You can
put this solution on YOUR website!Hi,
I don't know what definition of convex you're working with, but they shouldn't differ too much, so here's mine.
A set is convex, if for all pairs of elements of the set, the geodesic lies entirely within the set.
The geodesic is the shortest path, in
it is always a straight line.
So lets pick two arbitrary points
and
, and find the straight line between them.
=x_1+t(x_2-x_1)\;t\in\[0,1\])
Because
is a vector space, we can add and scalar multiply to our heart's content, without leaving
. Thus
\in\mathbb{R}^n\,\forall t)
And that's it, proved.
Hope that helps, if you have any problems, please email back.
Kev
Question from original poster
Thanks......I have my line segment as L=mX1+(1-m)X2 0<=m<=1
is this any different then the line that you had?
There are two ways to think about this. You can say that any two straight lines that go through the same two points must be the same.
My line passes through
when
and it passes through
when
. Your line passes through
when
and passes through
when
. So we have two straight lines that each pass through two points, therefore the must be the same.
The other much better(If you like geometry, which I do) way to think about it is this.
Think of a person travelling along my line, now for any
they are on the line. Lets consider a general function s(x), that maps the closed unit interval to its self.
. Because
\in\[0,1\])
then choosing
)
, we know that
))
is still the same line. Although the person moves along it differently. If we can choose an
)
that makes
)=L(m))
then the two lines are clearly the same(note that this technique works in general for any path, not just lines)
Let's choose
=1-m)
, so
)=x_1+(1-m)(x_2-x_1))
If you tidy this up you will find you get the same equation, thus the lines are the same.
Hope that helps,
Kev.
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