grunwald (01/27/83)
#N:uiucdcs:28200003:000:1448 uiucdcs!grunwald Jan 26 18:20:00 1983 I've got this little problem which I have been unable to put away for awhile, and yet I realise that there must be a trivial solution to the thing. Assume you have a measuring stick which can only measure the edges of a square. Now, you want to find the length of the diagonal of a unit square. So, for the first approximation, you measure the edges, sum them and say "length is 2". Well, to get a "better approximation" you divide the square into 4 even parts. You measure the inside path from the lower left corner to the upper right corner. Again, you get a length of 2. Now, you continue subdividing the unit square in to n sqaured sub-squares, and taking a path from lower-left to upper right. If we let the n -> infinity, then we would expect to eventally get a value of sqrt(2) for the length. But, as best as I have been able to determine, the limit is still 2. What's the flaw in the above argument? (the following diagrams may or may not make the problem easier to understand) 1: ------- square of unit area | X Path marked by X is of length 2 | X (well, not in this picture, but in real life...) XXXXXXX 2: ------- 4 squares on 1/4 unit area | | X path marked X is of length 2 ---XXXX | X | XXXX--- 4: ----xxx 16 squares, 1/16 unit area | | x | path marked x is of length 2 ----x-- | | x | --xxx-- | x | | --x---- | x | | xxx---- Carry this on till the area of each small square approachs 0.
ardis (01/27/83)
#R:uiucdcs:28200003:uiucdcs:28200004:000:493 uiucdcs!ardis Jan 27 07:27:00 1983 The "limit diagram" you describe is not a good approximation (or even a better approximation than the first one) for the diagonal. Consider the sides to be vectors. You always measure vectors that are either horizontal or vertical. So, you will always compute the same sum. In other words, you are factoring the diagonal vector into horizontal and vertical components. The limiting case of this still factors the vector, it just does so with an infinite number of segments of the vector.
mat (01/27/83)
Question is about measuring the length of a diagonal by zigzagging along lines parallel to the co-ordinate system ( sides of a square ). For some really good stuff along these lines, read B. Mandelbrot's book ``Fractals, Form, Chance, and Dimension''. But a quick description of what is happening: You are right in saying that the path length is always 2; it IS always 2. You are never actually measuring the diagonal. You are measuring a path which approximates the diagonal. The approximations are continually improving in their mean-square error, but their path lengths remain the same. -More than one way to measure the -goodness of fit of a cat skin. Mark Terribile
leichter (01/28/83)
There is nothing "wrong" with the process described, when you get right down to it, except that it gives the "wrong" answer! As soon as you go to any infinite or limiting process, you have no a priori reason to believe your intuition is correct. You've made an implicit claim here: The "limit", in some sense of the word, of your smaller and smaller subdivisions is the diagonal AND this notion of "limit" fits correctly with the notion of length. Well, unfortunately, this claim is incorrect - as your example shows. One has to be more subtle in defining "limit" for a series of geometric operations like this if one expects the limit of the areas to be the area of the limits. (In fact, if you want to define area for surfaces in three-space by the limit of smaller and smaller triangular patches covering the surface - and obvious notion - one finds that this just will not work at all - it's easy to find example of surfaces that have a finite area if you do the sub-division one way, but an infinite area if you do the subdivision another way. It MAY be possible - I don't remember, you should be able to find it in any good calculus text - to construct a figure for which one can get ANY answer above some lower limit as the area by an appropriate subdivision.) Another demonstration of why your intuition is just plain wrong for infinite cases: Consider the hyperbola y = 1/x for x>0. Spin it around the x axis to form an infinite horn. If you sit down and work it out - simple calculus - you will find that the VOLUME of the horn is finite - but the surface area is infinite! Alternatively: You can fill the horn with a finite amount of paint, but no amount of paint will cover the outside. Strange but true. -- Jerry
hickmott (01/28/83)
The mathematics system described by measuring distance as (x+y)
instead of (x**2 + y**2)**.5 is known as Lobachevskian geometry;
it is (as far as anyone has tested it and as far as I know)
completely self-consistent. It takes advantage of the fact, which you
may recall from 'way back in geometry I, that 'point', 'line', 'plane',
and 'distance' are all undefined terms. Although I'm only
vaguely familiar with it, I remember that it comes up with some
interesting things; e.g. a circle (the locus of all points
equidistant from a given point) looks like a square standing on
one corner. I'd be interested in hearing if it has any practical
application.
"And who deserves the credit, and who deserves the blame?
Nicolai Ivanovitch Lobachevsky is his name!"
Andy Hickmott
decvax!yale-comix!hickmottsibley (01/28/83)
The article uiucdcs.1389 suggests a way to determine the length of the diagonal of a square. It doesn't work, but the author wants to know why. First, it is true that it doesn't work. In the example given, the sum of all the lengths of the little vertical pieces is always 1, the original side length, and the same happens horizontally. That's why you get 2. This explains why the integral formula for arc length of a curve is so complicated. That is, if the suggested procedure worked, one could calculate arc length by integrating ( y' + 1 )dx instead of the usual (((y')**2 + 1 )**.5)dx. Note that the first (wrong) integral is actually easy to evaluate -- it is y(b) - y(a) + b - a if the integral is taken from a to b. The second (correct) one does not simplify. Dave Sibley Department of Mathematics Penn State University psuvax!sibley
dap1 (01/30/83)
#R:uiucdcs:28200003:ihlpb:6200008: 0:1126 ihlpb!dap1 Jan 29 16:29:00 1983 I believe that the paint in the horn paradox is just a little misleading. It implies ( to me anyway ) that although you can have a "layer" of paint "under" the horn, you can't have a layer of paint "on" the horn. Well, what is being referred to here as a "layer" is described in mathematics as a 2 dimensional manifold embedded in a 3 dimensional space. Such manifolds are intuitively "warped planes", that is, infinitely thin. The "paint" could cover the horn if it was spread out infinitely thin. That is to say, it can be proven that there is a many to one correspondence between the points contained in the horn and the points on the horn (in fact, between any three dimensional continuum and a two dimensional manifold). Therefore, the "thickness" of the layer of paint is the only thing that keeps it from covering the horn. Normally, it would take twice as much paint to cover a floor twice as large but this isn't the case if the paint is allowed to be placed on in negligibly thin layers. Sooooo, while the paint inside could not cover the horn physically, it can mathematically. Darrell Plank BTL-IH
ken (02/02/83)
I suspect that Andy Hickmott really means a distance measurement based on
D = | x | + | y |,
where | . | is the magnitude or absolute value.
Such a measure is known as the l1 norm. The Euclidean or l2 norm is the one
we normally use to interpret geometrical relationships. The other norm
widely used in mathematics is the l(infinity) norm, which is
D = max ( | x |, | y | )
In general, we can have the lp norm, defined as
p p 1/p
D = ( | x | + | y | )
but only the l1, l2, and l(infinity) norms are used in practice.
Such norms do indeed have numerous applications in areas such as optimization
and approximation. They are used to indicate closeness of two points in space
whether that space be 2-space, 3-space, n-space, or function space. The choice
of which norm to use usually depends on which is easiest to compute.
Computer graphics uses a modification of the l1 norm quite frequently. It is
D = max(|x|, |y|) + 1/2 * min(|x|, |y|)
and is much cheaper computationally than the l2 norm, but looks more like an
octagon than a circle. If using this to approximate the l1 norm, the RMS and
maximum errors are about 10%, whereas if the 1/2 above is replaced by 5/16
or 3/8, the errors are more around 5%.
Yes, Andy, such funny distance measurements do have their place in applied
mathematics.
Ken Turkowski
{ucbvax,decvax}!decwrl!turtlevax!kenleichter (02/02/83)
No, that's not right; the "paradox" is really there. A one-to-one correspon- dence between the points has nothing whatsoever to do with area; any two squares in the plane, or volumes in 3-space, have Aleph-1 points. Area is a much more sensitive measure. What the "paradox" shows is that the generalized definition of area, while it accords well with intuition for finite objects, breaks down for infinite objects - actually, of course, the problem is not with the defini- tion but with our intuition! If you really want to look at it as PHYSICAL paint, then consider: Paint is made of atoms of some finite size. Hence, when you pour paint into the horn, it cannot get "below" the point where the nozzle is narrower than an atom. On the other hand, a finite thickness paint will, after a certain point, cover the outside of the horn with a constant-diameter tube (two atoms across at least). Hence, even physically, painting the outside "as far as you can go" takes an infinite amount of paint, but filling "as far as you can go" takes only a finite amount. This last argument is a pretty picture, but has NOTHING to do with area or volume. Area and volume are mathematical constructs that are useful because when applied to real-world objects - which are always finite - they provide useful values - i.e. they predict how much paint you need to cover a wall. BTW, there is a error in my original description. I said to consider the curve y = 1/x for all x>0. There is an infinite area and volume, though, in any section of the resulting horn that gets arbitrarily close to 0, though, because 1/x is blowing up there. You have to cut the curve off at some particular positive x - say, x = 1; any value will do. -- Jerry decvax!yale-comix!leichter