Question

Calculo Numérico: Para resolver os exercícios a seguir, utilize o arredondamento de duas casas decimais para cada cálculo. Dada a Tabela 3.10, Determine o valor aproximado de f(11) por meio de interpolação polinomial.

Tabela 3.10 - Pares Ordenados f(x)
--x-----6-----8-----12
f(x)----2-----10----12

Answer 1

Temos três pontos, portanto o grau do polinômio interpolador será $n=2.$

$\begin{array}{ccc} x_{0}=6&\Rightarrow&f(x_{0})=2\\\\ x_{1}=8&\Rightarrow&f(x_{1})=10\\\\ x_{2}=12&\Rightarrow&f(x_{2})=12 \end{array}$


Vamos encontrar o polinômio interpolador pela forma de Lagrange:

$p_{2}(x)=f(x_{0})\,L_{0}(x)+f(x_{1})\,L_{1}(x)+f(x_{2})\,L_{2}(x)~~~~~~\mathbf{(i)}$


onde cada polinômio $L_{k}(x)$ é dado por

$L_{k}(x)=\displaystyle\prod_{i=0,\;i\ne k}^{n}{\dfrac{(x-x_{i})}{(x_{k}-x_{i})}}$


com $k=0,\;1,\;2.$

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Encontrando os polinômios de Lagrange em cada ponto:

$\bullet~~k=0:$

$L_{0}(x)=\dfrac{(x-x_{1})(x-x_{2})}{(x_{0}-x_{1})(x_{0}-x_{2})}\\\\\\ L_{0}(x)=\dfrac{(x-8)(x-12)}{(6-8)(6-12)}\\\\\\ L_{0}(x)=\dfrac{(x-8)(x-12)}{(-2)\cdot (-6)}\\\\\\ \boxed{\begin{array}{c} L_{0}(x)=\dfrac{(x-8)(x-12)}{12} \end{array}}$


$\bullet~~k=1:$

$L_{1}(x)=\dfrac{(x-x_{0})(x-x_{2})}{(x_{1}-x_{0})(x_{1}-x_{2})}\\\\\\ L_{1}(x)=\dfrac{(x-6)(x-12)}{(8-6)(8-12)}\\\\\\ L_{1}(x)=\dfrac{(x-6)(x-12)}{(2)\cdot (-4)}\\\\\\ \boxed{\begin{array}{c} L_{1}(x)=\dfrac{(x-6)(x-12)}{-8} \end{array}}$


$\bullet~~k=2:$

$L_{2}(x)=\dfrac{(x-x_{0})(x-x_{1})}{(x_{2}-x_{0})(x_{2}-x_{1})}\\\\\\ L_{2}(x)=\dfrac{(x-6)(x-8)}{(12-6)(12-8)}\\\\\\ L_{2}(x)=\dfrac{(x-6)(x-8)}{(6)\cdot (4)}\\\\\\ \boxed{\begin{array}{c} L_{2}(x)=\dfrac{(x-6)(x-8)}{24} \end{array}}$

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O polinômio interpolador na forma de Lagrange é

$p_{2}(x)=f(x_{0})\,L_{0}(x)+f(x_{1})\,L_{1}(x)+f(x_{2})\,L_{2}(x)\\\\\\ p_{2}(x)=f(6)\,L_{0}(x)+f(8)\,L_{1}(x)+f(12)\,L_{2}(x)\\\\\\ \boxed{\begin{array}{c} p_{2}(x)=2\cdot\dfrac{(x-8)(x-12)}{12}+10\cdot \dfrac{(x-6)(x-12)}{-8}+12\cdot \dfrac{(x-6)(x-8)}{24} \end{array}}$


Portanto.

$f(11)\approx p_{2}(11)\\\\\\ f(11)\approx 2\cdot\dfrac{(11-8)(11-12)}{12}+10\cdot \dfrac{(11-6)(11-12)}{-8}+12\cdot \dfrac{(11-6)(11-8)}{24}\\\\\\ f(11)\approx 2\cdot\dfrac{(3)\cdot (-1)}{12}+10\cdot \dfrac{(5)\cdot (-1)}{-8}+12\cdot \dfrac{(5)\cdot (3)}{24}\\\\\\ f(11)\approx 2\cdot\dfrac{-3}{12}+10\cdot \dfrac{-5}{-8}+12\cdot \dfrac{15}{24}\\\\\\ f(11)\approx -\,\dfrac{1}{2}+\dfrac{25}{4}+\dfrac{15}{2}\\\\\\ f(11)\approx -\,\dfrac{2}{4}+\dfrac{25}{4}+\dfrac{30}{4}$

$f(11)\approx \dfrac{-2+25+30}{4}\\\\\\ f(11)\approx \dfrac{53}{4}\\\\\\ \boxed{\begin{array}{c}f(11)\approx 13,25 \end{array}}$