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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,247

There could 3 intersections rather than 2.

**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. this is true. So the unknown variables will be six (the 3 intersection points)

So we need 6 cubics etc etc....

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,247

Yes, you will need 6 equations when there are 3 points of intersection.

**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

anonimnystefy,

do you agree with us?

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

How do you construct the polynomials?

MAthematica? I am not familiar with it..

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,247

There is a better tool for constructing them because it is dynamic. It is called Geogebra.

**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, and for solving the system you use Mathematica or the afore mentioned tool?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,247

Mathematica will solve the system and geogebra will construct it.

Here is a good one that I will provide all the data for and leave it as a challenge.

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

where is the data?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,247

Putting it together now. it will take some time to post it. Please hold.

a3 = 1 , point ( 4,18 )

a3 = -0.142857142857141, point ( -4 , 10 )

a3 = 0.017482517482521 point (14 , 40)

a3 = -0.018939393939392 point ( 9 , -10 )

a3 = 0.066137566137551 point ( 10 , 50 )

a3 = -1.111111111111111 point ( 4 , -20 )

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. Sorry!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,247

No problem, I am filling in post #135

A question about the formula in post #99. It has a P(xi) that the formula in post #98 did not have. What is 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

Its

the respective ofOffline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,247

I am getting for a3

you are saying that P(x3) is y3 which makes sense.

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 I think what you write in #139 is correct.

*Last edited by Herc11 (2013-06-21 06:36:05)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,247

Okay, I will see if I can solve the problem I posted in post 133 - 135.

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 think that it will be solved as you have 6 equations and 6 unknowns..x0---y3

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,247

Look at these two equations:

do you notice something different?

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

No.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,247

The a3 has coefficients in it, a0, a1, a2. While the formula for a2 does not have any coefficients in 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

Yes

So you have only to replace the coefficients with the are equals

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

So, if you now look a_3 you have six unknows. Using 6 cubics I think can leed to a soultion.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,247

Yes, if you wanted to solve for the a's too. But you can substitute for those a's with earlier terms can you not?

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?

It is the same thing. You have only to solve the set of equations of a3. \

Afterwards, you know the x0-y3. If you wish you can compute a0-a2, but there is no need.

Of course you can substitute. Either you solve for a0 to a2 or for x0-y3 its the same.

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

Hi guys

Sorry about not posting all day. has the interpolation worked to reduce the number of needed equations?

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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