| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 960.1 | | 38863::DERAMO | Daniel V. {AITG,ZFC}:: D'Eramo | Sun Oct 30 1988 23:41 | 24 |
| A new problem! Just what I was looking for. :-)
solution follows
Let g(x) = f(x) - f(-x)
f is continuous, and so are both f(-x) and g.
Since S1 is compact, the image f(S1) is compact, and
therefore the image has a maximum point fMAX [let x1 be
such that f(x1) = fMAX] and a minimum point fmin [let
x2 on the circle be such that f(x2) = fmin].
Now g(x1) = f(x1) - f(-x1) >= 0 because f(x1) is as high
as f goes. Likewise g(x2) = f(x2) - f(-x2) <= 0 because
f(x2) is as low as f goes.
Since g is a continuous function from the connected set
S1 its image is also connected, and so there is a point
x on the circle such that g(x) takes on the intermediate
value 0. For any such point x it must be the case that
f(x) = f(-x).
Dan
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| 960.2 | | 7413::XIA | | Mon Oct 31 1988 01:04 | 27 |
| re .1:
Well done, Dan. That proof is very clever. Now for a some what
harder problem:
3
Take the unit sphere S2 in R , i.e., the set of points
(x , x , x ) satisfying:
1 2 3
2 2 2
x + x + x = 1
1 2 3
2
Let R be the set of all the points on the plane . Let
2 2
f: S2-->R be a continuous function from S2 into R . Prove there
exists a point x in S2 such that:
f(x) = f(-x)
i.e., there is a pair of antipodal points which is mapped to the same
point.
Eugene
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| 960.3 | | AITG::DERAMO | Daniel V. {AITG,ZFC}:: D'Eramo | Mon Oct 31 1988 15:16 | 7 |
| Thanks. What was the unclever proof?
I'm working on the problem in .2 -- show there are two
points at opposite ends of the earth with both the same
temperature and humidity. :-)
Dan
|
| 960.4 | | HPSTEK::XIA | | Mon Oct 31 1988 16:47 | 18 |
| re .3
>Thanks. What was the unclever proof?
I am not aware of any other elementary proof of the theorem.
> I'm working on the problem in .2 -- show there are two
> points at opposite ends of the earth with both the same
> temperature and humidity. :-)
That is a very nice way of looking at it. Hmmm. It can also be
said about temperature and pressure. Wind speed and pressure.
Wow, theoretical math is useful after all :-) :-).
Eugene
As to the solution of .0. Does that mean if you draw a circle around
the earth, there exist two opposite ends with both the same
temperature? No wonder the North pole is as cold as the south
pole :-) :-).
|
| 960.5 | Translate that into space... | AKQJ10::YARBROUGH | I prefer Pi | Tue Nov 01 1988 19:38 | 14 |
| This discussion reminds me of a conjecture that was discovered during my
days as an analyst on the Apollo Project. The problem being attacked was
that of minimizing the energy required to transfer a satellite from one
elliptical orbit into another in the same plane; in other words, at what
point in the orbit do you fire the control rockets to maneuver into the new
orbit with minimum energy consumed? As I recall, there are always *two*
diametrically opposed points in the orbit with the same minimum. This
appears to be independent of any of the orbital parameters. We were never
able to find an analytic proof, but the concept is I believe still in use
without modification. The interesting thing to me is that the two points
may be at vastly different distances from the mass being orbitted, and the
speed of the vehicle at those points is quite different.
Lynn Yarbrough
|
| 960.6 | | ATLAST::FRAZER | Je suis prest! | Tue Nov 01 1988 19:59 | 4 |
| Or . . .
A square table with legs the same length can be made not to rock
on an uneven (continuous) floor.
|
| 960.7 | | HPSTEK::XIA | | Tue Nov 01 1988 21:03 | 25 |
| re .5, .6:
The theorem (as stated in .2) was proved a long time ago (maybe
50 years ago). Moreover, the proof is for the general case as stated
in the following:
---------------------------------------------------------------
Let Sn be the n-sphere, i.e, Sn is the set of all the points
n+1
x in R such that ||x|| = 1 where ||.|| is the Euclidean norm.
n
Let f: Sn-->R be a continuous. Then there exists x in Sn such that:
f(x) = f(-x)
i.e., there is a pair of antipodal points which is mapped to the same
point.
Eugene
P.S. The proof, however, is not elemetary. That is why I did not ask
for a proof for the general case. I am just looking for an elementary
proof of the case stated in .2 because the advanced proof for
the general case does not offer much intuition.
|
| 960.8 | re .2 | AITG::DERAMO | Daniel V. {AITG,ZFC}:: D'Eramo | Tue Nov 01 1988 21:11 | 53 |
| Now that you mention it, it was temperature and barometric
pressure, but I guess any pair of continuous functions will
do.
The problem in .2 is harder than the one in .0 in the
following sense. In .0 I used the facts that
the image of a compact set under a continuous function
is compact
the image of a connected set under a continuous
function is connected
compact subsets of the real line (of any Hausdorf
space) are closed and so contain all of their limit
points
etc.
and no one complained. Those will be used for .2, but are
not enough; to really solve the problem requires quoting
results that are essentially equivalent to it, or else being
called upon to supply the [long] proofs reducing them to
more well known results.
Anyway, here goes. Suppose the result in .2 is false, so
that there exists a continuous function f:S2 -> R2 such that
f(x) not= f(-x) for any pair of antipodal points x, -x on
the sphere. Let g be the function given by g(x) = f(x) - f(-x).
Then obviously g never takes on the value (0,0) [no that
isn't a face, it is an ordered pair of zeroes]. Write g as
g(x) = (g1(x), g2(x)) and define h by
g(x) g1(x) g2(x)
h(x) = ---------- = (----------------- , -----------------)
|| g(x) || sqrt(g1^2 + g2^2) sqrt(g1^2 + g2^2)
Since g1 and g2 are never simultaneously zero [by the
assumption on f] there will be no division by zero here, and
so h is well defined. Since f is continuous and the
arithmetic operations involved are continuous, both g and h
are also continuous. Now, h:S2 -> S1, i.e., h maps the
sphere into [actually onto] the circle x1^2 + x2^2 = 1.
Also, h(-x) = -h(x) because of how g and h were chosen.
So far this has shown that if there a continuous f:S2 -> R2
such that f(x) never equals f(-x), then there must also be a
continuous function h:S2 -> S1 such that always h(-x) = -h(x).
But it's a little known result that such a function h is
impossible. :-)
Dan [... to be continued]
|
| 960.9 | If you were a teacher, how would you grade this? :-) | AITG::DERAMO | Daniel V. {AITG,ZFC}:: D'Eramo | Sun Nov 13 1988 01:37 | 105 |
| re .8,
>> So far this has shown that if there a continuous f:S2 -> R2
>> such that f(x) never equals f(-x), then there must also be a
>> continuous function h:S2 -> S1 such that always h(-x) = -h(x).
>>
>> But it's a little known result that such a function h is
>> impossible. :-)
The proof of this involves a lot of stuff from topology.
Let X and Y be two topological spaces, f:X -> Y and g:X -> Y
be two continuous functions from X into Y, and let I be the
unit interval [0,1] of the real line. The following tries
to capture the intuitive concept of slowly deforming the
function f until it becomes the function g. The functions f
and g are said to be homotopic if there is a continuous
function H: X x I -> Y (that's the cross product of X and I)
such that for all x in X, H(x,0) = f(x) and H(x,1) = g(x).
A path (or curve) in a topological space Y is defined as a
continuous function from the unit interval I into Y.
Suppose in the above that X is the unit interval, i.e., that
f and g are paths in Y. Suppose further that f(0) = g(0) = a
and f(1) = g(1) = b, i.e., boths functions are paths from the
same starting point a to the same ending point b. Let H be
a homotopy between them as before. If H also satisfies the
condition that for every i in I, H(0,i) = a and H(1,i) = b
(i.e., as H continuously deforms f into g, each
"intermediate" stage is also a path from a to b), then f and
g are said to be path homotopic.
If a = b in the above, i.e., if f(0) = f(1), then the path f
is called a loop.
The spaces involved here -- the circle and the surface of a
sphere -- both have the property that any two points of the
space can be connected by a path f with f(0) one point and
f(1) the other. Such spaces are called path connected. Now
let a be a point of such a space and let f be a loop based
at a. Let g be the constant function from the unit interval
I given by g(i) = a for all i in I. Obviously g is the
simplest kind of loop possible. If f and this g are path
homotopic then that says something very interesting about f,
that it can be "deformed" into the constant function, or
stated another way the loop f can be "shrunk to a point."
If a path connected space has the property that every loop
based at some point a of the space is path homotopic to the
constant loop at a, the space is said to be simply
connected. [Which point a doesn't matter because the space
is path connected.]
Now, the surface of a sphere, S2, is simply connected. Any
loop on the surface can be slowly shrunk into a point
[i.e. into the constant function]. But the circle S1 is not
simply connected. For example, the loop f given by
f(i) = < cos 2 pi i, sin 2 pi i > for i in [0,1]
which "goes around the circle once" cannot be cointinuously
deformed into the constant function given by g(i) = <1, 0>.
It you go into the proof with enough detail every step can
be reduced to things out of a first course in topology or
real analysis.
The proof that
>> So far this has shown that if there a continuous f:S2 -> R2
>> such that f(x) never equals f(-x), then there must also be a
>> continuous function h:S2 -> S1 such that always h(-x) = -h(x).
>>
>> But it's a little known result that such a function h is
>> impossible. :-)
essentially comes down to showing that if such a function h
existed then there would be a path homotopy between two
loops in the circle that are known not to be path homotopic.
So let h be such a function. Let f:I -> S2 be a simple loop
around the equator of the sphere. Then g:I -> S1 given by
g(i) = h(f(i)) is a loop in the circle. Let x be the point
f(0) on the (equator of the) sphere, and the simple loop f
could be such that -x on the sphere is f(1/2), and in
general f(i + 1/2) = -f(i) for 0 <= i <= 1/2. So g(0) =
h(f(0)) = h(x) and g(1/2) = h(f(1/2)) = h(-x) = -h(x). Here
is where I will start the handwaving. The path g considered
as a function from [0,1/2] has gone around the circle "n + 1/2"
times for some integer n, and g(i + 1/2) = h(f(i + 1/2)) =
h(-f(i)) = -h(f(i)) = -g(i). So the loop g can't "unwind"
as i goes through [1/2,1]; that part of the path repeats the
[0,1/2] part of the path except g(i + 1/2) = -g(i) [it's
across the diameter from g(i)]. So the loop g "goes around
the circle 2(n + 1/2) = 2n + 1 times" -- and odd number of
times; such a g cannot be homotopic to the constant loop
which goes around the circle zero times, an even number.
But now (set handwave off) let H be a path homotopy between
the loop f in the sphere and the constant loop at f(0).
Then the composition of h and H will be a path homotopy between
the function g and the constant loop in the circle. But
this can't be; the handwavy part shows that g and the
constant loop are such that the stated-without-proof part
applies.
Dan
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