Results 1 to 6 of 6

Math Help - Polynomial roots

  1. #1
    Super Member
    Joined
    Mar 2010
    Posts
    715
    Thanks
    2

    Polynomial roots

    I've just noticed that every polynomial of degree 2k+1, with k \ne 0, has to at least have one real root (because complex numbers happen in pairs).
    This is true, right? It's shown in my A-level textbook that if z = x+iy is a root of a polynomial equation with real coefficients, then z = x-iy is
    also a root of the polynomial equation. But I doubt showing just that would be sufficient. How can it be shown?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by TheCoffeeMachine View Post
    I've just noticed that every polynomial of degree 2k+1, with k \ne 0, has to at least have one real root (because complex numbers happen in pairs).
    This is true, right? It's shown in my A-level textbook that if z = x+iy is a root of a polynomial equation with real coefficients, then z = x-iy is
    also a root of the polynomial equation. But I doubt showing just that would be sufficient. How can it be shown?
    2k+1 is odd; therefore, [iMATH]x^{2k+1}[/iMATH] must have at least one real solution. Actually induction may work here.

    For roots of the polynomial, we only have 2 options either x\in\mathbb{R} or x\in\mathbb{C}. If x\in\mathbb{C}, then x=a\pm b\mathbf{i}.
    Last edited by dwsmith; May 13th 2010 at 03:04 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Mar 2010
    Posts
    715
    Thanks
    2
    Quote Originally Posted by dwsmith View Post
    2k+1 is odd; therefore, x^{2k+1} must have at least one real solution.
    But your conclusion is exactly what I'm trying to prove. I've noticed it while sketching odd
    degree polynomials, but I have written it in that way (because I thought it might be better).
    And thanks, I don't know I was thinking when I put the restriction that  k\ne 0.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Since complex answers come in pairs, they are of the form 2p which is even.

    x^{2p+1}=x^{2p}x

    The roots of x^{2p} are complex.

    Since x is of degree 1, x must be a real solution.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by TheCoffeeMachine View Post
    I've just noticed that every polynomial of degree 2k+1, with k \ne 0, has to at least have one real root (because complex numbers happen in pairs).
    This is true, right? It's shown in my A-level textbook that if z = x+iy is a root of a polynomial equation with real coefficients, then z = x-iy is
    also a root of the polynomial equation. But I doubt showing just that would be sufficient. How can it be shown?
    Hello friend,

    Your hypothesis is true. Namely that if p(x)=\alpha_{2n+1} x^{2n+1}+\cdots+\alpha_0,\text{ }n\geqslant 1 then p(x)=0 for some x\in\mathbb{R}

    There are many proofs of this. One, as you mentioned is that by the fundamental theorem of algebra p:\mathbb{C}\to\mathbb{C} has exactly 2n+1 zeros (counting multiplicity), \alpha_0,\cdots,\alpha_{2n+1}\in\mathbb{R} then we know that p(x)=(x-\beta_1)\cdots(x-\beta_{2n+1}),\text{ }\beta_{1}\cdots,\beta_{2n+1}\in\mathbb{C}. Now, it is fairly easy to show that if all of the betas are non-real that p must have an imaginary coefficient contradictory to our hypothesis. Thus, p must have a real zero.

    Thinking more along the lines of analysis p:\mathbb{R}\to\mathbb{R} is continuous. But, clearly (why?) \lim_{x\to\infty}p(x)=-\lim_{x\to-\infty}p(x) and thus p(x)<0 at some point and p(x)>0 at some point and the intermediate value theorem implies the existence of a zero.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by TheCoffeeMachine View Post
    I've just noticed that every polynomial of degree 2k+1, with k \ne 0, has to at least have one real root (because complex numbers happen in pairs).
    This is true, right? It's shown in my A-level textbook that if z = x+iy is a root of a polynomial equation with real coefficients, then z = x-iy is
    also a root of the polynomial equation. But I doubt showing just that would be sufficient. How can it be shown?
    It is easily proved using the Intermediate Value Theorem.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. roots of i, polynomial
    Posted in the Algebra Forum
    Replies: 3
    Last Post: February 28th 2011, 06:08 AM
  2. Polynomial Roots
    Posted in the Algebra Forum
    Replies: 4
    Last Post: September 23rd 2010, 11:20 PM
  3. Polynomial Roots
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: March 1st 2010, 09:10 PM
  4. polynomial roots
    Posted in the Algebra Forum
    Replies: 4
    Last Post: September 1st 2009, 12:30 PM
  5. Roots of a polynomial.
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: November 6th 2008, 11:49 AM

Search Tags


/mathhelpforum @mathhelpforum