function NewtonMethod$()$

$\%$ Initial point

x$0=[10$; $10$; $10]$;

$\%$ Function handle for the objective function

f $=$ @$($x$)100*$x$(1)^2+$ x$(2)^2+0\mathrm{.}1*$x$(3)^2+$ x$(1)*$x$(2)+0\mathrm{.}1*$x$(2)*$x$(3)-6*$x$(1)+4*$x$(2)+6*$x$(3)+15$;

$\%$ Gradient of the objective function

grad$\_$f $=$ @$($x$)[200*$x$(1)+$ x$(2)-6$;

$2*$x$(2)+$ x$(1)+0\mathrm{.}1*$x$(3)+4$;

$0\mathrm{.}2*$x$(3)+0\mathrm{.}1*$x$(2)+6]$;

$\%$ Hessian of the objective function

hess$\_$f $=$ @$($x$)[200,1,0$ ;

$1,2,0\mathrm{.}1$ ;

$0,0\mathrm{.}1,0\mathrm{.}2]$;

$\%$ newton method

$[$x$\_$opt, f$\_$opt, grad$\_$opt, hess$\_$opt$]=$ Newton$\_$Method$($f$,$ grad$\_$f$,$ hess$\_$f$,$ x$0)$;

$\%$ Display results

disp$(\text{'}$Approximated Optimal point:$\text{'})$;

disp$($x$\_$opt$)$;

disp$(\text{'}$Approximated Function value at optimal point:$\text{'})$;

disp$($f$\_$opt$)$;

disp$(\text{'}$Approximated Gradient at optimal point:$\text{'})$;

disp$($grad$\_$opt$)$;

disp$(\text{'}$Approximated Hessian at optimal point:$\text{'})$;

disp$($hess$\_$opt$)$;

end

function $[$x$\_$opt, f$\_$opt, grad$\_$opt, hess$\_$opt$]=$ Newton$\_$Method$($f$,$ grad$\_$f$,$hess$\_$f$,$ x$0)$

$\%$ Parameters

tol $=1$e$-6$;

max$\_$iter $=1000$;

$\%$ Initialization

x $=$ x$0$;

g $=$ grad$\_$f$($x$)$;

h $=$ hess$\_$f$($x$)$;

d $=-$g;

k $=0$;

while norm$($g$)>$ tol && k $<$ max$\_$iter

$\%$ Line search to find optimal step size alpha

alpha $=$ fminbnd$($@$($a$)$ f$($x $+$ a$*$d$),0,1)$;

$\%$ Update the point

x $=$ x $+$ alpha $*$ d;

$\%$ Compute new gradient

g$\_$new $=$ grad$\_$f$($x$)$;

$\%$ Compute new Hessian

h$\_$new $=$ hess$\_$f$($x$)$;

$\%$ Update the direction

d $=-$h $\backslash$ g$\_$new;

$\%$ Update the gradient

g $=$ g$\_$new;

$\%$ Increment iteration counter

k $=$ k $+1$;

end

$\%$ Optimal values

x$\_$opt $=$ x;

f$\_$opt $=$ f$($x$\_$opt$)$;

grad$\_$opt $=$ g;

hess$\_$opt $=$ h;

end

Question : does my code correctly approximates the optimal solution, graident and hessian using newton method