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Solution to polynomial of unknown degree

  1. Dec 2, 2013 #1

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

  2. jcsd
  3. Dec 2, 2013 #2
    You have sevens unknowns and one equation.

    Could you solve 0 = a + b + c + d + e + f + g + h?
  4. Dec 2, 2013 #3
    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. Dec 2, 2013 #4


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    If it's a polynomial then a and b are natural.
  6. Dec 3, 2013 #5


    Staff: Mentor

    I count six: A, a, B, b, x, and y.
  7. Dec 3, 2013 #6
    omg, my question is simple!

    I would like to know if is possible to isolate the x variable in equation
  8. Dec 3, 2013 #7


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    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. Dec 3, 2013 #8
    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. Dec 3, 2013 #9


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    Staff Emeritus
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    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:

  11. Dec 3, 2013 #10


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    He's not looking to factor the polynomial but rather to find the inverse [itex]f^{-1}(x)[/itex], I think...
  12. Dec 3, 2013 #11
  13. Dec 4, 2013 #12
    Yes if the exponents are positive integers. Consider the general algebraic function, ##y(x)## written implicitly as:


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


    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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