Explaining Infinity

For the discussion of math. Duh.

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asimore
Posts: 10
Joined: Tue Oct 12, 2010 6:26 pm UTC

Explaining Infinity

Postby asimore » Wed Dec 01, 2010 9:45 am UTC

Hi,
first off im sorry for my bad english, i especially lack "specialised" vocabulary. I hope you can still guess what im tryin to ask :mrgreen:

So, i had a math-test yesterday, and one of the topics was curve sketching.
I think it was y=(x^2-10)/(x^2-3x-10)
I remember the limes being 3, and the poles being -2 and 5. So the curve was entering from the left above the limes, taking a left turn at the lime & left pole crossing into +infinite-y, then coming in from below, taking a turn back down (like an upside down U), and then came in again from top, taking a left at the right pole & limes crossing going to infinite-x.
Il'd draw it online somewhere but i think im not allowed to post links with the amounts of posts i have :S But i think you get the picture.

Anyways, we have to "describe" the different sides of the curve, for example:

Code: Select all

lim -> infinite
(x->3)


My question is..is there a negative infinite?
Like saying "curve comes from the left side, negative infinite x, taking a left going to a positive infinite y"?
Because from what i know, infinity can be explained like this:
If you take the number-line and draw a circle above it, every tangent relates to a number on the number-line, except for the ones that are parallel to it.
So, the infinity-tangents go from -infinite to +infinite. Does it even matter, or is it mathematically correct one way or the other, to say "-infinite" and "+infinite"? I mean, maybe if you have it in an equation and you can cancel it down, it would still leave -1...hm...
Anyways, if he strips some points because i didn't say "-infinite" and "+infinite" il'd like to be able to call bs on it (or know to just take it xD) :P

cheer,
-a :)

edit: Whoa..i just read about latex and tried to implement the \infty symbol..but that tore my text to shred, cuz it always centered it and stuff. So..sorry for not using LaTex, but i don't have the time to play around atm (@work :P)
"For an engineer, the glass is neither half full nor half empty. It is twice as big as it has to be!"

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skeptical scientist
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Re: Explaining Infinity

Postby skeptical scientist » Wed Dec 01, 2010 11:14 am UTC

asimore wrote:My question is..is there a negative infinite?

Yes. This is called the extended real number line, and consists of the real numbers together with \(+\infty\) (we sometimes just write \(\infty\) for \(+\infty\)) and \(-\infty\). This allows us to talk about infinite limits, and limits as x approaches infinity (or minus infinity). For example, for your function $$f(x)=\frac{x^2-10}{(x-5)(x+2)},$$ we can see that there are vertical asymptotes at 5 and -2. This is because $$\lim_{x \to 5^-} \, \frac{x^2-10}{(x-5)(x+2)}=-\infty,$$ which means that as x approaches 5 from the left, the value of the function eventually gets smaller than -1, then -10, then -100, then -1000, and in fact eventually it gets smaller than every negative number, no matter how large its magnitude. Informally, this is because if x is very close to 5 but a little smaller, then x^2-10 will be about 15, and x+2 will be about 7, and x-5 will be very close to zero (and negative), so putting this all together, we have $$f(x)=\frac{x^2-10}{(x-5)(x+2)}\approx \frac{15}{7\times \text{some tiny negative number}},$$ so f(x) is negative, and very large in magnitude, since the reciprocal of a tiny number has extremely large magnitude. (A more formal argument is hidden by the spoiler below, but you can skip it if you don't find such formal arguments useful.)
Spoiler:
If 5-e<x<5 (for e<1), then $$f(x)<\frac{25-10e+e^2-10}{(5-e-5)(5+e+2)}<\frac{5}{(-e)(8)}=-\frac{5}{8e}.$$ Given some large number M>1, by choosing e less than 5/(8M), if 5-e<x<5 (so x is very close to 5, and smaller than 5), since f(x) is smaller than -(5/8e), this means f(x)<-M, so f can be made arbitrarily negative.


We can also talk about horizontal asymptotes using the same framework. Your function has a horizontal asymptote at 1, which f(x) approaches as x approaches both positive infinity and negative infinity. (Formally, we write that \(\lim_{x \to +\infty} f(x)=1\), and similarly for \(x \to -\infty\).) This means that as x gets very large, f(x) is eventually always between 0 and 2, and then between .9 and 1.1, and then between .99 and 1.01, and so on, getting closer and closer to 1. To see this, we can rewrite f by dividing the top and bottom by x2, which gives us the same value of the function for x≠0 (and of course the value of the function at 0 doesn't affect the limit as x approaches ±infinity, which depend only on the longterm behavior of f). So we have $$f(x)=\frac{x^2-10}{x^2-3x+10}=\frac{1-\frac{10}{x^2}}{1-\frac{3}{x}+\frac{10}{x^2}}.$$ If x is extremely large, then 3/x and 10/x2 will be extremely close to 0, so for x very large, f(x) is very close to 1 (and as x gets bigger, 3/x and 10/x2 get even closer to 0, so the approximation gets more and more exact, meaning that the limit is 1). Again, you could prove this formally using an argument similar to the one I spoilered above, or by applying theorems about limits (such that the limit of a quotient is the quotient of limits, provided the limits of the numerator and denominator exist, and the limit of the denominator is nonzero).
I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

"With math, all things are possible." —Rebecca Watson

asimore
Posts: 10
Joined: Tue Oct 12, 2010 6:26 pm UTC

Re: Explaining Infinity

Postby asimore » Wed Dec 01, 2010 12:38 pm UTC

Thanks :D
Man, thats the only thing i missed..i hope the labeling on the drawing isn't costing me my grade ^^
"For an engineer, the glass is neither half full nor half empty. It is twice as big as it has to be!"

Suffusion of Yellow
Posts: 42
Joined: Thu Jan 01, 2009 6:15 pm UTC

Re: Explaining Infinity

Postby Suffusion of Yellow » Thu Dec 02, 2010 11:18 pm UTC

asimore wrote:Because from what i know, infinity can be explained like this:
If you take the number-line and draw a circle above it, every tangent relates to a number on the number-line, except for the ones that are parallel to it.
So, the infinity-tangents go from -infinite to +infinite. Does it even matter, or is it mathematically correct one way or the other, to say "-infinite" and "+infinite"? I mean, maybe if you have it in an equation and you can cancel it down, it would still leave -1...hm...


There are different definitions of "infinity" depending on what you're doing...what you're talking about is function theory idea of infinity (it's usually formulated on a sphere over the complex plane, but it's the same idea.) In that case there's only one infinity. In terms of analysis, which is what I think you're doing, there's two infinities, because in analysis, infinity doesn't mean a number, it means a limit (that your sequence gets arbitrarily small or arbitrarily big.)

So, in function theory: one infinity. In analysis, 2: positive and negative. (In set theory, infinite many infinities, but that's weird.)


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