w*******g
发帖数: 9932 | 1 This is STUPID.
just shift the curve alone line AB to the left or right by 1/n
draw a line 'l' parallel to AB from the intersection point 'p' of the old
and new curve
and if the intersection of 'l' with old curve is 'q', then 'pq' is of length
1/n.
for proof, just need to show that the new curve and old curve must intersect.
we can show by contradiction of course. piece of cake. | s**x
发帖数: 405 | 2
^^^^^^^^^^^^^ show it please...
i don't think it is so trivial
in particular, if you don't shift by 1/n, but by some value larger than
0.5 (for instance 0.7), then it is possible that two curves do not intersect
【在 w*******g 的大作中提到】
: This is STUPID.
: just shift the curve alone line AB to the left or right by 1/n
: draw a line 'l' parallel to AB from the intersection point 'p' of the old
: and new curve
: and if the intersection of 'l' with old curve is 'q', then 'pq' is of length
: 1/n.
: for proof, just need to show that the new curve and old curve must intersect.
: we can show by contradiction of course. piece of cake.
| w*******g
发帖数: 9932 | 3
let c be the original curve and c_1 be the curve shifted 1/n.
suppose there is no intersection.
Then, c_1 must be "above" or "below" c.
Suppose c_1 is "below" c.
Now shift c_1 by 1/n again and becoming c_2.
c_2 is also "below" c_1 by assumption.
Continue this n times. c_n and c are continuous but in between there are
n-1 curves however.
This is a contradiction.
【在 s**x 的大作中提到】
:
: ^^^^^^^^^^^^^ show it please...
: i don't think it is so trivial
: in particular, if you don't shift by 1/n, but by some value larger than
: 0.5 (for instance 0.7), then it is possible that two curves do not intersect
| s**x
发帖数: 405 | 4
I can see your idea, and I believe it is intuitively right.
However, I still feel something is missing in the proof.
Your above claim depends on how you define "above" and "below".
A simple definition would be:
c1 is "above" c iff for all points p in c, p1 in c1
s.t. p.x=p1.x, it is true that p.y
However, under this definition I don't see how to show that c1 must be
either "above" or "below" c. The problem is that curve c is not necessarily
expressable as a function in x (suppose AB
【在 w*******g 的大作中提到】
:
: let c be the original curve and c_1 be the curve shifted 1/n.
: suppose there is no intersection.
: Then, c_1 must be "above" or "below" c.
: Suppose c_1 is "below" c.
: Now shift c_1 by 1/n again and becoming c_2.
: c_2 is also "below" c_1 by assumption.
: Continue this n times. c_n and c are continuous but in between there are
: n-1 curves however.
: This is a contradiction.
| w*******g
发帖数: 9932 | 5 I see your point.
I can't give a precise definition of "above" or "below" based on the x-axis
value.
maybe c "above" c_1 means that from any point far below the x-axis, if it
emits an array, it must travel through c_1 before reaching c.
【在 s**x 的大作中提到】
:
: I can see your idea, and I believe it is intuitively right.
: However, I still feel something is missing in the proof.
: Your above claim depends on how you define "above" and "below".
: A simple definition would be:
: c1 is "above" c iff for all points p in c, p1 in c1
: s.t. p.x=p1.x, it is true that p.y: However, under this definition I don't see how to show that c1 must be
: either "above" or "below" c. The problem is that curve c is not necessarily
: expressable as a function in x (suppose AB
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