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Thread: Arbitrary polynomial: Algebraic impossibility?

  1. #1 Arbitrary polynomial: Algebraic impossibility? 
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    Why is it impossible for to derive the form ?

    I know , but I cannot calculate .

    I've tried Wolfram Alpha with several different assumed forms of input, but it was inconclusive. Why isn't it possible?


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  3. #2  
    Moderator Moderator Markus Hanke's Avatar
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    This is what is called a "transcendental equation" :

    Transcendental equation - Wikipedia, the free encyclopedia

    Many such equations cannot be solved analytically, but it may be possible to do it via numerical methods.


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    Quote Originally Posted by Markus Hanke View Post
    This is what is called a "transcendental equation" :

    Transcendental equation - Wikipedia, the free encyclopedia

    Many such equations cannot be solved analytically, but it may be possible to do it via numerical methods.
    Thanks Markus Hanke. The link helped me understand it much better.

    I know what a transcendental number is, but I'm not sure what transcendental function is. So it can't be expressed "in algebraic terms". What exactly does that mean? What makes not algebraic?
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    Quote Originally Posted by brody View Post
    I know what a transcendental number is, but I'm not sure what transcendental function is. So it can't be expressed "in algebraic terms". What exactly does that mean? What makes not algebraic?
    One way to think about it is to imagine that the only operations you are permitted to use are addition, multiplication and root-taking. If your function is expressible in terms of a finite number of any combination of those operations, the function is said to be algebraic. If you need other operations, or an infinity of any, then the function goes beyond ("transcends") algebra, hence the name "transcendental function."

    In the specific case you've presented, it's the exponential term that gets in the way if the variable y can take on any value. For some specific values of y (say, 2, for an easy example), the resulting equation is algebraic. But in general, it isn't.

    As Markus said, you need to resort to numerical or graphical methods in general to solve such equations.
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    Quote Originally Posted by tk421 View Post
    Quote Originally Posted by brody View Post
    I know what a transcendental number is, but I'm not sure what transcendental function is. So it can't be expressed "in algebraic terms". What exactly does that mean? What makes not algebraic?
    One way to think about it is to imagine that the only operations you are permitted to use are addition, multiplication and root-taking. If your function is expressible in terms of a finite number of any combination of those operations, the function is said to be algebraic. If you need other operations, or an infinity of any, then the function goes beyond ("transcends") algebra, hence the name "transcendental function."

    In the specific case you've presented, it's the exponential term that gets in the way if the variable y can take on any value. For some specific values of y (say, 2, for an easy example), the resulting equation is algebraic. But in general, it isn't.

    As Markus said, you need to resort to numerical or graphical methods in general to solve such equations.
    Thank you tk421. Great explanation.

    When I was introduced to trigonometry, I learned the right-angle approach to finding the sines, cosines, etc. of some number. And I wondered why you had to set up a geometric figure and look at other values instead of just using some operations directly on x.

    I was thinking there had to be some algebraic representation of the trigonometric functions like or something like that, other than what I later learned using the complex exponential functions. But now I understand what it means that the trigonometric functions are transcendental.
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    Forum Senior TheObserver's Avatar
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    If you are looking for the why, I think that you would want to study Galois Theory.
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    Quote Originally Posted by TheObserver View Post
    If you are looking for the why, I think that you would want to study Galois Theory.
    Thanks for the reference. But the concept seems to be a little above my level of understanding.
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    By the way, I've actually tried to find an algebraic representation for the trigonometric functions. The closest I've gotten was to the tangent function with a non-periodic approximation. This is how I arrived at it...

    If I were to graph the equation I would derive the unit circle, making it a rather interesting relation. And by taking it as a function for only the principal square root and differentiating the function, I would arrive at a function which looks similar to the tangent.

    Courtesy of Google:


    The first being and the second being .

    The derivative has the general look of the tangent function, and now I can take the opposite of that making it positive, and expand the domain to . And then stretch it ordinately by 2 and voila, something close to the tangent function. Although it's not periodic.



    The second one is .

    All in all, I realized this was a futile and pointless attempt. Thanks to transcendentiality. Argh!
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  10. #9  
    Forum Senior TheObserver's Avatar
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    Quote Originally Posted by brody View Post
    Quote Originally Posted by TheObserver View Post
    If you are looking for the why, I think that you would want to study Galois Theory.
    Thanks for the reference. But the concept seems to be a little above my level of understanding.
    Thats ok, mine too.
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  11. #10  
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    Ya know, it is not at all in the spirit of this subforum to answer genuine questions by posting internet links, least of all to Wikipedia - we all have internet access and (presumably) have all heard of Wikipedia. Especially if the responder doesn't really know what they are talking about

    So. A number is said to be algebraic if it is the zero of a polynomial with rational coefficients. Note that the zeros need not be drawn from the same field (more precisely commutative ring) as the coefficients..

    For consider with implied realcoefficient . The zeros are and .

    Now consider again with implied real coefficient. Then the zeros are and which are not real

    Now we need to distinguish between a polynomial form and a polynomial function.

    A polynomial form, say, does not require that the coefficients be drawn from the same field (rather ring, even rather integral domain) as the . A polynomial function DOES, as is easily seen........

    Consider the polynomial forms and over the quotient ring . (Actually it is a field, since is prime). Whatever. These two distinct forms refer to the same same polynomial function - namely the function that is identically zero.

    So. A function, a polynomial function in the above sense, is algebraic if it is the zero of a polynomial function with rational polynomial coefficients.

    Ummm.....as to what this means, specifically what is meant by a "rational polynomial" I need to think awhile. Plus I have a coupla texts here that may help us.

    You may want to stay tuned, but I suspect not............
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    Moderator Moderator Markus Hanke's Avatar
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    Quote Originally Posted by Guitarist View Post
    Ya know, it is not at all in the spirit of this subforum to answer genuine questions by posting internet links, least of all to Wikipedia
    Well, it seems to have done some good, considering the response it has solicited :

    "Thanks Markus Hanke. The link helped me understand it much better."

    Especially if the responder doesn't really know what they are talking about
    I would be interested to know how you arrived at this conclusion ?
    I am not a mathematician, in this you are correct. However, I do recognize a transcendantal function when I see one, so I merely pointed brody in the right direction.
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    Markus - don't take my last remark personally, it was not directed specifically at you.

    Look, this is your forum (collective you!), and, within limits, it is not for me to tell you how to use it. I am, however, entitled to my opinions, and these are:

    1. If a questioner needs someone else to do their Google search for them, one my reasonably ask whether they deserve a response at all;

    2. If a member has done their own search and still feels the need to ask a question, it seems reasonable to assume they haven't understood what they have read on the internet. Further links are unlikely to be of much help;

    3. There is a fantastic amount of mis-information out there; it requires a reasonable knowledge of the subject at hand to sort out the pearls from the swine

    4. The day this site becomes little more than an exchange of internet hyper-links will be the day I move on. It is meant to be a discussion board.
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  14. #13  
    Moderator Moderator Markus Hanke's Avatar
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    Quote Originally Posted by Guitarist View Post
    Markus - don't take my last remark personally, it was not directed specifically at you.
    That's alright, I understand.

    4. The day this site becomes little more than an exchange of internet hyper-links will be the day I move on. It is meant to be a discussion board.
    Never really thought about it this way, but I suppose you probably have a point here in general.

    In this particular case though one could argue that the OP genuinely was not familiar with the concept/term "transcendental function", and was thus unable to Google it. The pointer in the right direction could have just been what he needed, and he can go and do his own research now that he knows what to actually look for.
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