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  > An exponential curve looks locally the same at all points in time
This is true for any curve...

If your curve is continuous, it is locally linear.

There's no use in talking about the curve being locally similar without the context of your window. Without the window you can't differentiate an exponential from a sigmoid from a linear function.

Let's be careful with naive approximations. We don't know which direction things are going and we definitely shouldn't assume "best case scenario"



A curve isn't necessarily locally linear if it's continuous. Take f(x) = |x|, for example.


There may have been a discontinuity at the beginning of time... but there was nobody there to observe it. More seriously, the parent is saying that it always looks continuous linear when you're observing the last short period of time, whereas the OP (and many others) are constantly implying that there are recent discontinuities.


I think they read curve and didn't read continuous.

Which ends up making some beautiful irony. One small seemingly trivial point fucked everything up. Even a single word can drastically change everything. The importance of subtlety being my entire point ¯\_(ツ)_/¯


|x| is piece wise continuous, not absolutely continuous


For a function to be locally linear at a point, it needs to be differentiable at that point... |x| isn't differentiable at 0, so it isn't locally linear at 0... that's the entirety of what I'm saying. :-)


You're not wrong. But it has nothing to do with what I said. I think you missed an important word...

Btw, my point was all about how nuances make things hard. So ironically, thanks for making my point clearer.


Nothing to do with what you said?

  This is true for any curve...

  If your curve is continuous, it is locally linear.
Hmm...

Sometimes naive approximations are all you've got; and in fact, aren't naive at all. They're just basic. Don't overthink it.




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