Take some function such as $f(x)=x\sin x+3x-1$, or $g(x)=xe^x-2x+3$. These functions don't have a closed form inverse (I know this because in general we are told that these should be solved numerically, as there is no analytic way of solving equations such as $f(x)=0$).

Why do they not have an inverse? Is there any way of justifying this?


Briefly, because the equations you'd need to solve to get a closed form of this inverse are transcendental equations. You might also find the following article helpful.

Edit: Some more info - it might seem that if you end up with an equation of the form $x=h(x)$ where $h$ is a transcendental function, then you're out of luck and can't find a closed-form solution for $x$. However, this is not the case, for example $$\sin x = x$$ can be completely solved by eyeballing $x=0$ as the only solution (and this is admittedly a closed-form way to settle things!). The point is that transcendence is more correctly understood as preventing the existence of closed-form solutions for a class of equations instead of for a single equation. This is similar to the sense in which there is no general formula involving only the elementary algebraic operations and taking radicals to solve a polynomial equation of degree $5$ or more, and yet we can deal with many special cases (and indeed entire sub-classes of such polynomial equations).

So, briefly: transcendence is no magical excuse to stop thinking :) and with practice and over time you will develop a feeling for which equations are hopeless and which not so much.

  • $\begingroup$ How can we identify a particular equation as being transcendental? Is it simply if you can rearrange so that you get $x=h(x)$ where $h(x)$ is a transcendental function (from your second link)? I can see why my examples are transcendental if this is the case. $\endgroup$ – John Doe Apr 30 '17 at 3:41
  • $\begingroup$ thanks for asking - I think this actually lead to a much better answer (I just updated it!) $\endgroup$ – amakelov Apr 30 '17 at 3:52
  • $\begingroup$ Thanks, this clears some things up :) $\endgroup$ – John Doe Apr 30 '17 at 4:11

Yes, there is a way of justifying this.

"Closed form" means expressions of allowed functions (Wikipedia: Closed-form expression). If an equation is solvable in closed form depends therefore on the functions you allow.

For the elementary functions, there is a structure theorem that can help to decide if a given kind of equations of elementary functions can have solutions that can be expressed as function value of an elementary function.

The elementary functions are according to Liouville and Ritt those functions of one variable which are obtained in a finite number of steps by performing algebraic operations and taking exponentials and logarithms (Wikipedia: Elementary function).

The incomprehensibly unfortunately hardly noticed theorem of Joseph Fels Ritt in Ritt, J. F.: Elementary functions and their inverses. Trans. Amer. Math. Soc. 27 (1925) (1) 68-90 answers which kinds of Elementary functions can have an inverse which is an Elementary function. You can also take the method of Rosenlicht, M.: On the explicit solvability of certain transcendental equations. Publications mathématiques de l'IHÉS 36 (1969) 15-22.

For the functions $f$ you gave, no expln-form of Ritt exists that has the structure $A(f_{1}(x))$, where $f_{1}$ is a transcendental function and $A$ is an algebraic function of only one variable. Therefore both $f$ cannot have an elementary inverse, according to the theorem of Ritt.

If $f$ can be decomposed into compositions of algebraic functions and other known Standard functions than $\exp$ and $\ln$, an analog theorem to the theorem of Ritt of [Ritt 1925] could be applied. I hope to prove such a generalization of Ritt's theorem for this class of functions.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.