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P/q continuity

  1. Nov 15, 2008 #1
    1. The problem statement, all variables and given/known data
    Let p and q be a polynomial and x0 be a zero of q of multiplicity m. Prove that p/q can be assigned a value at x0 such that the function thus defined will be continuous there iff x0 is a zero of p of multiplicity greater than or equal to m.

    2. Relevant equations

    3. The attempt at a solution
    I'm not quite sure how to even get started. This question just confuses me in what it's asking.
  2. jcsd
  3. Nov 15, 2008 #2


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    If f(x) is a polynomial, and has a zero at x0 of degree m, then there exists a polynomial g(x) such that f(x) = g(x)*(x-x0)m

    As written, p/q (if q has a zero at x0) is not defined at x0, but is defined continuously in a ball around x0. A standard trick is to extend functions by finding what their limit is as you approach points that aren't in the original domain, and then defining a new function that equals the original function on the original domain, and if you're on one of the non-defined points, you define it to be the limit of the original function.

    So try writing p and q in the form I gave at the top of my post, and then use the tip to consider the relative degrees of the roots and figure out if the limit as x approaches x0 exists
  4. Nov 15, 2008 #3
    Ok so p: there exists a polynomial q such that p=q*(x-x0)m
    Is this right then do the same for q?
  5. Nov 16, 2008 #4
    or p=r*(x-x0)m?
  6. Nov 16, 2008 #5


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    Office Shredder said p(x)= g(x)(x- x0)m, NOT "q(x)(x- x0)m". You cannot use "q" to mean two different polynomials.

    Don't write "p(x)= q(x)(x- x0)m" or "p= r(x- x0)m). Do what Office Shredder suggested!
  7. Nov 19, 2008 #6
    Ok, so I have p=g(x)(x-[tex]x_{0}[/tex])[tex]^{m}[/tex]
  8. Nov 19, 2008 #7
    I'm just confused on where to go next... Like do I do a delta, epsilon proof?
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