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Math Help - Really basic parametric equations question

  1. #1
    MHF Contributor Mathstud28's Avatar
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    Really basic parametric equations question

    Hi, this is really simple but I just wanted to make sure that since it is so simple that I am not missing something,

    Consider the parametric equations

    x=f(\xi)

    and y=g(\xi)

    where x is of degree Nand y of degree n

    then since math \xi=f^{-1}(x) which I believ would have degree \frac{1}{N}

    and to get our equation into regular form with y as a dependent and x an independent variable we must take

    y=g(f^{-1}(x))

    will our final function in terms of x and y be of degree \frac{n}{N}?
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  2. #2
    Moo
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    Hello,

    Quote Originally Posted by Mathstud28 View Post
    Hi, this is really simple but I just wanted to make sure that since it is so simple that I am not missing something,

    Consider the parametric equations

    x=f(\xi)

    and y=g(\xi)

    where x is of degree Nand y of degree n

    then since math \xi=f^{-1}(x) which I believ would have degree \frac{1}{N}

    and to get our equation into regular form with y as a dependent and x an independent variable we must take

    y=g(f^{-1}(x))

    will our final function in terms of x and y be of degree \frac{n}{N}?
    Prove that t \mapsto f^{-1}(t) exists ^^


    But there's something disturbing me, because the degree of a polynomial is defined as being "the least integer such that the coefficient is null".

    x=f(\xi)=a_N\xi^N+a_{N-1}\xi^{N-1}+\dots+a_0 (assuming that x is a polynomial because you didn't mention it).


    How would you find f^{-1}(x) ? How can you know it's something like the N-th root of x ?
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    Hello,



    Prove that t \mapsto f^{-1}(t) exists ^^


    But there's something disturbing me, because the degree of a polynomial is defined as being "the least integer such that the coefficient is null".

    x=f(\xi)=a_N\xi^N+a_{N-1}\xi^{N-1}+\dots+a_0 (assuming that x is a polynomial because you didn't mention it).


    How would you find f^{-1}(x) ? How can you know it's something like the N-th root of x ?
    Thanks for responding Moo, this is not in any way a proof just an observation, i am saying that if you solved a quadratic for x you would get a squareroot, a cubic would yield a cubic root. I know this is informal but I think it works loosely. For example if you are given a few graphs and you need to match the parametric equations to it, you look for y(t) composed of x inverse of t which by my observation( which could totally be wrong) would be the degree of y over the degree of x.

    So for example if I was given

    x(t)=at+b
    and

    y(t)=cx^3+dx+e

    I would expect

    y(x) to be a cubic function
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  4. #4
    Moo
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    So for example if I was given

    x(t)=at+b
    and

    y(t)=cx^3+dx+e

    I would expect

    y(x) to be a cubic function
    It would be a quartic function, huh ?
    But yeah, this is true

    Quote Originally Posted by Mathstud28 View Post
    Thanks for responding Moo, this is not in any way a proof just an observation, i am saying that if you solved a quadratic for x you would get a squareroot, a cubic would yield a cubic root.
    While dealing with functions which are defined to be polynomials, yep.
    But since you were talking about f^{-1}, it just didn't make sense... for me at least..perhaps it is different for someone else >.<


    I know this is informal but I think it works loosely. For example if you are given a few graphs and you need to match the parametric equations to it, you look for y(t) composed of x inverse of t which by my observation( which could totally be wrong) would be the degree of y over the degree of x.
    I don't really agree with that... Because it would mean that the inverse function is also a polynomial... Prove it ? =)
    Plus, like I said, the degree of a polynomial is an integer (it was quoted from Wikipedia)

    Polynomial - Wikipedia, the free encyclopedia << they also give examples which are not polynomials.
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    It would be a quartic function, huh ?
    But yeah, this is true


    While dealing with functions which are defined to be polynomials, yep.
    But since you were talking about f^{-1}, it just didn't make sense... for me at least..perhaps it is different for someone else >.<



    I don't really agree with that... Because it would mean that the inverse function is also a polynomial... Prove it ? =)
    Plus, like I said, the degree of a polynomial is an integer (it was quoted from Wikipedia)

    Polynomial - Wikipedia, the free encyclopedia << they also give examples which are not polynomials.
    I never said the inverse was a polynomial, just an expression of degree 1/N
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  6. #6
    Moo
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    Quote Originally Posted by Mathstud28 View Post
    I never said the inverse was a polynomial, just an expression of degree 1/N
    Two things :
    - The term "degree" applies, as far as I know, to polynomials
    - The degree has to be an integer (I can't find any certain source for it, I read it in the wikipedia and Mathworld Wolfram doesn't mention it)
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    Two things :
    - The term "degree" applies, as far as I know, to polynomials
    - The degree has to be an integer (I can't find any certain source for it, I read it in the wikipedia and Mathworld Wolfram doesn't mention it)
    But, no offense I am not asking about the semantics of polynomial terminology (although oh so interesting as it is, that was not sarcastic), I am interested if my observation has any truth?
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  8. #8
    Moo
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    Quote Originally Posted by Mathstud28 View Post
    But, no offense I am not asking about the semantics of polynomial terminology (although oh so interesting as it is, that was not sarcastic), I am interested if my observation has any truth?
    Your observation isn't correct since you don't even know if f^{-1} is a polynomial
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