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

I am facing the following problem.

Lets consider 2 points that are not known

I know that from these 2 unknown points

For each of these polynomials I know one point

Is it possible to find the intersection points (i.e. the 2 unknown points) of the aforementioned polynomials?

*Last edited by Herc11 (2013-06-19 23:22:12)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,233

Hi Herc11;

Sounds like you have 3 points for each quadratic. That is enough to determine the two of them uniquely.

Wait you do not have 3 points. You have one point and the leading coefficient of each one. That may not be enough.

**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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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,335

hi Herc11

Welcome to the forum.

Someone will probably be able to answer this. There are members who are far brainier than me.

I'll edit your post into Latex to make the powers and subscripts easier to read. Please check in 5 minutes that I've got it right.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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

Hmmm.Thats the problem. I know the leading coefficient and one point from each of the polynomials.

But I m thinking that if I know the leading coef. and one point from four polynomials

I might be able to find a set of 4 equations to define the four variables x0 x1 y0 y1.

Is ti possible?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,233

Hi;

First of all taking one quadratic at at a time I cannot uniquely determine the other two coefficients just because I know the first one and a point.

If I told you that the the coefficient of x^2 was 2 and the point it passed through was (5,28) you would end up with this.

2x^2 + a*x + b = 28

50 + 5a + b = -22

5a + b = - 22

I am stuck. This has an infinite number of solutions for a and b.

The next quadratic will yield another 2 variables c and d. Coupled with the 4 other variables that define the unknown points we have maybe 4 equations and 8 unknowns. An overdetermined set. This spells doom for a unique solution.

I think I need more relationships between the points and the quadratics.

**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, That's right.

Let s use Newton Interpolation method to construct the known leading coefficient of the quadratic.

a_(2_known )=(y_known-(y_0+(y_1-y_(0 ))/(x_1-x_0 )(x_known-x_1))/((x_known-x_1)(x_known-x_0))

If I use the previous Equation for 4 quadratics, is it possible to define x0 x1 y0 y1 ?

P.S. How can I write efficiently the formulas??

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,233

Hi;

Hold on. There are many interpolation formulas, as a matter of fact in the old days you were not considered a mathematician unless you discovered one.

The fact is you are going to end up with variables a,b,c,d,x0,y0,x1,y1 and we know few facts about them. What relationships we do know will amount to 4 equations in 8 unknowns. Even if these were linear which they are not we would not be able to come up with a unique solution. Interpolation will not help us out. I can get the solutions using mathematica. They will be large because they will have a lot of variables left in them.

Is there anymore that you can provide me, no matter how tiny?

**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

Consider only coefficient a2.

The only unknowns are x0 x1 y0 y1.

If I have the coef. a2 of four other polynomials could I determine the x0 x1 y0 y1?

I might be cinfused...

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,233

Let me give you a concrete example here are two polynomials with the the leading coefficient known. Also each has one point known. Can you determine point A and B?

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...What I am thinking is:

a2=2= (7+ (y0- ( (y1-y0) / (x1-x0))*(1-x1)))/ ( (1-x1)*(1-x0)

similarly for the second polynomial

a2=-3= (1+ (y0- (( y1-y0)/(x1-x0))*(2-x1))/ (2-x0)*(2-x1)

If i had more polynomials could i find the x0 x1 y0 y1?

*Last edited by Herc11 (2013-06-20 00:41:59)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,233

If i had more polynomials could i find the x0 x1 y0 y1

That is what I am saying. You need more relationships between the variables. We call these equations. You have to make sure they are unique equations and not just the same as the others.

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

It seems that I used wrong a2 but the idea is still the same.

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

So, you mean that If I had 4 equations a2=.... I wouldn t be able to define x0,,y1?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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You want A, B in my drawing. Therefore you (x0,y0) and (x1,y1) that is 4 variables. To have a chance at getting those 4 as a number I need 4 equations. Then to get the 2 polynomials I need 4 more equations.

I need 8 equations to determine the uniquely. To give you answers that are numbers.

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

To have a chance at getting those 4 as a number I need 4 equations.

So If you have at your disposal two more polynomials passing from the same intersection points you were able to define x0,y1?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Only if you knew the equations of those polynomials. If you did not you would be just adding more variables!

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,507

I have a feeling that the intersections can be determined. You can set up three equations in x0,y0,x1,y1,a12, and a13. Then each three new equations add just 2 new variables, ai2 and ai3. So, with 12 equations and 12 variables, there is aa chance that the system will be determined.

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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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,233

With 8 and 8 it could be determined. He has more variables than relationships that is the problem.

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

a2=2= (7+ (y0- ( (y1-y0) / (x1-x0))*(1-x1)))/ ( (1-x1)*(1-x0)

a2=-3= (1+ (y0- (( y1-y0)/(x1-x0))*(2-x1))/ (2-x0)*(2-x1)

a2=8= (6+ (y0- (( y1-y0)/(x1-x0))*(2-x1))/ (2-x0)*(2-x1) polynomial with a2=8 and a point 2,6

a2=11= (12+ (y0- (( y1-y0)/(x1-x0))*(8-x1))/ (8-x0)*(8-x1) polynomial with a2=11 and a point 8,2

I can not solve this system of equations?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,233

Where is (2,6) and (8,2) coming from?

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

bobbym wrote:

Where is (2,6) and (8,2) coming from?

Consider that there are two more polynomials passing from the intersection points illustrated in your figure and I know their leading coefficient and one point form each of them i.e. 2,6 8,2.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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That will add more variables because you will not know two terms from each of them. You are adding 2 equations and 4 more unknowns. See post #4.

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

Which are the unknowns?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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You will have the x^2 coefficient but not the x and the constant term for both. You will need to know what the equation is for each.

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

Combinatorics is Algebra and Algebra is Combinatorics.

**Online**

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

a2=2= (7+ (y0- ( (y1-y0) / (x1-x0))*(1-x1)))/ ( (1-x1)*(1-x0)

If i use this equation for a2, which unknons does it introduce?

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