5 ms·
One big issue with that notation is that the log of the exponential is not the exponential of the log, so the order really does matter(for any complex valued ex
by melissalobos 4y ago
One big issue with that notation is that the log of the exponential is not the exponential of the log, so the order really does matter(for any complex valued expressions).
https://www.wolframalpha.com/input?i=log%28exp%28x%2Biy%29%29+-+exp%28log%28x%2Biy%29%29 https://www.wolframalpha.com/input?i=log%28exp%28x%2Biy%29%2...
So just having over and underbars loses some information.
- lupire 4y agopretty sure that's because log is not well defined over the complex numbers, it's just defined over branch cuts. similar to how sqrt(x^2) != (sqrt((x)^2) when x is Real but not Positive Real, because sqrt is not well defined over the full Reals. It's not a flaw of notation, it's a flaw in the attempted math.
- gjvnq 4y agoHow about we just accept that the square root is a relation inatead of a function?
- d_tr 4y agoI am not sure what you mean by "relation" and what the benefit would be. There is a definition for "relation" in mathematics. A function is a type of relation and a relation is a set. But you probably have something else in mind.
- ducttapecrown 4y agoMultivalued functions are types of relations. Relations are more general than multivalued functions because relations don't have to be defined on the entire domain.
- kergonath 4y agoWhat’s wrong with the square root, apart from the fact that it’s not defined over all of R? It’s a fairly well behaved function overall.
- moonchild 4y agoI think you mean sqrt(x^2) != (sqrt(x))^2