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Consider for example the function:

$$x^2 e^y + \log(x)y^2 = 0$$

I suspect that neither x nor y can be isolated in this function (that it, it cannot be written as either $x(y)$ or $y(x)$. However, I can't really prove that other than to say "it's difficult to do if not impossible".

Is there a method by which it can be proven that no finite number of algebraic operations will lead to such a simplification? (The above function was presented as an example, but I am looking for a general solution)

DMcMor
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Michael Stachowsky
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  • i think that is impossible – Dr. Sonnhard Graubner Apr 10 '17 at 18:25
  • I think it depends on what elementary operations you allow. Your original expression contains $log$ so apparently at least some "common" functions are allowed in your answer. Specifically, do you consider the Lambert W function permissible in the solution? In which case $y=2W(\frac{-\sqrt{log(x)}}{2x})$ is a solution. – Χpẘ Apr 10 '17 at 18:28
  • @Dr.SonnhardGraubner: Is it that you think the general method does not exist, or that it is not possible to isolate for any of the variables? – Michael Stachowsky Apr 10 '17 at 18:30
  • @user2460798: Overall, I would like to prove whether or not a method exists to write the function as either $x = f(y)$ or $y = f(x)$. The functions required to do so may be horrifying, but that's OK. – Michael Stachowsky Apr 10 '17 at 18:31
  • I understand that goal, but as my updated comment above shows, the answer is dependent on what functions you allow. To further elaborate, one could make up a function especially designed to make your implied function expressible as $y(x)$. You could call it the $MS$ function. But that isn't interesting. What is interesting is, based on some "vocabulary" of functions and operations, can you turn the function into an explicit form. – Χpẘ Apr 10 '17 at 18:33
  • Interesting and good point. I suppose, then, that I'd limit it this way: First, I'd limit it only to implicitly defined analytic functions. Then, I'd require that the isolation be in terms only of other analytic functions. Unless you can suggest a better limitation? – Michael Stachowsky Apr 10 '17 at 18:40
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    I'd suggest looking at http://math.stackexchange.com/questions/43263/if-a-function-can-only-be-defined-implicitly-does-it-have-to-be-multivalued, which doesn't directly address your question, but is relevant. My sense is to copy how explicit defined functions are usually presented: using "well known" functions, certainly, log, exp, trig, arctrig, hyperbolic trig, hyperbolic argtrig, etc. Power series also make sense. Beyond that I'm inclined to believe it depends on your audience, and tradition of the math discipline you're targeting - such as well known integrals, special functions, etc. – Χpẘ Apr 10 '17 at 18:53

1 Answers1

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By the way, this is obviously impossible using any kind of function iff there are two different solutions for the equation in x for any y and vice versa.

Considering some general implicit definition of the form $f(x,y) = 0$, this cannot be written as $x(y)$ if there exists a $y$ for which there are two $x_1, x_2$ with $f(x_1,y) = f(x_2,y) = 0$ and $x_1 \ne x_2$ as this would imply $x(y) = x_1$ and $x(y) = x_2$ in contradiction to $x_1 \ne x_2$, so $x(y)$ could not be a function. Uniqueness of the solutions of $f(x,y) = 0$ for any $y$ would therefore be necessary for $x(y)$ to be a function.

wave
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  • incorrect or just too obvious? – wave Apr 10 '17 at 18:55
  • I think the downvotes were because the answer was not rigorous enough or backed up well. Perhaps you could elaborate a bit and explain what you meant. Could you give an example? – Michael Stachowsky Apr 10 '17 at 19:09
  • I don't think the OP meant a function in the narrow sense - in the same way that the Lambert W function is not a function when argument $1/e<x<0$. I think he meant a relation specified by an implicit formula. – Χpẘ Apr 10 '17 at 19:35
  • @Rahul, either that's a deep philosophical insight like "Change is constant" or "I think therefore I am" or you left a word or more out of your comment. – Χpẘ Apr 10 '17 at 21:12
  • @user2460798 I am sure he meant that there always is such a "relation specified by an implicit formula", as the implicit formula already defines a relation. – wave Apr 11 '17 at 07:31