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Bump function

  1. Aug 23, 2007 #1
    This is not a homework question, just trying to understand the material in my differential geomety lecture.

    Consider the function

    [itex] f : \mathbb{R} \to [0,1][/itex]

    given by [itex] f(x) = \frac{1}{A}\int_{-\infty}^{\infty} a(t) a(1-t)dt[/itex]

    where a(x) is zero for x less than or equal to 0 and [itex]a(x) = e^{-x^{-1}}[/itex] for x > 0 and [itex]A = \int_{0}^{1}a(t)a(1-t)dt[/itex].

    The claim is that f(x) = 0 for x less than or equal to 0 and f(x) = 1 for x greater than or equal to 1.

    The first part is obvious. The second part I'm having trouble with.

    [itex] f(x > 1) = \frac{1}{A}\int_{-\infty}^{\infty} a(t) a(1-t)dt[/itex]
    [itex] f(x > 1) = \frac{1}{A}\left( \int_{-\infty}^0 + \int_{0}^{1} + \int_{1}^x \right) a(t) a(1-t)dt[/itex]
    [itex] f(x > 1) = 1 + \frac{1}{A} \int_{1}^x a(t) a(1-t)dt[/itex].

    So the question becomes, why does the second term vanish? My lecturer claims this is obvious but I just don't see it.
     
  2. jcsd
  3. Aug 25, 2007 #2
    There is a typo in the way I've defined f(x). It should be

    [itex] f(x) = \frac{1}{A}\int_{-\infty}^{x} a(t) a(1-t)dt[/itex].

    Does anyone have any ideas about this?
     
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