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Let $X$ be the set of all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ which are infinitely differentiable at $0$. Let us define an equivalence relation $\sim$ on $X$ by saying that $f\sim g$ if there exists a $\delta>0$ such that $f(x)=g(x)$ for all points in the interval $(-\delta,\delta)$. And let $Y$ be the set of equivalence classes of elements of $X$ under $\sim$.

My question is, what characterizes a given equivalence class in $Y$? The values of $f(0)$ and $f^{(n)}(0)$ for all $n$ aren't enough, because for any given analytic function $f$ there exists an non-analytic infinitely differentiable function $g$ such that $f(0)=g(0)$ and $f^{(n)}=g^{(n)}$ for all $n$ but where it's not the case that $f\sim g$.

So what is the minimum information needed to unambiguously specify an element of $Y$?

EDIT: It turns out my concept is an existing mathematical concept, known as a germ. So my question reduces to, what information uniquely characterizes the germ of a smooth function?

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    Maybe you want to look at this, but it does not contain an answer to your question: https://en.wikipedia.org/wiki/Germ_(mathematics) – Christian Blatter Aug 03 '18 at 11:34
  • @ChristianBlatter Wow, my concept is an existing mathematical concept! – Keshav Srinivasan Aug 03 '18 at 14:01
  • What do you mean "characterizes"? What you said in the first paragraph is the definition of germ, so one way to characterize it would be just use this definition. Do you want an alternative equivalent definition, or do you want the motivation behind this definition. – Xiao Aug 05 '18 at 18:39
  • @Xiao No, I mean, what is the minimum information needed to unambiguously specify one germ as opposed to all the other germs? The value of the function at zero along with the values of nth derivative of the function for all n isn’t enough to unambiguously specify a particular germ, because there are multiple germs which share that information. So what is the minimum information needed to specify a single germ? – Keshav Srinivasan Aug 05 '18 at 19:19
  • @Xiao I’ll tell you the motivation for my question. The higher-order derivative test is inconclusive if $f^{(n)}(0)=0$ for all $n$. So I’m wondering whether there’s some way to tell if a point is a local maximum or minimum in the case where the higher-order derivative fails, in a way that only depends on the germ of $f$. See my question here: https://math.stackexchange.com/q/2871719/71829 But in order to figure that out, I need to know what information about a function other than the values of its nth derivatives at the point is captured by the germ of the function. – Keshav Srinivasan Aug 05 '18 at 19:27
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    I somehow doubt there will be an answer that is satisfying. This question seems very similar to asking how you can tell whether two topological spaces are homeomorphic from information at a single point from either space. It's very possible the two spaces are locally homeomorphic around those two points. Some global information is going to be needed. There are several $C^\infty$ functions that are extremely flat at a point but not equivalently zero. How do you tell these apart without global information? –  Aug 07 '18 at 21:56

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Although I doubt this will have a satisfying answer, I still think it's an interesting question. Complex Analysis is precisely interesting because differentiability on a small neighborhood implies analyticity, which (in a certain sense) means that information at a point propagates to nontrivial local information. So investigating the real-analog is a terrific thing to be curious about and makes you appreciate Complex Analysis much more.

There are many ways in which Real Analysis is "ugly," and your question can exemplify one way in which this is the case.

Considering the difference of any two functions you're working with, your question is equivalent to asking

On an interval, just how "zero" can a function be without being equivalently zero?

Exploiting graph transformations we can focus on the case where $f:(-1, 1)\rightarrow \mathbb{R}$ is $C^\infty$.

Of course, if $f$ is zero on a dense subset of $(-1,1)$, then $f$ must be equivalently zero. This only exploits continuity.

In hopes of constructing a weird function off a weird subset, this might lead you to ask, "just how dense can something be without being dense?" The answer is "not much". If a subset is not dense, then there will be a point and a neighborhood of that point where the original wannabee-dense subset does not even have a single point in common with that neighborhood. In short, a subset that isn't dense doesn't show up in SOME open subset.

Turning back to the original problem here, a function that isn't zero on a neighborhood is pretty much not zero in my book.

But you might be comfortable with a function not being zero on a dense subset and still consider it "pretty much zero". Ways in which I can think of this being reasonable is a function which is zero on a closed nowhere dense set which has positive Lebesgue measure---e.g. a "fat" Cantor set. These are examples of subsets which are "small" topologically but "big" measure-theoretically. You can't quite construct a nowhere dense subset with full measure. You can get a full measure subset with empty interior, but this subset will be dense (and thus not interesting here).

For any closed set $F$, you can construct a function $f$ which is $C^\infty$ on $\mathbb{R}$ such that $f^{(n)}(x)=0$ for every $x$ in $F$ and any natural number $n$. This is one of the weird things about Real Analysis. Using this construction on something like a fat Cantor set would give you a pretty-much-zero function. So one half-way of answering your question is how would you tell such a function like this from the zero function? You would certainly need to know this function's value at some other point. The places where it is not zero is dense and open; however, you might have an arbitrarily small probability of picking a point where it evaluates to something else. Quite the conundrum. Depending on how "fat" this Cantor set was, the universe might die out before you have an appreciable probability of picking a point at random where it evaluates to not zero.

It seems you'd need to know how the function behaves on an entire interval, which isn't very satisfying.

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    By the way, this question is motivated by my question here, where I want to find whether $f$ has a local maximum or minimum at 0 in the case where the higher-order derivative test fails, i.e. $f^{(n)}(0)=0$ for all $n$, in a way that depends only on the germ of $f$. Do you think that's possible? – Keshav Srinivasan Aug 08 '18 at 15:47