8.4 Fórmulas de diferenças finitas

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Veremos nessa seção uma outra maneira de obter fórmulas de diferenças finitas para derivadas de qualquer ordem de tal forma que elas possuam alta ordem de precisão.

Dados $n+1$ pontos $\left\{x_1,x_2,\dots ,x_n\right\}$, queremos obter uma aproximação para a derivada de $f\left(x\right)$ calculada em $x^{\ast }$ do tipo $$\begin{array}{rcll}{f}^{\prime }\left(x^{\ast }\right)\approx c_{1}f\left(x_1\right)+c_{2}f\left(x_2\right)+\dots +c_{n}f\left(x_n\right)& & & \text{(8.62)}\end{array}$$

que seja exata para polinômios até ordem $n-1$.

Seja $q\left(x\right)=c_1{\varphi }_1\left(x\right)+c_2{\varphi }_2\left(x\right)+\dots +c_n{\varphi }_n\left(x\right)$ o polinômio de ordem $n$ que aproxima $f\left(x\right)$. Fixe a base ${\varphi }_k\left(x\right)=x^k$. Como a regra (8.62) deve ser exata para qualquer $q\left(x\right)$ até ordem $n-1$, então também deve ser exata para qualquer função da base. Substituindo $f\left(x\right)$ por ${\varphi }_1\left(x\right)=1$ em (8.62) obtemos $$\begin{array}{rcll}{\varphi }_1^{\prime }\left(x\right){|}_{x^{\ast }}& =& {\left(1\right)}^{\prime }{|}_{x^{\ast }}=& \text{(8.63)}\\ 0& =& c_1{\varphi }_1\left(x_1\right)+c_2{\varphi }_1\left(x_2\right)+\dots +c_n{\varphi }_1\left(x_n\right)& \text{(8.64)}\\ 0& =& c_1+c_2+\dots +c_n& \text{(8.65)}\end{array}$$

Da mesma forma para $k=1,\dots ,n-1$, obtemos $$\begin{array}{rcll}{\left(x\right)}_{x^{\ast }}^{\prime }=1& =& c_1{x}_1+c_2{x}_2+\dots +c_n{x}_n& \text{(8.66)}\\ {\left(x^2\right)}_{x^{\ast }}^{\prime }=2x^{\ast }& =& c_1{x}_1^2+c_2{x}_2^2+\dots +c_n{x}_n^2& \text{(8.67)}\\ {\left(x^3\right)}_{x^{\ast }}^{\prime }=3{\left(x^{\ast }\right)}^2& =& c_1{x}_1^3+c_2{x}_2^3+\dots +c_n{x}_n^3& \text{(8.68)}\\ ⋮& =& ⋮& \text{(8.69)}\\ {\left(x^{n-1}\right)}_{x^{\ast }}^{\prime }=\left(n-1\right){\left(x^{\ast }\right)}^{n-2}& =& c_1{x}_1^{n-1}+c_1{x}_1^{n-1}+\dots +c_n{x}_n^{n-1}& \text{(8.70)}\end{array}$$

que pode ser escrito na forma matricial $$\begin{array}{rcll}\left[\begin{array}{cccc}\hfill 1\hfill & \hfill 1\hfill & \hfill \dots \hfill & \hfill 1\hfill \\ \hfill x_1\hfill & \hfill x_2\hfill & \hfill \dots \hfill & \hfill x_n\hfill \\ \hfill x_1^2\hfill & \hfill x_2^2\hfill & \hfill \dots \hfill & \hfill x_n^2\hfill \\ \hfill ⋮\hfill & \hfill ⋮\hfill & \hfill \hfill & \hfill ⋮\hfill \\ \hfill x_1^{n-1}\hfill & \hfill x_2^{n-1}\hfill & \hfill \dots \hfill & \hfill x_n^{n-1}\hfill \end{array}\right]\left[\begin{array}{c}\hfill c_1\hfill \\ \hfill c_2\hfill \\ \hfill c_3\hfill \\ \hfill ⋮\hfill \\ \hfill c_n\hfill \end{array}\right]=\left[\begin{array}{c}\hfill 0\hfill \\ \hfill 1\hfill \\ \hfill 2x^{\ast }\hfill \\ \hfill ⋮\hfill \\ \hfill \left(n-1\right){\left(x^{\ast }\right)}^{n-2}\hfill \end{array}\right]& & & \text{(8.71)}\end{array}$$

Resolvendo o sistema, obtemos os coeficientes $c_k$ para a regra de diferenciação.

Solução. A regra terá a forma

$$f^{\prime }\left(x^{\ast }\right)\approx c_{1}f\left(x_1\right)+c_{2}f\left(x_2\right)+c_{3}f\left(x_3\right)$$

(8.72)

Considere a base polinomial $\left\{{\varphi }_1\left(x\right),{\varphi }_2\left(x\right),{\varphi }_3\left(x\right)\right\}=\left\{1,x,x^2\right\}$ e substitua $f\left(x\right)$ por ${\varphi }_k\left(x\right)$ obtendo $$\begin{array}{rcll}{\left(1\right)}_{x=0}^{\prime }=0& =c_1\left(1\right)+c_2\left(1\right)+c_3\left(1\right)& & \text{(8.73)}\\ {\left(x\right)}_{x=0}^{\prime }=1& =c_1\left(-h\right)+c_2\left(0\right)+c_3\left(h\right)& & \text{(8.74)}\\ {\left(x^2\right)}_{x=0}^{\prime }=0& =c_1{\left(-h\right)}^2+c_2{\left(0\right)}^2+c_3{\left(h\right)}^2& & \text{(8.75)}\end{array}$$

que pode ser escrito na forma matricial $$\begin{array}{rcll}\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill -h\hfill & \hfill 0\hfill & \hfill h\hfill \\ \hfill h^2\hfill & \hfill 0\hfill & \hfill h^2\hfill \end{array}\right]\left[\begin{array}{c}\hfill c_1\hfill \\ \hfill c_2\hfill \\ \hfill c_3\hfill \end{array}\right]=\left[\begin{array}{c}\hfill 0\hfill \\ \hfill 1\hfill \\ \hfill 0\hfill \end{array}\right]& & & \text{(8.76)}\end{array}$$

Resolvendo o sistema, obtemos $\left\{c_1,c_2,c_3\right\}=\left\{-\frac{1}{2h},0,\frac{1}{2h}\right\}$ fornecendo a regra $$\begin{array}{rcll}{f}^{\prime }{|}_{x=x_1}& \approx & -\frac{1}{2h}f\left(x_1\right)+\frac{1}{2h}f\left(x_3\right)& \text{(8.77)}\\ & \approx & \frac{f\left(x_3\right)-f\left(x_1\right)}{2h}& \text{(8.78)}\end{array}$$

Exercícios resolvidos

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Exercícios

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