Define the intersection points of polynomials

bobbym · 2013-06-20 01:25:28

But to get that you had to introduce new points.

Herc11 · 2013-06-20 01:37:52

x0 x1 y0 y1 are the same intersection points. the other one for each polynomial will be the known point.

bobbym · 2013-06-20 01:39:33

Yes, but that polynomial needs 3 known points to be a unique quadratic. You have 2 and one unknown point. You can not determine that quadratic knowing only two.

anonimnystefy · 2013-06-20 01:43:07

Hi Herc11

Do you have a test example we could try?

bobbym · 2013-06-20 01:43:59

What is wrong with post #9? Can anyone get A and B?

anonimnystefy · 2013-06-20 01:46:09

I need a of four quadratics.

Herc11 · 2013-06-20 01:47:31

Ok. You are right. But I am confused with the equation of defining a2.

In equation of a2 only the 4 variables x0 x1 y0 y1 exist.

And
if I have four equations for a2 (I do not know the 4 polynomials, only that they have the same 2 inters. points x0 y0 x1 y1 (unknowns),

I know

each polynomial's leading coef. i.e. a2

and one point of each of the polynomials.

Then from the set of 4 equations of a2, x0..y1 can be defined?

bobbym · 2013-06-20 01:48:08

Me too, and putting more points on the drawing does not give them to me.

I need more info about those quadratics!

anonimnystefy · 2013-06-20 01:49:29

I need you to give me the leading coefficient of two more quadratics through those two unknown points and a point on each of those.

Last edited by anonimnystefy (2013-06-20 01:50:27)

Herc11 · 2013-06-20 01:51:23

anonimnystefy wrote:

I need a of four quadratics.

So, if you have 4 quadratics plus a point of each of them you can solve it? i.e you can find x0...y1?

bobbym · 2013-06-20 01:52:13

For post #9? Yes any additional info would help. I have been asking the OP to state the full problem. It might be possible to get more info out of it.

anonimnystefy · 2013-06-20 01:56:27

Herc11 wrote:

anonimnystefy wrote:
I need a of four quadratics.
So, if you have 4 quadratics plus a point of each of them you can solve it? i.e you can find x0...y1?

I think so.

bobbym · 2013-06-20 01:57:52

From just the first coefficient?

anonimnystefy · 2013-06-20 01:59:16

First coefficient of four quadratics and a point on each.

bobbym · 2013-06-20 02:00:40

Want to use post #9 and I will provide the two more quadratics since I know what A and B is?

Herc11 · 2013-06-20 02:01:28

anonimnystefy

I think so.

I have the same opinion.

bobbym · 2013-06-20 02:02:39

It is more information than you provided in post #1 so I am willing to post the challenge. It is worth a shot.

Herc11 · 2013-06-20 02:03:35

bobbym wrote

Want to use post #9 and I will provide the two more quadratics since I know what A and B is?

What do you mean exactly?

bobbym · 2013-06-20 02:04:55

I know the answer for post #9 only because I created it. No one else can know it from that information ( maybe ). I can give 2 more quads.

anonimnystefy · 2013-06-20 02:05:01

Hi bobbym

Where does it say that we have only 2 quadratics in the first post??

bobbym · 2013-06-20 02:07:14

It does not, we just went from there. But so far no one has proved it is possible with n quadratics.

Herc11 · 2013-06-20 02:10:42

It does not, we just went from there. But so far no one has proved it is possible with n quadratics.

Thats why I am asking.

It seems to me that If there are 4 quad. you can define the 2 interesection points.

Similarly, if the polynomials were of degree 3, 6 polynomials of 3 degree might be needed...and so on...

But I am not sure

bobbym · 2013-06-20 02:12:27

I am going to say that with 4 quadratics or 176 of them you still can not define the two points. That is my opinion.

anonimnystefy · 2013-06-20 02:14:23

bobbym wrote:

I am going to say that with 4 quadratics or 176 of them you still can not define the two points. That is my opinion.

Okay, then give me the first coefficient of four quadratics and a point on each of them and I will try finding the 2 intersections.

bobbym · 2013-06-20 02:17:26

Yes, give some time. I am very interested in your result. Please hold.

Math Is Fun Forum

#26 2013-06-20 01:25:28

Re: Define the intersection points of polynomials

#27 2013-06-20 01:37:52

Re: Define the intersection points of polynomials

#28 2013-06-20 01:39:33

Re: Define the intersection points of polynomials

#29 2013-06-20 01:43:07

Re: Define the intersection points of polynomials

#30 2013-06-20 01:43:59

Re: Define the intersection points of polynomials

#31 2013-06-20 01:46:09

Re: Define the intersection points of polynomials

#32 2013-06-20 01:47:31

Re: Define the intersection points of polynomials

#33 2013-06-20 01:48:08

Re: Define the intersection points of polynomials

#34 2013-06-20 01:49:29

Re: Define the intersection points of polynomials

#35 2013-06-20 01:51:23

Re: Define the intersection points of polynomials

#36 2013-06-20 01:52:13

Re: Define the intersection points of polynomials

#37 2013-06-20 01:56:27

Re: Define the intersection points of polynomials

#38 2013-06-20 01:57:52

Re: Define the intersection points of polynomials

#39 2013-06-20 01:59:16

Re: Define the intersection points of polynomials

#40 2013-06-20 02:00:40

Re: Define the intersection points of polynomials

#41 2013-06-20 02:01:28

Re: Define the intersection points of polynomials

#42 2013-06-20 02:02:39

Re: Define the intersection points of polynomials

#43 2013-06-20 02:03:35

Re: Define the intersection points of polynomials

#44 2013-06-20 02:04:55

Re: Define the intersection points of polynomials

#45 2013-06-20 02:05:01

Re: Define the intersection points of polynomials

#46 2013-06-20 02:07:14

Re: Define the intersection points of polynomials

#47 2013-06-20 02:10:42

Re: Define the intersection points of polynomials

#48 2013-06-20 02:12:27

Re: Define the intersection points of polynomials

#49 2013-06-20 02:14:23

Re: Define the intersection points of polynomials

#50 2013-06-20 02:17:26

Re: Define the intersection points of polynomials

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