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bobbym, I think that i correct it #99. At Π i-1 and not i...I think
Ok.
So, from 4 quads that the only known info is a point of each of them and their leading coefficient we can find their intersection points?
Is that right?
I am thinking that the same will stand for cubics but we will need 6 cubics to solve the previous problem.
bobbym,
What do you mean exactly?
If I use the Eq. of post #98 or 99,
only the intersection points are the unknowns?
So, you think that for quadratics you can determine the intersection point If you know the leading coef of 4 quads and one point from each of them.
Now I am wondering, If I did not know the lead coef. and I had 8 quad. could I find again the intersection points?
I am confused.
Maybe it is easier to understand what I was saying...
I was saying that in Eq. of post 98 there are 4 unknowns i.e. x0...y1.
If I knew 3 more eq. like post #98, I think that the problem has a solution and the intersection points can be spotted.
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):
If I hadnt make any mistake this is the equation
Is there an option to choose office?
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..
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?
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?
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.
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..
Ok. that is why I need six polynomials.
The intersection points are three. i.e for each intersection point there are 2 unknown variables.
But the intersection points are three i.e. 6 unknowns
4????
please explain why is that...?
According to my post #76 the number of equations is equal to the number of unknowns,isnt it?
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.
Any suggestions, ideas etc?
anonimnystefy,
So, if the polynomials were of 3 degree in order to define the intersection points 6 degree 3 polynomials will be demanded?
Is that correct?
To be honest it was not.
Because until today,
I thought that the previous problem was intractable.
anonimnystefy
did you use matlab?
Haha!
So, It can be computed!
Is it right?