Hi,
I am learn how to calculate a fonction of parabole with 3 points given.
Me I do at hand pincil and paper, after I have resolve I obtain a system of 3 equations what I resolve with app linear solver. But I have to calculation by hand
the system as I say.
On internet I have see on a computer hand I dont' know the name a solution automatic, which I have tryied on Prime but no work. May be is a way to do same ?
In attachement I give what I do with hand and the other picture, may be you help me to do with Prime. Because I know how to do with hand ,it save me time for excercizes further if there is a solution automatic with Prime.
I hope it is clear because it is difficult for me to speek american.
1 - The quickest way might be to determine the Lagrange polynomial:
lagrange([1 -2 3], [2 1 4])
(then simplify the result).
2 - If you really want to solve the linear equations then you can do that as follows:
linsolve([a+b+c=2 4*a-2*b+c=1 9*a+3*b+c=4], [a b c])
3 - Or if you want to imitate the approach in your second screenshot:
f(x):=a*x^2+b*x+c
solve({f(1)=2,f(-2)=1,f(3)=4},[a b c])
(09-29-2017 11:56 AM)ggauny@live.fr Wrote: [ -> ]I am learn how to calculate a fonction of parabole with 3 points given.
Me I do at hand pincil and paper, after I have resolve I obtain a system of 3 equations what I resolve with app linear solver. But I have to calculation by hand
the system as I say.
I am sure there is a more elegant way on the Prime, but all you need is the linear equation solver. For a parabola the equation system is as follows:
Code:
| x1² x1 1 | → | y1 |
| x2² x2 1 | * p = | y2 |
| x3² x3 1 | | y3 |
So simply start the linear solver and fill the matrix.
Consider the example in your screenshot:
Enter the x-values into the center column and the y-values on the right-hand side:
Code:
| _ 1 _ | → | 2 |
| _ -2 _ | * p = | 1 |
| _ 3 _ | | 4 |
Fill the left column with the squares of the center column:
Code:
| 1 1 _ | → | 2 |
| 4 -2 _ | * p = | 1 |
| 9 3 _ | | 4 |
Fill the right column with 1s:
Code:
| 1 1 1 | → | 2 |
| 4 -2 1 | * p = | 1 |
| 9 3 1 | | 4 |
Now solve this system and you're done.
Dieter
(09-29-2017 12:52 PM)Gerson W. Barbosa Wrote: [ -> ]Interesting ideas in this old thread (others, not mine).
Thanks for that link.
And as you already suggested: Cramer's Rule is not very efficient in general and should probably be avoided for numerical calculations, especially when the dimension goes up.
And I am not 100% sure, but I seem to remember that the determinant calculations are also quite sensitive to rounding (not if all you care about are integer x values of course). Again, this becomes more of a problem as the dimension goes up.
Hi,
One again many thank you at all.
You are great teachers mathematician !
Have a good day.
(09-29-2017 12:32 PM)Dieter Wrote: [ -> ] (09-29-2017 11:56 AM)ggauny@live.fr Wrote: [ -> ]I am learn how to calculate a fonction of parabole with 3 points given.
Me I do at hand pincil and paper, after I have resolve I obtain a system of 3 equations what I resolve with app linear solver. But I have to calculation by hand
the system as I say.
I am sure there is a more elegant way on the Prime, but all you need is the linear equation solver. For a parabola the equation system is as follows:
Code:
| x1² x1 1 | → | y1 |
| x2² x2 1 | * p = | y2 |
| x3² x3 1 | | y3 |
So simply start the linear solver and fill the matrix.
Consider the example in your screenshot:
Enter the x-values into the center column and the y-values on the right-hand side:
Code:
| _ 1 _ | → | 2 |
| _ -2 _ | * p = | 1 |
| _ 3 _ | | 4 |
Fill the left column with the squares of the center column:
Code:
| 1 1 _ | → | 2 |
| 4 -2 _ | * p = | 1 |
| 9 3 _ | | 4 |
Fill the right column with 1s:
Code:
| 1 1 1 | → | 2 |
| 4 -2 1 | * p = | 1 |
| 9 3 1 | | 4 |
Now solve this system and you're done.
Dieter
[[2],[1],[4]]/vandermonde([1,-2,3])
for ill/weird/large/spare matrix: try
LSQ(vandermonde([1,-2,3]),[[2],[1],[4]])