You are not logged in.
It does not, we just went from there. But so far no one has proved it is possible with n quadratics.
Thats why I am asking.
It seems to me that If there are 4 quad. you can define the 2 interesection points.
Similarly, if the polynomials were of degree 3, 6 polynomials of 3 degree might be needed...and so on...
But I am not sure
bobbym wrote
Want to use post #9 and I will provide the two more quadratics since I know what A and B is?
What do you mean exactly?
anonimnystefy
I think so.
I have the same opinion.
anonimnystefy wrote:
I need a of four quadratics.
So, if you have 4 quadratics plus a point of each of them you can solve it? i.e you can find x0...y1?
Ok. You are right. But I am confused with the equation of defining a2.
In equation of a2 only the 4 variables x0 x1 y0 y1 exist.
And
if I have four equations for a2 (I do not know the 4 polynomials, only that they have the same 2 inters. points x0 y0 x1 y1 (unknowns),
I know
each polynomial's leading coef. i.e. a2
and one point of each of the polynomials.
Then from the set of 4 equations of a2, x0..y1 can be defined?
x0 x1 y0 y1 are the same intersection points. the other one for each polynomial will be the known point.
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?
Which are the unknowns?
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.
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?
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?
So, you mean that If I had 4 equations a2=.... I wouldn t be able to define x0,,y1?
It seems that I used wrong a2 but the idea is still the same.
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?
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...
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??
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?
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?