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

Why is that? 2 points?

I think that you only need one point per polynomial but six polynomials.

The reason that you need 6 polynomials is that the intersection points are 3 i.e. you must define 6 coordinates.

*Last edited by Herc11 (2013-06-20 04:13:49)*

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

Because we need the number of equations equal to the number of unknowns. If we have only one point, the equality of their numbers is unreachable.

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

According to my post #76 the number of equations is equal to the number of unknowns,isnt it?

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

That is true. And I have good news - 4 will be enough!

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

4????

please explain why is that...?

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

Well, that is when the number of unknown variables and equations will be the same.

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

But the intersection points are three i.e. 6 unknowns

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

No, there are more. For each cubic you introduce into the system, you get 3 new variables.

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

Ok. that is why I need six polynomials.

The intersection points are three. i.e for each intersection point there are 2 unknown variables.

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

Hm, you are right. It seems that for nth degree polynomials, you will need 2n quadratics. It isn't large for very, very small n, but in terms of system solving, 2n(n+1) variables can be a lot to handle, even for some smaller n, let alone larger ones.

Where does this problem come up?

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

In Newton Interpolation.

But, now why quadratic?

I think that you need you need 2n equationse e.g. when there are degree 3 polynomials, there are 6 unknowns so 6 equations are demanded/

degree:n=3---> 2n=6 equations

Similarly, n=10---->2n=20 equations..

*Last edited by Herc11 (2013-06-20 06:27:21)*

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

Not quadratics, sorry.

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

Ok. You think that I m right?

And

Can you send your code about the previous problem? If you want of course...

That one that you used for solving the problem of bobbym?

Thanks both of you.

*Last edited by Herc11 (2013-06-20 06:56:36)*

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

I'm on the phone now, so I am not able to post the code at the moment, but it's nothing spectacular anyway. I just used the built-in Solve function and entered the system of equations into it.

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

Ok no problem! Whenever you can, I am not familiar with Mathematica, thats why I am asking...

So, as far for the first part (6 equations, 6 unknowns)

do you think I am right?

And what equation did you use? Did you express a1 and a2 as x0 xi etc?

*Last edited by Herc11 (2013-06-20 08:30:41)*

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

Actually, there are 24 equations and unknowns, it's just that we can set them up from 6 cubics.

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

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?

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

Hi;

Could you please describe where the formulas in post #19 come from?

**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

Hi bobbym,

Yes I can but how can I write the equations in the forum, in order to be more intelligble?

By the way, the equations were from Newton Interpolation Method.

In specific, it was the formula for computing the leading coefficient of the polynomial (expressed in its Newton form). It is noted that only the leading

coefficient of the Newton's form of the polynomial coincides with the leading coeffeient of the polynomial..

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

If you want to latex them go here

**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

Is there an option to choose office?

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

Just type what you need into the box they provide. Or use the pull down menus. It will spit out latex underneath. Copy what is in the box and put it in here between the math tags.

**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 hadnt make any mistake this is the equation

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

The formula for recovering each Newton coefficient is (note also that except the leading coefficient Newton's coef are different than that of the polynomial):

*Last edited by Herc11 (2013-06-20 19:30:57)*

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

Hi;

Okay, I got it. Thank you.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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