Polinomio di Newton – 100

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Radici e zeri

Esercizio del libro di testo a pagina 100.
Calcola il valore approssimato per x=3 utilizzando il polinomio interpolatore di Newton passante per i punti dati.
Svolgi l’esercizio per 2 / 3 / 4 / 5 / 6 punti.

Otterrai gli stessi risultati del polinomio di Lagrange con gli stessi punti.

Punti = 2, Grado = 1

xyf_1
02-3
\displaystyle \frac{-3-15}{2-1} = -18
1115
p_1(x)= b_0 +b_1 (x-x_0)
= -3 -18 (x-2)
p_1(3)= -3 -18 (3-2)
= -21

Punti = 3, Grado = 2

xyf_1f_2
02-3
\displaystyle \frac{-3-15}{2-1} = -18
1115\displaystyle \frac{-18-\left(-\displaystyle\frac{13}{4}\right)}{2-5} = \displaystyle \frac{59}{12}
\displaystyle \frac{15-2}{1-5} = \displaystyle -\frac{13}{4}
252
p_2(x)= b_0 +b_1 (x-x_0) +b_2 (x-x_0)(x-x_1)
= \displaystyle -3 -18(x-2) + \frac{59}{12}(x-2)(x-1)
p_2(3)= \displaystyle -3 -18(3-2) + \frac{59}{12}(3-2)(3-1)
= \displaystyle -\frac{67}{6} = -11,166…

Punti = 4, Grado = 3

xyf_1f_2f_3
02-3
\displaystyle \frac{-3-15}{2-1} = -18
1115\displaystyle \frac{-18-\left(-\displaystyle\frac{13}{4}\right)}{2-5} = \displaystyle \frac{59}{12}
\displaystyle \frac{15-2}{1-5} = \displaystyle -\frac{13}{4}\displaystyle \frac{\displaystyle\frac{59}{12}-\displaystyle\frac{9}{20}}{2-6} = \displaystyle -\frac{67}{60}
252\displaystyle \frac{-\displaystyle\frac{13}{4}-(-1)}{1-6} = \displaystyle \frac{9}{20}
\displaystyle \frac{2-1}{5-6} = -1
361
p_3(x)= b_0 + b_1 (x-x_0) + b_2 (x-x_0)(x-x_1) + b_3 (x-x_0)(x-x_1)(x-x_2)
= \displaystyle -3 -18 (x-2) + \frac{59}{12}(x-2)(x-1)-\frac{67}{60}(x-2)(x-1)(x-5)
p_3(3)= \displaystyle -3 -18 (3-2) + \frac{59}{12}(3-2)(3-1)-\frac{67}{60}(3-2)(3-1)(3-5)
= \displaystyle -\frac{67}{10} = -6,7

Punti = 5, Grado = 4

xyf_1f_2f_3f_4
02-3
\displaystyle \frac{-3-15}{2-1} = -18
1115\displaystyle \frac{-18-\left(\displaystyle-\frac{13}{4}\right)}{2-5} = \displaystyle \frac{59}{12}
\displaystyle \frac{15-2}{1-5} = \displaystyle -\frac{13}{4}\displaystyle \frac{\displaystyle\frac{59}{12}-\displaystyle\frac{9}{20}}{2-6} = \displaystyle -\frac{67}{60}
252\displaystyle \frac{\displaystyle-\frac{13}{4}-(-1)}{1-6} = \displaystyle \frac{9}{20}\displaystyle \frac{-\displaystyle\frac{67}{60}-\left(-\displaystyle\frac{13}{20}\right)}{2-4} = \displaystyle \frac{7}{30}
\displaystyle \frac{2-1}{5-6} = -1\displaystyle \frac{\displaystyle\frac{9}{20}-\left(-\displaystyle\frac{3}{2}\right)}{1-4} = \displaystyle -\frac{13}{20}
361\displaystyle \frac{-1-\displaystyle\frac{1}{2}}{5-4} = \displaystyle -\frac{3}{2}
\displaystyle \frac{1-0}{6-4} = \displaystyle \frac{1}{2}
440
p_4(x)= b_0 + b_1(x-x_0) + b_2(x-x_0)(x-x_1) + b_3(x-x_0)(x-x_1)(x-x_2) + b_4(x-x_0)(x-x_1)(x-x_2)(x-x_3)
= \displaystyle -3 -18 (x-2) + \frac{59}{12}(x-2)(x-1)-\frac{67}{60}(x-2)(x-1)(x-5) + \displaystyle\frac{7}{30}(x-2)(x-1)(x-5)(x-6)
p_4(3)= \displaystyle -3 -18 (3-2) + \frac{59}{12}(3-2)(3-1)-\frac{67}{60}(3-2)(3-1)(3-5) + \displaystyle\frac{7}{30}(3-2)(3-1)(3-5)(3-6)
= \displaystyle -\frac{39}{10} = -3,9

Punti = 6, Grado = 5

xyf_1f_2f_3f_4f_5
02-3
\displaystyle \frac{-3-15}{2-1} = -18
1115\displaystyle \frac{-18-\left(\displaystyle-\frac{13}{4}\right)}{2-5} = \displaystyle \frac{59}{12}
\displaystyle \frac{15-2}{1-5} = \displaystyle -\frac{13}{4}\displaystyle \frac{\displaystyle\frac{59}{12}-\displaystyle\frac{9}{20}}{2-6} = \displaystyle -\frac{67}{60}
252\displaystyle \frac{\displaystyle-\frac{13}{4}-(-1)}{1-6} = \displaystyle \frac{9}{20}\displaystyle \frac{-\displaystyle\frac{67}{60}-\left(-\displaystyle\frac{13}{20}\right)}{2-4} = \displaystyle \frac{7}{30}
\displaystyle \frac{2-1}{5-6} = -1\displaystyle \frac{\displaystyle\frac{9}{20}-\left(-\displaystyle\frac{3}{2}\right)}{1-4} = \displaystyle -\frac{13}{20}\displaystyle \frac{\displaystyle \frac{7}{30}-\left(\displaystyle-\frac{17}{60}\right)}{2-0} = \displaystyle \frac{31}{120}
361\displaystyle \frac{-1-\displaystyle\frac{1}{2}}{5-4} = \displaystyle -\frac{3}{2}\displaystyle \frac{\displaystyle -\frac{13}{20}-\left(\displaystyle-\frac{11}{30}\right)}{1-0} = \displaystyle -\frac{17}{60}
\displaystyle \frac{1-0}{6-4} = \displaystyle \frac{1}{2}\displaystyle \frac{\displaystyle -\frac{3}{2}-\displaystyle\frac{1}{3}}{5-0} = \displaystyle -\frac{11}{30}
440\displaystyle \frac{\displaystyle\frac{1}{2}-\left(-\displaystyle\frac{3}{2}\right)}{6-0} = \displaystyle \frac{1}{3}
\displaystyle \frac{0-6}{4-0} = \displaystyle -\frac{3}{2}
506
p_5(x)= b_0 + b_1(x-x_0) + b_2(x-x_0)(x-x_1) + b_3(x-x_0)(x-x_1)(x-x_2) + b_4(x-x_0)(x-x_1)(x-x_2)(x-x_3)
+ b_5(x-x_0)(x-x_1)(x-x_2)(x-x_3)(x-x_4)
= \displaystyle -3 -18 (x-2) + \frac{59}{12}(x-2)(x-1)-\frac{67}{60}(x-2)(x-1)(x-5) + \displaystyle\frac{7}{30}(x-2)(x-1)(x-5)(x-6)
+ \displaystyle\frac{31}{120}(x-2)(x-1)(x-5)(x-6)(x-4)
p_5(3)= \displaystyle -3 -18 (3-2) + \frac{59}{12}(3-2)(3-1)-\frac{67}{60}(3-2)(3-1)(3-5) + \displaystyle\frac{7}{30}(3-2)(3-1)(3-5)(3-6)
+ \displaystyle\frac{31}{120}(3-2)(3-1)(3-5)(3-6)(3-4)
= -7

Ripeti i passi precedenti calcolando il polinomio interpolatore di Newton passante per 2 / 3 / 4 / 5 / 6 punti

Punti = 2, Grado = 1

xyf_1
02-3
\displaystyle \frac{-3-15}{2-1} = -18
1115
p_1(x)= b_0 +b_1 (x-x_0)
= -3 -18 (x-2)
= -18 x +33

Punti = 3, Grado = 2

xyf_1f_2
02-3
\displaystyle \frac{-3-15}{2-1} = -18
1115\displaystyle \frac{-18-\left(-\displaystyle\frac{13}{4}\right)}{2-5} = \displaystyle \frac{59}{12}
\displaystyle \frac{15-2}{1-5} = \displaystyle -\frac{13}{4}
252
p_2(x)= b_0 +b_1 (x-x_0) +b_2 (x-x_0)(x-x_1)
= \displaystyle -3 -18(x-2) + \frac{59}{12}(x-2)(x-1)
= \displaystyle \frac{59}{12} x^2- \frac{131}{4}x +\frac{257}{6}

Punti = 4, Grado = 3

xyf_1f_2f_3
02-3
\displaystyle \frac{-3-15}{2-1} = -18
1115\displaystyle \frac{-18-\left(-\displaystyle\frac{13}{4}\right)}{2-5} = \displaystyle \frac{59}{12}
\displaystyle \frac{15-2}{1-5} = \displaystyle -\frac{13}{4}\displaystyle \frac{\displaystyle\frac{59}{12}-\displaystyle\frac{9}{20}}{2-6} = \displaystyle -\frac{67}{60}
252\displaystyle \frac{-\displaystyle\frac{13}{4}-(-1)}{1-6} = \displaystyle \frac{9}{20}
\displaystyle \frac{2-1}{5-6} = -1
361
p_3(x)= b_0 + b_1 (x-x_0) + b_2 (x-x_0)(x-x_1) + b_3 (x-x_0)(x-x_1)(x-x_2)
= \displaystyle -3 -18 (x-2) + \frac{59}{12}(x-2)(x-1)-\frac{67}{60}(x-2)(x-1)(x-5)
= \displaystyle -\frac{67}{60} x^3+ \frac{277}{20}x^2- \frac{776}{15}x +54

Punti = 5, Grado = 4

xyf_1f_2f_3f_4
02-3
\displaystyle \frac{-3-15}{2-1} = -18
1115\displaystyle \frac{-18-\left(\displaystyle-\frac{13}{4}\right)}{2-5} = \displaystyle \frac{59}{12}
\displaystyle \frac{15-2}{1-5} = \displaystyle -\frac{13}{4}\displaystyle \frac{\displaystyle\frac{59}{12}-\displaystyle\frac{9}{20}}{2-6} = \displaystyle -\frac{67}{60}
252\displaystyle \frac{\displaystyle-\frac{13}{4}-(-1)}{1-6} = \displaystyle \frac{9}{20}\displaystyle \frac{-\displaystyle\frac{67}{60}-\left(-\displaystyle\frac{13}{20}\right)}{2-4} = \displaystyle \frac{7}{30}
\displaystyle \frac{2-1}{5-6} = -1\displaystyle \frac{\displaystyle\frac{9}{20}-\left(-\displaystyle\frac{3}{2}\right)}{1-4} = \displaystyle -\frac{13}{20}
361\displaystyle \frac{-1-\displaystyle\frac{1}{2}}{5-4} = \displaystyle -\frac{3}{2}
\displaystyle \frac{1-0}{6-4} = \displaystyle \frac{1}{2}
440
p_4(x)= b_0 + b_1(x-x_0) + b_2(x-x_0)(x-x_1) + b_3(x-x_0)(x-x_1)(x-x_2) + b_4(x-x_0)(x-x_1)(x-x_2)(x-x_3)
= \displaystyle -3 -18 (x-2) + \frac{59}{12}(x-2)(x-1)-\frac{67}{60}(x-2)(x-1)(x-5) + \displaystyle\frac{7}{30}(x-2)(x-1)(x-5)(x-6)
= \displaystyle \frac{7}{30} x^4 -\frac{263}{60} x^3 +\frac{1741}{60} x^2 -\frac{1168}{15} x + 68

Punti = 6, Grado = 5

xyf_1f_2f_3f_4f_5
02-3
\displaystyle \frac{-3-15}{2-1} = -18
1115\displaystyle \frac{-18-\left(\displaystyle-\frac{13}{4}\right)}{2-5} = \displaystyle \frac{59}{12}
\displaystyle \frac{15-2}{1-5} = \displaystyle -\frac{13}{4}\displaystyle \frac{\displaystyle\frac{59}{12}-\displaystyle\frac{9}{20}}{2-6} = \displaystyle -\frac{67}{60}
252\displaystyle \frac{\displaystyle-\frac{13}{4}-(-1)}{1-6} = \displaystyle \frac{9}{20}\displaystyle \frac{-\displaystyle\frac{67}{60}-\left(-\displaystyle\frac{13}{20}\right)}{2-4} = \displaystyle \frac{7}{30}
\displaystyle \frac{2-1}{5-6} = -1\displaystyle \frac{\displaystyle\frac{9}{20}-\left(-\displaystyle\frac{3}{2}\right)}{1-4} = \displaystyle -\frac{13}{20}\displaystyle \frac{\displaystyle \frac{7}{30}-\left(\displaystyle-\frac{17}{60}\right)}{2-0} = \displaystyle \frac{31}{120}
361\displaystyle \frac{-1-\displaystyle\frac{1}{2}}{5-4} = \displaystyle -\frac{3}{2}\displaystyle \frac{\displaystyle -\frac{13}{20}-\left(\displaystyle-\frac{11}{30}\right)}{1-0} = \displaystyle -\frac{17}{60}
\displaystyle \frac{1-0}{6-4} = \displaystyle \frac{1}{2}\displaystyle \frac{\displaystyle -\frac{3}{2}-\displaystyle\frac{1}{3}}{5-0} = \displaystyle -\frac{11}{30}
440\displaystyle \frac{\displaystyle\frac{1}{2}-\left(-\displaystyle\frac{3}{2}\right)}{6-0} = \displaystyle \frac{1}{3}
\displaystyle \frac{0-6}{4-0} = \displaystyle -\frac{3}{2}
506
p_5(x)= b_0 + b_1(x-x_0) + b_2(x-x_0)(x-x_1) + b_3(x-x_0)(x-x_1)(x-x_2) + b_4(x-x_0)(x-x_1)(x-x_2)(x-x_3)
+ b_5(x-x_0)(x-x_1)(x-x_2)(x-x_3)(x-x_4)
= \displaystyle -3 -18 (x-2) + \frac{59}{12}(x-2)(x-1)-\frac{67}{60}(x-2)(x-1)(x-5) + \displaystyle\frac{7}{30}(x-2)(x-1)(x-5)(x-6)
+ \displaystyle\frac{31}{120}(x-2)(x-1)(x-5)(x-6)(x-4)
= \displaystyle \frac{31}{120}\ x^5 - \frac{53}{12}\ x^4 + \frac{215}{8}\ x^3 -\frac{805}{12}\ x^2 + \frac{1601}{30}\ x + 6

Osserva i risultati successivi

PuntiPolinomiox=3
2p_1(x) = -18x+33-21
3p_2(x) = \displaystyle \frac{59}{12} x^2 - \frac{131}{4}x + \frac{257}{6}-11,166…
4p_3(x) = \displaystyle -\frac{67}{60} x^3 + \frac{277}{20}x^2 - \frac{776}{15}x+54-6,7
5p_4(x) = \displaystyle \frac{7}{30} x^4 -\frac{263}{60} x^3 +\frac{1741}{60} x^2 -\frac{1168}{15} x + 68-3,9
6p_5(x) = \displaystyle \frac{31}{120}\ x^5 - \frac{53}{12}\ x^4 + \frac{215}{8}\ x^3 -\frac{805}{12}\ x^2 + \frac{1601}{30}\ x + 6-7