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Math Help - functions differing by a constant

  1. #1
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    functions differing by a constant

    Let D be a set of nonzero real numbers. Suppose that the functions g : D \rightarrow R and h : D \rightarrow R are differentiable and that g'(x) = h'(x) for all x in D. Do the functions differ by a constant???



    so we can say that:

    If D is an open interval then, we can define f = g-h : D\rightarrow R . then we have

    f'(x) = g'(x) -h'(x) = 0 for all x in D

    So, f : D \rightarrow R is constant iff g and h differ by a constant.

    however, what i wanted to know is how can we assure that D is an (open) interval?
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  2. #2
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    The answer is no, as you have apprehended. Take D:=(-2,-1)\cup(1,2) and set g to be the indicator on (-2,-1) and h to be zero. Then they're derivatives are equal but they do not differ by a constant. You will need to insist that D be connected if you want this to hold.
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  3. #3
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    Quote Originally Posted by maddas View Post
    The answer is no, as you have apprehended. Take D:=(-2,-1)\cup(1,2) and set g to be the indicator on (-2,-1) and h to be zero. Then they're derivatives are equal but they do not differ by a constant. You will need to insist that D be connected if you want this to hold.
    Hi Maddas, thank you for your continual help.

    I could not completely understand this reasoning. Could you please elaborate a little?

    thank you
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  4. #4
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    Maybe I misunderstood you. Let D:=(-2,-1)\cup(1,2). Define

    g(x) = \begin{cases} 1, & \mbox{if }  x \in (-2,-1)  \\ 0,  & \mbox{otherwise}. \end{cases}

    and h(x)=0. Then you can easily see that g'=h'=0 on D, but g-h=g, which is not a constant.

    In general, if D is the disjoint union of several connected components, then you can treat each one "independantly", bumping a function up by a different constant on each piece without changing the derivative.
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