1. How to prove an expression is irreducible?

Discussion in 'Physics & Math' started by Secret, Apr 24, 2013.

  1. Secret Registered Senior Member

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    299
    Some months ago I came across the "solving equation with indeterminate answers" and this bit interested me:

    So is it true that for an expression to be reducible, it must be a ratio of polynomials ( where x, and y can be a function) or is there more subtle cases?
    If there is more subtle cases, then in general, how to prove that an arbitrary expression f(x)/g(x) is irreducible?
     
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  3. Fednis48 Registered Senior Member

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    It doesn't have to be a ratio of polynomials. You just have to be able to decompose the numerator into a product of the denominator and something else. Tach's example that you quoted works because \(x^2-y^2=(x-y)(x+y)\) and \(x-y\) is the denominator. A non-polynomial example would be \(\frac{\tan(x)}{\csc(x)}=\frac{\cos(x)\csc(x)}{\csc(x)}=\cos(x)\). (In case you've forgotten, \(\csc(x)=\frac{1}{\sin(x)}\); I wrote it in this form to make it clearer what's in the numerator and what's in the denominator.) I don't know of an algorithmic way to decide whether an expression is irreducible, but if you have a polynomial in one part of the fraction and an exponential in the other, that's generally a bad sign.

    Also, I don't know if this was mentioned in the thread you quoted, but the irreducible example can still be solved by taking the limit as x approaches zero and applying L'Hospital's rule.
     
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  5. Secret Registered Senior Member

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    ok, so I guess another user might be able to tackle the "an algorithmic way to decide whether an expression is irreducible". I'll also take note about the "but if you have a polynomial in one part of the fraction and an exponential in the other, that's generally a bad sign." for my future studies

    As for the "Also, I don't know if this was mentioned in the thread you quoted, but the irreducible example can still be solved by taking the limit as x approaches zero and applying L'Hospital's rule", yes I knew of that, and yes it has mentioned in that thread

    Maybe I should have been more clear in my OP. By reducible, I mean "it is possible to simplify the expression, as long the stuff being cancelled does not involve division by zero"

    So to refine my question it is as follows:
    Q: In general, is there a way to check whether an arbitrary expression, given division by zero is not involved, can be simplified

    P.S. Solving this question might make me understand more clearly why expressions such as x=sin(x) (i.e. x-sin(x)=0) cannot be solved symbolically
     
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  7. rpenner Fully Wired Valued Senior Member

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    You need a concept of a class of all functions that meet a certain property, like \(C^{\tiny \infty}\).

    Then f(x)/g(x) is "reducible" if there is a unique admissible function h(x) such that whenever g(x) is not zero f(x) = h(x) g(x) and whenever g(x) is zero, then f(x) is also zero and the limit of f(x)/g(x) at the zeros of g(x) is equal to the limit of h(x).

    \(\textrm{divides}(g,f) \leftrightarrow \exists ! h \left[ \left( \forall x f(x) = h(x) g(x) \right) \; \textrm{and} \; \forall x \left( g(x) = 0 \rightarrow \left( f(x) = 0 \; \textrm{and} \; \lim_{y\to x} \frac{f(y)}{g(y)} = \lim_{y\to x} h(y) \right) \right) \right] \)

    So if the class of "all functions" are polynomials in \(\mathbb{C}\) this conforms to the standard definition of polynomial division.

    Likewise if the class is "all smooth functions in \(\mathbb{R}\) with isolated singularities" we have "sin(2x)/cos(x) = 2 sin(x)". And "tan(x)/sec(x) = sin(x)". But not "sin(x)/cos(x) = tan(x)", because cos(x) has zeros where sin(x) is finite, so the domain of f is not the domain of h.
     

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