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Math Help - Differentiating a function with a Heaviside Step Function

  1. #1
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    Differentiating a function with a Heaviside Step Function

    The function H(x) defined by
    H(x) = 0 , x < 0
    H(x) = 1/2 , x = 0
    H(x) = 1 , x > 0

    F(x) = f(x)H(x - b)

    Find the conditions on f(x) that make F(x)
    1. Continuous at x = b
    2. Differentiable at x = b
    Last edited by glio; March 31st 2008 at 04:05 AM.
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  2. #2
    Junior Member roy_zhang's Avatar
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    Quote Originally Posted by glio View Post
    The function H(x) defined by
    H(x) = 0 , x < 0
    H(x) = 1/2 , x = 0
    H(x) = 1 , x > 0

    F(x) = f(x)H(x - c)

    Find the conditions on f(x) that make F(x)
    1. Continuous at x = c
    2. Differentiable at x = c
    1. In order to make F(x) continuous at x = c, the function F(x) needs to satisfy the following conditions:

    (1) F(c) is defined;
    (2) \lim\limits_{x\rightarrow c} F(x) exists; and
    (3) \lim\limits_{x\rightarrow c} F(x)=F(c)

    To satisfy (1), consider F(c)=f(c)H(c-c)=f(c)H(0)=\frac{1}{2}f(c), we need f(c) is defined.
    To satisfy (2), we need to make sure both \lim\limits_{x\rightarrow c^-} F(x) and \lim\limits_{x\rightarrow c^+} F(x) exist, also \lim\limits_{x\rightarrow c^-} F(x) = \lim\limits_{x\rightarrow c^+} F(x).
    Here we have \lim\limits_{x\rightarrow c^+} F(x)=f(c)\left(\lim\limits_{x\rightarrow c^+}H(x-c)\right)=f(c)(1)=f(c) and \lim\limits_{x\rightarrow c^-} F(x)=f(c)\left(\lim\limits_{x\rightarrow c^-}H(x-c)\right)=f(c)(0).
    So we need |f(c)|<\infty and f(c)=f(c)(0)\Rightarrow f(c)=0.
    To satisfy (3), realize that from above, if \lim\limits_{x\rightarrow c} F(x) exists then \lim\limits_{x\rightarrow c} F(x)=0. Also we have F(c)=\frac{1}{2}f(c), hence \frac{1}{2}f(c)=0\Rightarrow f(c)=0.

    In summary, in order to make F(x) continuous at x = c, function f(x) has to satisfy f(c)=0.


    2. Given that F^\prime(c)=\lim\limits_{x\rightarrow c}\frac{F(x)-F(c)}{x-c}, so in order to make F(x) differentiable at x = c, we need to show the limit \lim\limits_{x\rightarrow c}\frac{F(x)-F(c)}{x-c} exists. More specifically, we need to show the one-sided limits \lim\limits_{x\rightarrow c^-}\frac{F(x)-F(c)}{x-c} and \lim\limits_{x\rightarrow c^+}\frac{F(x)-F(c)}{x-c} exist and are equal. Can you pick up from here?

    Roy
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  3. #3
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    Thanks man.

    I finally have ideas how to appreach this

    Hmmm... It looks like I'll need more help with this.
    Last edited by glio; March 30th 2008 at 03:44 PM. Reason: Changed my mind
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