| ▲ | ajkjk 5 hours ago | ||||||||||||||||||||||
One announcement from a company or government can suddenly change the derivative discontinuously. Derivatives irl do not follow the rules of calculus that you learn in class because they don't have to be continuous. (you could quibble that if you zoom in enough it can be regarded as continuous.. But you don't gain anything from doing that, it really does behave discontinuous) | |||||||||||||||||||||||
| ▲ | lucianbr 2 hours ago | parent | next [-] | ||||||||||||||||||||||
Person who draws comparison from current situation to derivatives points out that derivatives rules don't apply to current situation. Awesome stuff. | |||||||||||||||||||||||
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| ▲ | Neywiny 4 hours ago | parent | prev | next [-] | ||||||||||||||||||||||
Not sure what kinda calculus you took at least here in the states it's very standard to learn about such functions in class, and yes there is a difference between discontinuous and the slope being really large (though finite) for a brief period of time | |||||||||||||||||||||||
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| ▲ | alwa 3 hours ago | parent | prev | next [-] | ||||||||||||||||||||||
As seems to have sort of happened between March and April of this year, at least from the Ramp chart in TFA. I wonder what that was about. | |||||||||||||||||||||||
| ▲ | umanwizard 4 hours ago | parent | prev [-] | ||||||||||||||||||||||
Derivatives in actual calculus don’t have to be continuous either. Consider the function defined by f(x) = x^2 sin(1/x) for x != 0; f(0) = 0. The derivative at 0 exists and is 0, because lim h-> 0 (h^2 sin(1/h))/h = lim h-> 0 (h sin(1/h)), which equals 0 because the sin function is bounded. When x !=0, the derivative is given by the product and chain rules as 2x sin(1/x) - cos(1/x), which obviously approaches no limit as x-> 0, and so the derivative exists but is discontinuous. | |||||||||||||||||||||||