Define the intersection points of polynomials

Herc11 · 2013-06-20 18:20:24

Maybe it is easier to understand what I was saying...

I was saying that in Eq. of post 98 there are 4 unknowns i.e. x0...y1.

If I knew 3 more eq. like post #98, I think that the problem has a solution and the intersection points can be spotted.

Herc11 · 2013-06-20 18:23:30

So, you think that for quadratics you can determine the intersection point If you know the leading coef of 4 quads and one point from each of them.

Now I am wondering, If I did not know the lead coef. and I had 8 quad. could I find again the intersection points?

I am confused.

anonimnystefy · 2013-06-20 18:25:37

Herc11 wrote:

24??? Can you explain please?
When I count the unknowns, I count only the missing intersection points.
It is considered that the cubics intercept at 3 points. Each intersction point has two coordinates which are unknown i.e six unknowns.
From each cubic I know one point and its leading coefficient. (****Now I am confused and I m starting thinking that I dont even need
the lead. coef.)
So, in order to find the 6 unkowns I need 6 cubics and the respective lead. coefs and one point from each cubic.
*** It is possible if I do not know the leading coefficient to finde the interscetion points only by using more cubics??
I am confused...
How easy or not is to specify the n-1 intersection points of all the degree n polynomials which pass from the intersection points?

The coordinates of intersection points aren't the only unknowns. What about the 2nd, the 3rd and the 4th coefficients of each polynomial?

bobbym · 2013-06-20 18:25:46

I am still working with your divided difference - Newton formula.

Herc11 · 2013-06-20 18:30:52

If I use the Eq. of post #98 or 99,

only the intersection points are the unknowns?

Last edited by Herc11 (2013-06-20 18:31:35)

bobbym · 2013-06-20 18:32:09

Formula in post #98 checks out.

Herc11 · 2013-06-20 18:33:41

bobbym,
What do you mean exactly?

bobbym · 2013-06-20 18:37:22

I just found the intersection points of a new problem using it and it worked fine. You wrote that you were worried you made a mistake in writing it.

Herc11 · 2013-06-20 18:40:41

Ok.
So, from 4 quads that the only known info is a point of each of them and their leading coefficient we can find their intersection points?
Is that right?

I am thinking that the same will stand for cubics but we will need 6 cubics to solve the previous problem.

bobbym · 2013-06-20 18:48:49

Hold on please we have the time to go slow. I am having trouble with post #99's formula. Please check it.

Herc11 · 2013-06-20 18:57:37

bobbym, I think that i correct it #99. At Π i-1 and not i...I think

bobbym · 2013-06-20 19:06:34

That was not the problem. The denominator is always 0. The upper limit on the product can not be k.

Herc11 · 2013-06-20 19:31:36

Ok. I correct the denom..

Hi anonimnystefy,
Did you see my post #105?

Last edited by Herc11 (2013-06-20 19:33:00)

anonimnystefy · 2013-06-20 19:47:44

Hm, I am not sure. maybe it could work. I am not sure at all, now.

Herc11 · 2013-06-20 19:48:50

When you solved the problem of bobbym what Equations did you use?

anonimnystefy · 2013-06-20 19:54:44

I used a set of 12 equations. Each one was basically each point's coordinates inserted into the appropriate polynomial.

Herc11 · 2013-06-20 19:56:43

12? you had only the lead coefficient, and a point..

It would be helpful if you could post the algorithm.

Last edited by Herc11 (2013-06-20 20:16:07)

bobbym · 2013-06-20 21:27:53

Only 4 equations for my problem are necessary.

Herc11 · 2013-06-20 21:29:56

Ok. That is what I am thinking too. Did you find the solution?

We can suppose that for a cubic 6 equations are demanded?

bobbym · 2013-06-20 21:33:52

The formula in post #99 is now working.

We can suppose that for a cubic 6 equations are demanded?

To test that would require a model problem. I do not have one yet.

Herc11 · 2013-06-20 21:40:40

But it seems that at the formula #, when a3, there will be six unknown variables x0y0...x2y2. There fore, 6 equations will be needed

so 6 cubics plus their leading coefficient and one point of each...

bobbym · 2013-06-20 21:44:01

It seems that is correct but a test always helps.

Herc11 · 2013-06-20 21:50:27

Yes.
Thats the only fact!:)

bobbym · 2013-06-20 21:51:20

If we have cubics there is another difference.

Herc11 · 2013-06-20 21:57:14

what difference?

Math Is Fun Forum

#101 2013-06-20 18:20:24

Re: Define the intersection points of polynomials

#102 2013-06-20 18:23:30

Re: Define the intersection points of polynomials

#103 2013-06-20 18:25:37

Re: Define the intersection points of polynomials

#104 2013-06-20 18:25:46

Re: Define the intersection points of polynomials

#105 2013-06-20 18:30:52

Re: Define the intersection points of polynomials

#106 2013-06-20 18:32:09

Re: Define the intersection points of polynomials

#107 2013-06-20 18:33:41

Re: Define the intersection points of polynomials

#108 2013-06-20 18:37:22

Re: Define the intersection points of polynomials

#109 2013-06-20 18:40:41

Re: Define the intersection points of polynomials

#110 2013-06-20 18:48:49

Re: Define the intersection points of polynomials

#111 2013-06-20 18:57:37

Re: Define the intersection points of polynomials

#112 2013-06-20 19:06:34

Re: Define the intersection points of polynomials

#113 2013-06-20 19:31:36

Re: Define the intersection points of polynomials

#114 2013-06-20 19:47:44

Re: Define the intersection points of polynomials

#115 2013-06-20 19:48:50

Re: Define the intersection points of polynomials

#116 2013-06-20 19:54:44

Re: Define the intersection points of polynomials

#117 2013-06-20 19:56:43

Re: Define the intersection points of polynomials

#118 2013-06-20 21:27:53

Re: Define the intersection points of polynomials

#119 2013-06-20 21:29:56

Re: Define the intersection points of polynomials

#120 2013-06-20 21:33:52

Re: Define the intersection points of polynomials

#121 2013-06-20 21:40:40

Re: Define the intersection points of polynomials

#122 2013-06-20 21:44:01

Re: Define the intersection points of polynomials

#123 2013-06-20 21:50:27

Re: Define the intersection points of polynomials

#124 2013-06-20 21:51:20

Re: Define the intersection points of polynomials

#125 2013-06-20 21:57:14

Re: Define the intersection points of polynomials

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