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        riddles >> general problem-solving / chatting / whatever >> infinity by zero
        
(Message started by: Naomi Rockwood on Jun 3^rd, 2003, 5:00am)

Title: infinity by zero
Post by Naomi Rockwood on Jun 3^rd, 2003, 5:00am

Tell me, do you know what infinity by zero is?
now just think for a mo...
all right. firstly, it should be zero, because zero lots of infinity is zero and an infinite amount of lots of zero is also zero, riiiiggghhhhttt?
nope.
think about this. a point has zero length. a line is made up of an infinite amount of points. a line has infinite length
therefore, infinity times zero is infinity...
there is a similar scenario about time and instants, but that goes into the fourth dimension, so i'm not sure how accurate it is.
think of this...(x/0) = infinity. (0/1) = zero. but (x/0)(0/1) = (0/0) now, is that zero or infinity?
well, i don't know
???
goode lucke

Title: Re: infinity by zero
Post by Icarus on Jun 3^rd, 2003, 6:27pm

Infinity times zero is indeterminant. The value of an apparent "infinity times zero" situation will depend on other aspects of the situation.

For cardinal numbers, 0*[infty] = 0.

lim (x)(1/x) = 1
x[to][subinfty]

lim (x)(2/x) = 2
x[to]0

lim (x)(1/x²) = 0
x[to][subinfty]

lim (x²)(1/x) = [infty]
x[to][subinfty]

Title: Re: infinity by zero
Post by aero_guy on Jun 6^th, 2003, 7:17am

Basically what he is saying is that it depends on how you are getting to that zero and how you are getting to that infinity. I believe they beat the hell out of this point in first (or maybe second) semester college calculus. I use to tutor that stuff, almost none of the students who finish the course actually understand it.

Title: Re: infinity by zero
Post by Justin on Apr 28^th, 2004, 11:23am

Check l'Hopital's rule in calculus. f(x)*g(x)=0 as lim(f(x))--> infinity and lim(g(x))-->0. Infinity times zero is zero.

Title: Re: infinity by zero
Post by william wu on Apr 28^th, 2004, 12:47pm

L'Hospital's rule doesn't say that. As Icarus's examples show, the result depends on f and g.

Title: Re: infinity by zero
Post by Icarus on Apr 28^th, 2004, 7:45pm

Indeed L'Hospital's rule does not! L'Hospital's rule only applies to the two indeterminant cases 0/0 and [infty]/[infty] in the first place (if you have a different indeterminant case, you need to convert it to one of these two before you can apply the rule).

And given such an indeterminant case, it DOES NOT say that the answer is always 0 or [infty] or anything else! If such were always the answer, the case would NOT be indeterminant, and L'Hospital's rule would not apply.

Just to be sure we are talking about the same thing, this is L'Hospital's rule:

If f and g are functions defined and differentiable on a neighborhood of a, and if either ( lim_x[to]a f(x) = 0 and lim_x[to]a g(x) = 0 ) or ( lim_x[to]a f(x) = [infty] and lim_x[to]a g(x) = [infty] ) and if lim_x[to]a f'(x)/g'(x) converges (or diverges to [pm][infty]), then

lim_x[to]a f(x)/g(x) = lim_x[to]a f'(x)/g'(x).

The answer you get for your indeterminant case will depend on the derivatives.

Title: Re: infinity by zero
Post by Benoit_Mandelbrot on May 3^rd, 2004, 6:06am

Actually, 1/0 could be ±[infty], right? It depends on how you get to this limit. lim_n->0(1/n^2)=[infty], while lim_n->0(1/n)=±[infty]. The interesting thing is that ±1/[infty]=0.

Title: Re: infinity by zero
Post by Icarus on May 4^th, 2004, 6:57pm

Actually you could split 0 in two, so you have a +0 and a -0. Then 1/+0 = +[infty] and 1/-0 = -[infty], 1/+[infty] = +0, 1/-[infty] = -0.

Another trick, which unlike that one you will find used on occasions as mathematicians have need, is to have just one infinity which exists at both ends of the number line. To see how to do this, set [bbr] as the x-axis in the plane and consider the unit circle at the origin. We identify each point on the circle other than (0,1) with a real number by projection: for each point on the circle, draw the line through that point and (0,1). The value where the line crosses the x-axis([bbr]) is the real number assigned to the point. Conversely, if we are given a real number, we can find the corresponding point on the circle by drawing the line from (0,1) through the point on the x-axis and seeing where it intersects the circle.

So by this procedure, instead of a "real number line", we have a "real number circle", except that one point is missing. (0,1) has no corresponding real number. If we allow our real numbers to increase without bound, we see that they move around the circle in a counterclockwise fashion, approaching (0,1) from the right as they increase. Conversely, if we let them decrease without bound, they move around the circle in a clockwise fashion and approach (0,1) from the left. It should be obvious therefore that we can use (0,1) to stand in for a single [infty] that lies at both ends of the number line.

As such, we now have 1/0 = [infty], and 1/[infty] = 0, and neither is signed (-0 = 0, -[infty] is undefined since no real number nor [infty] itself produces 0 when added to [infty]).

Neither of these definitions changes the fundamental problem here: 0[cdot][infty], 0/0, [infty]/[infty], [infty]-[infty], 0⁰, 1^[subinfty], [infty]⁰ remain indeterminant.

Even if we defined values for them, the forms would still be indeterminant. 0⁰ = 1, but lim f^g may take on any value from 0 to 1 or fail to converge when lim f = lim g = 0.