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Re: [2013-03-26] 2927

Posted: Wed Mar 27, 2013 3:48 pm
by Jazzbo
I agree that it depends on the definition of emptiness. If emptiness is some measure of how much space the graph takes compared to the whole, like the area of the curve over the area of the whole, then the answer depends on your perspective. Ideally, the graph is 1 dimensional and thus has no area. So it would be zero the whole way.

If you consider the actual physical artifact instead of the idealized concept it represents, then the curve, of course, is not actually 1-dimensional but 2-d since any real-world line has thickness. And if you think not in terms of the area of the whole plane but instead the area of the rectangle before you and consider the ratio of those, then it is not zero. If you consider a straight line, it's area scales with length assuming constant thickness (an idealization). If you merely change the width of the graph and not the height, then it's area also scales by the length. And so you have a horizontal line again, just not at zero. That constant is dependent on the thickness of the line and the height of the graph.

Now I was wondering if one could have a curve as a solution instead of a horizontal line. Before I do that, I will briefly explore why a line has to be horizontal from a different point of view.

First consider a diagonal line with thickness w and length l at an angle theta going across a graph of with width x and height h:

h w
| /
|l/
|/theta
|___ x

x/l = cos theta
thus l = x / (cos theta)
Area of line over area of graph is lw/(hx) = wx/(hx cos theta) = w/(h cos theta)
Since w, h, and theta are constant, this is a constant which means it must be a horizontal line as argued before. The graph would be y = w/h.

Now let us look at a general curve. We are proposing that the area of the thick line after horizontal distance x over the area of the rectangle of the graph up to width x equals the value of the function at x.

(integral x'=0->x [w*dl])/(hx) = y, where h is height of the graph and x is the current width of the graph and dl is the differential length of the curve (construed here has a function of x) and w is the width of the curve.

So (w/h) integral x'=0->x [dl] = xy

Take derivative of each side by x.

(w/h)dl/dx = x*dy/dx + y
(w/h)sqrt( (dy/dx)^2 + 1) = x*dy/dx + y

Now, let us consider boundary conditions. The highest ratio possible between the areas is 1. Of course it is going to be less than one. The graph will not take up the whole area, so there must be a horizontal asymptote. That means that as x goes to infinity, dy/dx goes to zero. dy/dx should go to zero much faster than x goes to infinity, so the above will give you,

w/h = y when x approaches infinity.

In other words, the asymptote is the horizontal line discussed earlier. We also know that any deviation of the curve from a horizontal line lengthens the line and hence increases the area. So the curve cannot go down. It can only go horizontally or up. Thus if there is a solution to the differential equation that is not a horizontal line, it must start below the asymptote. But the initial behavior of the curve must mirror the behavior of the tangent line. Any line will have a value of w/(h cos theta) which is always greater than or equal to w/h. Since it cannot be greater, the horizontal line is the only possible solution.

Re: [2013-03-26] 2927

Posted: Wed Mar 27, 2013 7:28 pm
by Lethal Interjection
ThePeople wrote:You people need to look at the picture a lot more closely. You don't seem to understand it.
Clearly they aren't spending enough time analysing it. The meaning/solution requires more thought.

Re: [2013-03-26] 2927

Posted: Wed Mar 27, 2013 8:44 pm
by Guest
You could just label the y axis in reverse, so it's 100% where the x axis crosses, and 0% near the top. Job done.

Re: [2013-03-26] 2927

Posted: Wed Mar 27, 2013 10:33 pm
by Oldrac the Chitinous
ThePeople wrote:You people need to look at the picture a lot more closely. You don't seem to understand it.
I think you'd better post it again.

Re: [2013-03-26] 2927

Posted: Thu Mar 28, 2013 12:25 am
by ThePeople
I take no joy in this

Image

Re: [2013-03-26] 2927

Posted: Thu Mar 28, 2013 12:44 am
by rjsfox911
A horizontal line at y=.5 (shaded below, of course) would be valid, no? If the graph's "width" is only the part of the horizontal which has been graphed (which seems logical), then for any given width, half is filled and half is not.

Re: [2013-03-26] 2927

Posted: Thu Mar 28, 2013 1:18 am
by Edminster
maybe the problem is that people aren't seeing the full majesty because the forum software cuts off the image

maybe flipping the image will help

Image


maze1125 i know you reported the first image but this is necessary

maybe in time you will come to understand

Re: [2013-03-26] 2927

Posted: Thu Mar 28, 2013 10:20 am
by ThePeople
I think the orientation is important, just as in english we read from left to right, the eye "reads" along the pig resting finally on the poo on his balls.


But perhaps a smaller picture will deal with the problem.

Image

Re: [2013-03-26] 2927

Posted: Fri Mar 29, 2013 1:55 am
by john greyman
ok, so, I think the joke is that, the entirety of the emptiness is determined by how full the graph is, in regard to the graph's width. So, you should just make a line, near the top, which goes strait across. Since the line is homogenous, and fills the graph at the same rate, as it's width grows, then the amount of empiness actually doesn't change as the width grows. So it's always the same amount of empty, so long as teh line you're drawing is very very regular.

Anyway, that's my two cents.

My other two cents is that if their speaking about the totality of the graph, regardless of the point on the x-axis. Then you should make the y axis negative, with increasing negativity. Then you leave it blank. Because it's empty, and thus, you shouldn't decrease the emptiness... or something, don't know if that breaks some graphing rule or not, simply how i'd chose to orient my graph.

Re: [2013-03-26] 2927

Posted: Fri Mar 29, 2013 2:31 am
by Lethal Interjection
john greyman wrote:ok, so, I think the joke is that...
Not a good way to start a thread.

Also, interestingly, not a good way to finish a joke...

Re: [2013-03-26] 2927

Posted: Fri Mar 29, 2013 2:35 am
by Edminster
i think maybe the smaller picture removes the symbolic weight of the feces resting upon the hale and hearty pig's scrotum

i may have found a way to preserve most of the image as well as providing a similar traveling of the pig towards the ultimate destination of pooballs

.jpg

Image

Re: [2013-03-26] 2927

Posted: Fri Mar 29, 2013 4:38 am
by ThePeople
It seems to be working, good job ed, I love you

USER WAS BANNED FOR THIS POST

Re: [2013-03-26] 2927

Posted: Fri Mar 29, 2013 4:53 am
by DonRetrasado
too much love.

Emptiness of this Graph (Problem Solved)

Posted: Fri Mar 29, 2013 1:24 pm
by Warchon
Image
That is all.

Re: [2013-03-26] 2927

Posted: Mon Apr 01, 2013 5:32 am
by Apocalyptus
I'm sorry guys, but I think the message still isn't getting across quite strongly enough. Here, I've attempted a post-modern distillation of the artistic essence of the pooballs.

Image