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**Herc11****Member**- Registered: 2013-06-19
- Posts: 169

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.

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**Herc11****Member**- Registered: 2013-06-19
- Posts: 169

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.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,544

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?

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,271

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

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**Herc11****Member**- Registered: 2013-06-19
- Posts: 169

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)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,271

Formula in post #98 checks out.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**Herc11****Member**- Registered: 2013-06-19
- Posts: 169

bobbym,

What do you mean exactly?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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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.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**Herc11****Member**- Registered: 2013-06-19
- Posts: 169

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.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,271

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

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Herc11****Member**- Registered: 2013-06-19
- Posts: 169

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,271

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

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Herc11****Member**- Registered: 2013-06-19
- Posts: 169

Ok. I correct the denom..

Hi anonimnystefy,

Did you see my post #105?

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

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,544

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

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**Herc11****Member**- Registered: 2013-06-19
- Posts: 169

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

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,544

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

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**Herc11****Member**- Registered: 2013-06-19
- Posts: 169

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)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Only 4 equations for my problem are necessary.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Herc11****Member**- Registered: 2013-06-19
- Posts: 169

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

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,271

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.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Herc11****Member**- Registered: 2013-06-19
- Posts: 169

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...

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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It seems that is correct but a test always helps.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Herc11****Member**- Registered: 2013-06-19
- Posts: 169

Yes.

Thats the only fact!:)

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,271

If we have cubics there is another difference.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Herc11****Member**- Registered: 2013-06-19
- Posts: 169

what difference?

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