You are not logged in.
ok ok.
I think that anonimnystefy can solve it as he solved the quadratics...
Ok:)
Yes they are coeffients.
But see equation in #146.
Then suppose you have 6 equations like #146.
You know a3 y3 x3, and you dont know x0x1x2 y0y1y2.
By a set of 6 equations isn t the set of equations solvable?
Our aim is to define the intersection points i.e. x0x1x2 y0y1y2.
Once you solve the set of the 6 equations you will have discovered the x0x1x2 y0y1y2.
If you d like you can then define a0=y0 a1=... and a2=...
Why to eliminate?
Simply substitute the a0 a1 and a2 with x0-y2.
See post #146.
Dont use a0..etc if it confuses you...
You have only to solve for x0..y2.
And six equations like that in #146 are enough
I think it has not to do with interpolation. See in #146 post
the formula for computing a_3
if the unknowns are x0 y0 x1 y1 x2 y2
and known the x3 y3 and a3
I think that by having a set of 6 equations you can define x0-y2.
Also, bobbym
posted another one problem
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.
So, if you now look a_3 you have six unknows. Using 6 cubics I think can leed to a soultion.
Yes
So you have only to replace the coefficients with the are equals
I think that it will be solved as you have 6 equations and 6 unknowns..x0---y3
Yes I think what you write in #139 is correct.
Its
the respective ofOk. Sorry!
where is the data?
Ok, and for solving the system you use Mathematica or the afore mentioned tool?
How do you construct the polynomials?
MAthematica? I am not familiar with it..
anonimnystefy,
do you agree with us?
Yes. this is true. So the unknown variables will be six (the 3 intersection points)
So we need 6 cubics etc etc....
what difference?
Yes.
Thats the only fact!:)
But it seems that at the formula #, when a3, there will be six unknown variables x0y0...x2y2. There fore, 6 equations will be needed
so 6 cubics plus their leading coefficient and one point of each...
Ok. That is what I am thinking too. Did you find the solution?
We can suppose that for a cubic 6 equations are demanded?
12? you had only the lead coefficient, and a point..
It would be helpful if you could post the algorithm.
When you solved the problem of bobbym what Equations did you use?
Ok. I correct the denom..
Hi anonimnystefy,
Did you see my post #105?