:45 | history | edited | Alexandre Eremenko | CC BY-SA 3.0 |
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| Mar 17, 2018 at 17:34 | history | edited | Alexandre Eremenko | CC BY-SA 3.0 |
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| Mar 17, 2018 at 16:45 | history | edited | Alexandre Eremenko | CC BY-SA 3.0 |
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| Mar 17, 2018 at 15:57 | history | edited | Alexandre Eremenko | CC BY-SA 3.0 |
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| Mar 17, 2018 at 15:49 | history | answered | Alexandre Eremenko | CC BY-SA 3.0 |