Lamé curve

Yes...Most, if not all, of those examples are flawed in that way.
And you CAN use absolute value in ellipses, without changing the end result at all:
|x/a|^2+|y/b|^2=c^2
The reason being that x^2 is always positive if x is a real number, whether x is positive or negative. So making x positive before squaring it does nothing at all. If x is not a real number, then it's not an ellipse (right? please correct me if I'm wrong...) In fact, I'm fairly sure this is one of those things that's taught 50/50 in schools (with absolute value and without).
You HAVE to use absolute value for Lame' curves or else they become open curves, as you were saying (undefined in the third quadrant). Because when n is not integral, and is not a multiple of the reciprocal of an odd number, then if x or y is negative, x^n or y^n is imaginary.

(Another thing wrong with your trying to prove to me that n/0 is undefined: I never said 0/0 was a defined value. Your proof demonstrates that 0/0 is not defined, but not that n/0 where n!=0 is undefined 🙂 )

hmmmm... Im lost in all this math.
Some questions:
Jheriko: Isnt elipse defined by apsolute value |Z|=1 whith
Z=x/a+(y/b)*i?
Atero: How can n!=0?????And what has the ! thing have to do whith deviation by 0???

!= is the not-equal-to operator.

Originally posted by nixa
Jheriko: Isnt elipse defined by apsolute value |Z|=1 whith
Z=x/a+(y/b)*i?

x^2/a^2+y^2/b^2=1

EDIT: I just thought of a rough proof by contradiction:

if 1/x is a continuous function than it is defined for all points (including x=0). as x tends towards zero 1/x tends towards +/- infinity, therefore at x=0 there is a vertical line x=0 which joins the y = +/- infinity. As a result 1/x can be any number on this line, therefore it can not be defined. This logically applies to n/x which always has a discontinuity at x=0.

Still flawed. The 'vertical line' at n=0 is an asymptote, not an actual line, just a line there to clarify where a curve is tending to. You ought to know that by know, Jheriko 😉
And who said that 1/0 is +/- infinity? Personally I feel there is a difference between 'the reciprocal of x' and '1/x' in this respect: n/0=positive infinity and -n/0=negative infinity (where n is any positive real number). The tangent of multiples of pi/2 (90) is natural infinity. The reciprocal of 0 is also natural infinity. (natural meaning it has no polarity/sign).

Or at least that's MY view on infinity and it's related definitions....

Atero: you're right to differentiate between 1/(0+), 1/(0-) and 1/(0) but you're wrong to distinguish between n/x and tan(n): they both have asymptotes.

And note that an asymptote can be 1/(0+) instead of 1/0, for example 1/(x^2).

Originally posted by Atero
Still flawed. The 'vertical line' at n=0 is an asymptote, not an actual line, just a line there to clarify where a curve is tending to. You ought to know that by know, Jheriko 😉

That was the key point of the rough proof. If the function is continuous then 1/x where x=0 can not be multi-valued, but, if it is continuous then the asymptote IS a part of the curve and it is multi-valued. You can't define 1/x because if you could it could be any number on the line x=0. Its a proof by contradiction.

Originally posted by Jheriko
as x tends towards zero 1/x tends towards +/- infinity

I never mentioned the sign of x, I was being lazy and rather than saying 'as +x tends to zero 1/x tends to +infinity and as -x tends to zero 1/x tends to -infinity' I said what I did, sorry for the confusion.

Taking 1/x as infinite isn't something which can be relied on always. Sometimes it is useful and at other times it can just screw things up.

Erm...my differentiation was that n/0 can equal +/-/natural infinity, but that tan(90n) equals natural infinity. Not that one had asymptotes and the other didn't...I don't know where you got that from 🙂

Whoever said that the curve(s) had to be continuous?

And look, all I'm saying is that I feel that today's ideas of basic mathematic principles are severely flawed; especially those ideas which are taught in basic math classes.
But you've sort of gotten off track with this...weren't you trying to tell me that the absolute value sign is evil incarnate?

(btw, nixa, if n == 0, then n! == 0, doesn't it?)

Originally posted by Atero
Whoever said that the curve(s) had to be continuous?

You implied it by saying n/0=something. if n/0=something, then since every other point on the curve y=n/x is defined then the curve is continuous.

Originally posted by Atero
But you've sort of gotten off track with this...weren't you trying to tell me that the absolute value sign is evil incarnate?

Because it has a point singularity (and isn't differentiable) at x=0, because 0/0 is undefined. Also because you don't always want the positive value in geometry (although I think you understand this).

Originally posted by Atero
(btw, nixa, if n == 0, then n! == 0, doesn't it?)

0! = 1 (as in 'factorial 0' = 1)

Originally posted by Atero
And look, all I'm saying is that I feel that today's ideas of basic mathematic principles are severely flawed; especially those ideas which are taught in basic math classes.

I think that they teach things in the wrong way too. If I had it my way schools would use Euclid's style of textbook as a basis for teaching, where by every piece of knowledge is derived from basic axioms, since even a very young child can understand the basic axioms and definitions of a mathematical system.

Anyway, I'm going to drop this whole thing, your Lamé curve formula generates something more useful for AVS than the real one anyway.

I said n/0 equals something. I didn't say it was a real number, though. Just like y=(x+1)^2 has imaginary roots; but the curve never touches the x-axis, y=1/x has a surreal y-intercept, even though it never touches the y-axis.

General mathematical question: What is the (psuedo-)exact definition of a factorial? Cus I thought it was just n*(n-1)*(n-2)...*1=n!, which means that if n=0, it's 0*1=n!, right?
Another general mathematical question: Is there a real mathematical definition of the absolute value function, or is it just |x| = +x? And if I'm not mistaken it does have an inverse funciton: f(x)=|x|; f^-1(x)=+-x (the plus-or-minus sign which I can't write in arial). Which is why conics and Lame' curves are usually written with absolute value signs - when inverted, you use the positive-or-negative square root, instead of just the square root (which is always negative when written by itself).

What do you mean by 'the real thing?' If you're talking about the real formula, that IS the real formula, just re-arranged to fit scripting 🙂

Originally posted by Atero
I said n/0 equals something.

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.

Originally posted by Atero
General mathematical question: What is the (psuedo-)exact definition of a factorial? Cus I thought it was just n*(n-1)*(n-2)...*1=n!, which means that if n=0, it's 0*1=n!, right?

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).

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)

Originally posted by Atero
Another general mathematical question: Is there a real mathematical definition of the absolute value function, or is it just |x| = +x?

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.

Originally posted by Atero
And if I'm not mistaken it does have an inverse funciton: f(x)=|x|; f^-1(x)=+-x

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.

Originally posted by Atero
Which is why conics and Lame' curves are usually written with absolute value signs - when inverted, you use the positive-or-negative square root, instead of just the square root (which is always negative when written by itself).

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.

Originally posted by Atero
What do you mean by 'the real thing?' If you're talking about the real formula, that IS the real formula, just re-arranged to fit scripting 🙂

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

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? 😛

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.

Originally posted by Atero
+/-/natural infinity are not real numbers, or else they could 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.

Originally posted by Atero
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.

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:

And if I'm not mistaken it does have an inverse funciton: f(x)=|x|; f^-1(x)=+-x

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