Use o método de Broyden para calcular \mathbf{x}^{(2)} para cada um dos sistemas não lineares.
\operatorname{sen}\left(4 \pi x_{1} x_{2}\right)-2 x_{2}-x_{1}=0, \left(\frac{4 \pi-1}{4 \pi}\right)\left(e^{2 x_{1}}-e\right)+4 e x_{2}^{2}-2 e x_{1}=0. Use \mathbf{x}^{(0)}=(0,0)^{t}.

Passo 1

Temos o sistema
\begin{cases} f_{1}(x_{1},x_{2})=\sin\bigl(4\pi x_{1}x_{2}\bigr)-2x_{2}-x_{1}=0,\\\ f_{2}(x_{1},x_{2})=\frac{4\pi-1}{4\pi}\bigl(e^{2x_{1}}-e\bigr)+4e\,x_{2}^{2}-2e\,x_{1}=0, \end{cases}
e queremos aplicar o método de Broyden com \mathbf{x}^{(0)}=(0,0)^{t},
onde as fórmulas de atualização são
\mathbf{x}^{(n+1)}=\mathbf{x}^{(n)}-B_{n}^{-1}f\bigl(\mathbf{x}^{(n)}\bigr), \qquad B_{n+1}=B_{n}+\frac{\bigl[\Delta f_{n}-B_{n}\,\Delta x_{n}\bigr]\Delta x_{n}^{t}}{\Delta x_{n}^{t}\,\Delta x_{n}},
com \Delta x_{n}=\mathbf{x}^{(n+1)}-\mathbf{x}^{(n)} e \Delta f_{n}=f(\mathbf{x}^{(n+1)})-f(\mathbf{x}^{(n)}).

Passo 2

Calculemos B_{0}=Jf(\mathbf{x}^{(0)}). Chamando \displaystyle a=\frac{4\pi-1}{4\pi},
obtemos
Jf(x_{1},x_{2})= \begin{pmatrix} \partial_{x_{1}}f_{1}&\partial_{x_{2}}f_{1}\\\ \partial_{x_{1}}f_{2}&\partial_{x_{2}}f_{2} \end{pmatrix} = \begin{pmatrix} \cos(4\pi x_{1}x_{2})\cdot4\pi x_{2}-1 & \cos(4\pi x_{1}x_{2})\cdot4\pi x_{1}-2\\[6pt] 2a\,e^{2x_{1}}-2e & 8e\,x_{2} \end{pmatrix}.
Em \mathbf{x}^{(0)}=(0,0) isto vale
B_{0}= \begin{pmatrix} -1 & -2\\\\ 2-\tfrac1{2\pi}-2e & 0 \end{pmatrix}, \qquad f\bigl(\mathbf{x}^{(0)}\bigr)= \begin{pmatrix}0\\[3pt]a\,(1-e)\end{pmatrix}.
Resolvendo \mathbf{x}^{(1)}=\mathbf{x}^{(0)}-B_{0}^{-1}f(\mathbf{x}^{(0)})
encontra-se
\mathbf{x}^{(1)} \approx \begin{pmatrix}-0,44002\\[3pt]0,21993\end{pmatrix}.]

Passo 3

Primeiro avaliamos \displaystyle f\bigl(\mathbf{x}^{(1)}\bigr)\approx\begin{pmatrix}-0,93744\\[3pt]0,79890\end{pmatrix}.
Calculamos \displaystyle \Delta x_{0}=\mathbf{x}^{(1)}-\mathbf{x}^{(0)}
e \displaystyle \Delta f_{0}=f(\mathbf{x}^{(1)})-f(\mathbf{x}^{(0)}),
logo
\Delta x_{0}\approx \begin{pmatrix}-0,44002\\[3pt]0,21993\end{pmatrix}, \quad \Delta f_{0}\approx \begin{pmatrix}-0,93744\\[3pt]2,38094\end{pmatrix}.
O correção de Broyden é
B_{1}=B_{0} +\frac{\bigl[\Delta f_{0}-B_{0}\,\Delta x_{0}\bigr]\, (\Delta x_{0})^{t}} {(\Delta x_{0})^{t}\,\Delta x_{0}} \approx \begin{pmatrix} 0,7055 & -2,8522\\\\ -5,0483 & 0,7258 \end{pmatrix}.]

Passo 4

Para \mathbf{x}^{(2)} resolvemos
B_{1}\,d = f\bigl(\mathbf{x}^{(1)}\bigr), \qquad \mathbf{x}^{(2)}=\mathbf{x}^{(1)}-d.
Isso dá \displaystyle d\approx\begin{pmatrix}-0,11496\\[3pt]0,30030\end{pmatrix}
e portanto
\mathbf{x}^{(2)} \approx \begin{pmatrix} -0,44002\\[3pt]0,21993 \end{pmatrix} - \begin{pmatrix} -0,11496\\[3pt]0,30030 \end{pmatrix} = \begin{pmatrix} -0,32506\\[3pt]-0,08037 \end{pmatrix}.]

Resposta

\mathbf{x}^{(2)}\approx \begin{pmatrix} -0,3251\\[3pt]-0,0804 \end{pmatrix}.]