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Homework Help: Derivative of a function at a isolated point

  1. Mar 9, 2012 #1
    This might be a very miscellaneous problem, but it really makes me concern. This problem is about existence and definability.

    1. The problem statement, all variables and given/known data
    Suppose [itex]f[/itex] is a real-valued function defined on [itex][a,a]=\{a\}[/itex].

    (i) Then can we use the definition below to this function?

    (ii) If it can be used, then is [itex]f'(a)[/itex] is undefined?

    (iii) If it is undefined, why is it that? Is it because the [itex]\phi[/itex] function was undefinable in the first place?

    2. Relevant equations
    Definition (Rudin, p. 103). Let [itex]f[/itex] be defined (and real-valued) on [itex][a,b][/itex]. For any [itex]x\in[a,b][/itex], form the quotient [itex]{\displaystyle \phi(t)=\frac{f(t)-f(x)}{t-x}\quad(a<t<b,t\neq x)}[/itex], and define [itex]f'(x)=\lim_{t\to x}\phi(t)[/itex], provided this limit exists (in [itex]\mathbb{R}[/itex]).

    3. The attempt at a solution

    (i) It seems so, but only the author knows the truth: I don't know whether he (Rudin) allows me to use this kind of interval for the definition. (When he proposes the definition of segment and interval by notation [itex](a,b), [a,b][/itex] in the preceding pages, he does not specify what [itex]a,b[/itex] are.)

    (ii) Undefined.

    (iii) And the reason is seemingly that I cannot define [itex]\phi[/itex] as stated in the definition. Otherwise, no idea.

    Note: I just feel so confused about these notions of 'define', 'form', 'construct', such and such because I can't translate properly these things to the first order language (logic). Please help me.
  2. jcsd
  3. Mar 9, 2012 #2
    I may be wrong on this, but I think it's undefined. Look at your definition. You want x = a in this case, but what is t? There's nowhere else in the interval for t to go. Since t != x, but f is only defined for t = a, I think it's undefined.

    As an example, think about a piecewise function; say, f(x) = x, for x != 0, and f(0) = 1. That's just the graph of f(x) = x with a hole at x = 0, where the value of the function at that point is 1. What would you say the value of f'(0) is?
  4. Mar 9, 2012 #3


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    The derivative of a function gives it rate of change. If, for example, the function is constant over some range, its derivative is 0. But, here, there is no "range". Since we cannot even vary "x", there is no derivative.
  5. Mar 9, 2012 #4
    Can't say I'm convinced. Any function from a singleton domain is a constant function, whose derivative is zero. The difference quotient is perfectly well defined:

    (f(x) - f(x+h))/h = 0 for any nonzero h; hence the limit as h->0 is defined and equal to 0. So I'd say such a function is differentiable, for essentially the same reason any constant function is differentiable. The function f(x) = 5 doesn't vary whether its domain is the entire real numbers or just some particular real number. Can't see how it makes a difference.
  6. Mar 9, 2012 #5

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    The derivative is defined using the definition of a limit.
    The definition of a limit (see for instance wiki) specifically requires t ≠ x (0 < |t - x| < δ).

    In other words, the derivative cannot be defined in this case and as such it does not exist, cannot be formed, and cannot be constructed.
  7. Mar 9, 2012 #6
    I understand your point, but you are in the position of stating that under some circumstances a constant function may not be differentiable. To me that seems worse than what I said, which is that the difference quotient is well-defined.

    As a warm-up exercise, you WOULD agree that a constant function is always continuous, right? Even if the domain is a singleton.

    For my understanding, are you objecting to my forming the difference quotient? Or do you admit the legitimacy of taking the difference quotient, but object to taking the limit? There's a lot of subtle stuff going on in this example. You are claiming that if a math-oriented maniac (think the Unabomber) comes up to you in the street, points a gun to your head, and says: "For your life: True or false? A constant function is differentiable," you are going to shout FALSE.

    Is that your claim? Are you certain? I'm just saying I'm not so certain that there's not a suitable definition of limit floating around somewhere that would clarify this case. To me, a function that does not change is a constant function, whose derivative is zero. And I believe I proved that above, when I took the limit of the difference quotient. My two cents. When the Unabomber comes up to me in the street, I'm going to stand by my claim that a constant function is differentiable. Period.

    (pps) Have you seen the formal derivative of a polynomial?


    Polynomials behave properly with respect to derivatives even if you have no definition of limit. A constant function is a polynomial; and the derivative of a constant polynomial is zero.
    Last edited: Mar 9, 2012
  8. Mar 9, 2012 #7

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    There needs to be an interval in which the function is defined before you can say it's continuous at a point that is part of the interval.

    It's not a matter of what's reasonable, it's a matter of mathematical definition.
  9. Mar 9, 2012 #8

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    For my life... I'm certain (for real-valued functions defined on a real domain, that is)! ;)

    In math there is no middle ground.
    Select the system of axioms, and stuff follows without ambiguity from the definitions and the axioms.
    If you want, I can quote from the definitions of derivative and limit.

    Yep. That's a "formal" derivative - specifically defined in a polynomial ring.
    But then, that is not about a "a real-valued function", which is defined on a subset of R.
  10. Mar 9, 2012 #9
    In general topology, a function is continuous iff the inverse image of an open set is open. Any function on a singleton domain is continuous. This is beyond dispute. Any function whose domain is a discrete topological space is continuous, since in a discrete space, every subset is open.

    If you don't understand why a function on a singleton is continuous, you are not in a position to tackle differentiability. That's why I mentioned it. Any function on a discrete space is continuous. That's why continuity is a good warmup exercise for differentiability. You need to really understand what a continuous function is, for starters. "Intervals" have nothing to do with it at all.
  11. Mar 9, 2012 #10
    That would be extremely helpful. Because I already used the definition of the derivative to show that the OP's function is in fact differentiable. So if I'm wrong, I'd appreciate technical specifics.

    You need to look at this in the most general sense. Talking about intervals of real numbers is not helpful if the domain is a discrete space, let alone a singleton, where the definition of derivative is clearly tricky.

    Do you at least agree that the OP's function is continuous? If not, you need to think about discrete spaces and the general definition of continuity.

    Do you object to my forming the difference quotient? Or do you object to my taking the limit? It would help for me to understand exactly why you think I'm wrong.



    "A 0-dimensional manifold (or differentiable or analytical manifold) is nothing but a discrete topological space. We can therefore view any discrete group as a 0-dimensional Lie group."

    Now I'm not really sure if the above quote tells us that a map on a singleton is differentiable. But it does show that in discrete spaces, you get limits "for free" as it were. Because every subset is open. So there's a lot of continuity floating around.
  12. Mar 9, 2012 #11


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    How do you arrive at this? At least one of f(x) and f(x+h) is not defined so you cannot form this quotient at all.

  13. Mar 9, 2012 #12


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    How can you say that the difference quotient is perfectly well defined, when f(x + h) for h ≠ 0 is itself undefined?
    I doubt that many people would agree that a function that maps a single number to some other number fits the description of a constant function.

    Your example below does fit this description, as it maps every real number to 5.
  14. Mar 9, 2012 #13
    What else could you possibly call it?

    If you have a singleton set, say {x}, no matter what x is, EVERY function with domain {x} is a constant function.

    If I defined f:{3} -> R by f(3) = 5, is that not a constant function? You disagree that's a constant function?

    I admit to some lack of certainty over the question of differentiability. It depends how you choose to interpret the difference quotient.

    But still, clear thinking's going to be helpful here. Any function on a discrete space is continuous. And any function on a singleton space is constant. At least let's agree on the easy stuff.

    Because once you agree to those two points, you then have to be willing to claim that a continuous, constant function might not be differentiable.
  15. Mar 9, 2012 #14
    I agree with that. However, in order for the derivative to exist, you still must have values for f(x-δ) and f(x+δ), which you would not have in a singleton domain.
  16. Mar 9, 2012 #15
    Okay guys. As some of you guys said, I totally get that

    the derivative of this function both in an intuitive level and

    formalistic level does not exist, and also that the point has to be a limit point

    of the domain of f for the derivative to be defined at the point.

    But you guys kinda didn't yet clearly

    answer my question. Let me restate a bit my question for the sake of clearance.

    (i) Can I use the Rudin's definition for this function that I defined? I'm not sure about this because his definition of interval is kinda unclear.

    (i') If I cannot use it, then please let me know a better and stronger and more rigorous version of this definition.

    (ii) As you guys said, if I use this definition, I get the result that f'(a) is undefined. So please show me how this is true in a very formal and rigorous presentation.

    And actually the problem that I'm more concerned with is posted on here: especially #4


    which is about definability and existence.
  17. Mar 9, 2012 #16
    Okay, I've tried to clarify the Rudin's definition by using the first order language as much as I can. Please check if this is correct:

    Definition (Rudin, p. 103; revised by me). Suppose [itex]f[/itex] is defined (and real-valued) on [itex][a,b][/itex]. Suppose [itex]x\in [a,a][/itex]. Suppose there exists a function [itex]\phi[/itex] defined by [itex]{\displaystyle \phi(t)=\frac{f(t)-f(x)}{t-x}\quad(a<t<b,t\neq x)}[/itex]. If [itex]\lim_{t\to x}\phi(t)[/itex] exists (in [itex]\mathbb{R}[/itex]) then we say [itex]f[/itex] is differentiable at [itex]x[/itex]. Suppose the limit exists for some [itex]x \in [a,b][/itex]. Then, in this case, we define a function [itex]f'[/itex], naming it the derivative of [itex]f[/itex], by [itex]f'(x)=\lim_{t\to x}\phi(t)[/itex] for all [itex]x \in [a,b][/itex] such that the limit exists.

    If this revised definition is correct (and the revision is correct), then we cannot use this definition for the function that I defined in #1 because such [itex]\phi[/itex] function does not exist (it can be proved by contradiction), and therefore it's not that the derivative does not exist but that it is undefined.
    Last edited: Mar 10, 2012
  18. Mar 10, 2012 #17

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    Looks correct to me.
  19. Mar 10, 2012 #18
    Thanks. But after reconsidering my definition, I've come to a thought that this definition is still somewhat wrong. As for the function defined on [itex][a,a][/itex], since [itex]t\notin \{t \in \mathbb{R} : a<t<a, t\neq x \}[/itex], the notation [itex] \phi (t) [/itex] is undefined. Thus the function [itex]\phi[/itex], even before it does not exist, is undefined. Therefore, in this case I cannot use the revised definition.
  20. Mar 10, 2012 #19


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    I don't see what the problem there is.
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