Interpolation: We consider the function f (x) = (sin (x)) ^ 2...

tibad582

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Hello. I block at question 3 of the following exercise:We consider the function f (x) = (sin (x)) ^ 2

1) Give in a single line, using the Lagrange method, the expression factored from the interpolation polynomial of f at the points 1,2,3,5/2.

2) What happens to the interpolation polynomial if we add the 0 point to the others previous points?

3) What points can be added to have an interpolation polynomial of f that is odd and of degree seven? Give then the expression of such a polynomial, knowing that its dominant coefficients equal to - 128/10395.

4) Find the value of the estimate of the actual error committed at point x = 2/5 for interpolation at the points ± 1 and 1/4.
 
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Hello. I block at question 3 of the following exercise:We consider the function f (x) = (sin (x)) ^ 2

1) Give in a single line, using the Lagrange method, the expression factored from the interpolation polynomial of f at the points 1,2,3,5/2.

2) What happens to the interpolation polynomial if we add the 0 point to the others previous points?

3) What points can be added to have an interpolation polynomial of f that is odd and of degree seven? Give then the expression of such a polynomial, knowing that its dominant coefficients equal to - 128/10395.

4) Find the value of the estimate of the actual error committed at point x = 2/5 for interpolation at the points ± 1 and 1/4.
What are your thoughts regarding the assignment?

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Interpolation

I am sorry , I had to show you prealably what I did .
I made a mistake , the function is f ( x ) = (sin ( pi x) ) ^ 2
My anwer is :

For question 1 / I had P3 (x) = (-8/3) (x-1) (x-2) (x-3)For the second:p 4 (x) = P 3 (x) + a (x - 1) (x -2) (x -3) or a = y (1, 2, 3, 5/2, 0)We have P4 (0) = 0 then a = - 16/15therefore P4 (x) = (-16/15) x (x-1) (x-2) (x-3)
For question 1 / I had *P3 (x) = (-8/3) (x-1) (x-2) (x-3)For the second:p 4 (x) = P 3 (x) + a (x - 1) (x -2) (x -3) or a = y (1, 2, 3, 5/2, 0)We have P4 (0) = 0 then a = - 16/15therefore P4 (x) = (-16/15) x (x-1) (x-2) (x-3)
 
The Lagrange interpolation polynomial through the set of points \(\displaystyle (x_0, y_0)\), \(\displaystyle (x_1, y_1)\), ..., \(\displaystyle (x_{n-1}, y_{n-1})\) is \(\displaystyle \sum_{i=0}^{n-1} y_i \frac{(x-x_0)(x- x_1)\cdot\cdot\cdot(x- x_{i-1})(x- x_{i+1}\cdot\cdot\cdot(x- x_n)}{(x_i- x_0)(x_i-x_1)\cdot\cdot\cdot(x_i- x_n)}\), an n-1 degree polynomial for n data points.

Given the four points \(\displaystyle (1, f(1))\), \(\displaystyle (2, f(2))\), \(\displaystyle (5/2, f(5/2))\), \(\displaystyle (3, f(3))\), that would be \(\displaystyle f(1)\frac{(x- 2)(x- 5/2)(x- 3)}{(1-2)(1-5/2)(1-3)}+ f(2)\frac{(x- 1)(x- 5/2)(x- 3)}{(2-1)(2-5/2)(2-3)}+ f(5/2)\frac{(x- 1)(x- 2)(x- 3)}{(5/2-1)(5/2-5/2)(5/2-3)}+ f(3)\frac{(x- 1)(x- 2)(x- 5/2)}{(3-1)(3-2)(3-5/2)}\)
 
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