Prev: Elastic Numbers as a Model for DNS Expansion
Next: Theorem: The maximum cut problem
From: Thomas Nordhaus on 13 Apr 2010 03:35
Robert H. Lewis schrieb:
>> (Problem): Given ellipses e1: x^2/a^2 + y^2/b2 = 1
>> and e2: x^2/A^2 + y^2/B^2 =1. Compute the minimal distance D between
>> points P, Q lying in e1, e2 resp.
>> ...
>> Here is what I found out:
>>
>> (Result): Provided that
>> (1) (A/a) < (A^2-B^2)/(a^2-b^2) < B/a
>>
>> the minimal distance D between e1 and e2 is given by:
>>
>> (2) D = (a^2*B^2 - b^2*A^2)*(1/(a^2-b^2) - 1/(A^2-B^2))
>>
>> For example let a=2, b=1, A=2.5, B=1.5 then D ~=
>> 0.47871. Compare that to (A-a)=(B-b)=0.5. ...
>
> I think (1) was supposed to be (A/a) < (A^2-B^2)/(a^2-b^2) < B/b.

Oh yes, of course - I'm sorry. These equations should be symmetric under
mutual permutatations of the symbols. Maybe that explained the confusion
with Narasimhan in an older follow-up... I hope he's still reading this
thread.


>
> This is really cool, thanks for sharing.
>
> I thought of generalizing it. Let f1 be an ellipse centered at the origin and f2 any conic. We can write their equations as
> f1 := a1^2*x1^2 + b1^2*y1^2 - 1;
> f2 := a2*x2^2 + b2*y2^2 + c2*x2 + d2*y2 + e2*x2*y2 + ff;
>
> Then there are four equations, based on the equations and partial derivatives:
>
> f1 = 0
> f2 = 0
> Deriv(f1,x1,1)*Deriv(f2,y2,1) - Deriv(f1,y1,1)*Deriv(f2,x2,1) = 0
> (y2 - y1)*Deriv(f1,x1,1) - (x2 - x1)*Deriv(f1,y1,1) = 0
>
> A resultant can be computed for each of x1, x2, y1, y2 in turn. Numerical values could then be plugged in for the parameters, and the various possibilities checked.
>
> Using the Dixon resultant, I got the resultant for x1 to have 1529 terms and the one for x2 to have 10400 terms. Each takes a few seconds to compute.
>
> Alternatively, one can work with ds =distance between the points (x1,y1) and (x2,y2). There are then five equations, and one can eliminate x1, y1, x2, y2 and have a resultant for ds in terms of the parameters. That is a lot harder, and blows out my RAM.

Oh boy! - I will not go along that path. At the moment the ellipses are
tough enough for me. ;-)

--
Thomas Nordhaus
From: gudi on 13 Apr 2010 14:01
On Apr 10, 1:46 pm, Thomas Nordhaus <thnord2...(a)yahoo.de> wrote:
> gudi schrieb:
>
> > On Apr 9, 3:12 pm, Thomas Nordhaus <thnord2...(a)yahoo.de> wrote:
>
> >> Thomas Nordhaus schrieb:
>
> >>> (Result): Provided that
> >>> (1) (A/a) < (A^2-B^2)/(a^2-b^2) < B/a
>
> > Trying to follow this. Letting c^2 = a^2 - b^2 and C^2 = A^2 - B^2.
>
> OK.
>
>
>
> > For the special case of smaller circle c = 0 , B/a > Infinity
> > and the other special case of bigger circle C = 0 ,  A/a < 0
>
> > OK? !
>
> The assumption was that a>b, A>b. But if you fix A,B and b and let a->b
> then the middle-term of the inequality becomes bigger then B/a and the
> inequalty becomes void which means the formula for D cannot be applied.
> In that case the minimum for D will either be A-a or B-b.
>
> >>> the minimal distance D between e1 and e2 is given by:
> >>> (2)  D = (a^2*B^2 - b^2*A^2)*(1/(a^2-b^2) - 1/(A^2-B^2))
> >> Correction:
>
> >>  D^2 = (a^2*B^2 - b^2*A^2)*(1/(a^2-b^2) - 1/(A^2-B^2))
>
> > Needs one more correction to D^6, to tally equation dimension?
>
> I dont understand - the dimensions are correct: D^2 is of order two. The
> r.h.s is of order 2*2/2 = 2. That checks.

> > If either c or C is zero, then the circle should touch the ellipse?
> > Please check.
>
> Mhhm - I think you're overcomplicating matters. The case you stated
> isn't covered by the assumptions. The formula for D doesn't work  in
> this case(isn't even defined). What is true is the following: If the
> assumptions on a,b,A,B are'nt met, then

I am so sorry, please ignore this. I found and removed my error in
posting
a few minutes after posting but the server was down, before it got
working
once again you have already noticed it.

Narasimham

> --
> Thomas Nordhaus

From: gudi on 14 Apr 2010 07:04
On Apr 8, 4:47 pm, Thomas Nordhaus <thnord2...(a)yahoo.de> wrote:
> Hi,
> (Problem): Given ellipses e1: x^2/a^2 + y^2/b2 = 1 and e2: x^2/A^2 +
> y^2/B^2 =1. Compute the minimal distance D between points P, Q lying in
> e1, e2 resp.
>
> W.l.o.g assume that a>b, A>a, B>b. So e2 lies on the outside of e1.
> Candidates for points P and Q where this minimum is attained are
> adjacent vertices of the ellipses. That would mean D = A-a or D=B-b
> whatever is smallest. However this is not always true. In some cases the
> minimal distance is attained "in between" vertices. Here is what I found
> out:
>
> (Result): Provided that
> (1) (A/a) < (A^2-B^2)/(a^2-b^2) < B/a
>
> the minimal distance D between e1 and e2 is given by:
>
> (2)  D = (a^2*B^2 - b^2*A^2)*(1/(a^2-b^2) - 1/(A^2-B^2))
>
> For example let a=2, b=1, A=2.5, B=1.5 then D ~= 0.47871. Compare that
> to (A-a)=(B-b)=0.5. One can easily see from (1) that for example similar
> and confocal ellipses do not meet condition 1. Given a and b (1) yields
> a tongueshaped region in (A,B) space.
>
> Question: Is there a geometric proof to this without all these messy
> computations?
>
> Thomas Nordhaus

Hi,
What are coordinates of P and Q calculated/obtained in each case in
terms of a,b,A and B?
Narasimham



From: Thomas Nordhaus on 14 Apr 2010 09:44
gudi schrieb:
> On Apr 8, 4:47 pm, Thomas Nordhaus <thnord2...(a)yahoo.de> wrote:
>> Hi,
>> (Problem): Given ellipses e1: x^2/a^2 + y^2/b2 = 1 and e2: x^2/A^2 +
>> y^2/B^2 =1. Compute the minimal distance D between points P, Q lying in
>> e1, e2 resp.
>>
>> W.l.o.g assume that a>b, A>a, B>b. So e2 lies on the outside of e1.
>> Candidates for points P and Q where this minimum is attained are
>> adjacent vertices of the ellipses. That would mean D = A-a or D=B-b
>> whatever is smallest. However this is not always true. In some cases the
>> minimal distance is attained "in between" vertices. Here is what I found
>> out:
>>
>> (Result): Provided that
>> (1) (A/a) < (A^2-B^2)/(a^2-b^2) < B/a
>>
>> the minimal distance D between e1 and e2 is given by:
>>
>> (2) D = (a^2*B^2 - b^2*A^2)*(1/(a^2-b^2) - 1/(A^2-B^2))
>>
>> For example let a=2, b=1, A=2.5, B=1.5 then D ~= 0.47871. Compare that
>> to (A-a)=(B-b)=0.5. One can easily see from (1) that for example similar
>> and confocal ellipses do not meet condition 1. Given a and b (1) yields
>> a tongueshaped region in (A,B) space.
>>
>> Question: Is there a geometric proof to this without all these messy
>> computations?
>>
>> Thomas Nordhaus
>
> Hi,
> What are coordinates of P and Q calculated/obtained in each case in
> terms of a,b,A and B?
> Narasimham

Hi Narasimhan,
first of all there is a "slight" error in condition (1). It should read:

(1) A/a < (A^2-B^2)/(a^2-b^2) < B/b

If P = (xP,yP), Q =(xQ,yQ) then

(2) xP^2 = f(a,b,A,B), yP^2 = f(b,a,B,A)
xQ^2 = f(A,B,a,b), yQ^2 = f(B,A,b,a)

where


B^2*(a^2-b^2)^2 - b^2*(A^2-B^2)^2
(3) f(a,b,A,B) = a^4 -----------------------------------
(a^2-b^2)^2*(B^2*a^2 - A^2*b^2)

I used these eqns to derive (1).

BTW: To derive the Dmin-formula one doesn't need to know these
coordinates. We found a new proof where one just minimizes the distance
between mutual tangents whether they have a common normal or not. The
minimal tangent-distance is then attained at the points with common
normals (as it should be), but one doesnt' have to compute these points
before hand. So I guess, your hint went in this direction?

Again, here is the formula for Dmin:


(4) Dmin = (a^2*B^2 - b^2*A^2)*(1/(a^2-b^2) - 1/(A^2-B^2))

P.S. The new proof is much simpler. Basically one derives at equations
for the parallel tangents t1, t2 of the form

t1: cos(t)*x + sin(t)*y = d1
t2: cos(t)*x + sin(t)*y = d2

where d1 and d2 are the distance of t1, t2 from the origin and t is a
parameter giving the slope of the tangent. The distance D between the
tangents is then simpy given by D = d2-d1.

Furthermore d1, d2 are given by:

d1 = a^2*cos^2(t) + b^2*sin^2(t)
d2 = A^2*cos^2(t) + B^2*sin^2(t)

Minimizing d2-d1 (or rather (d2-d1)^2) w.r.t. t isn't too difficult and
doesn't involve very complicated algebra. Still I'm trying to fit that
into a "geometric picture".

--
Thomas Nordhaus
From: gudi on 24 Apr 2010 15:43
On Apr 8, 4:47 pm, Thomas Nordhaus <thnord2...(a)yahoo.de> wrote:
> Hi,
> I'm looking for an "elegant" geometric proof of of a result I and others
> worked out completely algebraically and by using CAS. The equations
> turned real messy quick but at the end everything turned out nice. So I
> suspect some deeper lying and geometric arguments exist.

> Question: Is there a geometric proof to this without all these messy
> computations?
>
> --
> Thomas Nordhaus

Hi Thomas Nordhaus,


If we ask for intersection points between two non-intersecting curves
in CAS
like Mathematica, we get complex roots.The minimum distance in case of
real
roots is zero and in case of complex roots in some way related to the
function
derivatives at the minimum distance points.The treatment had been in
earlier "classical"
text books books one gathers.Hope someone would shed some light on
that topic now.

To find minimum distance between x-axis and ellipses etc. I had posted
in NG
geometry.puzzles earlier, and mention it fwiw, due to the same same
technique
needed I believe, to get at a general solution:

geometric method to find Complex Roots when an Ellipse“cuts” x –
axis.

The present question is much more general than the above particular
case.

Just find the general method or procedure to get minimum distance c -
b^2/4a
for a parabola y = ax^2 + bx + c at min point got by putting dy/dx = 2
a x + b
= 0 and apply it over here :) ...

MfG

Narasimham


First  |  Prev  |  Next  |  Last
Pages: 1 2 3 4
Prev: Elastic Numbers as a Model for DNS Expansion
Next: Theorem: The maximum cut problem