Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

When you $\tt{diff}$erence discrete observations of a function $f$ you lose one observation each time you apply $\tt{diff}$.

When you $\partial$ifferentiate a $\mathcal{C}^n \ni$ function $f: \mathbb{R} \to \mathbb{R}$, losing an observation would make the cardinality of the domain $\aleph_1-1$ which does nothing since $\aleph_1-1 = \aleph_1$. So we require $f' \in \mathcal{C}^{n-1}$ instead. In an epsilon-delta sense this is like saying $$\forall \varepsilon > 0, \ \exists \delta \text{ such that } \lim_{h \downarrow 0} { f((x + \delta)-h) - f(x + \delta) \over h } < \lim_{h \downarrow 0} {f(x+h) - f(x) \over h} \pm \varepsilon.$$

I've also heard the saying "Calculus is topology" because of the similarity of the boundary operator $\partial$ to the derivative operator $\partial$, as could be seen in Green's theorem.

Is there a deeper connection between "lose one observation" and "lose a continuity"? If so, what are the next steps to understand it?

share|cite|improve this question
up vote 2 down vote accepted

Let's think in terms of Taylor series; the derivative of $\sum_{n=0}^\infty a_n x^n$ is $\sum_{n=1}^\infty na_nx^{n-1}$. Note that we can recover all of the $a_n$'s from the latter series -- except for $a_0$, the constant term. So in a sense, the derivative is an operation that "forgets" the constant term.

Now suppose you have a discrete sequence of numbers $b_n$. Consider the sequence $b_n' = b_n + C$ where $C$ is some constant. Now note that $b_n' - b_{n-1}' = b_n - b_{n-1}$. In other words, the extra constant term that we introduced disappears -- it no longer plays any role -- once we consider successive differences.

So maybe instead of thinking of it in terms of "losing an observation", you should think of it in terms of "forgetting the constant term". In the discrete case, these two things are equivalent (use $C = -b_0$). Then the analogue in the continuous world is also "forgetting the constant term".

share|cite|improve this answer
Thanks, makes sense. So in what sense does losing the constant term move me from $\mathcal{C^n}$ to $\mathcal{C^{n-1}}$? – isomorphismes Mar 12 '14 at 19:29
Well, I suppose you could say that the constant term is forgotten, the quadratic term is shifted down to the linear term, the cubic term is shifted down to the quadratic term, etc... If the "non-differentiability" lives above the $n$th term, now it lives above the $(n-1)$st term. It's just a shifting operation. – Kevin H. Lin Mar 12 '14 at 19:37
Riiight. Ok. But ∞−1=∞. So…how does it work with that? – isomorphismes Mar 12 '14 at 19:51
Oh wait. ∞ in the Taylor series would mean $\mathcal{C}^\infty$. I get it! Thanks Kevin!! – isomorphismes Mar 12 '14 at 19:53
Your answer gives an interesting way to think about $\mathcal{C}^\infty$ versus $\mathcal{C}^\omega$ $\ni$ functions as well. – isomorphismes Mar 12 '14 at 20:49

I don't think the connection is as strong as I hoped. Reason being, $\mathcal{C}^0$ functions can be jumpy everywhere, not just have one discontinuity. So it's not just a small difference between $\mathcal{C}^n$ and $\mathcal{C}^{n-1}$, more like $\aleph_0$ many singularities in between the smaller class and the bigger class.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.