determine the polynomial F of degree 2 that interpolates. f at points (0;1) (2;5) (4;6). calculate F(0.8). Note: Using the polynomial expression with difference operator.



Answer to a math question determine the polynomial F of degree 2 that interpolates. f at points (0;1) (2;5) (4;6). calculate F(0.8). Note: Using the polynomial expression with difference operator.

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Para determinar o polinômio F de grau 2 que interpola f nos pontos (0,1), (2,5) e (4,6), usaremos a expressão de polinômio com operador diferença.

Primeiro, vamos calcular as diferenças divididas de ordem 1:

\Delta y_0 = y_1 - y_0 = 5 - 1 = 4
\Delta y_1 = y_2 - y_1 = 6 - 5 = 1

Agora, calculamos as diferenças divididas de ordem 2:

\Delta^2 y_0 = \Delta y_1 - \Delta y_0 = 1 - 4 = -3

A partir dessas diferenças divididas, podemos construir o polinômio interpolador de Newton:

F(x) = y_0 + \Delta y_0 \cdot (x - x_0) + \Delta^2 y_0 \cdot (x - x_0) \cdot (x - x_1)

Substituindo os valores conhecidos:

F(x) = 1 + 4x - 3x^2

Agora, para calcular F(0,8), substituímos x por 0,8 na expressão de F(x):

F(0,8) = 1 + 4 \cdot 0,8 - 3 \cdot (0,8)^2


F(0,8) = 1 + 3,2 - 1,92

F(0,8) = 2,28

Portanto, o valor de F(0,8) é igual a 2,28.
Answer: F(0,8) = 2,28

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