i.e. n/0 is defined, therefore 1/x is continuous.

Also, +infinity and -infinity are real numbers (Sup(R) and Inf(R)). I'm not sure about 'natural infinity' since it seems not to fit in anywhere, or even to make sense in terms of real numbers.

If that definition went down to zero then every factorial would be zero right? Anyway, what you are looking for is the gamma function. It is defined to be the extension of the factorial function into the field of complex numbers (and therefore reals, rationals and (the ring of) integers too). There are several ways to define and evaluate the function, for real numbers there are some fairly straightforward methods (especially for x>0) mostly integrals and infinite products if I can remember correctly. Here are a couple that I can remember:

I even bothered to make it look correct for you (shame that there is no subscript and superscript).

code:




for x>0.

        1

G(x) = ò (-log t)^(x-1) dt

        0



        ¥

G(x) = ò t^(x-1) e^-t dt

        0

Those two are actually the same thing just in a different form, the bottom one can often be easier to evaluate even though it involves infinity. (i'll explain integrals if you want but preferably not on the forums, it would require a colossal post)


z=a+ib
|z|=sqrt(a*a+b*b)

The modulus function is mainly for complex numbers. If b=0 then z is real, so |z| is real.

No single valued inverse, that isn't really what I would call an inverse function, multi-valued functions like that are useless since you will never know which value you need without already knowing it.

I'd never seen a conic or any curve using |x| until we had this discussion. Its probably just convention, but it would definately cause problems when dealing with a general conic, for instance, where the curve is off centre and rotated, which is probably why I've never seen it since I was introduced to conics in the general form and then shown the simpler (specific) equations as a consequence of the general form.

I'm just assuming (like you are) that I am correct. :p

The definition of n! is:

n! = n*((n-1)!)
0! = 1

The reason 0! = 1, is because it's useful in statistical analysis. Related to the factorial is the binomial coefficient (A B) (vertical notation).

Erm, Jheriko, what I meant was is the series 0! starts with 0, therefore it will equal 0, riiiiight? :p

Also, x^n where n is even (and integral) has no single-valued inverse function. But I know for a fact that's not a 'shunned' function in mathematics.

+/-/natural infinity are not real numbers, or else they could be graphed. Infinity (and so on) is a surreal number. Natural infinity is infinity without a sign, like zero has no sign (and is a 'natural' number).
Just because infinity equals something doesn't mean that the curve n/x is continuous. Think of curves using imaginary numbers. I can't think of an example right now, but my point is that the curve isn't necessarily continuous, but every value on the curve (or off, as the case may be) has a value - it just can't be graphed.

That has nothing to do with whether or not they are real. Natural infinity is definately not real, +/- infinity are, they are the first and last members of the field of real numbers. How else would you define them?

As I said before, they're surreal numbers...I'll dig up some information on them for you if you'd like.

I also found an absolutely excellent little way to disprove your ideas today:

y=1/x is algebraically equivalent, and therefore graphically equivalent, to x=1/y. In x=1/y, at y=0, as you have said there is a line upon the x-axis where any value could be defined. However there is not a line at y=infinity to y=-infinity, but there is in y=1/x. But they are graphically equivalent.
Let's assume there is a line on both axes in this graph. Then you are saying that infinity must equal -infinity if the function is continuous. Algebraically: x=-x, then 0=-0, which means that infinity=0, which is absurd; you said yourself that +/-infinity are the ends of the field of real numbers.

And a proof that 1/0 must equal infinity: Divide 1 into an infinite number of parts. Each part must be infinitely small. Mathematically speaking, 1/infinity=0. Thus, 1/0=infinity.



/me does a victory dance

Umm, here's what I think:

Did you miss the point of my 'proof'. The whole point of it was that this exact sort of thing results from assuming that 1/x is a defined value. The other problem with this is that I never said that infinity=-infinity all i said was that the curve tends towards them at x=0.

What you have proved is not 1/infinity=0 or 1/0=infinity, but rather you have just shown that 1/infinity tends towards 0 and 1/0 tends towards infinity.


I'm going to give up on this discussion right now with the hopes that one day you will find out for yourself that 1/0 is not always going to behave how you expect it to.

Just give it a rest... Atero: I thought once like you too, trying to think of a 'unified theory of infinity'. It works nicely as long as you stick to asymptotes... for example, suppose you think of R as a ring (at one end, it meets at x=0, at the other end it meets at x=+/- infinity). Instead of graphing on a plane, you could graph on a cilinder of infinite size. You could use something like atan(x) to map R to a finite interval. Then the function would look like a spiral around the cilinder.

However, if it was this easy, don't you think mathematicians would've formalised this thinking into a common set of rules? Fact is, infinity is not as easy as you think it is. Once you delve into advanced maths (integrals and such) you'll encounter much more difficult situations with infinity which cannot be solved by stating infinity is something concrete.

By the way, one big error:

A function is only a function if and only if there is only one image for every point in its domain.

For every x, there is a |x|. But for every |x| there are 2 x's. Thus, any description of the relationship between |x| and x would not be a function, unless it covers only R- or R+.

So for the domain R, |x| has no inverse function, just an inverse relationship.

Two final notes:
I didn't say you said infinity=-infinity, I merely said that what you were arguing implied that infinity=-infinity.
Also, if what UnConeD pointed out is true (not that I'm saying it isn't), then x^n where n is even (and integral) doesn't have an inverse function either, which was my point in my earlier post.

Anyway, I hereby declare this discussion closed :)

Yup Atero... the function sqrt(x) is not the inverse function of x^2 for R, only for R+. Similarly -sqrt(x) is the inverse function x^2 for R-.

Same way (x) is the inverse function for |x| only in R+, and -(x) is the inverse function of |x| in R-, which was what I said before - I thought I made that clear, guess I didn't, sry

Geez this thread is LAME














Geez that pun was LAME