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

I think it has not to do with interpolation. See in #146 post

the formula for computing a_3

if the unknowns are x0 y0 x1 y1 x2 y2

and known the x3 y3 and a3

I think that by having a set of 6 equations you can define x0-y2.

Also, bobbym

posted another one problem

*Last edited by Herc11 (2013-06-21 08:34:00)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,676

Unless you eliminate the a0, a1 and a2 you will have more than 6 variables. You will have 9 variables. This means 9 cubics.

Fortunately I have a way around that.

**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

Why to eliminate?

Simply substitute the a0 a1 and a2 with x0-y2.

See post #146.

Dont use a0..etc if it confuses you...

You have only to solve for x0..y2.

And six equations like that in #146 are enough

*Last edited by Herc11 (2013-06-21 08:43:29)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,676

Aren't the a0, a1, a2, a3 the coefficients of the cubic?

**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

Yes they are coeffients.

But see equation in #146.

Then suppose you have 6 equations like #146.

You know a3 y3 x3, and you dont know x0x1x2 y0y1y2.

By a set of 6 equations isn t the set of equations solvable?

Our aim is to define the intersection points i.e. x0x1x2 y0y1y2.

Once you solve the set of the 6 equations you will have discovered the x0x1x2 y0y1y2.

If you d like you can then define a0=y0 a1=... and a2=...

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,676

Hi;

Yes, I see that. I am agreeing with you. They are replaced as you did in post #146, so did I. Eventually all of them are replaced and just the 6 variables we want remain.

I did not understand you were saying the same thing I was.

**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:)

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,676

Hi;

I am currently busy answering a question. I will need some sleep right after. I will work out the problem I posed and get the solution as soon as I wake up. I am really beat.

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 ok.

I think that anonimnystefy can solve it as he solved the quadratics...

*Last edited by Herc11 (2013-06-21 09:17:08)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,676

He is sleeping by now, I think. I need some rest. see you later and thanks for the problems.

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

Thans both of you!

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

By the way are you students or college students?

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,525

Hi guys

I haven't had time to post here. Have you got the interpolation thing to work?

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

Hi anonimnystefy,

See the posts #133,135 and 146,147.

The aim is to solve a set of 6 equations of a_3 with the information provided in post 133 and 135.

I thinh that you can with mathematica.

You have 6 equations, with 6 unknown variables. I cant see why this is not sovlable.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,676

Hi;

I have solved the problem given above using just 6 equations.

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

Thats nice!

So I think one can say that If you have n intesection points, and someone provides you with the lead coeficient and a point of 2n polynomials you can define the intersection point.

Scool student?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,676

Hi;

Yes, I would say so but the actually calculation becomes more and more difficult as n increases.

I am not a student.

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 do you mean "difficult"?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,676

Numerical difficulties. Although A,B and C in my problem are integers the equations have enough instability in them to run off to the complex plane. Already Geogebra cannot create the points accurately enough and it has 16 digits of precision.

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, if the operations were over a galois field will have any difference?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,676

You mean a set consisting of {1,2,3,4,...p} where p is primes? If so, that is what I did.

You noticed the ai's I provided for the problem? They are floating point numbers. Geogebra lost 4 digits in computing them. That suggests a condition number for the problem of around 10000.

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

Something like that...

The problem is that when you work over Finite fields,

addition multiplication and division are not equal to standard operations. operations are executed modulo the prime number etc..etc..

*Last edited by Herc11 (2013-06-21 20:06:32)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,676

Yes, I know my grasp of theory is horrendous.

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

I am familiar with Galois Fields and the operations executed over GFs but it is not ot easy to explain...

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,676

A small thing to improve performance of solving that would be to have the coefficient of the constant term rather than the leading coefficient.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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