How to know if I can't solve an equation with "standard" methods?

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Solution 1

The problem you pose stems from an implicit false dichotomy introduced from the various meanings of "solve".

Any equation can be solved, whether polynomial, linear or transcendental. The algebraic process of solving usually involves algebraic transformations which preserve or do not preserve the equivalence of the solutions to the original one. When it does not preserve it, care is taken to examine superfluous or extraneous solution.

The last step, not always but usually, involves inverting a given function. The problem is that the domain of functions which are algebraically invertible (i.e. the inverse has a closed form which allows direct calculations) is somewhat more limited than the domain of functions which are generally invertible.

Suppose, for example, I have reached the step and need to solve:

$$e^z+z^2+1/z+z^8=0$$

Solving for $z$ amounts to inverting $f(z)=e^z+z^2+1/z+z^8$. The existence of such an inverse is always guaranteed by the Lagrange Inversion Theorem, so away from critical points, there always exists an $f^{-1}(w)$, such that:

$$w=f(z)\Leftrightarrow$$

$$z=f^{-1}(w)$$

Call this $f^{-1}$ "myCrazyInverse(w)". It exists and is well defined in a neighborhood of a good solution, provided the solution is not a critical point of $f$.

This function can now be approximated, using numerical methods, provided it is fairly smooth locally and therefore the original equation can be solved to the user's $\epsilon$ satisfaction.

The inverse of this particular example either has a closed form, or it doesn't. If it does, we dump this example to the list of "alebraically" solvable examples and are done with, since the inversion can be done manually. If it doesn't, it may take a while for mathematicians to hold-on to this function in the mathematical memory of things, depending on usage.

For example, it's fairly useless to classify the inverse of this $f$ in the official archives of mathematical functions, because its use has been very rare. Have you seen it before? Chances are nil. It is a very rare kind of function (the inverse) and functions are usually classified in order of some use.

The most notable (and fairly late) example you may probably already be familiar with, is the Lambert $W$ function. Its use has become somewhat "standard" lately, not because it just "solves" the transcendental equation $z\exp(z)=w$, rather because mathematicians have encountered it often-enough in various branches of science, such as astronomy, quantum physics and optics, away from mathematics themselves, which guaranteed a fairly frequent usage.

If you didn't know its properties, it would be fairly useless, too. The following implication (as an algebraic step) is actually valid:

$$z\cdot \exp(z)=y\Leftrightarrow z=W(y)$$

but away from knowing actual ways to calculate values $W(y)$, it is as moot a step as the inversion above with "mycrazyinverse".

It just so happens that this function is of use in many fields, hence the step of the inversion of $z\exp(z)=y$ has been included in the allowed algebraic transformations which solve all the related transcendental equations and provide a "closed form" solution.

Even under the above notes, many mathematicians still doubt the inclusion of this inversion step, given the multivalued nature of (most crazy) inverses and I actually sympathize with them. In view of the multivalued nature of some maps, it can be argued that such steps are valid algebraically only under specific qualifications, which preclude semantic mathematical nonsense, such as:

$$z\cdot\exp(z)=y\Leftrightarrow z=W(y)$$

The correct step is qualified by the branch index only and denotes an equivalence of sets:

$$z\cdot\exp(z)=y\Leftrightarrow z=W(k,y),\,\,\,k\in\mathbb{Z}$$

You can wittness the resulting confusion in answers which use the Lambert $W$ function to solve equations, as a wrong answer: $x=W(whatever)$. Whenever this function solves an equation, it always specifies infinitely many solutions, whether real or complex, because of the branch index. It's never a single value, whether the teacher wants only real solutions or not.

Addednum (for clarification for #2)

For some reason I didn't see that last question, so I am adding a reply here.

The Lambert $W$ function enjoys certain algebraic properties, which allow it to solve many transcendental equations, such as close variants of the original $z\cdot \exp(z)=y$, like $z^3\cdot \exp(z)=y$.

All these are exactly the equations which through some valid algebraic manipulations can be eventually be brought to the form:

$$f(z)\cdot\exp(f(z))=y\,\,\,\,(1)$$

for some $f$ and some $y$, since the solutions can then be described as:

$$f_k(z)=W(k,y)\Leftrightarrow z_k=f^{-1}(W(k,y)),\,\,\,\,k\in\mathbb{Z}$$

Maple contains an algorithm which does exactly that in each given case, but the algorithm's success alone is not a proof, because the algorithm has to stop sometime.

Fortunately, correct algebraic syntax makes for a solvable algebra/group (by definition), so you can be sure that if some equation can be factored into form (1), it will be factored in $O(g(n))$ time, where the latter is the factoring time of the algorithm and this gives you deterministic certainty with a very small fail rate $\epsilon(g(n))>0$.

Now push $\epsilon\to 0$ and you've proved it.

Solution 2

"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.

A general method for solving a given equation $H(x)=0$ is to apply the compositional inverse $H^{-1}$ of $H$: $x=H^{-1}(0)$. In general, $H$ and $H^{-1}$ are correspondences. But often it is possible to split the problem into subproblems where $H$ and $H^{-1}$ are functions. For applying this method, $H$ and $H^{-1}$ have to be known. That means they have to be in closed form.

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 1925] 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.
Risch also proved this theorem in Risch, R. H.: Algebraic Properties of the Elementary Functions of Analysis. Amer. J. Math 101 (1979) (4) 743-759.

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.

Two methods for simpler transcendental elementary equations are given in [Lin 1983] Ferng-Ching Lin: Schanuel's Conjecture Implies Ritt's Conjectures. Chin. J. Math. 11 (1983) (1) 41-50 and in [Chow 1999] Chow, T.: What is a closed-form number. Am. Math. Monthly 106 (1999) (5) 440-448.

If a given function 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.

If you can find a representation of your given function $H$ that is a composition of functions that doesn't contain a multivariate algebraic function and if the inverses of all member functions of this composition representation are known, you can apply the single inverses in opposite order and gets the whole inverse in closed form.

For equations solvable by Lambert W, see my answer here: Algebraic solution to natural logarithm equations like $1-x+x\ln(-x)=0$ and my comment here: https://mathoverflow.net/questions/289248/what-are-the-most-general-methods-for-solving-equations-in-closed-form-with-lamb.

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Updated on July 07, 2020

Comments

  • Renato Faraone
    Renato Faraone over 3 years

    I'm particularly fascinated by transcendental equations whose posses closed form solutions and when I pose some of them to my friends or teachers I heard a lot of "You can't solve this in closed form" or "You can't solve this with ordinary functions" and I always thought: when and how can I be sure that this is the case ?

    In other words suppose that I have an equation in one variable involving some exponentials and polynomials or even trigonometric functions etc. 99% of the time if the solution is nontrivial no one can get to an explicit formula but only to an approximation. But shouldn't we actually have to prove that ordinary functions is not enough to solve such an equation ? I mean may there is a solution which is 3 pages long but contains only common functions.

    So what I want to know is:

    1) If I have an equation how do I know that I can't solve it using any finite composition of polynomials, exponentials, trigonometrics and their inverses ?

    2) If I'm in the case in which I can't and I decide to add one new function to the list (take Lambert's as an example) can I apply a similar reasoning to know when I can use this new function?

    clarification for 2) take as an example the equation $xe^x=1$ and let the Lambert's function represent its solution: $x=W(1)$ if then I want to solve $x^2e^x=1$ I don't need to introduce another function to my list, the solution still can be expressed in terms of Lambert's function: $x=2W(\frac 12)$ but if I then approach $xe^x+x=1$ I need another function? How can I prove/disprove this?

    • Asinomás
      Asinomás almost 8 years
      There is a simple algorithm: try to solve it with standard methods, fail, lie down, try not to cry, cry a lot.
    • David K
      David K almost 8 years
      If you do not even have transcendental functions, only a polynomial, you still might not have a closed-form solution in radicals. The usual explanation goes something like this: "Galois theory mumble mumble." Even the criteria to tell which polynomials above degree 4 have closed-form solutions in radicals and which don't are complicated. Is the situation better if we allow a few transcendental functions to be part of "closed form" but open the possible equations to the same kinds of transcendental functions? I don't know, but it doesn't sound good.
    • mweiss
      mweiss almost 8 years
      @DavidK "Galois theory mumble mumble" is a wonderful phrase and I am pretty sure I have said it a bunch of times.
    • Renato Faraone
      Renato Faraone almost 8 years
      @DavidK you pretty got the point, actually I heard that there is a new method to solve equation with something called "ultra-radicals" or "hyper-radicals" sry but I don't remember the exact name.
    • Peter
      Peter almost 8 years
      I am not even sure whether the given problem is decideable in general.