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Question on polynomial orders

  1. Jul 23, 2015 #1
    I am trying to use a numerical polynomial root finding method, but I am unsure of the order of an expression. For example, if I have something that looks like
    x2+5x √(x2+3)+x+1=0
    what is the coefficient of the second order (and potentially even the first order) term? Is the entire 5x√... term second order, or does it have lower order components as well?
     
  2. jcsd
  3. Jul 23, 2015 #2
    I'm thinking, and I'm not an expert on polynomial mathematics, but that the equation you listed isn't actually one. I make that statement with the thought in mind that a term cannot include an irreducible radical as part of the product.

    See more about the definition of a polynomial here: http://www.purplemath.com/modules/polydefs.htm
     
    Last edited: Jul 23, 2015
  4. Jul 23, 2015 #3
    If not a polynomial, then what kind of equation is it? This is asked in an effort to determine a numerical method to use.
     
  5. Jul 23, 2015 #4
    And you can always square both sides to get a polynomial.
     
  6. Jul 23, 2015 #5
    You can square both sides to get a PE, that is true, but that doesn't necessarily make the original equation a PE. Think about it this way. You can square an imaginary number to get a real one, but the operation fundamentally undoes the operation that created the imaginary number in the first place. If you're looking for a root, than I suspect that you've already found your answer to find a solution. Square it and solve. For PE, I know with certainty that changing it's form by multiplying by a constant doesn't change the roots, so I suspect the same is true in this case. For instance, the roots of Q(x) are the same for A*Q(x) where A is in the reals. I suspect squaring both sides, simplifying and solving for the roots will give you the answer. If you want to verify (since you're using numerical methods), you could simply graph the two equations with Geogebra and see if they intersect... actually... I'll pull out my laptop and give that a try myself... you've got me curious.
     
  7. Jul 23, 2015 #6
    Oooh. Bad news... the radical doesn't disappear when you square it.

    LHS2 = x4 + 26x4 + 2x3(5x√x2+3) + 2x2(5x√x2+3) + 2x(5x√x2+3) + 1

    This means that strategy's out the door.
     
    Last edited: Jul 23, 2015
  8. Jul 23, 2015 #7

    mathman

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    Change to [itex]x^2+x+1=-5x\sqrt{x^2+3}[/itex]

    Now square both sides to get a polynomial equation.
     
  9. Jul 23, 2015 #8
    Did it quickly and got:

    [tex]24x^4-2x^3+72x^2-2x-1 = 0[/tex]

    What about using Newton's Method on it? The derivative a polynomial is trivial to compute?
     
  10. Jul 23, 2015 #9
    Thanks, all. Would the roots of the squared equation in the last post be the same as the roots of the starting equation? Multiplying by a constant preserves the roots, but does squaring? If not, what is the transformation from these solutions back to those of the original problem?
     
  11. Jul 23, 2015 #10

    micromass

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    Squaring preserves the roots, but does not reflect the roots. That is, if you have an equation ##f(x) = g(x)## then any ##x## that satisfies this, will also satisfy ##f(x)^2 = g(x)^2##. But not every ##x## that satisfies ##f(x)^2 = g(x)^2## will satisfy ##f(x) = g(x)##. That is: you may have added some roots by squaring. No harm though: just find the roots of ##f(x)^2 = g(x)^2## and check for each root whether it satisfies ##f(x) = g(x)##.
     
  12. Jul 23, 2015 #11
    Thanks everyone!
     
  13. Jul 23, 2015 #12
    Here's a graphical representation of the two equations:
    Two Functions.png
     
  14. Jul 23, 2015 #13
    That's nice to see, a graphical representation always helps think about things. Sadly, this equation was just a thing that I pulled out of thin air as a simpler example for the true problem at hand. The "isolating the root and squaring" method should still hold in my case, though.
     
    Last edited: Jul 23, 2015
  15. Jul 23, 2015 #14
    Despite it, you should be cautious whenever you square the both sides of an equation, for it tends to give more roots that don't meet the origin one.
     
  16. Jul 24, 2015 #15

    julian

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    The solutions of ##f^2(x) = g^2(x)## obviously give the solutions of both ##f(x) =g(x)## and ##f(x) = - g(x)##.

    In any case, doesn't the Newton-Raphson method apply to arbitrary functions you can calculate the derivative of, not just polynomials? If you can obtain a computer generated sketch of the function you can plug in the approx values for the zeroes of the function and iterate with N-R.
     
    Last edited: Jul 24, 2015
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