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This is my first StackExchange question:

What is the reason that equations such as $\tan x = 2x$, $\cos x = x$, $\sin(x) = x^2$ and other questions that involve the same variable within a trigonometric function and outside the trigonometric function can only be solved with computer algorithms and have extremely complex closed forms; such as the Dottie number, the root of $\cos x = x$, having the mind-blowing closed form $\sqrt{1-\left(2I_{\frac{1}{2}}^{-1}\left(\frac{1}{2},\frac{3}{2}\right)-1\right)^2}\,,$ which is really fascinating to me.

For $\cos x = x$, I tried solving the equation using the complex definition of $\cos x$ (aka. the cooler version of cosine), $\frac{e^{ix}+e^{-ix}}{2} = x$ but after using a quadratic formula to solve for $e^{ix}$, I found that I just ended up with $x$ being equal to the complex definition of inverse cosine (arccos) in the formula $x = -i\ln(x + \sqrt{x^2-1})$ and I had gone around in circles, I thought "Is this really an impossible equation to solve for $x$?" The sum inside the natural logarithm, the $i$ being present in the equation even though the Dottie number is a real number approximately $0.739085$... and the $\sqrt{x^2-1}$, I had seen that thing everywhere in Pythagoras and trigonometric calculus. It's as if mathematics had put as many barriers around the $x$ as possible to prevent you from solving for $x$. And I had a similiar problem with $\sin x = x$, I knew for a fact that the solution was definitely $0$ and yet the "solution" I got was total garbage that looked nothing like $0$. And apparently this equation also had infinitely many complex solutions as well, which I don't grasp at all.

These equations look so simple from first glance and yet are mathematically impossible for a human to solve for an exact form, but why is that? is it something to do with "transcendental numbers"? Does it have any applications in trigonometry and calculus?

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    There's just a very specific set of functions that count as "closed forms" and most interesting questions can't be answered exactly in terms of them. In a typical mathematics education this fact is thoroughly hidden from you because teachers are careful to only show you questions which do have closed form answers. "Random" mathematical questions simply don't. – Qiaochu Yuan Oct 28 '22 at 20:28
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    The solution can't have a closed form when the equation is composed of different kind of functions such $x e^x = 1$, $x + \ln x = 2$, $\sin x + e^x = x^2$ ... – Essaidi Oct 28 '22 at 20:34
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    The answer by Qiaochu Yuan is excellent. I will add that the little society of "elite functions" accept to integrate newcomers having a sufficiently good Vitae. For example "cardinal sine" function which is $\operatorname{sinc}(x)=\frac{\sin(x)}{x}$ (with the exceptional value $\operatorname{sinc}(0)=1$ encountered in equations like $\sin x = 0.1 x$ has been "ennobled" around 100 years ago (in particular for its role in Fourier Transform). Same thing for Lambert function $W$, for a certain number of "special functions" like Airy, etc. discovered mainly in the 19th century. – Jean Marie Oct 28 '22 at 20:41
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    @Essaidi Not only. For instance polynomials of degree $5$ or more are generally not solvable so for such a polynomials $P(x)=2x$ likely too, yet both sides are the same types (i.e. polynomials). The impossibility of finding a closed form is not limited to the so called transcendental equations. – zwim Oct 28 '22 at 20:41
  • See here for the inverse of $2\tan(x)-x$, a similar problem – Тyma Gaidash Oct 28 '22 at 22:26
  • When you want to address a comment to some one, in the body, add @user_name. If you ask the question about why $\frac \pi 4$, I chose it because it is the simplest angle close to the solution and which gives nice coefficients in th first expansion. – Claude Leibovici Oct 29 '22 at 09:24
  • Glibly: Undoing things is harder than doing things. – JonathanZ Oct 29 '22 at 21:22
  • @Essaidi obviously, $x e^x = 1$ has solution $W(1)$, $x + \ln x = 2$ has solution $W(e^2)$, you can verify it using Wolfram Alpha. – Anixx May 20 '23 at 20:47
  • I'm talking about expressing solution with elementary functions. Other way, one can always write: the solution of $a = f(x)$ is $x = f^{-1}(a)$ when $f$ is bijective. – Essaidi May 24 '23 at 01:22

3 Answers3

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I love your question, and while I won't be able to put into words a complete answer that does it justice, I hope to provide enough intuition to get you thinking and find your own satisfying answer.

First of all, I bring to your attention that working with "complex closed form" expressions is not new to you, and in fact you've been working with them for most of your mathematical life. Indeed, $\cos x$, $\sin x$, $e^x$, $\log x$, and even something as "elementary" as $\sqrt{2}$ are "complex closed form" expressions. They are just expressions that appear so often, and that we use so much, and that we know so many properties about, that we just consider them as part of the "not complex" repertoire. Two thousand or so years ago, a mathematician just like you would look at $x^2 = 2$ and say that it can only be solved with a "computer" algorithm that approximates the answer and has an extremely complex closed form with a weird v-shaped symbol. What you are experiencing is not the exception, it is the rule!

More generally, some objects of interest appear sufficiently often in math that it is simpler, more elegant, and more efficient to give them their own symbol and add them to our "fundamental repertoire". As an analogy, say that you and I develop a language with only two letters, say $a$ and $b$, and when we communicate, the word $ababababa$ keeps appearing all over the place. It is only natural to decide after some time that rather than repeating that monstrosity every time, we just invent a new letter, call it $c$, and just refer to $c$ from now on. If you are familiar with information theory, this is exactly what compression theory aims to achieve! And notation such as $\cos x$ and $\sqrt{2}$ are just convenient "letters" we added to our alphabet. The solution to $cos x = x$ simply does not appear often enough for mathematicians to bother developing syntax around it. Most of math starts with very few extremely basic symbols, such as $1,2,+,x^2$ etc then develops problems and solutions with those symbols, then develops new symbols to make doing math more efficient, for example by associating the symbol $\sqrt{2}$ to the chain of symbols $x^2=2$.

J.G.
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Carl Chaanin
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  • Very nice answer! In the beginning there was only 0 and the successor function and we built an empire out of these humble beginnings. – Arkady Oct 29 '22 at 06:27
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    Perfect answer with perhaps just one thing missing. You focus essentially on beauty - although that word is not approved of by everyone. What needs to be added is the faith of the mathematician - which is that “Beauty is truth, truth beauty”. We don’t just say $sin$ because it’s a short way of saying “sine”. We say it because the abbreviation, as well as being beautiful, is true and leads to truth. (The thing that everyone believes but nobody is allowed to say is that there is an element of “because of being beautiful” in that sentence). – Martin Kochanski Oct 29 '22 at 08:17
  • You could extend the definition of $a^b$ to fractions and ignore the square root notation since $\sqrt 2=2^\frac12$ – Тyma Gaidash Oct 29 '22 at 22:16
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If we want a complete and mathematical correct answer, we have to translate the question into mathematical terms. What you mean is: What is the reason that such equations cannot be solved by applying only finite numbers of elementary functions/operations?

It's a kind of closed-form problems. That means, solutions in closed form are wanted.

Note that I'm reflecting only my personal findings here.

The term $\mathbb{C}$-algebraic means "algebraic over $\mathbb{C}$".

The problem of solving a given equation can be split into two mathematical sub-problems. Let $F$ be a function, $x$ a variable and $c$ a constant. Let the equation $F(x)=c$ be given. If we find the inverse relation $F^{-1}$ of $F$ (that means the appropriate partial inverses of $F$), the solutions of the equation can be found by applying $F^{-1}$: $F^{-1}(F(x))=F^{-1}(c)$, $x=F^{-1}(c)$.
The second mathematical sub-problem is to decide if a solution of the equation can be represented by a closed-form number.

You are asking for solutions that can be represented by elementary expressions (means function terms of elementary functions) or by elementary numbers.

According to Liouville, the elementary functions are generated by applying only finite numbers of only $\exp$, $\ln$ and/or unary or multiary $\mathbb{C}$-algebraic functions. Each elementary standard function (i.e. the trigonometric functions, the hyperbolic functions, the arcus functions, the inverse hyperbolic functions) can be represented in that form.

1) Applying the inverse relation

1 a) Ritt's theorem

We have the main theorem of [Ritt 1925], that's also proved in [Risch 1979], for deciding if a given elementary function can have partial inverses with non-discrete domains that are elementary functions. I call it structure theorem about elementary invertible elementary functions. With help of Ritt's theorem, we can conclude that the elementary functions that have elementary partial inverses with non-discrete domains are generated by applying only finite numbers of only $\exp$, $\ln$ and/or unary $\mathbb{C}$-algebraic functions.

But your equations are multiary $\mathbb{C}$-algebraic equations. Your functions $F$ are multiary $\mathbb{C}$-algebraic functions therefore. That means, your functions $F$ don't have partial inverses with non-discrete domains that are elementary functions.

The following is my own conclusion: That means your kind of equations cannot be solved by rearranging them by applying only finite numbers of elementary functions/operations we can read from the equation.

[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

1 b) Khovanskii's Topological Galois theory

[Khovanskii 2014]:
"Vladimir Igorevich Arnold discovered that many classical questions in mathematics are unsolvable for topological reasons. In particular, he showed that a generic algebraic equation of degree 5 or higher is unsolvable by radicals precisely for topological reasons. Developing Arnold’s approach, I constructed in the early 1970s a one-dimensional version of topological Galois theory. According to this theory, the way the Riemann surface of an analytic function covers the plane of complex numbers can obstruct the representability of this function by explicit formulas. The strongest known results on the unexpressibility of functions by explicit formulas have been obtained in this way.
...
The monodromy group of an algebraic function is isomorphic to the Galois group of the associated extension of the field of rational functions. Therefore, the monodromy group is responsible for the representability of an algebraic function by radicals. However, not only algebraic functions have a monodromy group. It is defined for the logarithm, arctangent, and many other functions for which the Galois group does not make sense. It is thus natural to try using the monodromy group for these functions instead of the Galois group to prove that they do not belong to a certain Liouville class. This particular approach is implemented in one-dimensional topological Galois theory ...
...
Nevertheless, the set of singular points of an elementary function is at most countable, and its monodromy group is solvable. If a function does not satisfy these restrictions, then it cannot be elementary."

[Khovanskii 2014] Khovanskii, A.: Topological Galois Theory - Solvability and Unsolvability of Equations in Finite Terms. Springer 2014

[Khovanskii 2019] Khovanskii, A.: One dimensional topological Galois theory. 2019

[Khovanskii 2021] Topological Galois Theory - Slides 2021, University Toronto

Khovanskii's publications

[Belov-Kanel/Malistov/Zaytsev 2020] Belov-Kanel, A.; Malistov, A.; Zaytsev, R.: Solvability of equations in elementary functions. Journal of Knot Theory and Its Ramifications 29 (2020) (2) 204-205
$\ $

see also:

Conditions to be able to write down the inverse function as a closed-form expression

How do we know if a function has an elementary inverse?

Proving a Problem has a Closed Form Solution

Resolution of symbolic equations

Why some inverse functions do not have a closed form

2) Solutions in the Elementary numbers

A given equation can have solutions that are representable by an elementary number even if the function $F$ doesn't have an elementary partial inverse with non-discret domain. But we cannot find this representation by only rearranging the equation by only applying finite numbers of elementary functions.

Your equations can be rearranged to irreducible polynomial equations over $\overline{\mathbb{Q}}$ in dependence of $x$ and $e^x$. Therefore the main theorems in [Lin 1983] and [Chow 1999] can be used. They state that those kinds of equations don't have solutions except $0$ in the Elementary numbers or Explicit elementary numbers respectively.

And it follows with my theorem in Proof Check: Non-existence of the inverse function in a given class of functions that the function $F$ of these kind of equations cannot have partial inverses with non-discrete domains that are elementary functions.

[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

see also Trigonometric/polynomial equations and the algebraic nature of trig functions

3) Non-applicability of Lambert W

Some kinds of $\mathbb{C}$-algebraic equations that contain simultaneously $x$ and $e^x$ can be solved in terms of Lambert W. Lambert W isn't an elementary function, but solving equations by applying Lambert W can be done manually.

Your kind of equations are non-symmetrical $\mathbb{C}$-algebraic equations that depend simultaneously on $x$ and $e^{ix}$ wherein the degree of $e^{ix}$ is greater than $1$. This kind of equations cannot be solved in terms of Lambert W.

IV_
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Congratulations for a so well-posed first question on this site to which I welcome you !

As said in comments, very few transcendental equations have a closed form and when they have, they express in terms of special functions.

@Tyma Gaidash (in particular) gave a few of them; they are nice, beautiful and interesting. For example (have a look here) Laplace's limit constant corresponds to the solution of

$$ e^{-2x}=\frac{x-1}{x+1}\quad \implies \quad x=\frac{1}{2}W\left({+2\atop -2};1\right)$$ this is an exact solution in terms of the generalized Lambert function but for computing it $$x=1+2\sum_{n=1}^\infty\frac{L_{n-1}^{(1)}(4n)}{n}e^{-2n}$$ which is an infinite summation.

$\color{red}{\text{Playing the role of the devil's advocate}}$, let me consider, with a more than simplistic approach, the case of the zero of the simple function $$f(x)=x-\cos(x)$$ which write $$f(x)=\frac{\pi -2 \sqrt{2}}{4}+\left(1+\frac{1}{\sqrt{2}}\right)\left(x-\frac{\pi }{4}\right) +\sum_{n=2}^\infty \frac{\sin \left(\frac{\pi n}{2}\right)-\cos \left(\frac{\pi n}{2}\right)}{\sqrt{2} n!}\left(x-\frac{\pi }{4}\right)^n$$ In this form, it can be inversed leading to $$x=\frac \pi 4+\sum_{n=1}^\infty a_n\,t^n\qquad \text{where}\qquad t=\frac{2 \sqrt{2}-\pi}{4}$$ where we know all the $a_n$ (have look here). This is then (at leat to me) an exact definition of Dottie number.

To give an idea, truncating the infinite series to $O(t^{13})$ gives an absolute error of $1.14\times 10^{-19}$

Claude Leibovici
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  • Your answer is very helpful. Why for the function $f(x)=x-\cos(x)$ , you make a development at $x=\frac{\pi}{4}$, and not at $x=1$ for example?$$$$ And do you think we could apply the method that you used for $f(x)=x-\cos(x)$ for another function, for example $f(x)=x-\sin(x)$ and find the $a_n$? – NN2 Oct 29 '22 at 09:09