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From: mike3 on 22 Jun 2010 22:29
On Jun 22, 8:03 pm, Robert Israel
<isr...(a)math.MyUniversitysInitials.ca> wrote:
> mike3 <mike4...(a)yahoo.com> writes:
> > Hi.
>
> > Is it true that if one has a continuous curve connecting two points,
> > makes a copy of it and slides the copy along the direction between
> > those points a distance less than 1/2 the distance between the points,
> > then the original curve and translated copy will intersect at at least
> > one point? If not, can you provide me a counterexample, and if so, how
> > about either a proof or just a hint at the proof, if the proof isn't
> > too heavily sophisticated?
>
> Counterexample: points are (0,0) and (1,1), curve is y = sin(9/2 pi x) - x
> for 0 <= x <= 1, slide right by distance 0.45.

Uh, slide "right"? I mentioned to "slide the copy along the direction
between
the two points". That is, along the direction of the _line_ through
the two
points. I know, I didn't state that quite clearly enough, sorry. This
example
fails to refute the hypothesis, since a translation by 0.45 along the
line
through (0, 0) and (1,1) leads to intersections.
From: mike3 on 22 Jun 2010 22:30
On Jun 22, 8:11 pm, Tim Little <t...(a)little-possums.net> wrote:
> On 2010-06-23, mike3 <mike4...(a)yahoo.com> wrote:
>
> > Is it true that if one has a continuous curve connecting two points,
> > makes a copy of it and slides the copy along the direction between
> > those points a distance less than 1/2 the distance between the points,
> > then the original curve and translated copy will intersect at at least
> > one point?
>
> No.
>
> > If not, can you provide me a counterexample
>
> A helix.
>

That's not 2D (i.e. I'm asking about a _plane_ curve.). I should've
mentioned that...
From: mike3 on 22 Jun 2010 22:30
On Jun 22, 8:22 pm, Tim Little <t...(a)little-possums.net> wrote:
> On 2010-06-23, Robert Israel <isr...(a)math.MyUniversitysInitials.ca> wrote:
>
> > Counterexample: points are (0,0) and (1,1), curve is y = sin(9/2 pi x) - x
> > for 0 <= x <= 1, slide right by distance 0.45.
>
> Nice counterexample, though I suppose you really mean (0,0) and (1,0).
>
> - Tim

Oh, now I see... it does work! Thank you...
From: Bart Goddard on 22 Jun 2010 22:32
mike3 <mike4ty4(a)yahoo.com> wrote in news:705b3ff9-66e8-4e4b-887a-
1ec7ba0f7497(a)u7g2000yqm.googlegroups.com:

> Hi.
>
> Is it true that if one has a continuous curve connecting two points,
> makes a copy of it and slides the copy along the direction between
> those points a distance less than 1/2 the distance between the points,
> then the original curve and translated copy will intersect at at least
> one point? If not, can you provide me a counterexample, and if so, how
> about either a proof or just a hint at the proof, if the proof isn't
> too heavily sophisticated?

There must be a slicker way, but:

First translate and rotate so that the points are on
the x-axis and the continuous function f has f(a)=f(b)=0.
Then slide the curve along the x-axis a distance p where
p < (b-a)/2. Call this function g(x) = f(x-p).
Consider h(x) = f(x-p)-f(x), which is continuous and
defined on the interval [a+p,b]. Then h(a+p)=-f(a+p)
and h(b)=f(b-p). If these two values have opposite
sign, you're done, since h(q) = 0 for some q in the
interval [a+p,b-p]. If not, then f(a+p) and f(b-p)
have opposite signs, so there exists p1 in the
interval [a+p,b-p] such that f(p1)=0. Then h(p1)=
f(p1-p). If this is of the opposite sign of h(a+p)
and h(b-p), you're done. Else repeat to get p2, p3,...
which is a bounded sequence and must have a limit point
q. Since h is continuous, h(q)=0 and you're done.

--
Cheerfully resisting change since 1959.
From: Tim Little on 22 Jun 2010 23:03
On 2010-06-23, mike3 <mike4ty4(a)yahoo.com> wrote:
> On Jun 22, 8:03 pm, Robert Israel
>> Counterexample: points are (0,0) and (1,1), curve is y = sin(9/2 pi x) - x
>> for 0 <= x <= 1, slide right by distance 0.45.
>
> Uh, slide "right"?

Yes. The (1,1) should be (1,0).


- Tim
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