dementedwombat

About the original problem; how about the graph of the function resulting when lim.(Z->∞) sin(Z·X) is graphed over {x|0<X<1,X∈R}?

One heck of a fast and loose equation, but it works (at least in my mind). Because the function is compressed a just-less-than infinite amount [lim.(Z->∞)], and so has a just-less-than infinite amount of maxima (and minima). This should be no different from an infinite amount of maxima, which is uncountable. Of course the whole thing most likely falls apart because lim(Z->∞) is equivalent to ∞; thus rendering the entire function non-continuous.

Unfortunately my school isn't very math oriented. That up there is the ramblings of a dangerously uninformed high school senior with one quarter of AP Calculus under his belt.

P.S. lim.(Z->∞) is used above as a quick and dirty way of getting a number "just less than" infinity. I'm sure there's a better way (made more difficult because ∞-1=∞) but right now I'm too lazy to care.

unitofobscurity

I'm pretty sure lim(z -> ∞) (k∞) is an aleph-naught infinity, if we're talking about cardinality, and is thus a 'countable' infinity. (Aleph number - Wikipedia, the free encyclopedia)

To be honest, Cantor's distinction between countable and uncountable infinities seems so obtuse that I honestly can't imagine having a non-cardinal (ie non-ordered field) number of maxima over [0,1] assuming that we're sticking to a purely functional expression.

unitofobscurity

Wait - how about z = (sin(1/(x + yi))) ? where x and y are both in the interval [0, 1]? That seems pretty uncountable, and seems to resemble an ordinal infinity a little bit more than sin(1/x).

dementedwombat

Can't comment on your addition as I've never properly learned the orders of infinity. I'll grill my Calc teacher on it and get back to you tomorrow.

BTW, are you using "i" there to refer to "Square root of -1" or is there another i that I don't know about?

EDIT: That equation also has the unfortunate tendency to divide by zero at one end point.

unitofobscurity

No, it's √-1 . Other uses of i just annoy me, almost as much as using e as a variable.

And the discontinuity at x + yi = 0 shouldn't be a problem, because it still has uncountably infinite local maxima (?).

But yeah, my math education's been a bit wishy-washy as well, so I'm definitely not 100% sure on this one.

dementedwombat

My math teacher said the equation I put up should work fine since the real numbers are uncountably infinite (when you include decimals). One person already solved it with y=1, since every point is a maxima, but I like ours better.

theZcd

Weierstrass function anyone?

And for the record, i != Sqrt[-1] because this definition creates certain problems in commutativity.
It's defined as i²=-1

unitofobscurity

Whoas! I think theZcd wins. That's really a mind-bending function.