Exercise 6.5.4

We have $\nabla <!--\; FUNCTION\; APPLICATION\; -->F=\left[\begin{array}{c}\hfill -2(y-x)\hfill \\ \hfill 4y-2x\hfill \end{array}\right]$, at $(x_0,y_0)=(1,1)$, we have $\nabla <!--\; FUNCTION\; APPLICATION\; -->F=\left[\begin{array}{c}\hfill 0\hfill \\ \hfill 2\hfill \end{array}\right]$

Then we have $\left[\begin{array}{c}\hfill x_1\hfill \\ \hfill y_1\hfill \end{array}\right]=\left[\begin{array}{c}\hfill x_0\hfill \\ \hfill y_0\hfill \end{array}\right]-s_1\nabla <!--\; FUNCTION\; APPLICATION\; -->F(x_0,y_0)=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 1\hfill \end{array}\right]-\frac{1}{2}\left[\begin{array}{c}\hfill 0\hfill \\ \hfill 2\hfill \end{array}\right]=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 0\hfill \end{array}\right]$