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From this answer to how to get a get a nice “cosine looking” curve following the y=x direction?

It is not possible to invert $x=t+\cos t$ analytically.

I'm sure it's true and I wouldn't know how to try, but how can the impossibility be shown?

uhoh
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    In this case it can be shown by the inverse function theorem. Since $(t+\cos t)' = 1-\sin t$ is zero at $t=2\pi n + \pi/2$ with $n \in \mathbb{Z}$, in a neighborhood of these points the function is not analytically invertible. These are in fact the points of vertical tangent (infinite slope) in the graph from the linked answer. – pregunton Feb 23 '20 at 07:47
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    @pregunton no I've undone my caveat because it may make this a different question, or at least different than what the linked answer was getting at. I don't know if I should ask about inverting $x = 2t + \cos t$ separately or not. – uhoh Feb 23 '20 at 08:08
  • @pregunton oh, I've stopped back here just now and realize that since that is the simple and obvious answer, your comment can simply be reposted as an answer which I can accept and we can call this "case closed". – uhoh Jun 07 '21 at 05:23
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    Okay, it's done! – pregunton Jun 07 '21 at 07:16
  • @uhoh: Variables in transcendental functions cannot be inverted. – Narasimham Jun 07 '21 at 07:48
  • @Narasimham if there's a more general and supportable answer to "How do we know..." such your comment, please consider posting it as an answer for the benefit of future readers! The goal as always in Stack Exchange is to generate the best answer post. Thanks! – uhoh Jun 07 '21 at 07:50
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    Let me do that fwiw.. – Narasimham Jun 07 '21 at 08:13

3 Answers3

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For the purpose of the question, inversion of $x = t + \cos t$ analytically is formally defined as whether or not there exists a representation, in terms of elementary operations, of $t$ purely as a function of $x$.

The space of mathematical expressions can be represented as a syntax tree. As well, algebraic equivalences can be represented as modifications on that syntax tree. The problem then becomes, given this set of possible transitions and replacements, is it possible to arrive at a syntax tree which obeys this set of rules, in this case, a syntax tree where the LHS is $t$ and the RHS does not contain $x$. Proving this can likely be done using graph theory, and is the general approach of Mathematica, which does imply, but does not assert, that this representation does not exist

Nij
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mp-
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  • You need to surround your hyperlink with curved brackets () for it to take effect. – Sam Feb 23 '20 at 05:06
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    Yes, in theory - but I suspect the implied decision problem is non-trivial at least (i.e. NP-Hard), if not even undecidable, in general: I remember hearing a theorem suggesting that the similarly-flavored problem of deciding equality between arbitrary expressions for real numbers that are formed using only arithmetic, rational numbers, and sine, is undecidable. So we might need some more constraints to prove this by computer, if that is so. Though I could also be missing something. – The_Sympathizer Feb 23 '20 at 06:35
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In this case it can be shown by the inverse function theorem. Since $(t+\cos t)' = 1-\sin t$ is zero at $t=2\pi n + \pi/2$ with $n \in \mathbb{Z}$, in a neighborhood of these points the function is not analytically invertible. These are in fact the points of vertical tangent (infinite slope) in the graph from the linked answer.

pregunton
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    Note that the assumptions of the inverse function theorem include that the inverse is differentiable. If we relax this assumption, it is possible to algebraically invert some functions with points of zero slope. For example, $f(x) = x^3$ has an inverse of $f^{-1}(x) = x^{1/3}$. – Michael Seifert Dec 22 '21 at 12:53
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It can be shown by simple rearranging the equation, that the equation is not in a form that is solvable in terms of Lambert W.

The equation is an equation of elementary functions. You are looking for the inverse (means the inverse function) of the elementary function $t\mapsto t+\cos(t)$ over domains where the function is bijective.

The main theorem in [Ritt 1925], that's proven also in [Risch 1979], lets suggest that the elementary function above cannot have an elementary inverse. The main theorems in [Lin 1983] and [Chow 1999] imply, assuming Schanuel's conjecture is true, that your equation cannot have solutions that are elementary numbers or explicit elementary numbers if $x$ is an elementary number. This implies, assuming Schanuel's conjecture is true, that the elementary function above over non-discrete domains cannot have an elementary inverse therefore.
$\ $

[Chow 1999] Chow, T.: What is a closed-form number. Am. Math. Monthly 106 (1999) (5) 440-448

[Lin 1983] Ferng-Ching Lin: Schanuel's Conjecture Implies Ritt's Conjectures. Chin. J. Math. 11 (1983) (1) 41-50

[Risch 1979] Risch, R. H.: Algebraic Properties of the Elementary Functions of Analysis. Amer. J. Math. 101 (1979) (4) 743-759

[Ritt 1925] Ritt, J. F.: Elementary functions and their inverses. Trans. Amer. Math. Soc. 27 (1925) (1) 68-90

IV_
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