What would YOU call this?

[Chuck]

Last week a few friends and I discussed this "function":


f(x)=ln(1-x)+ln(x-2)


It was seen in a classroom, to which the instructor had written, "the domain is empty, therefore this is NOT a function". On further thinking, though, it does not violate any of the parts of the definition of a function. Sooooo, what would you call it? I have my own idea(s).

p.s. Stay out of the complex's. :)

[Valentin Albillo]

Hi, Chuck:

I guess you want this:

    f(x) = ln(1-x)+ln(x-2)

         = ln((1-x)*(x-2))

         = ln(-2+3*x-x2)

    where the argument of ln() is positive for x in (1,2),

    so f(x) is defined and real for x in that range, i.e

    between 1 and 2, both excluded, with extrema at x=1.5

Best regards from V.

[Chuck]

Ahh, but your first step is not allowed; you have drastically changed the function. The original function f(x) = ln(1-x)+ln(x-2)
cannot be evluated at x = 1.5 without delving into the complexes. It's like saying Log[x^2] = 2Log[x]. The graphs of these are clearly not identical (for x>0 they are, but for x<0 they are not). Soooo, is the original a function or not?

[John Gustaf Stebbins]

Looking at a couple definitions of "function" I guess you could say that it is a function on the empty set, assuming you consider the empty set to be a valid domain. Doesn't seem proper, but I don't see where the laws of mathematics would fall apart.

[Crawl]

I'd call it a function, because "stay out of the complex" is an artifical human requirement, while analytic continuity is mathematically natural. I'd also say x / x = 1 at x = 0 (not that it's undefined), and 1 - 1 + 1 - 1 + 1 -... = 1/2, though.

If you want to have a function that has no domain, why not define a function that really has no domain? Or at least something weirder than the logarithm, which is a perfectly normal function, except that it's multi-valued. How about

f(x) = 1^x + 1^(2*x) + 1^(3*x) + 1^(4*x) + 1^(5*x) + ...

which diverges to infinity for all x. Or even use something non-mathematical.

          x, if x loves 0

f(x) =    x^2, if x doesn't love 0

f(x) could take values, but since it's hard to establish if one number loves another, it might be undefined for all x.