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Is this function continous?

  1. Nov 15, 2006 #1
    Is

    f(t)= {0, -2<t<0
    {t, 0<=t<2

    continous everywhere in -2<t<2. I am a bit concerned at t=0

    I think yes, after applying the epsilon delta argument but I rememer that a similar function

    f(t) = {-t, -2<t<0
    {t, 0<=t<2

    is not differentiable hence not continous at t=0.

    Have I made a mistake somwhere?
     
    Last edited: Nov 15, 2006
  2. jcsd
  3. Nov 15, 2006 #2

    matt grime

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    It is perfectly possible for a function to be continuous and not differentiable.
     
  4. Nov 15, 2006 #3
    So

    f(t)= {0, -2<t<0
    {t, 0<=t<2

    is continous at t=0?
     
  5. Nov 15, 2006 #4

    HallsofIvy

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    What is [itex]\lim_{t\rightarrow0}f(t)[/itex]?
     
  6. Nov 15, 2006 #5
    I have to say it is 0 because from the left, it must be 0 since the function is 0. From the right, it is 0 since the function is t. f(0)=0 is defined so it should be continous. But is it differentiable at t=0?

    This means
    f(t) = {-t, -2<t<0
    {t, 0<=t<2
    is continous as well. But it is not differentiable at t=0 is it? How come?
     
  7. Nov 15, 2006 #6

    HallsofIvy

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    Yes, the limit from both left and right is 0 so the limit is 0. Since the limit there exists and is equal to the value of the function, the function is continuous at 0.

    Your f(t) is, of course, |t| between -2 and 2. It is well know that |x|, while continuous, does not have a derivative at x= 0. Essentially, it is not "smooth": there is a corner at x= 0 so the tangent line there is not well defined. More precisely, if h< 0 then f(0+h)= f(h)= -h so
    (f(0+h)- f(0))/h= -h/h= -1 while if h> 0, (f(0+h)- f(0))/h= h/h= 1. The two one-sided limits are different so the limit itself, and therefore the derivative at t=0, does not exist.

    It's easy to see that the derivative of f is 1 for t> 0 and -1 for t< 0. While the derivative of a function is not necessarily continuous, it does satisfy the "intermediate value property" so the derivative cannot exist at 0.
     
  8. Nov 15, 2006 #7
    So |x| is everywhere continous but not differentiable at x=0 because the derivative function is undefined at x=0.

    Back to
    f(t)= {0, -2<t<0
    {t, 0<=t<2

    If I wanted to represent this function as a fourier series than I am assured that the fourier series is uniformly convergent in -2<t<2 since f(t) is continous everywhere in this domain as have been shown. Or could this not happen because if an infinite series of functions is uniformly convergent than the function it represents is continous everywhere in the domain. But the function being continous everywhere dosen't always mean uniform convergence of the series representing it. In other words is it 'if' or 'if and only if'?
     
    Last edited: Nov 15, 2006
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