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How to show some systems of equations do not have a closed form solution?

for example I was once given something similar to ( this might not be the exact problem but i am just using it as an example to convey what type of problems I am talking about):

$$xy^2+y^2x=10 \tag{I}$$ and $$\frac{1}{x}+\frac{1}{y}=15 \tag{II}$$

What are the numerical values that would satisfy both I and II.

(again I reiterate that solving the above is not the main issue, if somebody can give a simpler example where it can be shown that it can NOT be explicitly solved for x and y they are welcome to edit this post), maybe the fact is that a solution does not exist (easy way to look at their graphs).

But is there a way to not end up wasting time trying to explicitly come up with solutions?

Is there a simple example similar to above that can be shown that it can NOT be solved explicitly? I am aware of a computational algebraic result to show that some function do NOT have closed form integrals but nothing to suggest the similar for simple systems of non-linear equations being explicitly solvable.

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@QED : yes,now solve that for y :) the point is NOT solution to this problem either, but to show that some problems can not be explicitly solved, they might look as simple as this one. –  Arjang Dec 30 '11 at 12:08
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Theoretically, as long as your system's entirely algebraic, you can reduce to the solution of some polynomial. When transcendental functions are involved, all bets are off. –  J. M. Dec 30 '11 at 12:20
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At the heart of the matter: What is a closed form solution? In Abel-Ruffini only radicals and basic arithmetic operations are allowed, but if hypergeometric (or theta) series are included, then we have "closed form solutions". Will you allow Bessel-functions? Lambert? Where do you draw the line? Many an interesting equation (or a DE) defines a function as a solution (by implicit function theorem or another relatively high-powered result). How familiar do we need to be with a function to call it "closed form"? In this sense the original question is unfortunately ill-defined. –  Jyrki Lahtonen Dec 30 '11 at 12:52
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That's beautifully written, @Jyrki. I think that's admissible as an answer here. –  J. M. Dec 30 '11 at 12:57
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Well, I had Gröbner in mind when I said "you can reduce to the solution of some polynomial". As long as you have an algebraic system, I don't see why the triangularization isn't possible. –  J. M. Dec 30 '11 at 13:11
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up vote 4 down vote accepted

The notion that some polynomial equations in one variable can't be solved with just radicals - nth roots - requires Galois theory. Specifically, Galois theory shows that there are some 5th degree polynomials which cannot be solved this way.

As others have pointed out, it depends on what you mean by closed form.

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