Derivation of the exponential distribution - that infinitesimal

AI Thread Summary
The discussion focuses on understanding the role of the little-o notation, o(Δt), in the derivation of the exponential function. It clarifies that o(Δt) represents all small terms that are not significant for the derivation, such as higher-order terms like (Δt)² and (Δt)³. Participants explain that this notation simplifies the process by allowing the use of a first-order Taylor approximation, which neglects higher-order terms. The conversation also touches on the Taylor expansion of the exponential function, specifically in the context of e^x and its behavior with small increments. Overall, the discussion aims to demystify the use of little-o notation in mathematical derivations.
thomas49th
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Hello,

I've been looking at the derivation of the exponential function, here
http://www.statlect.com/ucdexp1.htm
amongst other places, but I don't get how, why or what the o(delta t) really does. How does it help?

It's really confusing me, and all the literature I've looked at just seems to quickly dance over it

Thanks
 
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##o(\Delta t)## contains all small terms the derivation does not want to (and does not have to) care about. Something like ##(\Delta t)^2##, ##(\Delta t)^3## and so on.

Little-o notation
 
why do those terms exist? was there some kind of taylor expansion?
 
Exactly. It is a taylor approximation to first order (linear in Δt) and higher orders are neglected.
 
what the taylor expansion of? e^x ?
 
Should be ##e^x - e^{x+Δt}## with some prefactors I did not check.
 
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