3 ms·
Great. Now I'm going to spend the rest of the morning trying to figure out whether you can really do it that way. [Edit: Ah yes. The formalism goes like thi
by Dove 17y ago
Great. Now I'm going to spend the rest of the morning trying to figure out whether you can really do it that way.
[Edit: Ah yes. The formalism goes like this: Consider the above series. By the above argument, it differentiates to itself. QED. ]
- tome 17y agoAh, but how do you know that e^x is the only thing that differentiates to itself? :)
- philh 17y agoYou don't need to. The uniqueness part of the definition is never used in that argument. (In fact, a*e^x also differentiates to itself, for any a; but that's a trivial case.) Uniqueness almost follows from that argument. It's now easy to see that exp is the only analytic function satisfying exp' = exp and exp(0) = 1: if you have another one, by the same argument, it has the same Maclaurin expansion, hence is the same function. However, I don't know how to prove uniqueness over all functions, not just analytic ones.