Diferencias divididas
lida.caleroPráctica o problema18 de Agosto de 2021
3.053 Palabras (13 Páginas)129 Visitas
PROBLEMA 1:
[pic 1] | X | f ()[pic 2] |
[pic 3] | 0 | -1 |
[pic 4] | 1 | 6 |
[pic 5] | 2 | 31 |
[pic 6] | 3 | 18 |
Mediante la fórmula de newton para diferencias divididas
= f(x) + f(x)(x-) + f(x)(x-)(x-) + f(x)(x-)(x-) (x-)[pic 7][pic 8][pic 9][pic 10][pic 11][pic 12][pic 13][pic 14][pic 15][pic 16]
Calculo de coeficientes:
- f(x):[pic 17]
f(,) = = = 7[pic 18][pic 19][pic 20][pic 21]
f(,) = = = 25[pic 22][pic 23][pic 24][pic 25]
f(,) = = = -13[pic 26][pic 27][pic 28][pic 29]
- f(x)[pic 30]
f(,,) = = = 9[pic 31][pic 32][pic 33][pic 34][pic 35]
f(,,) = = = -19[pic 36][pic 37][pic 38][pic 39][pic 40]
- f(x)[pic 41]
f (,,,) = = = -9.33[pic 42][pic 43][pic 44][pic 45][pic 46][pic 47]
Reemplazando valores a la tabla de Diferencias Divididas
[pic 48] | f(x): | f(x):[pic 49] | f(x)[pic 50] | f(x)[pic 51] |
[pic 52] | -1 | 7 | 9 | -9.33 |
[pic 53] | 6 | 25 | -19 | |
[pic 54] | 31 | 13 | ||
[pic 55] | 18 |
Reemplazando valores al polinomio:
= -1+ 7(x-0) + 9(x-0) (x-1) + -9.33(x-0) (x-1) (x-2)[pic 56]
= + - – 1[pic 57][pic 58][pic 59][pic 60]
Comprobación:
X=0
= + - – 1 = -1[pic 61][pic 62][pic 63][pic 64]
X=1
= + - – 1 = 6[pic 65][pic 66][pic 67][pic 68]
X=2
= + - – 2 = 31[pic 69][pic 70][pic 71][pic 72]
X=3
= + - – 3 = 18[pic 73][pic 74][pic 75][pic 76]
PROBLEMA 2:
[pic 77] | X | f ()[pic 78] |
[pic 79] | 1 | 120 |
[pic 80] | 2 | 94 |
[pic 81] | 3 | 75 |
[pic 82] | 4 | 62 |
Mediante la fórmula de newton para diferencias divididas
= f(x) + f(x) (x-) + f(x) (x-) (x-) + f(x) (x-) (x-) (x-)[pic 83][pic 84][pic 85][pic 86][pic 87][pic 88][pic 89][pic 90][pic 91][pic 92]
Calculo de coeficientes:
- f(x):[pic 93]
f (,) = = = -26[pic 94][pic 95][pic 96][pic 97]
f (,) = = = -19[pic 98][pic 99][pic 100][pic 101]
f (,) = = = -13[pic 102][pic 103][pic 104][pic 105]
- f(x)[pic 106]
f(,,) = = = 3.5[pic 107][pic 108][pic 109][pic 110][pic 111]
f(,,) = = = 3[pic 112][pic 113][pic 114][pic 115][pic 116]
- f(x)[pic 117]
f (,,,) = = = -0.167[pic 118][pic 119][pic 120][pic 121][pic 122][pic 123]
Reemplazando valores a la tabla de Diferencias Divididas
[pic 124] | f(x): | f(x):[pic 125] | f(x)[pic 126] | f(x)[pic 127] |
[pic 128] | 120 | -26 | 3.5 | -0.167 |
[pic 129] | 94 | -19 | 3 | |
[pic 130] | 75 | -13 | ||
[pic 131] | 62 |
Reemplazando valores al polinomio:
= 120 - 26(x-1) + 3.5(x-1) (x-2) – 0.167(x-1) (x-2) (x-3)[pic 132]
= - + + + 154[pic 133][pic 134][pic 135][pic 136]
Comprobación:
X=1
= - + + + 154 = 120[pic 137][pic 138][pic 139][pic 140]
X=2
= - + + + 154 = 94[pic 141][pic 142][pic 143][pic 144]
X=3
= - + + + 154 = 75[pic 145][pic 146][pic 147][pic 148]
X=4
= - + + + 154 = 62[pic 149][pic 150][pic 151][pic 152]
- Para x = 3.5
= - + + + 154[pic 153][pic 154][pic 155][pic 156]
= 67.8125[pic 157]
PROBLEMA 3:
[pic 158] | X | f ()[pic 159] |
[pic 160] | -1 | 3 |
[pic 161] | 0 | 0 |
[pic 162] | 1 | -1 |
[pic 163] | 2 | 1 |
[pic 164] | 3 | 2 |
Mediante la fórmula de newton para diferencias divididas
= f(x) + f(x) (x-) + f(x) (x-) (x-) + f(x) (x-) (x-) (x-) + f(x) (x-) (x-) (x-) (x-)[pic 165][pic 166][pic 167][pic 168][pic 169][pic 170][pic 171][pic 172][pic 173][pic 174][pic 175][pic 176][pic 177][pic 178][pic 179]
Calculo de coeficientes:
- f(x):[pic 180]
f (,) = = = -3[pic 181][pic 182][pic 183][pic 184]
f (,) = = = -1[pic 185][pic 186][pic 187][pic 188]
f (,) = = = 2[pic 189][pic 190][pic 191][pic 192]
f (,) = = = 1[pic 193][pic 194][pic 195][pic 196]
- f(x)[pic 197]
f (,,) = = = 1[pic 198][pic 199][pic 200][pic 201][pic 202]
...