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Suppose we have a differential ring $(R,+,\cdot)$ with derivation $\partial: R\to R$ which is linear $$\partial(f+g)=\partial f+\partial g$$ and obeys the Leibniz rule: $$\partial(f\cdot g)=(\partial f) \cdot g + f\cdot (\partial g)$$ and we assume that the elements of $R$ are some kind of function.

Now we want to get the nice linear approximation property $$f(x+\epsilon) \approx f(x)+ \epsilon\cdot\partial f(x)$$ where $\approx$ means something like "minimises error" or is "is equal for sufficiently small $\epsilon$". This is clearly true for nice spaces like $C^1(\mathbb R)$, but I can't prove it with just the above. What structure needs to be added/how can it be proved?

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    $\begingroup$ For one, if you want the problem as stated to make sense, you need $R$ to be a ring of functions from one metric space to another. $\endgroup$ – Arthur Sep 13 '17 at 10:54
  • $\begingroup$ I agree that you need to have functions, but can you prove having a metric is necessary or sufficient? For example it looks like you could get away with just an order. $\endgroup$ – Sean D Sep 13 '17 at 11:18
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    $\begingroup$ You're right, an order should work. But I don't think "get away with just" is the right phrase to use. Many metric spaces cannot be ordered in a nice way, and many ordered spaces cannot me metricised. It's just two different things. $\endgroup$ – Arthur Sep 13 '17 at 11:21
  • $\begingroup$ I mean there are trivial ways to do this. For example you could simply define the term "linear approximation" as being the same as a derivation. For anything substantially different, you need more assumptions. For example if you want "error minimising", you need to define what kind of "minimising" you mean. That's why Arthur is asking are about metric spaces. Your question of "what one can get away with" also needs more assumptions: Get away with while doing what ? What do you need that for? Do you want to prove something? If so, what exactly are you trying to do? $\endgroup$ – Johannes Hahn Feb 14 '18 at 12:42

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