Is Multi-Valued Functions a Misnomer?

[Gib Z]

I'm just checking here, but Wolfram and Wikipedia (my 2 most common and highly ranked internet mathematical encyclopedias) are really correct to use the term "MultiValued Function" are they?

I commonly see on Wolfram "f(x) is a multi valued function, and hence requires a branch cut..." but I have learned that if it has 2 output values for a single input, then it is not a function, *by definition*. This abuse of notation is really annoying me because it could put into confusion the actual definition of a function.

The perfect example of an oxymoron =]

 

[HallsofIvy]

In terms of functions of real numbers, yes, a function must be single-valued. But in complex numbers, that becomes much too restrictive. For example ln(z), a very useful (single-valued) function in the real numbers is multi-valued in the complex numbers. Rather than give that up as a function,we relax the definition of function to allow "multi-valued" functions. We still will separate the areas in the range (by the "branch cuts") so that the function is single valued on each "branch".

 

[Gib Z]

I don't understand, wouldn't it be more correct then to just say that the log is a relation in the complex numbers, and only *after* a branch cut is taken, becomes a function?

 

[matt grime]

Who is saying log is a function on C? I see people saying log is a multivalued function on C. Don't just throw out the word 'multivalued'.

Given the genuine abuses of 'functions' propogated throughout introductory courses in mathematics, I would have thought this was incredibly minor. I mean, every single question that asks 'find the domain if this function' is completely an utterly mathematically unsound, for instance.

 

[D H]

You are thinking of "multivalued" as a qualifier, as if multivalued functions are a subset of all functions. Think of it the other way around: Multivalued functions are a generalization of the concept of a function rather than a specialization.

Similar abuses of language notation occurs elsewhere in math. One that comes to mind is "Cartesian tensor". Any tensor is a Cartesian tensor, but a Cartesian tensor is not necessarily a tensor. The rules regarding what constitutes a Cartesian tensor are looser than the rules regarding what constitutes a tensor. Similarly, the rules regarding what constitutes a multivalued function are looser than the rules regarding what constitutes a function.

 

[Gib Z]

So basically, it is just a very minor abuse of notation that shouldn't cause any problems =] And:

every single question that asks 'find the domain if this function' is completely an utterly mathematically unsound

It was only from you matt grime that I learned this, I hear teachers at school giving these questions all the time. I now know that the domain is part of the original definition of the function.

Thank you all for your replies!

 

[Dragonfall]

Since we are being overly pedantic here, remember that there is no "set of all functions."

 

[D H]

I was being a bit loose and shouldn't have used the term set when I didn't mean "set".

OTOH, isn't $\beth_2$, the power set of the continuum, of the same cardinality as the set of all mappings (i.e., functions) from R to R?

 

[CRGreathouse]

OTOH, isn't $\beth_2$, the power set of the continuum, of the same cardinality as the set of all mappings (i.e., functions) from R to R?

Yeah... and wouldn't that make $\beth_3$ the cardinality of the set of multivalued functions in one variable? That is, for each point in the domain there are $\beth_2$ possibilities for a multivalued function, so
$$\mathfrak{c}^{\beth_2}=\aleph_0^{\beth_2}=\beth_3$$
is the number of multivalued functions in one real variable. Right?

Wow, that's almost hard to think about -- every point has its own one-dimensional fractal. Shudder.

 

[uart]

Originally Posted by matt grime
"every single question that asks 'find the domain if this function' is completely an utterly mathematically unsound"

It was only from you matt grime that I learned this, I hear teachers at school giving these questions all the time. I now know that the domain is part of the original definition of the function.

Thank you all for your replies!

About this point, while I do understand what Matt is getting at here you do have to take into account that when students are first introduced to the concept of domains that they generally know nothing about anything other than real numbers. So the concept of a functions domain being determined by the functions natural limitations of existence over the reals seems like a resonable thing to do.

Personally when the concept of domain is introduced to students in this manner I like to qualify it as the "natural domain" and at least give some other examples where the domain is specified as part of the function definition and qualify it as the "defined domain". To illustrate "defined domain" I usually like to give a nice simple example of a piecewise function like,

$$y = \left\{ \begin{array} {cc} x + 1 & : x \le -1 \\ x - 1 & : x \ge +1 \\ \end{array}$$

Really is that so terrible?

 

[matt grime]

So the concept of a functions domain being determined by the functions natural limitations of existence over the reals seems like a resonable thing to do.

No, it isn't/doesn't (depending on how one corrects that fragment so it is a sentence). It is just poor mathematics, with no excuse.

 

[Dragonfall]

You can ask questions like "where in $$\mathbb{R}$$ is $$f(x)=\sqrt{x}$$ defined so that it is real-valued?" To many students this is the "domain" of a function; and to an extent this is true if you define a limited "universe of discourse".