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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 98,200

There could 3 intersections rather than 2.

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

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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: 98,200

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.**

**If it ain't broke, fix it until it is.**

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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: 98,200

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.**

**If it ain't broke, fix it until it is.**

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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: 98,200

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.

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

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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: 98,200

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 )

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

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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: 98,200

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?

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

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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: 98,200

I am getting for a3

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

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

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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: 98,200

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

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

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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: 98,200

Look at these two equations:

do you notice something different?

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

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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: 98,200

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

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

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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: 98,200

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?

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

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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: Harlan's World
- Registered: 2011-05-23
- Posts: 15,954

Hi guys

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

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