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Solving non-elementary functions

  1. Apr 2, 2009 #1

    Mentallic

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    I'm curious to understand why there are such equations that cannot be solved with elementary methods, but rather only numerically.

    e.g. [tex]2^x=x^2[/tex]

    Is there a proof that claims that elementary methods cannot be used to solve such an equation, or is it that there could quite possibly be methods to solve these, but they are too complicated and yet to be discovered?
     
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  3. Apr 2, 2009 #2

    MathematicalPhysicist

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    I am not aware of any such proofs, but it would be tremendous if there were.

    But I guess because no matter what operation you will do on the equation you will still get an iterative equation which you cannot go outside it.

    I mean usually in numerical analysis, for example your equation we may take log_2 on the equation and recieve the equation: [tex]x=2log_2 (x)[/tex] which can only be solved by iterations.
    I guess a proof should show that you cannot differentiate the unique solution(s) from the equation, providing that both of them are defined in the same interval, and indeed both of the functions intersect, if I'm not mistaken at both ends (negative and plus).
     
  4. Apr 2, 2009 #3

    Mentallic

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    Yes I would've also found it tremendous that any such legitimate proofs exist but while I always see the words "it can't be solved with elementary functions" I'd also like to see a proof to support this statement. Can't it be solved because mathematicians have yet to find any method of doing so, or is it impossible for some reason, such as how you've said it can only be solved by iterations?

    Looking back in history, there would've been a time when linear equations [tex]ax+b=0[/tex] could be solved, however [tex]ax^2+bx+c=0[/tex] were supposedly unsolvable, until the method of completing the square was discovered.
    Could we just be in a primitive version of mathematical history that have yet to find links between [tex]a^x-x^a=0[/tex] and many other examples?
     
  5. Apr 2, 2009 #4

    MathematicalPhysicist

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    Type transcedental equation, and you will find your answer apparently they have closed form solutions but via special functions which is still an approximation.
     
  6. Apr 2, 2009 #5

    CRGreathouse

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    Solving a univariate linear equation requires subtraction and division. Solving a univariate quadratic equation requires in addition root extraction.

    Similarly, solving a^x = x^a requires Lambert's W (or similar functions). As we gain reasons to solve equations like this support for and knowledge of these special functions will increase.
     
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