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I'm studying "Smoothness".

If a function is once differentiable for all x's, shouldn't it be considered smooth? Because it does "look smooth" for all f(x), there's no way it will have sharp corners or cusps because it's differentiable. Then why does it have to be differentiable way more times than once (and actually has to be differentiable infinite times) to be considered smooth?

Or unless this is not about "looking smooth" but smooth in other meaning?

Any help is greatly appreciated!

Edit: Sorry. Please use a bit less formal math language so I can understand. I not very good at it.

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    $\begingroup$ "smooth" does not have a unique meaning. In differential geometry it is ofen used to denote $C^1$ while in distributions theory it is $C^\infty$. It depends on how much regularity you need. Hence "smooth" means that the function is regular enough, depending on the context. $\endgroup$
    – nicomezi
    Commented Apr 26, 2022 at 4:41
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    $\begingroup$ @nicomezi In differential geometry, I have personally never heard smooth mean anything other than $C^{\infty}$. A smooth manifold is one where the charts are $C^{\infty}$. $\endgroup$
    – Arthur
    Commented Apr 26, 2022 at 4:55
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    $\begingroup$ Smoothness is a strong statement utilized in analysis, and especially in differential geometry. For a function to be continuous, it must only satisfy the epsilon-delta defintion. Smoothness, implies continuity, but not the other way around. When you say it "looks smooth," I think you mean it looks continuous. Why do we care about smoothness? There are many applications in various fields of math, and each one has their own reason why we want "smooth" functions. Even stronger, smooth mappings. $\endgroup$
    – help
    Commented Apr 26, 2022 at 4:58
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    $\begingroup$ @IanAmbrose Math smoothness is certainly not about "looks". The functions $f(x)=x^3$ and $g(x)=\sqrt[3]{x}$ "look" the same, since one is the reflection of the other, but only the former is $C^1$ on $\mathbb R$, let alone $C^\infty$. $\endgroup$
    – dxiv
    Commented Apr 26, 2022 at 5:20
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    $\begingroup$ May be I was over enthusiastic when writing "often". On my side I had a whole class in differential geometry where smooth was $C^1$ because it was constructed in a way that this regularity was the "baseline". If more regularity was needed, it was explicitely written. @Arthur $\endgroup$
    – nicomezi
    Commented Apr 26, 2022 at 5:47

3 Answers 3

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It seems like your question is about how the term "smooth" in math corresponds to the term "smooth" as it's used in English. A once-differentiable function might "look smooth" if you graph it, but if it's not smooth in the $C^\infty$ sense, it can still have "wild" behavior that we don't want when you look at the higher-order derivatives, and we don't want to call such a thing "smooth".

For example, imagine you're walking along a 100-meter cliff until you fall off it after 5 seconds. If you let $f(t)$ be your height at time $t$, you'd have $f(t) = 100$ for $t < 5$, and $f(t) = 100 - 9.81/2(t-5)^2$ for $t \geq 5$. You can check that this function is differentiable. However, it's not smooth in the sense of $C^\infty$, because it is not twice-differentiable. The lack of smoothness is reflected in the sudden change in acceleration as gravity takes effect: $f''(t) = 0$ for $ t < 5$ while $f''(t) = -9.81$ for $t > 5$. The curve $y=f(t)$ might look smooth if you graph it, but the situation would certainly not feel smooth! The point is that higher-order derivatives reveal more information about "general smoothiness" of the curve that might not be apparent if you only look at the graph of the curve, and this "general smoothiness" is what we are interested in when we define term smooth in the mathematical sense.

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  • $\begingroup$ Thank you for the very detailed answer. You said: "The point is that higher-order derivatives reveal more information ... you only look at the graph of the curve" - Where can I read more about this? What's the name of the topic that contains this idea? Thank you! <I will mark this as the answer in the next few hours> $\endgroup$
    – Orange Cat
    Commented Apr 26, 2022 at 7:14
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    $\begingroup$ There's not a particular place I have in mind where you can read about that idea. It's just there in the definition of "smooth": that derivatives to every order exist. Or are you asking what the applications are of higher-order derivatives? Well, there are many... for example, Taylor series approximate a function by creating a polynomial that agrees with the first $n$-th derivatives at a point. Cheers $\endgroup$
    – Jair Taylor
    Commented Apr 26, 2022 at 15:24
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I don't know whether this corresponds with the original coinage of the term, but I think it's fair to say that in this usage mathematicians have not sat down and thought, "OK, we want a mathematical theory of things that are not rough or angular. What tools do we need?". Rather they have sat down and thought, "OK, here's this mathematical property of being infinitely differentiable: can anyone think of a catchier name?"

So yes, you're right, "smoothness" here is not to do with trying to model river-washed pebbles or whatever: it's a nice name for a mathematical property that is not wholly unrelated to the real-life concept of "smoothness".

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  • $\begingroup$ Thank you! This is supposed to be the accepted answer but user Tanner Swett made a more in-depth version of it. $\endgroup$
    – Orange Cat
    Commented Apr 27, 2022 at 0:09
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Most likely, it's because the concept came first and the name came afterward.

It sounds like you're imagining that some mathematicians were thinking about the English word "smooth," and they were trying to figure out how the word could be mathematically defined, and they decided that it should be defined as "differentiable arbitrarily many times." You're asking why they decided that it should be defined as "differentiable arbitrarily many times" instead of "differentiable (once)."

However, that's almost certainly not what actually happened. Most likely, mathematicians were thinking about the concept of a function that's differentiable arbitrarily many times, and they decided that they wanted to have a single word referring to this concept. They thought about the various words they could use for that concept, and they decided to use the word "smooth."

That's something that happens a lot in mathematics: mathematicians start thinking about a concept, and then they choose an existing English word to name that concept. As a result, a lot of words have meanings in mathematics that are very different from their meanings in everyday English. For example, in everyday English, the words "loop," "ring," and "circle" all mean more or less the same thing, but in mathematical English, these words have very, very different meanings.

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  • $\begingroup$ Thank you. This is the perfect answer. $\endgroup$
    – Orange Cat
    Commented Apr 27, 2022 at 0:08

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