Solution to polynomial of unknown degree

  1. Dear!

    Is possible to solution a polynomial of kind y = Ax^a + Bx^b ?

    Thx!
     
  2. jcsd
  3. You have sevens unknowns and one equation.

    Could you solve 0 = a + b + c + d + e + f + g + h?
     
  4. I assume you are asking if it is possible to find the zeroes of y(x) = Ax^a + Bx^b, where a,b are real numbers?
     
  5. pwsnafu

    pwsnafu 933
    Science Advisor

    If it's a polynomial then a and b are natural.
     
  6. Mark44

    Staff: Mentor

    I count six: A, a, B, b, x, and y.
     
  7. omg, my question is simple!

    I would like to know if is possible to isolate the x variable in equation
    [tex]f(x)=ax^{\alpha}+bx^{\beta}[/tex]
     
  8. Mentallic

    Mentallic 3,749
    Homework Helper

    In general, no. If [itex]\alpha[/itex] and [itex]\beta[/itex] are both less than 5, then yes. If they're both less than three, then it's a quadratic or simpler and hence easily done by the quadratic formula. In most other cases it's either difficult or impossible.
     
  9. Yes, it's possible, and quite easy. Let us suppose that [itex]\alpha > \beta[/itex]. Let [itex]\zeta = e^{2\pi i/(\alpha - \beta)}[/itex] be a primitive root of unity. Then the polynomial factors as:
    [tex]ax^\beta \prod_{k=1}^{\alpha - \beta} (x - \left( \frac{b}{a} \right)^{1/(\alpha - \beta)} \zeta^k)[/tex]
     
  10. SteamKing

    SteamKing 9,602
    Staff Emeritus
    Science Advisor
    Homework Helper

    In mathematics, as in other areas of science, just because a question is 'simple', it does not necessarily follow that the solution will be 'simple'.

    For example, consider Fermat's Last Theorem:

    http://en.wikipedia.org/wiki/Fermat's_Last_Theorem
     
  11. Mentallic

    Mentallic 3,749
    Homework Helper

    He's not looking to factor the polynomial but rather to find the inverse [itex]f^{-1}(x)[/itex], I think...
     
  12. Yeah!
     
  13. Yes if the exponents are positive integers. Consider the general algebraic function, ##y(x)## written implicitly as:

    $$f(x,y)=a_1(x)+a_2(x)y+a_3(x)y^2+\cdots+a_n(x)y^n=0$$

    with ##a_i(x)## polynomials. In your case we would simply have:

    $$f(x,y)=x-ay^{\alpha}-by^{\beta}=0$$

    Then by Newton-Puiseux's Theorem, we can compute power series representations of the various branches (solutions) of ##y(x)## having the form:

    $$y_d(x)=\sum_{n=-p}^{\infty} c_n\left(x^{1/d}\right)^n$$

    with radii of convergences extending at least to the nearest singular point of ##f(x,y)## and often further than that.

    Do a search for "Newton Polygon" if you're interested in knowing how to compute these "Puiseux" series.
     
    Last edited: Dec 4, 2013
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