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Math Help - Continuity and Integration by Partial Fractions

  1. #1
    Junior Member qspeechc's Avatar
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    Continuity and Integration by Partial Fractions

    The problem is from Stewart, Appendix G, A58, no.45.

    Suppose that F, G, and Q are polynomials, and:
    F(x)/Q(x) = G(x)/Q(x)
    for all x except when Q(x) = 0. Prove that F(x) = G(x) for all x. [Hint: Use Continuity]

    I thought the statement was obvious, but ofcourse it isn't if they want you to prove it.

    This is under the section "Integration of Rational Functions by Partial Fractions", so I guess the proof must somehow use this technique.

    I guess we know it is true for real numbers a, b, c, that if a/c = b/c then a = b. We just haven't proved it for functions, and I think that is what this is about- am I correct? So the proof needs to be based on the fact that it is true for real numbers- is this correct?
    That in mind, I decided to take the limit as x approaches a (except if Q(a) = 0) on both sides (sorry about not using Latex- I am in a rush):
    lim [F(x)/Q(x)] = lim [G(x)/Q(x)]
    lim[F(x)]/lim[Q(x)] = lim[G(x)]/lim[Q(x)]
    And because they are all polynomials:
    F(a)/Q(a) = G(a)/Q(a)
    Now these are all real numbers, and thus F(a) = G(a).

    This seems a little dodgy to me, because it does not hold for a when Q(a) = 0, and the questions says prove for all a. Also, we say, when we have:
    F(a)/Q(a) = G(a)/Q(a)
    that F(a), Q(a), G(a) are all real numbers, but the answer claims they are functions all-of-sudden!
    Help is much appreciated.
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  2. #2
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    Quote Originally Posted by qspeechc View Post
    The problem is from Stewart, Appendix G, A58, no.45.

    Suppose that F, G, and Q are polynomials, and:
    F(x)/Q(x) = G(x)/Q(x)
    for all x except when Q(x) = 0. Prove that F(x) = G(x) for all x. [Hint: Use Continuity]
    Multiply both sides by Q(x) to get F(x) = G(x) for all x such that Q(x)!=0. There is only a finite # of such x (if Q is a non-zero polynomial) hence we have two polynomials agreeing together for an infinite number of different values thus it must be the same polynomial. And thus F(x) = G(x) for any x.
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  3. #3
    Junior Member qspeechc's Avatar
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    Hmm, I think your proof assumes to much on behalf of my schooling. I know nothing of the theorem:

    two polynomials agreeing together for an infinite number of different values thus it must be the same polynomial
    I guess you are correct, because you know more than me, but is there not a proof with integration by partial fractions, because as I said, I have not been taught that theorem.
    Thank-you for your answer.
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