{"id":51760,"date":"2025-01-12T22:23:22","date_gmt":"2025-01-12T21:23:22","guid":{"rendered":"https:\/\/www.valcon.it\/pp\/?p=51760"},"modified":"2026-09-19T10:31:50","modified_gmt":"2026-09-19T08:31:50","slug":"polinomio-di-lagrange-100","status":"publish","type":"post","link":"https:\/\/www.valcon.it\/pp\/polinomio-di-lagrange-100\/","title":{"rendered":"Polinomio di Lagrange – 100"},"content":{"rendered":"\n
\n
Interpolazione e regressione<\/a><\/div>\n<\/div>\n\n\n\n

<\/p>\n\n\n\n

Esercizio del libro di testo a pagina 100.<\/mark><\/em><\/p>\n\n\n\n

Utilizza il polinomio interpolatore di Lagrange<\/strong> passante per i punti dati per calcolare il valore approssimato per x=3<\/strong>.<\/mark><\/em>
Svolgi l’esercizio per 2 \/ 3 \/ 4 \/ 5 \/ 6 punti.<\/mark><\/em><\/p>\n\n\n\n

\n\n\n\n

Punti = 2 (grado = 1)<\/h2>\n\n\n\n
\n
\n
<\/th>x<\/th>y<\/th><\/tr><\/thead>
0<\/td>2<\/td>-3<\/td><\/tr>
1<\/td>1<\/td>15<\/td><\/tr><\/tbody><\/table><\/figure>\n<\/div>\n\n\n\n
\n
\"\"<\/figure>\n<\/div>\n<\/div>\n\n\n\n
\"p_1(x)\"<\/td>= \"y_0 \cdot L_0(x) +y_1\cdot L_1(x)\"<\/td><\/tr>
<\/td>= \"y_0 \cdot \displaystyle \frac{x-x_1}{x_0-x_1} +y_1\cdot \displaystyle \frac{x-x_0}{x_1-x_0}\"<\/td><\/tr>
<\/td>= \"-3 \cdot \displaystyle \frac{x-1}{2-1} +15\cdot \displaystyle \frac{x-2}{1-2}\"<\/td><\/tr>
<\/td><\/td><\/tr>
\"p_1(3)\"<\/td>= \"-3 \cdot \displaystyle \frac{3-1}{2-1} +15\cdot \displaystyle \frac{3-2}{1-2}\"<\/td><\/tr>
<\/td> = \"-3 \cdot \displaystyle \frac{2}{1} +15\cdot \displaystyle \frac{1}{-1}\" = \"-6-15\" = -21<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Punti = 3 (grado = 2)<\/h2>\n\n\n\n
\n
\n
<\/th>x<\/th>y<\/th><\/tr><\/thead>
0<\/td>2<\/td>-3<\/td><\/tr>
1<\/td>1<\/td>15<\/td><\/tr>
2<\/td>5<\/td>2<\/td><\/tr><\/tbody><\/table><\/figure>\n<\/div>\n\n\n\n
\n
\"\"<\/figure>\n<\/div>\n<\/div>\n\n\n\n
\"p_2(x)\"<\/td>= \"y_0 \cdot L_0(x) + y_1\cdot L_1(x) + y_2\cdot L_2(x)\"<\/td><\/tr>
<\/td>= \"y_0 \cdot \displaystyle \frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)} + y_1\cdot \displaystyle \frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)} + y_2\cdot \displaystyle \frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}\"<\/td><\/tr>
<\/td>= \"-3 \cdot \displaystyle \frac{(x-1)(x-5)}{(2-1)(2-5)} +15\cdot \displaystyle \frac{(x-2)(x-5)}{(1-2)(1-5)} +2 \cdot \displaystyle \frac{(x-2)(x-1)}{(5-2)(5-1)}\"<\/td><\/tr>
<\/td><\/td><\/tr>
\"p_2(3)\"<\/td>= \"-3 \cdot \displaystyle \frac{(3-1)(3-5)}{(2-1)(2-5)} +15\cdot \displaystyle \frac{(3-2)(3-5)}{(1-2)(1-5)} +2 \cdot \displaystyle \frac{(3-2)(3-1)}{(5-2)(5-1)}\" <\/td><\/tr>
<\/td>= \"-3 \cdot \displaystyle \frac{(2)(-2)}{(1)(-3)} +15\cdot \displaystyle \frac{(1)(-2)}{(-1)(-4)} +2 \cdot \displaystyle \frac{(1)(2)}{(3)(4)}\"<\/td><\/tr>
<\/td>= \"-3 \cdot \displaystyle \frac{-4}{-3} +15\cdot \displaystyle \frac{-2}{4} +2 \cdot \displaystyle \frac{2}{12}\" = \"\displaystyle -4-\frac{15}{2}+\frac{1}{3}\" = \"\displaystyle -\frac{67}{6}\" = -11,166…<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
\n\n\n\n

Punti = 4 (grado = 3)<\/h2>\n\n\n\n
\n
\n
<\/th>x<\/th>y<\/th><\/tr><\/thead>
0<\/td>2<\/td>-3<\/td><\/tr>
1<\/td>1<\/td>15<\/td><\/tr>
2<\/td>5<\/td>2<\/td><\/tr>
3<\/td>6<\/td>1<\/td><\/tr><\/tbody><\/table><\/figure>\n<\/div>\n\n\n\n
\n
\"\"<\/figure>\n<\/div>\n<\/div>\n\n\n\n
\"p_3(x)\"<\/td>= \"y_0 \cdot L_0(x) + y_1\cdot L_1(x) + y_2\cdot L_2(x) + y_3\cdot L_3(x)\"<\/td><\/tr>
<\/td>= \"y_0 \cdot \displaystyle \frac{(x-x_1)(x-x_2)(x-x_3)}{(x_0-x_1)(x_0-x_2)(x_0-x_3)}\" \"+ y_1\cdot \displaystyle \frac{(x-x_0)(x-x_2)(x-x_3)}{(x_1-x_0)(x_1-x_2)(x_1-x_3)}\" \"+ y_2\cdot \displaystyle \frac{(x-x_0)(x-x_1)(x-x_3)}{(x_2-x_0)(x_2-x_1)(x_2-x_3)}\"
\"+ y_3\cdot \displaystyle \frac{(x-x_0)(x-x_1)(x-x_2)}{(x_2-x_0)(x_2-x_1)(x_2-x_2)}\"<\/td><\/tr>
<\/td>= \"-3 \cdot \displaystyle \frac{(x-1)(x-5)(x-6)}{(2-1)(2-5)(2-6)} +15\cdot \displaystyle \frac{(x-2)(x-5)(x-6)}{(1-2)(1-5)(1-6)}\" \"+2 \cdot \displaystyle \frac{(x-2)(x-1)(x-6)}{(5-2)(5-1)(5-6)}\"
\"+1 \cdot \displaystyle \frac{(x-2)(x-1)(x-5)}{(6-2)(6-1)(6-5)}\"<\/td><\/tr>
<\/td><\/td><\/tr>
\"p_3(3)\"<\/td>= \"-3 \cdot \displaystyle \frac{(3-1)(3-5)(3-6)}{(2-1)(2-5)(2-6)}\" \"+15\cdot \displaystyle \frac{(3-2)(3-5)(3-6)}{(1-2)(1-5)(1-6)}\" \"+2 \cdot \displaystyle \frac{(3-2)(3-1)(3-6)}{(5-2)(5-1)(5-6)}\"
\"+1 \cdot \displaystyle \frac{(3-2)(3-1)(3-5)}{(6-2)(6-1)(6-5)}\"<\/td><\/tr>
<\/td>= \"-3 \cdot \displaystyle \frac{(2)(-2)(-3)}{(1)(-3)(-4)} +15\cdot \displaystyle \frac{(1)(-2)(-3)}{(-1)(-4)(-5)}\" \"+2 \cdot \displaystyle \frac{(1)(2)(-3)}{(3)(4)(-1)} +1 \cdot \displaystyle \frac{(1)(2)(-2)}{(4)(5)(1)}\"<\/td><\/tr>
<\/td>= \"-3 \cdot \displaystyle \frac{12}{12} +15\cdot \displaystyle \frac{6}{-20}\" \"+2 \cdot \displaystyle \frac{-6}{-12} +1 \cdot \displaystyle \frac{-4}{20}\" = \"\displaystyle -3-\frac{9}{2}+1-\frac{1}{5}\" = \"\displaystyle -\frac{67}{10}\" = -6,7<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
\n\n\n\n

Punti = 5 (grado = 4)<\/h2>\n\n\n\n
\n
\n
<\/th>x<\/th>y<\/th><\/tr><\/thead>
0<\/td>2<\/td>-3<\/td><\/tr>
1<\/td>1<\/td>15<\/td><\/tr>
2<\/td>5<\/td>2<\/td><\/tr>
3<\/td>6<\/td>1<\/td><\/tr>
4<\/td>4<\/td>0<\/td><\/tr><\/tbody><\/table><\/figure>\n<\/div>\n\n\n\n
\n
\"\"<\/figure>\n<\/div>\n<\/div>\n\n\n\n
\"p_4(x)\"<\/td>= \"y_0 \cdot L_0(x) + y_1\cdot L_1(x) + y_2\cdot L_2(x) + y_3\cdot L_3(x) + y_4\cdot L_4(x)\"<\/td><\/tr>
<\/td>= \"y_0 \cdot \displaystyle \frac{(x-x_1)(x-x_2)(x-x_3)(x-x_4)}{(x_0-x_1)(x_0-x_2)(x_0-x_3)(x_0-x_4)}\" \"+y_1\cdot \displaystyle \frac{(x-x_0)(x-x_2)(x-x_3)(x-x_4)}{(x_1-x_0)(x_1-x_2)(x_1-x_3)(x_1-x_4)}\"
\"+ y_2\cdot \displaystyle \frac{(x-x_0)(x-x_1)(x-x_3)(x-x_4)}{(x_2-x_0)(x_2-x_1)(x_2-x_3)(x_2-x_4)}\" \"+y_3\cdot \displaystyle \frac{(x-x_0)(x-x_1)(x-x_2)(x-x_4)}{(x_3-x_0)(x_3-x_1)(x_3-x_2)(x_3-x_4)}\"
\"+y_4\cdot \displaystyle \frac{(x-x_0)(x-x_1)(x-x_2)(x-x_3)}{(x_4-x_0)(x_4-x_1)(x_4-x_2)(x_4-x_3)}\"<\/td><\/tr>
<\/td>= \"-3 \cdot \displaystyle \frac{(x-1)(x-5)(x-6)(x-4)}{(2-1)(2-5)(2-6)(2-4)}\" \"+15 \cdot \displaystyle \frac{(x-2)(x-5)(x-6)(x-4)}{(1-2)(1-5)(1-6)(1-4)}\"
\"+2 \cdot \displaystyle \frac{(x-2)(x-1)(x-6)(x-4)}{(5-2)(5-1)(5-6)(5-4)}\" \"+ 1 \cdot \displaystyle \frac{(x-2)(x-1)(x-5)(x-4)}{(6-2)(6-1)(6-5)(6-4)}\"
\"+ 0 \cdot \displaystyle \frac{(x-2)(x-1)(x-5)(x-6)}{(4-2)(4-1)(4-5)(4-6)}\"<\/td><\/tr>
<\/td><\/td><\/tr>
\"p_4(3)\"<\/td>= \"-3 \cdot \displaystyle \frac{(3-1)(3-5)(3-6)(3-4)}{(2-1)(2-5)(2-6)(2-4)}\" \"+15 \cdot \displaystyle \frac{(3-2)(3-5)(3-6)(3-4)}{(1-2)(1-5)(1-6)(1-4)}\"
\"+2 \cdot \displaystyle \frac{(3-2)(3-1)(3-6)(3-4)}{(5-2)(5-1)(5-6)(5-4)}\" \"+ 1 \cdot \displaystyle \frac{(3-2)(3-1)(3-5)(3-4)}{(6-2)(6-1)(6-5)(6-4)}\"
\"+ 0 \cdot \displaystyle \frac{(3-2)(3-1)(3-5)(3-6)}{(4-2)(4-1)(4-5)(4-6)}\"<\/td><\/tr>
<\/td>= \"-3 \cdot \displaystyle \frac{(2)(-2)(-3)(-1)}{(1)(-3)(-4)(-2)}\" \"+15 \cdot \displaystyle \frac{(1)(-2)(-3)(-1)}{(-1)(-4)(-5)(-3)}\" \"+2 \cdot \displaystyle \frac{(1)(2)(-3)(-1)}{(3)(4)(-1)(1)}\" \"+ 1 \cdot \displaystyle \frac{(1)(2)(-2)(-1)}{(4)(5)(1)(2)}\" \"+ 0 \cdot \displaystyle \frac{(1)(2)(-2)(-3)}{(2)(3)(-1)(-2)}\"<\/td><\/tr>
<\/td>= \"-3 \cdot \displaystyle \frac{-12}{-24}\" \"+15 \cdot \displaystyle \frac{-6}{60}\" \"+2 \cdot \displaystyle \frac{6}{-12}\" \"+ 1 \cdot \displaystyle \frac{4}{40}\" \"+ 0 \cdot \displaystyle \frac{12}{12}\" = \"-\displaystyle \frac{3}{2} - \displaystyle \frac{3}{2}-1 + \displaystyle \frac{1}{10}+ 0\" = \"\displaystyle -\frac{39}{10}\" = -3,9<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
\n\n\n\n

Punti = 6 (grado = 5)<\/h2>\n\n\n\n
\n
\n
<\/th>x<\/th>y<\/th><\/tr><\/thead>
0<\/td>2<\/td>-3<\/td><\/tr>
1<\/td>1<\/td>15<\/td><\/tr>
2<\/td>5<\/td>2<\/td><\/tr>
3<\/td>6<\/td>1<\/td><\/tr>
4<\/td>4<\/td>0<\/td><\/tr>
5<\/td>0<\/td>6<\/td><\/tr><\/tbody><\/table><\/figure>\n<\/div>\n\n\n\n
\n
\"\"<\/figure>\n<\/div>\n<\/div>\n\n\n\n
\"p_5(x)\"<\/td>= \"y_0 \cdot L_0(x) + y_1\cdot L_1(x) + y_2\cdot L_2(x) + y_3\cdot L_3(x) + y_4\cdot L_4(x) + y_5\cdot L_5(x)\"<\/td><\/tr>
<\/td>= \"y_0 \cdot \displaystyle \frac{(x-x_1)(x-x_2)(x-x_3)(x-x_4)(x-x_5)}{(x_0-x_1)(x_0-x_2)(x_0-x_3)(x_0-x_4)(x_0-x_5)}\" \"+y_1\cdot \displaystyle \frac{(x-x_0)(x-x_2)(x-x_3)(x-x_4)(x-x_5)}{(x_1-x_0)(x_1-x_2)(x_1-x_3)(x_1-x_4)(x_1-x_5)}\"
\"+ y_2\cdot \displaystyle \frac{(x-x_0)(x-x_1)(x-x_3)(x-x_4)(x-x_5)}{(x_2-x_0)(x_2-x_1)(x_2-x_3)(x_2-x_4)(x_2-x_5)}\" \"+y_3\cdot \displaystyle \frac{(x-x_0)(x-x_1)(x-x_2)(x-x_4)(x-x_5)}{(x_3-x_0)(x_3-x_1)(x_3-x_2)(x_3-x_4)(x_3-x_5)}\"
\"+y_4\cdot \displaystyle \frac{(x-x_0)(x-x_1)(x-x_2)(x-x_3)(x-x_5)}{(x_4-x_0)(x_4-x_1)(x_4-x_2)(x_4-x_3)(x_4-x_5)}\" \"+y_5\cdot \displaystyle \frac{(x-x_0)(x-x_1)(x-x_2)(x-x_3)(x-x_4)}{(x_5-x_0)(x_5-x_1)(x_5-x_2)(x_5-x_3)(x_5-x_4)}\"<\/td><\/tr>
<\/td>= \"-3 \cdot \displaystyle \frac{(x-1)(x-5)(x-6)(x-4)(x-0)}{(2-1)(2-5)(2-6)(2-4)(2-0)}\" \"+15\cdot \displaystyle \frac{(3-2)(3-5)(3-6)(3-4)(3-0)}{(1-2)(1-5)(1-6)(1-4)(1-0)}\"
\"+2\cdot \displaystyle \frac{(x-2)(x-1)(x-6)(x-4)(x-0)}{(5-2)(5-1)(5-6)(5-4)(5-0)}\" \"+1\cdot \displaystyle \frac{(x-2)(x-1)(x-5)(x-4)(x-0)}{(6-2)(6-1)(6-5)(6-4)(6-0)}\"
\"+0\cdot \displaystyle \frac{(x-2)(x-1)(x-5)(x-6)(x-0)}{(4-2)(4-1)(4-5)(4-6)(4-0)}\" \"+6\cdot \displaystyle \frac{(x-2)(x-1)(x-5)(x-6)(x-4)}{(0-2)(0-1)(0-5)(0-6)(0-4)}\"<\/td><\/tr>
<\/td><\/td><\/tr>
\"p_5(3)\"<\/td>= \"-3 \cdot \displaystyle \frac{(3-1)(3-5)(3-6)(3-4)(3-0)}{(2-1)(2-5)(2-6)(2-4)(2-0)}\" \"+15\cdot \displaystyle \frac{(3-2)(3-5)(3-6)(3-4)(3-0)}{(1-2)(1-5)(1-6)(1-4)(1-0)}\"
\"+2\cdot \displaystyle \frac{(3-2)(3-1)(3-6)(3-4)(3-0)}{(5-2)(5-1)(5-6)(5-4)(5-0)}\" \"+1\cdot \displaystyle \frac{(3-2)(3-1)(3-5)(3-4)(3-0)}{(6-2)(6-1)(6-5)(6-4)(6-0)}\"
\"+0\cdot \displaystyle \frac{(3-2)(3-1)(3-5)(3-6)(3-0)}{(4-2)(4-1)(4-5)(4-6)(4-0)}\" \"+6\cdot \displaystyle \frac{(3-2)(3-1)(3-5)(3-6)(3-4)}{(0-2)(0-1)(0-5)(0-6)(0-4)}\"<\/td><\/tr>
<\/td>= \"-3 \cdot \displaystyle \frac{(2)(-2)(-3)(-1)(3)}{(2)(-3)(-4)(-2)(2)}\" \"+15\cdot \displaystyle \frac{(1)(-2)(-3)(-1)(3)}{(-1)(-4)(-5)(-3)(1)}\" \"+2\cdot \displaystyle \frac{(1)(2)(-3)(-1)(3)}{(3)(4)(-1)(1)(5)}\"
\"+1\cdot \displaystyle \frac{(1)(2)(-2)(-1)(3)}{(4)(5)(1)(2)(6)}\" \"+0\cdot \displaystyle \frac{(1)(2)(-2)(-3)(3)}{(2)(3)(-1)(-2)(4)}\" \"+6\cdot \displaystyle \frac{(1)(2)(-2)(-3)(-1)}{(-2)(-1)(-5)(-6)(-4)}\"<\/td><\/tr>
<\/td>= \"-3 \cdot \displaystyle \frac{-36}{-48} +15\cdot \displaystyle \frac{-18}{60} +2\cdot \displaystyle \frac{18}{-60}+1\cdot \displaystyle \frac{12}{240}\" \"+0\cdot \displaystyle \frac{36}{48}+6\cdot \displaystyle \frac{-12}{-240}\" = \"\displaystyle -\frac{9}{4}-\frac{9}{2}-\frac{3}{5}+\frac{1}{20}+0+\frac{3}{10}\" = -7,0<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
\n\n\n\n

Ripeti gli esercizi calcolando il polinomio interpolatore<\/strong> per 2 \/ 3 \/ 4 \/ 5 \/ 6 punti <\/p>\n\n\n\n

\n\n\n\n
\n
\n

Punti = 2 (grado = 1)<\/h2>\n\n\n\n
<\/th>x<\/th>y<\/th><\/tr><\/thead>
0<\/td>2<\/td>-3<\/td><\/tr>
1<\/td>1<\/td>15<\/td><\/tr><\/tbody><\/table><\/figure>\n<\/div>\n\n\n\n
\n
\"\"<\/figure>\n<\/div><\/div>\n<\/div>\n\n\n\n
\"L_0(x)\"<\/td>= \"\displaystyle \frac{x-x_1}{x_0-x_1}\" = \"\displaystyle \frac{x-1}{2-1}\" = \"\displaystyle x-1\"<\/td><\/tr>
\"L_1(x)\"<\/td>= \"\displaystyle \frac{x-x_0}{x_1-x_0}\" = \"\displaystyle \frac{x-2}{1-2}\" = \"\displaystyle -(x-2)\"<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
\"p_1(x)\"<\/td>= \"y_0 \cdot L_0(x) +y_1\cdot L_1(x)\"<\/td><\/tr>
<\/td>= \"-3 (x-1) +15\left(-(x-2)\right)\"<\/td><\/tr>
<\/td>= \"-18x+33\"<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
\n\n\n\n
\n
\n

Punti = 3 (grado = 2)<\/h2>\n\n\n\n
<\/th>x<\/th>y<\/th><\/tr><\/thead>
0<\/td>2<\/td>-3<\/td><\/tr>
1<\/td>1<\/td>15<\/td><\/tr>
2<\/td>5<\/td>2<\/td><\/tr><\/tbody><\/table><\/figure>\n<\/div>\n\n\n\n
\n
\"\"<\/figure>\n<\/div><\/div>\n<\/div>\n\n\n\n
\"L_0(x)\"<\/td>= \"\displaystyle \frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}\" = \"\displaystyle \frac{(x-1)(x-5)}{(2-1)(2-5)}\" = \"\displaystyle \frac{x^2-6x+5}{-3}\" = \"\displaystyle -\frac{1}{3}x^2+2x-\frac{5}{3}\"<\/td><\/tr>
\"L_1(x)\"<\/td>= \"\displaystyle \frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)}\" = \"\displaystyle \frac{(x-2)(x-5)}{(1-2)(1-5)}\" = \"\displaystyle \frac{x^2-7x+10}{4}\" = \"\displaystyle \frac{1}{4}x^2-\frac{7}{4}x+\frac{5}{2}\"<\/td><\/tr>
\"L_2(x)\"<\/td>= \"\displaystyle \frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}\" = \"\displaystyle \frac{(x-2)(x-1)}{(5-2)(5-1)}\" = \"\displaystyle \frac{x^2-3x+2}{12}\" = \"\displaystyle \frac{1}{12}x^2-\frac{1}{4}x+\frac{1}{6}\"<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
\"p_2(x)\"<\/td>= \"y_0 \cdot L_0(x) +y_1\cdot L_1(x) +y_2\cdot L_2(x)\"<\/td><\/tr>
<\/td>= \"-3\left(\displaystyle -\frac{1}{3}x^2+2x-\frac{5}{3}\right)\" \"+15\left(\displaystyle \frac{1}{4}x^2-\frac{7}{4}x+\frac{5}{2} \right)\" \"+2 \left(\displaystyle \frac{1}{12}x^2-\frac{1}{4}x+\frac{1}{6} \right)\"<\/td><\/tr>
\ufeff<\/td>= \"\displaystyle \frac{59}{12} x^2 - \frac{131}{4}x + \frac{257}{6}\"<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
\n\n\n\n
\n
\n

Punti = 4 (grado = 3)<\/h2>\n\n\n\n
<\/th>x<\/th>y<\/th><\/tr><\/thead>
0<\/td>2<\/td>-3<\/td><\/tr>
1<\/td>1<\/td>15<\/td><\/tr>
2<\/td>5<\/td>2<\/td><\/tr>
3<\/td>6<\/td>1<\/td><\/tr><\/tbody><\/table><\/figure>\n<\/div>\n\n\n\n
\n
\"\"<\/figure>\n<\/div><\/div>\n<\/div>\n\n\n\n
\"L_0(x)\"<\/td>= \"\displaystyle \frac{(x-x_1)(x-x_2)(x-x_3)}{(x_0-x_1)(x_0-x_2)(x_0-x_3)}\"<\/td>= \"\displaystyle \frac{(x-1)(x-5)(x-6)}{(2-1)(2-5)(2-6)}\"<\/td><\/tr>
<\/td>…<\/td>= \"\displaystyle \frac{1}{12}\ x^3 -x^2 + \frac{41}{12}x-\frac{5}{2}\"<\/td><\/tr>
\"L_1(x)\"<\/td>= \"\displaystyle \frac{(x-x_0)(x-x_2)(x-x_3)}{(x_1-x_0)(x_1-x_2)(x_1-x_3)}\"<\/td>= \"\displaystyle \frac{(x-2)(x-5)(x-6)}{(1-2)(1-5)(1-6)}\"<\/td><\/tr>
<\/td>…<\/td>= \"\displaystyle -\frac{1}{20} x^3 + \frac{13}{20}x^2 - \frac{13}{5}x+3\"<\/td><\/tr>
\"L_2(x)\"<\/td>= \"\displaystyle \frac{(x-x_0)(x-x_1)(x-x_3)}{(x_2-x_0)(x_2-x_1)(x_2-x_3)}\"<\/td>= \"\displaystyle \frac{(x-2)(x-1)(x-6)}{(5-2)(5-1)(5-6)}\"<\/td><\/tr>
<\/td>…<\/td>= \"\displaystyle -\frac{1}{12} x^3+ \frac{3}{4}x^2 - \frac{5}{3}x+1\"<\/td><\/tr>
\"L_3(x)\"<\/td>= \"\displaystyle \frac{(x-x_0)(x-x_1)(x-x_2)}{(x_3-x_0)(x_3-x_1)(x_3-x_2)}\"<\/td>= \"\displaystyle \frac{(x-2)(x-1)(x-5)}{(6-2)(6-1)(6-5)}\"<\/td><\/tr>
<\/td>…<\/td>= \"\displaystyle \frac{1}{20} x^3 - \frac{2}{5}x^2 + \frac{17}{20}x- \frac{1}{2}\"<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
\"p_3(x)\"<\/td>= \"y_0 \cdot L_0(x) + y_1 \cdot L_1(x) + y_2 \cdot L_2(x) + y_3 \cdot L_3(x)\"<\/td><\/tr>
<\/td>= \"-3 \left(\displaystyle \frac{1}{12} x^3 -x^2 + \frac{41}{12}x-\frac{5}{2} \right)\" \"+15 \left(\displaystyle -\frac{1}{20} x^3 + \frac{13}{20}x^2 - \frac{13}{5}x+3 \right)\" \"+2 \left(\displaystyle -\frac{1}{12}\ x^3+ \frac{3}{4}x^2 - \frac{5}{3}x+1 \right)\"
\"+ \left(\displaystyle \frac{1}{20} x^3 - \frac{2}{5}x^2 + \frac{17}{20}x- \frac{1}{2} \right)\"<\/td><\/tr>
\ufeff<\/td>= \"\displaystyle -\frac{67}{60} x^3 + \frac{277}{20}x^2 - \frac{776}{15}x+54\"<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
\n\n\n\n
\n
\n

Punti = 5 (grado = 4)<\/h2>\n\n\n\n
<\/th>x<\/th>y<\/th><\/tr><\/thead>
0<\/td>2<\/td>-3<\/td><\/tr>
1<\/td>1<\/td>15<\/td><\/tr>
2<\/td>5<\/td>2<\/td><\/tr>
3<\/td>6<\/td>1<\/td><\/tr>
4<\/td>4<\/td>0<\/td><\/tr><\/tbody><\/table><\/figure>\n<\/div>\n\n\n\n
\n
\"\"<\/figure>\n<\/div><\/div>\n<\/div>\n\n\n\n
\"L_0(x)\"<\/td>= \"\displaystyle \frac{(x-x_1)(x-x_2)(x-x_3)(x-x_4)}{(x_0-x_1)(x_0-x_2)(x_0-x_3)(x_0-x_4)}\"<\/td>= \"\displaystyle \frac{(x-1)(x-5)(x-6)(x-4)}{(2-1)(2-5)(2-6)(2-4)}\"<\/td><\/tr>
<\/td>= …<\/td>= \"\displaystyle -\frac{1}{24} x^4 +\frac{2}{3} x^3 -\frac{89}{24} x^2 +\frac{97}{12} x - 5\"<\/td><\/tr>
\"L_1(x)\"<\/td>= \"\displaystyle \frac{(x-x_0)(x-x_2)(x-x_3)(x-x_4)}{(x_1-x_0)(x_1-x_2)(x_1-x_3)(x_1-x_4)}\"<\/td>= \"\displaystyle \frac{(x-2)(x-5)(x-6)(x-4)}{(1-2)(1-5)(1-6)(1-4)}\"<\/td><\/tr>
<\/td>= …<\/td>= \"\displaystyle \frac{1}{60} x^4 -\frac{17}{60} x^3 +\frac{26}{15} x^2 -\frac{67}{15} x + 4\"<\/td><\/tr>
\"L_2(x)\"<\/td>= \"\displaystyle \frac{(x-x_0)(x-x_1)(x-x_3)(x-x_4)}{(x_2-x_0)(x_2-x_1)(x_2-x_3)(x_2-x_4)}\"<\/td>= \"\displaystyle \frac{(x-2)(x-1)(x-6)(x-4)}{(5-2)(5-1)(5-6)(5-4)}\"<\/td><\/tr>
<\/td>= …<\/td>= \"\displaystyle -\frac{1}{12} x^4 +\frac{13}{12} x^3 -\frac{14}{3} x^2 +\frac{23}{3} x - 4\"<\/td><\/tr>
\"L_3(x)\"<\/td>= \"\displaystyle \frac{(x-x_0)(x-x_1)(x-x_2)(x-x_4)}{(x_3-x_0)(x_3-x_1)(x_3-x_2)(x_3-x_4)}\"<\/td>= \"\displaystyle \frac{(x-2)(x-1)(x-5)(x-4)}{(6-2)(6-1)(6-5)(6-4)}\"<\/td><\/tr>
<\/td>= …<\/td>= \"\displaystyle \frac{1}{40} x^4 -\frac{3}{10} x^3 +\frac{49}{40} x^2 -\frac{39}{20} x + 1\"<\/td><\/tr>
\"L_4(x)\"<\/td>= \"\displaystyle \frac{(x-x_0)(x-x_1)(x-x_2)(x-x_3)}{(x_4-x_0)(x_4-x_1)(x_4-x_2)(x_4-x_3)}\"<\/td>= \"\displaystyle \frac{(x-2)(x-1)(x-5)(x-6)}{(4-2)(4-1)(4-5)(4-6)}\"<\/td><\/tr>
<\/td>= …<\/td>= \"\displaystyle \frac{1}{12} x^4 -\frac{7}{6} x^3 +\frac{65}{12} x^2 -\frac{28}{3} x + 5\"<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
\"p_4(x)\"<\/td>= \"y_0 \cdot L_0(x) + y_1 \cdot L_1(x) + y_2 \cdot L_2(x) + y_3 \cdot L_3(x) + y_4 \cdot L_4(x)\"<\/td><\/tr>
<\/td>= \"-3 \left(\displaystyle -\frac{1}{24} x^4 +\frac{2}{3} x^3 -\frac{89}{24} x^2 +\frac{97}{12} x - 5 \right)\" \"+15 \left(\displaystyle \frac{1}{60} x^4 -\frac{17}{60} x^3 +\frac{26}{15} x^2 -\frac{67}{15} x + 4 \right)\"
\"+2 \left(\displaystyle -\frac{1}{12} x^4 +\frac{13}{12} x^3 -\frac{14}{3} x^2 +\frac{23}{3} x - 4 \right)\" \"+ \left( \displaystyle \frac{1}{40} x^4 -\frac{3}{10} x^3 +\frac{49}{40} x^2 -\frac{39}{20} x + 1 \right)\"
\"+0 \left(\displaystyle \frac{1}{12} x^4 -\frac{7}{6} x^3 +\frac{65}{12} x^2 -\frac{28}{3} x + 5 \right)\"<\/td><\/tr>
<\/td>= \"\displaystyle \frac{7}{30} x^4 -\frac{263}{60} x^3 +\frac{1741}{60} x^2 -\frac{1168}{15} x + 68\"<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
\n\n\n\n
\n
\n

Punti = 6 (grado = 5)<\/h2>\n\n\n\n
<\/th>x<\/th>y<\/th><\/tr><\/thead>
0<\/td>2<\/td>-3<\/td><\/tr>
1<\/td>1<\/td>15<\/td><\/tr>
2<\/td>5<\/td>2<\/td><\/tr>
3<\/td>6<\/td>1<\/td><\/tr>
4<\/td>4<\/td>0<\/td><\/tr>
5<\/td>0<\/td>6<\/td><\/tr><\/tbody><\/table><\/figure>\n<\/div>\n\n\n\n
\n
\"\"<\/figure>\n<\/div><\/div>\n<\/div>\n\n\n\n
\"L_0(x)\"<\/td>= \"\displaystyle \frac{(x-x_1)(x-x_2)(x-x_3)(x-x_4)(x-x_5)}{(x_0-x_1)(x_0-x_2)(x_0-x_3)(x_0-x_4)(x_0-x_5)}\"<\/td>= \"\displaystyle \frac{(x-1)(x-5)(x-6)(x-4)(x-0)}{(2-1)(2-5)(2-6)(2-4)(2-0)}\"<\/td><\/tr>
<\/td>= \"\displaystyle \frac{x^5-16x^4 +89x^3 +120x}{-48}\"<\/td>= \"\displaystyle -\frac{1}{48}x^5 +\frac{1}{3}x^4 -\frac{89}{48}x^3 +\frac{97}{24}x^2 -\frac{5}{2}x\"<\/td><\/tr>
\"L_1(x)\"<\/td>= \"\displaystyle \frac{(x-x_0)(x-x_2)(x-x_3)(x-x_4)(x-x_5)}{(x_1-x_0)(x_1-x_2)(x_1-x_3)(x_1-x_4)(x_1-x_5)}\"<\/td>= \"\displaystyle \frac{(x-2)(x-5)(x-6)(x-4)(x-0)}{(1-2)(1-5)(1-6)(1-4)(1-0)}\"<\/td><\/tr>
<\/td>= \"\displaystyle \frac{x^5-17x^4 +104x^3 -268x^2+240x}{60}\"<\/td>= \"\displaystyle \frac{1}{60}x^5 -\frac{17}{60}x^4 +\frac{26}{15}x^3 -\frac{67}{15}x^2 +4 x\"<\/td><\/tr>
\"L_2(x)\"<\/td>= \"\displaystyle \frac{(x-x_0)(x-x_1)(x-x_3)(x-x_4)(x-x_5)}{(x_2-x_0)(x_2-x_1)(x_2-x_3)(x_2-x_4)(x_2-x_5)}\"<\/td>= \"\displaystyle \frac{(x-2)(x-1)(x-6)(x-4)(x-0)}{(5-2)(5-1)(5-6)(5-4)(5-0)}\"<\/td><\/tr>
<\/td>= \"\displaystyle \frac{x^5-13x^4 +56x^3 -92x^2+48x}{-60}\"<\/td>= \"\displaystyle -\frac{1}{60}x^5 +\frac{13}{60}x^4 -\frac{14}{15}x^3 +\frac{23}{15}x^2 -\frac{4}{5}x\"<\/td><\/tr>
\"L_3(x)\"<\/td>= \"\displaystyle \frac{(x-x_0)(x-x_1)(x-x_2)(x-x_4)(x-x_5)}{(x_3-x_0)(x_3-x_1)(x_3-x_2)(x_3-x_4)(x_3-x_5)}\"<\/td>= \"\displaystyle \frac{(x-2)(x-1)(x-5)(x-4)(x-0)}{(6-2)(6-1)(6-5)(6-4)(6-0)}\"<\/td><\/tr>
<\/td>= \"\displaystyle \frac{x^5-12x^4 +49x^3 -78x^2+40x}{240}\"<\/td>= \"\displaystyle \frac{1}{240}x^5 -\frac{1}{20}x^4 +\frac{49}{240}x^3 -\frac{13}{40}x^2 +\frac{1}{6}x\"<\/td><\/tr>
\"L_4(x)\"<\/td>= \"\displaystyle \frac{(x-x_0)(x-x_1)(x-x_2)(x-x_3)(x-x_5)}{(x_4-x_0)(x_4-x_1)(x_4-x_2)(x_4-x_3)(x_4-x_5)}\"<\/td>= \"\displaystyle \frac{(x-2)(x-1)(x-5)(x-6)(x-0)}{(4-2)(4-1)(4-5)(4-6)(4-0)}\"<\/td><\/tr>
<\/td>= \"\displaystyle \frac{x^5-14x^4 +65x^3 +122x^2+60x}{48}\"<\/td>= \"\displaystyle \frac{1}{48}x^5 -\frac{7}{24}x^4 +\frac{65}{48}x^3 -\frac{7}{3}x^2 +\frac{5}{4}x\"<\/td><\/tr>
\"L_5(x)\"<\/td>= \"\displaystyle \frac{(x-x_0)(x-x_1)(x-x_2)(x-x_3)(x-x_4)}{(x_5-x_0)(x_5-x_1)(x_5-x_2)(x_5-x_3)(x_5-x_4)}\"<\/td>= \"\displaystyle \frac{(x-2)(x-1)(x-5)(x-6)(x-4)}{(0-2)(0-1)(0-5)(0-6)(0-4)}\"<\/td><\/tr>
<\/td>= \"\displaystyle \frac{x^5-18x^4 +121x^3 -372x^2+508x-240}{-240}\"<\/td>= \"\displaystyle -\frac{1}{240}x^5 +\frac{3}{40}x^4 -\frac{121}{240}x^3 +\frac{31}{20}x^2 -\frac{127}{60}x+1\"<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
\"p_5(x)\"<\/td>= \"y_0 \cdot L_0(x) + y_1 \cdot L_1(x) + y_2 \cdot L_2(x) + y_3 \cdot L_3(x) + y_4 \cdot L_4(x) + y_5 \cdot L_5(x)\"<\/td><\/tr>
<\/td>= \"-3 \left( \displaystyle -\frac{1}{48}x^5 +\frac{1}{3}x^4 -\frac{89}{48}x^3 +\frac{97}{24}x^2 -\frac{5}{2}x \right)\" \"+15 \left(\displaystyle \frac{1}{60}x^5 -\frac{17}{60}x^4 +\frac{26}{15}x^3 -\frac{67}{15}x^2 +4 x\right)\"
\"+2 \left(\displaystyle -\frac{1}{60}x^5 +\frac{13}{60}x^4 -\frac{14}{15}x^3 +\frac{23}{15}x^2 -\frac{4}{5}x\right)\" \"+ \left(\displaystyle \frac{1}{240}x^5 -\frac{1}{20}x^4 +\frac{49}{240}x^3 -\frac{13}{40}x^2 +\frac{1}{6}x\right)\"
\"+0 \left(\displaystyle \frac{1}{48}x^5 -\frac{7}{24}x^4 +\frac{65}{48}x^3 -\frac{7}{3}x^2 +\frac{5}{4}x\right)\" \"+6 \left(\displaystyle -\frac{1}{240}x^5 +\frac{3}{40}x^4 -\frac{121}{240}x^3 +\frac{31}{20}x^2 -\frac{127}{60}x+1\right)\"<\/td><\/tr>
<\/td>= \"\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\"<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
\n\n\n\n

Osserva i risultati successivi<\/p>\n\n\n\n

Punti<\/th>Polinomio<\/th>x=3<\/th><\/tr><\/thead>
2<\/td>\"p_1(x) = -18x+33\"<\/td>-21<\/strong><\/td><\/tr>
3<\/td>\"p_2(x) = \displaystyle \frac{59}{12} x^2 - \frac{131}{4}x + \frac{257}{6}\"<\/td>-11<\/strong>,166…<\/td><\/tr>
4<\/td>\"p_3(x) = \displaystyle -\frac{67}{60} x^3 + \frac{277}{20}x^2 - \frac{776}{15}x+54\"<\/td>-6,7<\/strong><\/td><\/tr>
5<\/td>\"p_4(x) = \displaystyle \frac{7}{30} x^4 -\frac{263}{60} x^3 +\frac{1741}{60} x^2 -\frac{1168}{15} x + 68\"<\/td>-3,9<\/strong><\/td><\/tr>
6<\/td>\"p_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\"<\/td>-7<\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
\n\n\n\n
\n
Python<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Esercizio del libro di testo a pagina 100. Utilizza il polinomio interpolatore di Lagrange passante per i punti dati per calcolare il valore approssimato per x=3.Svolgi l’esercizio per 2 \/ 3 \/ 4 \/ 5 \/ 6 punti. Punti = 2 (grado = 1) x y 0 2 -3 1 1 15 = = = … Leggi tutto<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[114],"tags":[140],"class_list":["post-51760","post","type-post","status-publish","format-standard","hentry","category-calcolo","tag-lagrange"],"_links":{"self":[{"href":"https:\/\/www.valcon.it\/pp\/wp-json\/wp\/v2\/posts\/51760","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.valcon.it\/pp\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.valcon.it\/pp\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.valcon.it\/pp\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.valcon.it\/pp\/wp-json\/wp\/v2\/comments?post=51760"}],"version-history":[{"count":304,"href":"https:\/\/www.valcon.it\/pp\/wp-json\/wp\/v2\/posts\/51760\/revisions"}],"predecessor-version":[{"id":64546,"href":"https:\/\/www.valcon.it\/pp\/wp-json\/wp\/v2\/posts\/51760\/revisions\/64546"}],"wp:attachment":[{"href":"https:\/\/www.valcon.it\/pp\/wp-json\/wp\/v2\/media?parent=51760"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.valcon.it\/pp\/wp-json\/wp\/v2\/categories?post=51760"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.valcon.it\/pp\/wp-json\/wp\/v2\/tags?post=51760"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}