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Rigorous definition of derivative

  1. Jan 21, 2012 #1
    Hi, I have a question about the definition of derivative.

    As far as I know, for a real valued function f defined on a subset of R, the derivative of f at a is

    (f(x+h)-f(x))/h as h → 0.

    And if it exists it's said f is differentiable at x.

    What if I define f : Q → R as follows,

    f(x) = sin(x) (the domain of f is all rational numbers.)

    Then it satisfies every condition to be a differentiable function, thus it's a differentiable function on its domain.

    Is this a valid argument?

    Or what about f : [0,1]U[2,3] → R defined as follows,

    f(x) = 1 when x belongs in the first closed interval
    2 when x belongs in the second closed interval

    f'(x) at points 1 and 2 seem to exist. But do they really?

    And one small question; when you have f'(x) = ∞, it implies f isn't defined at x. Is it even possible to have such an equality? If f(x) isn't defined, how can f'(x) can be calculated in the first place when the definition of f'(x) requires the existence of f(x)?

    Thank you in advance.
     
  2. jcsd
  3. Jan 21, 2012 #2
    As far as I know, differentiability is defined for maps whose domain is an open interval (a,b), or a union of such intervals.

    f'(1) and f'(2) are considered one-sided derivatives. For example, for f'(1) the difference quotient [itex] \frac{f(1+h) - f(1)}{h} [/itex] cannot be formed if h > 0, so we would calculate f'(1) by evaluating the limit [itex] \lim_{h \rightarrow 0^{-}} \frac{f(1+h) - f(1)}{h} [/itex].

    However, more formally, if [itex] f: S \rightarrow R [/itex] is a map where [itex] S \subset \mathbb{R} [/itex] and S isn't necessarily an open interval, we say that f is differentiable if we can find a map [itex] g: T \rightarrow \mathbb{R} [/itex] such that:

    1) [itex] S \subset T [/itex]
    2) [itex] g = f [/itex] on [itex] S \cap T [/itex]
    3) T is an open interval.

    In this case we say that we have extended f to a map g.


    Saying that [itex] f'(x) = \infty [/itex] for some number x doesn't really make sense. What you are really trying to say is that [itex] \lim_{x \rightarrow a} f'(x) = \infty [/itex].

    For example, look at the graph of [itex] f(x) = \frac{1}{x} [/itex]. If you look at the graph, you can see that as x approaches 0, f'(x) will approach infinity since the tangent lines will become increasingly steeper. To algebraically verify this just note that [itex] \frac{d}{dx} (\frac{1}{x}) = \frac{-1}{x^2} [/itex] and that upon letting [itex] x \rightarrow 0 [/itex] we see that [itex] \frac{d}{dx} (\frac{1}{x}) [/itex] becomes infinite.
     
  4. Jan 21, 2012 #3

    pwsnafu

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    Consider f(x) = x^{1/3} when x=0.
     
  5. Jan 21, 2012 #4
    Thank you very much!!

    There's one thing I'm still not sure about.

    If it says f'(a) exists, can I take it that f is defined on some open interval that includes a?
    Or does it cover the case where 'a' can be an end point of some close interval such as [a,b] on which f is defined?
     
  6. Jan 21, 2012 #5
    Well consider the map [itex] f: [a,b] \rightarrow R [/itex] where [itex] f(x) = x^2 [/itex]. One would surely say that f'(a) exists, however f isn't defined on an open interval containing 'a'. One could of course extend f to make this possible, but you aren't asking about the extended map, you're asking about f and we have defined it so its domain is [a,b]!.
     
  7. Jan 21, 2012 #6

    Fredrik

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    The statement "f is differentiable on [a,b]" is sometimes defined to mean that the domain of f is an open set that contains [a,b], and for some ε>0, f is differentiable on (a-ε,b+ε). I think I saw something like this in "Introduction to smooth manifolds", by John M. Lee.

    "f is differentiable" usually means that "the limit that defines f'(x) exists for all x in the domain of f". It only makes sense to talk about that limit when x is an interior point of the domain, so if we are to use phrases like "f is differentiable" without fancy definitions like the one above, the domain must be an open set.
     
  8. Jan 21, 2012 #7

    jgens

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    As a general note, it is possible to define the derivative of mappings between normed vector spaces using bounded linear operators. I think that this is generally introduced in functional analysis type courses so I am guessing it is not what the OP is looking for.

    This is a fairly common interpretation. Another interpretation just requires the left- or right-hand limits to exist at the end points.
     
  9. Jan 22, 2012 #8
    You only need an accumulation point to define a limit. So, why not on the set of rational numbers?

    The only problem... maybe it's useless, at least for physical purposes. But you can take the limits in such rational-real function
     
  10. Jan 22, 2012 #9

    Fredrik

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    Because the limit we're talking about is the limit of
    $$\frac{f(x+h)-f(x)}{h}$$ as ##h\to 0##, and the numerator (specifically "f(x)") makes sense only when x is in the domain.
     
  11. Jan 22, 2012 #10
    Of course, so let x be the rational (an accumulation point) and h run over the neighborhood of x;

    you only need an acc point to define the limit
     
  12. Jan 22, 2012 #11
    Why have you chosen this and what does your book say about continuity?
     
  13. Jan 22, 2012 #12
    The weird examples from mathematicians... :D

    Just to ask whether it is differentiable.

    I think yes: you can take the derivative in the whole domain (all points are accumulation points, so you can take the limit), they exist so they define the derivative.

    This would not be possible provided that the domain was the set of integers: they are not accumulation points.
     
  14. Jan 22, 2012 #13

    Fredrik

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    An accumulation point of what? I would have guessed that you meant an accumulation point of the domain of f, but then you agreed that x needs to be a member of the domain.

    ##f(x)\to y## as ##x\to a## if for all ##\varepsilon>0##, there's a ##\delta>0## such that for all x in the domain, ##|x-a|<\delta\ \Rightarrow\ 0<|f(x)-y|<\varepsilon##. Note that the entire interval ##(a-\delta,a+\delta)## is mapped into ##\{z\in\mathbb R:0<|z-y|<\varepsilon\}##, so f must be defined on that entire interval. This is why this definition doesn't work for functions defined on e.g. ##[0,1]\cap\mathbb Q##.

    To define f'(x), x must be an interior point of an open subset of the domain of f. It certainly isn't enough that x is an accumulation point of the domain, if that's what you meant.
     
  15. Jan 22, 2012 #14
    Is f(x) = {sin(x) such that x is rational} a continuous function?

    What about all the non rational x for which sin(x) are not defined?
     
  16. Jan 22, 2012 #15

    Fredrik

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    I assume that you're asking about the function ##f:\mathbb Q\to\mathbb R## defined by f(x)=sin x for all ##x\in\mathbb Q##. Since it's not defined on any open subset of ##\mathbb R##, it can't satisfy the definition of "continuous at x" for any x.

    So it's not continuous with respect to the topology of ##\mathbb R##. It is however continuous with respect to the topology of ##\mathbb Q##.
     
  17. Jan 22, 2012 #16
    I'm just trying to clarify why the OP specified a sine function with lots of 'gaps' in it, by restricting the domain to Q.

    For instance if we consider an epsilon-delta argument should the values of epsilon and delta be restricted to rational numbers, and if so why?
     
  18. Jan 22, 2012 #17
    Why not? You can take a limit as well. You are too restrictive with your real-analysis definition.
     
  19. Jan 23, 2012 #18

    Fredrik

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    I already answered that. If x isn't in the domain, "f(x)" doesn't make sense, and we're talking about the limit of
    $$\frac{f(x+h)-f(x)}{h}$$ as ##h\to 0##.
     
  20. Jan 23, 2012 #19
    Not only x but also h?
     
  21. Jan 23, 2012 #20
    Ah, of course. But on the domain intersection accumulation points of the domain (any rational in this case). It's not necessary rigorously to have an interior point to take the limit in a topological sense, right?
     
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