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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: Harlan's World
- Registered: 2011-05-23
- Posts: 16,037

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?

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

The knowledge of some things as a function of age is a delta function.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

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

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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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****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Formula in post #98 checks out.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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

bobbym,

What do you mean exactly?

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

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.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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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****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

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

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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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****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

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

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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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: Harlan's World
- Registered: 2011-05-23
- Posts: 16,037

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

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

The knowledge of some things as a function of age is a delta function.

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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: Harlan's World
- Registered: 2011-05-23
- Posts: 16,037

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

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

The knowledge of some things as a function of age is a delta function.

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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****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Only 4 equations for my problem are necessary.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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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****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

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.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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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****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

It seems that is correct but a test always helps.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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

Yes.

Thats the only fact!:)

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

If we have cubics there is another difference.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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

what difference?

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