Define the intersection points of polynomials

Herc11 · 2013-06-20 04:08:11

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)

anonimnystefy · 2013-06-20 04:12:35

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.

Herc11 · 2013-06-20 04:16:17

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

anonimnystefy · 2013-06-20 05:45:21

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

Herc11 · 2013-06-20 05:51:55

4????
please explain why is that...?

anonimnystefy · 2013-06-20 05:54:34

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

Herc11 · 2013-06-20 06:06:30

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

anonimnystefy · 2013-06-20 06:09:20

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

Herc11 · 2013-06-20 06:11:55

Ok. that is why I need six polynomials.

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

anonimnystefy · 2013-06-20 06:21:02

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?

Herc11 · 2013-06-20 06:26:34

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)

anonimnystefy · 2013-06-20 06:29:31

Not quadratics, sorry.

Herc11 · 2013-06-20 06:37:50

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)

anonimnystefy · 2013-06-20 07:03:22

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.

Herc11 · 2013-06-20 07:08:22

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)

anonimnystefy · 2013-06-20 12:08:04

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

Herc11 · 2013-06-20 17:23:30

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?

bobbym · 2013-06-20 17:29:18

Hi;

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

Herc11 · 2013-06-20 17:36:56

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..

bobbym · 2013-06-20 17:46:17

If you want to latex them go here

http://latex.codecogs.com/editor.php

Herc11 · 2013-06-20 17:56:28

Is there an option to choose office?

bobbym · 2013-06-20 17:58:05

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.

Herc11 · 2013-06-20 18:06:53

If I hadnt make any mistake this is the equation

Herc11 · 2013-06-20 18:13:28

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)

bobbym · 2013-06-20 18:18:57

Hi;

Okay, I got it. Thank you.

Math Is Fun Forum

#76 2013-06-20 04:08:11

Re: Define the intersection points of polynomials

#77 2013-06-20 04:12:35

Re: Define the intersection points of polynomials

#78 2013-06-20 04:16:17

Re: Define the intersection points of polynomials

#79 2013-06-20 05:45:21

Re: Define the intersection points of polynomials

#80 2013-06-20 05:51:55

Re: Define the intersection points of polynomials

#81 2013-06-20 05:54:34

Re: Define the intersection points of polynomials

#82 2013-06-20 06:06:30

Re: Define the intersection points of polynomials

#83 2013-06-20 06:09:20

Re: Define the intersection points of polynomials

#84 2013-06-20 06:11:55

Re: Define the intersection points of polynomials

#85 2013-06-20 06:21:02

Re: Define the intersection points of polynomials

#86 2013-06-20 06:26:34

Re: Define the intersection points of polynomials

#87 2013-06-20 06:29:31

Re: Define the intersection points of polynomials

#88 2013-06-20 06:37:50

Re: Define the intersection points of polynomials

#89 2013-06-20 07:03:22

Re: Define the intersection points of polynomials

#90 2013-06-20 07:08:22

Re: Define the intersection points of polynomials

#91 2013-06-20 12:08:04

Re: Define the intersection points of polynomials

#92 2013-06-20 17:23:30

Re: Define the intersection points of polynomials

#93 2013-06-20 17:29:18

Re: Define the intersection points of polynomials

#94 2013-06-20 17:36:56

Re: Define the intersection points of polynomials

#95 2013-06-20 17:46:17

Re: Define the intersection points of polynomials

#96 2013-06-20 17:56:28

Re: Define the intersection points of polynomials

#97 2013-06-20 17:58:05

Re: Define the intersection points of polynomials

#98 2013-06-20 18:06:53

Re: Define the intersection points of polynomials

#99 2013-06-20 18:13:28

Re: Define the intersection points of polynomials

#100 2013-06-20 18:18:57

Re: Define the intersection points of polynomials

Board footer