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The equation $x\sin x=a$ can be computed numerically, but are there methods to solve it algebraically?

Note: A similar equation in that form, $xe^x=a$ can be solved using lambert W function.

KKZiomek
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    Not solvable algebraically + infinitely many solutions because $\sin x$ is periodic. – Simply Beautiful Art Jun 10 '17 at 23:31
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    @TobyMak: How does that help? – copper.hat Jun 10 '17 at 23:31
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    @SimplyBeautifulArt is there a proof that it's not solvable algebraically? – KKZiomek Jun 10 '17 at 23:44
  • The basic idea is outlined here: https://math.stackexchange.com/questions/2082103/trigonometric-polynomial-equations-and-the-algebraic-nature-of-trig-functions – Simply Beautiful Art Jun 10 '17 at 23:45
  • @copper.hat It might help... though not in the way I expected. See my answer for details. – Simply Beautiful Art Jun 11 '17 at 00:00
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    The Lambert W function is not a magic wand $-$ it's just a name for the inverse of the function $xe^x$. You still have to evaluate it numerically. If you want, you can define a Ziomek Z function that is the inverse of $x\sin x$; and now you have an 'algebraic' solution to your problem! – TonyK Jun 11 '17 at 00:03
  • @TonyK Mmm... but the point is to reduce the problem down to a more specific case. One could define $\sqrt2$ in terms of $\sin(\pi/4)$ and $\cos(\pi/4)$, but we prefer to do so in the other manner. The Greeks didn't believe in square roots, but now we do, despite still calculating it numerically most of the time... who's to say the Lambert W function won't be considered so elementary in the near future! – Simply Beautiful Art Jun 11 '17 at 00:07

2 Answers2

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I think there are two things that shouldn't be confused here.

1 : algebraic solution

First there is the notion of algebraic functions, which can be defined as roots of polynomials with rational coefficients.

But this is a very limited set of functions, for instance $\exp,\ln, \sin,...$ would not be considered.

If we consider the problem $f(x)=a$ for $a\in\mathbb Q$ for a certain function $f$, we can wonder whether $x$ should be an algebraic number or not.

The explicit form would be to find an algebraic function $\phi$ such that $x=\phi(a)$.

The implicit form would be to find a polynomial $P$ with rational coefficients such that $P(x)=a$ (at least for some $a$).

[Simply Beautiful Art just addressed this very question in his answer].

2 : algebraic combination of known functions

And then, there is the different problem of finding a closed formula for $x=f^{-1}(a)$ in term of an algebraic combination of known functions (meaning using $+,-,\times,/$ and powers to a rational).

But this statement is quite fuzzy, since it all depends of your initial set of so-called known functions. Does it includes $\ln,\exp$, does it includes trig functions, reciprocal of trig functions, etc...

Before Lambert W function was invented, there was no closed formula according to this rule for $xe^x=a$. So if I define today the Ziomek function $Z$ such that $Z(a)$ is solution of $x\sin(x)=a$, now I have a closed formula for $x$.

Of course, it only becomes to be interesting when this $Z$ function can be used for other kind of equations, if we can find interesting properties for it, if we can develop it in power series, or find algorithms to compute it quickly and accurately, or if it simply has any kind of theoretic interest.

This is the reason why our limited set of known functions has been extended to erf, Bessel, Gamma, polylogs and other creatures there were initially not even considered.

Simply Beautiful Art
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zwim
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  • @TonyK seems I had the same idea, while writing my answer. – zwim Jun 11 '17 at 00:32
  • It may be of interest to wonder if your $Z(a)$ is related to the Lambert W function or any other special functions, which would be the most interesting. (+1) for a good overview of the problem at hand. – Simply Beautiful Art Jun 11 '17 at 00:34
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If $x=f(a)$ is solved for $a\in\mathbb Q$ and $f(a)$ is continuous, then $x=f(a)$ for $a\notin\mathbb Q$.

Next, we see that if $a\ne0,a\in\mathbb Q$, then $x$ is not algebraic. If $x$ were algebraic, then $\sin(x)$ would be transcendental, and thus $x\sin(x)$ would be transcendental, a contradiction to $x\sin(x)=a\in\mathbb Q$.

Likewise, we show that $x$ is not a rational multiple of $\pi$. If it were, then $\sin(x)$ would be algebraic, but then $x\sin(x)$ would be transcendental.

I believe that more or less covers the basis of what is algebraically solvable, excluding the Lambert W function. One is able to solve it using the Lambert W function though.

$$x\sin(x)=\Im(xe^{ix})=a\implies xe^{ix}=b+ai$$

$$\implies x=\frac1iW_k(bi-a)$$

But this holds only if $x$ is real, which occurs when $W_k(bi-a)$ is imaginary, where $W_k$ is the multi-valued Lambert W function. Likely put, there is probably no closed form solution for $b$, unless someone's discovered a way to dissect the Lambert W function into real and imaginary parts, something I have yet to see.

Solving for $b$ is as hard as solving for $x\cos(x)=b$, literally...

Simply Beautiful Art
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