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Analysis of a polynomial (induction)

  1. Feb 26, 2009 #1
    1. The problem statement, all variables and given/known data
    Consider the polynomial

    [tex]p(x) = a_0 + a_1x + \cdots + a_nx^n[/tex]

    where each of [tex]a_0,...,a_n[/tex] is an integer (some of them may be non-positive), and [tex]a_n \geq 1[/tex]

    a) Show that there is [tex]k_0 \in \aleph[/tex] such that [tex]p(k) \geq 2 for each k \geq k_0[/tex]

    b) Show that there is an integer [tex]k \geq k_0[/tex] such that p(k) is not a prime number.


    2. Relevant equations
    Hint: Consider a polynomial [tex]q(y) = p(k_0 + y)[/tex]


    3. The attempt at a solution
    I've attempted (a)... but (b) I'm clueless. I tried (a) by induction...
    Want to show [tex]p(k_0) \geq 2[/tex]
    Base: x=2
    Since [tex]a_n \geq 1[/tex], take the lowest possible value it could be, ie, [tex]p(k) = a_0 + a_1(x) = 0 +1(x)[/tex]
    Then [tex]p(2) = 0 + 1(2) = 2[/tex]
    So the base case holds, [tex]p(k_0) \geq 2[/tex]

    Induction
    Assume [tex]p(k_0)[/tex] true.
    [tex]p(k_0 + 1) = a_0 + a_1(k_0+1)[/tex]
    [tex]=[a_0 + a_1(k_0)]+a_1 [/tex]
    [tex]=[ 2 ] + 1[/tex] by induction hypothesis, and since [tex]a_n \geq 1[/tex]
    [tex]=3 \geq 2 \[/tex]
    So [tex]p(k_0+1) \geq 2 \forall k \geq k_0[/tex]

    First, is this legitimate reasoning? I'm still trying to get the hang of induction/analysis/etc. For example, [tex]a_0,\cdots,a_n[/tex] - does this mean that if a1 =1, then a2 =2, a3=3 and so on, just so long as [tex]a_n \geq 1[/tex] ?

    Second, I'm not even sure where to start on (b). I'm assuming the hint will be useful... but where should I take it?

    Thank you! (ps, I also just recently learned basic LaTeX... super useful that this forum has it! Also, takes some damn time getting used to writing things like that, haha! How do I do a new line? Just a backslash ( \ ) won't work...)
     
    Last edited: Feb 26, 2009
  2. jcsd
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