ACOS logarithmic form

[ljubo]

(04-30-2016 07:01 AM)Ángel Martin Wrote:  Interesting, I never looked at those diagrams as showing mappings between domain and results regions but as defining the ranges of applicability, i.e. showing where will the results be; yet what you're saying appears to be accurate.

Well, there is no other definition of single-valued inverse function (restricted to the principal branch). Without diagram they would need some curly brackets and different equations depending of Re(z) and Im(z) >1, <-1, etc.

(04-30-2016 07:01 AM)Ángel Martin Wrote:  What's intriguing is the sentence in pg. 62: "The principal branches in the last four graphs above are obtained from the equations shown, but don't necessarily use the principal branches of ln(z) and sqr(z)"

They have introduced a clear notation and are sticking to it: "In the discussion that follows, the single-valued inverse function (restricted to the principal branch) is denoted by uppercase letters-such as COS−1(z)—to distinguish it from the multivalued inverse—cos−1(z)." (page 59).
When reading one needs to differentiate between functions in uppercase and in lowercase in equations.

(04-30-2016 07:01 AM)Ángel Martin Wrote:  So this leads us to believe that in some cases they had another criteria for choosing the principal branches, like keeping symmetries or other properties to make them more analogous to the behavior in the real domain.

Definitely, on the page 60 they are writing: "The principal branches used by the HP-15C were carefully chosen. First, they are analytic in the regions where the arguments of the real-valued inverse functions are defined. That is, the branch cut occurs where its corresponding real-valued inverse function is undefined. Second, most of the important symmetries are preserved. For example, SIN−1(−z) = -SIN−1(z) for all z."

Interesting question is why are they choosing Im(ARCCOS(z))>0 for Re(z)>1 and Im(z)=0, maybe it is related to ARCCOS(-x) = pi - ARCCOS(x), but I don't see it yet. It needs to be consistent with arccos(x) = pi/2 - arcsin(x) - so in a way it is consequence of arcsin principal branch.

(04-30-2016 07:01 AM)Ángel Martin Wrote:  Perhaps a mathematician in the audience could stop our poking the beast and provide a more rigorous clarification?

I'm a physicist - so almost a mathematician :-), but yes, it is interesting question why they have chosen this exact principal branch - or in other words, what would be broken or "ugly" if principal branch would be different, especially if they would took Im(ARCCOS(z))<0 for Re(z)>1 and Im(z)=0.