Function discontinuity

Discussion in 'Undergraduate Math' started by Mat, Jan 14, 2005.

  1. Mat

    Mat Guest

    Hello,

    Say we have a piecewise function f: [a, b] -> |R. Is this function
    continuous at a and b? How I see it, it's not because limits don't exist at
    a and b? (We have only one-sided limits at those two points, don't we?)
    Could you please clarify this to me? Thanks.
     
    Mat, Jan 14, 2005
    #1
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  2. What does "piecewise function" mean here?

    Under the usual calculus definition of "continuous at x", you are
    correct: the function cannot be continuous at a point unless it
    satisfies a number of conditions, one of which being that the function
    has to be defined on an open interval that contains the point. This is
    not the case with a and b here.

    ->HOWEVER<-, there are other common definitions of "continuous at x",
    which do not require this. For example, we can say that a real valued
    function of real variable f is continuous at x if and only if:

    (i) f is defined at x; and
    (ii) For every e>0 there exists d>0 such that, for every y in the
    domain of f, if |x-y|<d then |f(x)-f(y)|<e.

    Under this definition, it is certainly possible for your function f to
    be continuous at a or at b.

    MORAL: It depends on what your ->exact<- definition of "f is
    continuous at x" is!
    You'll have to clarify what the definition of "continuous" that you
    are using is.

    --
    ======================================================================
    "It's not denial. I'm just very selective about
    what I accept as reality."
    --- Calvin ("Calvin and Hobbes")
    ======================================================================

    Arturo Magidin
     
    Arturo Magidin, Jan 14, 2005
    #2
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  3. A function f(t) is said to be piecewise continuous on a finite interval
    [a,b] if f(t) is continuous at every point in [a,b], except possibly for a
    finite number of points at which f(t) has a jump discontinuity.

    for Example :
    f(t) = { t when 0 <= t < 1; 2 when 1< t <= 2, 0 when
    2<= t <= 3}
    in this example f(t) is not continuous on [0,3], but it is piecewise
    continuous ( it has a jump discontinuity at t = 2 and at 3)

    Jump discontinuity at a point means that the one-sided limits at this point
    exist and are finite.
     
    Someonekicked, Jan 15, 2005
    #3
  4. Mat

    Mat Guest

    You'll have to clarify what the definition of "continuous" that you
    Actually, I wasn't aware that there are other definitions of a continuous
    function. I've read that a function is a continuous at a point if it has a
    limit at that point that is equal to the function value at that point. So by
    that definition, I assumed that the function is discontinuous at the points
    a and b, because limits don't exist at those two points. We only have
    left-sided limit at b, and right-sided limit at a.

    By piecewise function I meant exactly what Someonekicked posted below.


    Thanks for helping me out ;)
     
    Mat, Jan 15, 2005
    #4
  5. By that definition, the function is not continuous at a and not
    continuous at b. I am afraid I have to tell you that whether or not
    this is the same as "discontinuous" also depends on your meaning of
    that term (though, I assume, in your experience "not continuous" and
    "discontinuous" mean that same thing; however, sometimes
    "discontinuous" is used to mean a specific kind of non-continuity).
    Correct. By that definition, the function cannot be continuous at a
    nor at b.
    Then you goofed: "Someonekicked" gave the definition of "piecewise
    CONTINUOUS". Leaving out such an important word ('continuous') is a
    major mistake. You did it again here.



    --
    ======================================================================
    "It's not denial. I'm just very selective about
    what I accept as reality."
    --- Calvin ("Calvin and Hobbes")
    ======================================================================

    Arturo Magidin
     
    Arturo Magidin, Jan 15, 2005
    #5
  6. Mat

    Mat Guest

    Could you elaborate? I'm still not seeing what I wrote wrong.

    I'd also like to note that I'm not a native English speaker. I'm almost sure
    that "discontinous" and "not continuous" mean the same thing in my language.
     
    Mat, Jan 16, 2005
    #6
  7. You wrote:

    "f is a piecewise function."

    Someonekicked wrote

    "f is a piecewise CONTINUOUS function."

    You left out the adjective "continuous", which was extremely
    important.

    "Piecewise" just means "in pieces". A function may be piecewise
    defined, piecewise continuous, piecewise monotone, piecewise
    integrable, piecewise anything.

    What someonekicked said was "f is a function which can be broken into
    a finite number of pieces, each of which is continuous."

    What you sayd was "f is a function which is broken into pieces."

    What you said meant nothing.
    Neither am I. All the more reason to be doubly careful with
    definitions, lest your mental translation lead you astray, or lest you
    confuse two terms which might mean the same thing in your native
    language but mean different things in English.
    Doesn't matter what they mean in your (or my) native language. What's
    important is how they are defined in the context you are using
    them. As I said, it is most likely that, given your definition of
    continuous, they ->do<- mean the same thing in your text. But it is
    not ->certain<-, so you should just be sure they do, or else not use
    them interchangeably. When ->I<- learned calculus (in Spanish), we
    used "not continuous" for a point where the function that failed
    ->any<- of the requirements for continuity (for example, where it was
    not defined, etc.); we used "discontinuous" exclusively only for
    points where the function was defined but the value of the function
    did not match the (adequate) limit.

    --
    ======================================================================
    "It's not denial. I'm just very selective about
    what I accept as reality."
    --- Calvin ("Calvin and Hobbes")
    ======================================================================

    Arturo Magidin
     
    Arturo Magidin, Jan 16, 2005
    #7
  8. Mat

    Mat Guest

    I see the error, thanks.

    Thanks for clearing it out.
     
    Mat, Jan 16, 2005
    #8
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