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Math Help - Writing a polynomial as a product of it's roots

  1. #1
    Ant
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    Writing a polynomial as a product of it's roots

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

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

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

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

    Next, in my notes it says "Clearly, |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; April 22nd 2013 at 07:10 AM.
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  2. #2
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    Re: Writing a polynomial as a product of it's roots

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

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

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

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

    Next, in my notes it says "Clearly, |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:

    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, z_i \leq 1?
    Thanks from Ant
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  3. #3
    Ant
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    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 \mathbb{C}
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  4. #4
    Ant
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    Re: Writing a polynomial as a product of it's roots

    Okay, I think I might have it.

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

    Then, I think the inequality follows? Could anyone confirm this?
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    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 n is the degree of our polynomial P(z) and so P_1(z) is a polynomial of degree n-m_1.

    Then, I think the inequality follows? Could anyone confirm this?
    Since 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,

    |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, |z_i|\leq 1 then

    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; April 22nd 2013 at 10:41 PM.
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    Re: Writing a polynomial as a product of it's roots

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

    let z^* = \max\{|z_1|,|z_2|,...\} then 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 n = (k-1) \log{(|z|+z^*)}/\log{(|z|+1)}
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