Results 1 to 6 of 6
Like Tree1Thanks
  • 1 Post By majamin

Thread: Writing a polynomial as a product of it's roots

  1. #1
    Ant
    Ant is offline
    Member
    Joined
    Apr 2008
    Posts
    145
    Thanks
    4

    Writing a polynomial as a product of it's roots

    Let $\displaystyle P(z)$ be a polynomial. We're trying to show that we can write:

    $\displaystyle P(z)=c(z-z_1)^{m_1}(z-z_2)^{m_2}...(z-z_{k})^{m_k}$.

    If $\displaystyle P(z)$ has a root $\displaystyle z_1$ (and it does by the FTA) then we can write:

    $\displaystyle P(z)=(z-z_1)^{m_1}P_1(z)$ for some entire function $\displaystyle P_1(z)$.

    Next, in my notes it says "Clearly, $\displaystyle |P_1(z) \leq C(|z|+1)^n $

    I really don't see how that is "clear", could anyone help?

    Thanks.
    Last edited by Ant; Apr 22nd 2013 at 06:10 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Mar 2013
    From
    BC, Canada
    Posts
    106
    Thanks
    19

    Re: Writing a polynomial as a product of it's roots

    Quote Originally Posted by Ant View Post
    Let $\displaystyle P(z)$ be a polynomial. We're trying to show that we can write:

    $\displaystyle P(z)=c(z-z_1)^{m_1}(z-z_2)^{m_2}...(z-z_{m_k})^{m_k}$.

    If $\displaystyle P(z)$ has a root $\displaystyle z_1$ (and it does by the FTA) then we can write:

    $\displaystyle P(z)=(z-z_1)^{m_1}P_1(z)$ for some entire function $\displaystyle P_1(z)$.

    Next, in my notes it says "Clearly, $\displaystyle |P_1(z) \leq C(|z|+1)^n $

    I really don't see how that is "clear", could anyone help?

    Thanks.
    Did you mean to have subscript of k instead of m_k in the last factor? As in:

    $\displaystyle P(z)=c(z-z_1)^{m_1}(z-z_2)^{m_2}...(z-z_{k})^{m_k}$

    Are there further details for the z_i's? For instance, $\displaystyle z_i \leq 1$?
    Thanks from Ant
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Ant
    Ant is offline
    Member
    Joined
    Apr 2008
    Posts
    145
    Thanks
    4

    Re: Writing a polynomial as a product of it's roots

    yeah I did, I'll edit that now. Thanks.

    There are no further restrictions on the z_i's, they are the roots distinct roots of P(Z) in $\displaystyle \mathbb{C}$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Ant
    Ant is offline
    Member
    Joined
    Apr 2008
    Posts
    145
    Thanks
    4

    Re: Writing a polynomial as a product of it's roots

    Okay, I think I might have it.

    I think $\displaystyle n$ is the degree of our polynomial $\displaystyle P(z)$ and so $\displaystyle P_1(z)$ is a polynomial of degree $\displaystyle n-m_1$.

    Then, I think the inequality follows? Could anyone confirm this?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Mar 2013
    From
    BC, Canada
    Posts
    106
    Thanks
    19

    Re: Writing a polynomial as a product of it's roots

    Quote Originally Posted by Ant View Post
    Okay, I think I might have it.

    I think $\displaystyle n$ is the degree of our polynomial $\displaystyle P(z)$ and so $\displaystyle P_1(z)$ is a polynomial of degree $\displaystyle n-m_1$.

    Then, I think the inequality follows? Could anyone confirm this?
    Since $\displaystyle P(z)=c(z-z_1)..(z-z_k)$, wouldn't it make sense that deg(P(z))=k? Not sure how "n" fits into this. And I still think that would be too "handwavy" to say it follows thereafter. Regardless,

    $\displaystyle |P(z)|=|c(z-z_1) ... (z-z_k)| \leq |c(z-z_1)||(z-z_2)...(z-z_k)| \leq C |(z-z_2)|...|(z-z_k)|$

    The reason why I asked about the z_i earlier was because it would follow nicely that if for each z_i, $\displaystyle |z_i|\leq 1$ then

    $\displaystyle C |(z-z_2)|...|(z-z_k)| \leq C (|z|+|z_2|)...(|z|+|z_k|) \leq C (|z| +1) ... (|z| +1) = C(|z|+1)^{k-1}$

    Otherwise, if this restriction on z_i isn't as above, then more will have to be done to bound the polynomial. Thoughts?
    Last edited by majamin; Apr 22nd 2013 at 09:41 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Mar 2013
    From
    BC, Canada
    Posts
    106
    Thanks
    19

    Re: Writing a polynomial as a product of it's roots

    Another attempt here, bypassing the need for $\displaystyle |z_i|\leq 1$:

    let $\displaystyle z^* = \max\{|z_1|,|z_2|,...\}$ then $\displaystyle P_1(z) \leq C |z+z^*|...|z+z^*| = C|z-z^*|^{k-1} \leq C (|z|+|z^*|)^{k-1} \leq C (|z|+1)^n$

    where $\displaystyle n = (k-1) \log{(|z|+z^*)}/\log{(|z|+1)}$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Writing Polynomial Systems in Matrix Form
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Nov 19th 2012, 11:22 AM
  2. Writing number as a product
    Posted in the Number Theory Forum
    Replies: 0
    Last Post: May 13th 2011, 03:58 AM
  3. Replies: 6
    Last Post: Apr 24th 2010, 11:34 PM
  4. Writing Polynomial functions
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Feb 11th 2010, 06:29 PM
  5. re-writing a quartic polynomial
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Nov 2nd 2009, 06:40 AM

Search Tags


/mathhelpforum @mathhelpforum