Diferencias Divididas

Sesión 5
Ejercicios: Diferencias Divididas

2. Realiza los ejercicios que te presento a continuación.

1. a información de la tabla se obtuvo del polinomio; 
x | f(x) |
-2 | -18 |
-1 | -5 |
0 | -2 |
2 | -2 |
3 | 7 |
6 | 142 |

A partir de ésta tabla elaborar una tabla de diferencias divididas e interpolar x=1, x=0.5 y x=5.
Xi | f(Xi) | f(Xi,Xi+1) | f(Xi,Xi+1,Xi+2) |f(Xi,Xi+1,Xi+2,Xi+3) | f(Xi,Xi+1,Xi+2,Xi+3,Xi+4) | f(Xi,Xi+1,Xi+2,Xi+3,Xi+4,Xi+5) |
-2 | -18 | | | | | |
-1 | -5 | -5--18-1--2=13 | | | | |
0 | -2 | -2--50--1=3 | 3-130--2=-5 | | | |
2 | -2 | -2--22-0=0 | 0-32--1=-1 | -1--52--2=1 | | |
3 | 7 | 7--23-2=9 | 9-03-0=3 | 3--13--1=1 | 1-13--2=0 | |
6 | 142 | 142-76-3=45 | 45-96-2=9 | 9-36-0=1 | 1-16--1=0 | 0 |

Xk=1
f(Xk)=-2+(1-0) (0) + (1-2) (1-0) (3) + (1-3) (1-2) (1-0) (1) = -3
Valor Exacto
(1)ᶟ - 2(1)² - 2 = - 3
Xk=0.5
f(Xk)= -2+(0.5-0) (0) + (0.5-2) (0.5-0) (3) + (0.5-3) (0.5-2) (0.5-0) (1) = - 2.375
Valor Exacto
(0.5)ᶟ - 2(0.5)² - 2 = - 2.375
Xk= 5
f(Xk)= 142 + (5-6) (45) + (5-6) (5-3) (9) + (5-6) (5-3) (5-2) (1)= 73
Valor Exacto
(5)ᶟ - 2(5)² - 2 = 73

2. a)Elabore una tabla de diferenciasdivididas para los siguientes datos:
x | f(x) |
0.1 | .99750 |
0.2 | 0.99002 |
0.4 | 0.96040 |
0.7 | 0.88120 |
1.0 | 0.76520 |
1.2 | 0.67113 |
1.3 | 0.62009 |

x | f(x) | f(X0,X1) | f(X0,.,X2) | f(X0,..X3) | f(X0,..X4) | f(X0,..X5) | f(X0,..X6) |
0.1 | .99750 | | | | | | |
0.2 | 0.99002 | -0.0748 | | | | | |
0.4 | 0.96040 | -0.1483 | -0.245 | | | | |
0.7 |0.88120 | -0.264 | -0.2314 | 0.026667 | | | |
1.0 | 0.76520 | -0.386667 | -0.204445 | 0.03369375 | 0.0078075 | | |
1.2 | 0.67113 | -0.47035 | -0.167366 | 0.04634875 | 0.012655 | 0.004896464 | |
1.3 | 0.62009 | -0.5104 | -0.135 | 0.053943333 | 0.008438425 | -0.00383325 | -0.007274761 |

b) Evalúe los polinomios de interpolación en x=0.3 y x=0.55.
X=0.3f(x)=0.99002+[(0.1)(-0.1483)]+[(-0.1)(0.1)(-0.2314)]+[(-0.4)(-0.1)(0.1)(0.04634875)]+ [(-0.7)(-0.4)(-0.1)(0.1)(0.008438425)] = 0.977665768
X= 0.55
f(x)= 0.96040+[(0.15)(-0.264)]+[(-0.15)(0.15)(-0.204445)]+[(-0.45)(-0.15)(0.15)(0.53943333)] = 0.930861774

3) con los siguientes datos obtener 4 diferencias
Xk=1905, 1968, 1974, 1995
x | f(x) |
1900 | 13.6 |
1910 | 15.2 |
1920 | 14.3 |1930 | 16.6 |
1940 | 19.6 |
1950 | 25.8 |
1960 | 34.9 |
1970 | 48.2 |
1980 | 66.8 |
1990 | 81.2 |
2000 | 97.5 |
Xk=1905
x | f(x) |   |   |   |   |   |   |
1900 | 13.6 |   |   |   |   |   |   |
  |   | 0.16 |   |   |   |   |   |
1910 | 15.2 |   | -0.0125 |   |   |   |   |
  |   | -0.09 |   | 0.0009500 |   |   |   |
1920 | 14.3 |   | 0.0160 |   | -0.0000341667 |   |   |  |   | 0.23 |   | -0.0004167 |   | 0.00000110 |   |
1930 | 16.6 |   | 0.0035 |   | 0.0000208333 |   | -0.0000000292 |
  |   | 0.3 |   | 0.0004167 |   | -0.00000065 |   |
1940 | 19.6 |   | 0.0160 |   | -0.0000116667 |   | 0.0000000169 |
  |   | 0.62 |   | -0.0000500 |   | 0.00000037 |   |
1950 | 25.8 |   | 0.0145 |   | 0.0000066667 |   | -0.0000000086 |
  |   | 0.91 |   | 0.0002167 |  | -0.00000015 |   |
1960 | 34.9 |   | 0.0210 |   | -0.0000008333 |   | -0.0000000119 |
  |   | 1.33 |   | 0.0001833 |   | -0.00000087 |   |
1970 | 48.2 |   | 0.0265 |   | -0.0000441667 |   | 0.0000000508 |
  |   | 1.86 |   | -0.0015833 |   | 0.00000218 |   |
1980 | 66.8 |   | -0.0210 |   | 0.0000650000 |   |   |
  |   | 1.44 |   | 0.0010167 |   |   |   |
1990 | 81.2 |   | 0.0095 |  |   |   |   |
  |   | 1.63 |   |   |   |   |   |
2000 | 97.5 |   |   |   |   |   |   |
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0.000000000659 |   |   |   |
  | -0.0000000000128 |   |   |
-0.000000000365 |   | 0.000000000000186 |   |
  | 0.0000000000040 |   | -0.000000000000001 |...