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Continuity and Integration by Partial Fractions

  1. Feb 4, 2008 #1
    1. The problem statement, all variables and given/known data
    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]

    3. The attempt at a solution

    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.
     
  2. jcsd
  3. Feb 4, 2008 #2
    It does not need to hold for when Q(a)=0, since you have at the very beggining that this must hold true besides for those x-is, this also icludes x=a such that Q(a)=0.
     
  4. Feb 4, 2008 #3
    I'm sorry, I don't quite get that?
     
  5. Feb 4, 2008 #4

    Dick

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    The trouble with the limits argument applied to the quotients F(x)/Q(x) and G(x)/Q(x) is that those limits may not exist. Argue by contradiction. Assume F(a) is not equal to G(a). Then you can write F(a)=G(a)+c for some non-zero c. Since F and G are continuous there is a value of e>0 such that |F(x)-F(a)|<|c|/2 and |G(x)-G(a)|<|c|/2 for |x-a|<e. But F(x)=G(x) if x is not equal to a. Do you see the contradiction?
     
  6. Feb 5, 2008 #5
    Shouldn't we say:
    F(x) = G(x) + P(x)? Where P(x) is some polynomial.
    I'm not quite sure I have learnt the stuff about limits you use in your proof, but I vaguely get it, I think. Stewart relegates this stuff to an appendix (rigorous treatment of limits). I'll go over it and get back to you.
    Thank-you for helping.
     
  7. Feb 5, 2008 #6

    Dick

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    The point is just that if F(a) and G(a) are different then if x is close enough to a, F(x)=G(x) CAN'T be close to BOTH F(a) and G(a). The rest of the words are just an attempt to express that using the limit definition.
     
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