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Simple math

  1. Nov 6, 2006 #1
    if i have two functions |g(x)|<e, |f(x)|<=1, and i perform the following operation |g(x)|*|f(x)| will the result be |f(x)||g(x)|<e or |f(x)||g(x)|<=e

    how do i even go about proving such things?
     
  2. jcsd
  3. Nov 6, 2006 #2

    HallsofIvy

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    If x< a, y< b, all numbers positive, then, multiplying on both sides of the first inequality by y, xy< ay. Multiplying on both sides of the second inequality by a, ay< ab. By the transitive law, then, ax< ab.
     
  4. Nov 6, 2006 #3
    Sorry, i forgot to put a question mark on the general question.
    if i understand what you are saying correctly, the result of the multiplication of |f(x)|<=1 by |g(x)|<e will be |f(x)||g(x)|<e
    not |f(x)|g(x)|<=e ?
     
  5. Nov 6, 2006 #4

    Office_Shredder

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    Suppose |f(x)|=1, the maximum it can be. Then no matter what g(x) is, you can't quite get to |f(x)||g(x)| = e

    So it's a strong inequality. But you can use the weak one if you want, since it certainly is true too
     
  6. Nov 6, 2006 #5

    HallsofIvy

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    If |f(x)|<= 1 and |g(x)|< e, then |f(x)||g(x)|<= |g(x)|< e.
    |f(x)||g(x)|< e.
     
  7. Nov 6, 2006 #6
    thank you
    for your help, all of you
    i really feel stupid for not seeing it before, i usually go off even if it takes me days to work out problems, but my mind has been awafully loaded lately
    i am really sorry for bothering you with such pettiness
    again thank you.
     
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