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

Passo 1

Resumindo, queremos aplicar o método de Broyden ao sistema
\begin{cases} f_1(x_1,x_2)=\ln\bigl(x_1^2+x_2^2\bigr)-\sin(x_1x_2)-\ln2-\ln\pi=0,\\\ f_2(x_1,x_2)=e^{\,x_1 - x_2}+\cos(x_1x_2)=0, \end{cases}
com chute inicial \mathbf x^{(0)}=(2,2)^T.
Para isso vamos usar a fórmula do Broyden “bom” (first method):
\mathbf x^{(k+1)}=\mathbf x^{(k)}-\bigl[B_k\bigr]^{-1}f\bigl(\mathbf x^{(k)}\bigr),\quad B_{k+1}=B_k+\frac{\bigl[y_k - B_k s_k\bigr]\,s_k^T}{s_k^T s_k},
onde s_k=\mathbf x^{(k+1)}-\mathbf x^{(k)} e y_k=f(\mathbf x^{(k+1)})-f(\mathbf x^{(k)}).
Vamos pintar dois passos: de k=0 a k=1 e de k=1 a k=2.

Passo 2

Cálculo de f(\mathbf x^{(0)}) e da Jacobiana exata J(\mathbf x^{(0)}). \mathbf x^{(0)}=(2,2)\implies x_1^2+x_2^2=8,\;x_1x_2=4.
f(\mathbf x^{(0)})=\begin{pmatrix} \ln8-\sin4-\ln2-\ln\pi\\[6pt] e^{2-2}+\cos4 \end{pmatrix} \approx\begin{pmatrix}0.99836\\\ 0.34636\end{pmatrix}.
As derivadas parciais em \mathbf x^{(0)} são
\frac{\partial f_1}{\partial x_1}=\frac{2x_1}{x_1^2+x_2^2}-x_2\cos(x_1x_2),\quad \frac{\partial f_1}{\partial x_2}=\frac{2x_2}{x_1^2+x_2^2}-x_1\cos(x_1x_2),
\frac{\partial f_2}{\partial x_1}=e^{x_1-x_2}-x_2\sin(x_1x_2),\quad \frac{\partial f_2}{\partial x_2}=-e^{x_1-x_2}-x_1\sin(x_1x_2),
avaliando em (2,2)
J(\mathbf x^{(0)})\approx \begin{pmatrix} 1.80728 & 1.80728\\\\ 2.51360 & 0.51360 \end{pmatrix} \equiv B_0.]

Passo 3

Resolver B_0\,\delta_0=-f(\mathbf x^{(0)}).
Numericamente encontra-se \delta_0\approx(-0.03134,\;0.32605)^T.
Logo
\mathbf x^{(1)}=\mathbf x^{(0)}+\delta_0\approx(1.9687,\;2.3261)^T.

Passo 4

Cálculo de f(\mathbf x^{(1)}).
Para \mathbf x^{(1)}\approx(1.9687,2.3261) obtemos
f(\mathbf x^{(1)})\approx\begin{pmatrix}1.3813\\\ 0.5679\end{pmatrix}.
Definimos
s_0=\mathbf x^{(1)}-\mathbf x^{(0)}\approx\begin{pmatrix}-0.03134\\\ 0.32605\end{pmatrix},\quad y_0=f(\mathbf x^{(1)})-f(\mathbf x^{(0)})\approx\begin{pmatrix}0.38294\\\ 0.22154\end{pmatrix}.

Passo 5

Atualização de B_1 via Broyden:
B_1=B_0+\frac{(y_0-B_0\,s_0)\,s_0^T}{s_0^T s_0} \approx \begin{pmatrix} 1.85105 & 1.35143\\\\ 2.47481 & 0.91690 \end{pmatrix}.

Passo 6

Resolver B_1\,\delta_1=-f(\mathbf x^{(1)}).
Encontra-se \delta_1\approx(0.3027,\,-1.4370)^T.
Portanto
\mathbf x^{(2)}=\mathbf x^{(1)}+\delta_1 \approx(2.2714,\;0.8891)^T.

Resposta

\mathbf x^{(2)}\approx \begin{pmatrix} 2.2714\\\\ 0.8891 \end{pmatrix}.

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