Consider the following trust$-$region algorithm: Specify some $x_0$ as an

initial guess. Let the constants ${}_1,{}_{2}in(0,1)$ are given. Typical values

are ${}_1=\frac{1}{4},{}_2=\frac{3}{4}.$

For $k=0,1,$dots

If $x_k$ is optimal, then stop.

Compute

${}_k=\frac{f(x_k)-f(x_k+p_k)}{f(x_k)-{}_k(p_k)}$

where

${}_k(p_k)=f(x_k)+$gradf$(x_k{)}^T{p}_k+\frac{1}{2}{p}_k^{T}gra{d}^{2}f(x_k)p_k$

with $\{$:$p_k=-(gra{d}^{2}f(x_k)+I{)}^{-1}$gradf$(x_k)).$

if ${}_k<msub><mrow><mi></mi></mrow><mrow><mn>1</mn></mrow></msub>$ then the step is failed: $x_{k+1}$:$=x_k,$:$=2.$

if then the step is predicted: $x_{k+1}$:$=x_k+p_k,$:$=.$

if ${}_k>{}_2$ then the step is very good: $x_{k+1}$:$=x_k+p_k,$:$=\frac{1}{2}.$

Compute the trust$-$region radius ${}_k=||p_k()||.$

To minimize the function $f[x_1,x_2)=e^{x_1+x_2-2}+(x_1-x_2{)}^2$

$($a$)$ Let $x_0=(1,1{)}^T.$ Apply the full Newton step to give $x_1.$

$($b$)$ Let $x_0=(1,1{)}^T.$ Calculate the trust$-$region search direction with

initial value $=1.$ Would you accept this step in the trust region

algorithm above or $$ should be changed?