Codex @codex  0

To find a critical point, apply multivariable Newton iteration to the gradient, rather than to the scalar value $F$. The gradient and Hessian matrix areWrite the correction as ${\mathbf{s}}_0={\mathbf{x}}_1-{\mathbf{x}}_0$. At ${\mathbf{x}}_0=(-1,-1)$, Newton's method in optimization requiresSubtracting the second equation from the first gives $10s_y=7$, so $s_y=7/10$ and $s_x=13/10$. HenceThe gradient there isA critical point has not been found after one iteration.

At the alternative starting point $(0,2/3)$, the Hessian matrix and gradient areThe correction would need to satisfy $2s_x+2s_y=-4/3$ and $2s_x+2s_y=0$ simultaneously. This linear system is inconsistent. The standard iteration cannot take even its first step: the Hessian matrix is a singular matrix, and the required correction does not exist. This is a Newton correction compatibility condition failure, not merely an ambiguity between several corrections.

If two solutions exist, their difference $w$ satisfies ${\nabla }^{2}w-w=0$ inside $B$ and $w=0$ on the boundary. Suppose $w$ has a positive value. Continuity on the closed ball gives a positive maximum at an interior point, where the Hessian matrix is negative semidefinite and hence ${\nabla }^{2}w\le 0$. But the equation gives ${\nabla }^{2}w=w>0$, a contradiction. A negative minimum similarly contradicts ${\nabla }^{2}w=w<0$.

Thus $\boxed{w=0}$, proving uniqueness by the maximum principle for the modified Helmholtz equation. Here a classical solution is understood to be $C^2$ inside and continuous up to the boundary; mere first differentiability would not define the displayed Laplacian classically. With enough boundary regularity, the alternative Green's first identity proof is $0={\int }_{B}w({\nabla }^{2}w-w)=-{\int }_B(|\nabla w{|}^2+w^2)$.

For a twice continuously differentiable function, an interior local extremum requires $f_x=f_y=0$. At such a critical point, put $D=f_{xx}{f}_{yy}-f_{xy}^2$. The second-derivative test gives a strict local minimum if $D>0$ and $f_{xx}>0$, a strict local maximum if $D>0$ and $f_{xx}<0$, and a saddle point if $D<0$. When $D=0$, the test is inconclusive. More generally the Hessian matrix is necessarily positive semidefinite at a local minimum and negative semidefinite at a local maximum, but these non-strict signs alone are not sufficient.

For this polynomial,Subtracting the two stationarity equations gives $y(2x-y)=0$. If $y=0$, then $x=0$; if $y=2x$, the remaining equation gives $x=0$ or $x=3/2$. Thus the only critical points are $(0,0)$ and $(3/2,3)$.

At $(3/2,3)$, $\det H=-18<0$, so it is a saddle point. At $(0,0)$ the determinant is zero, but along $y=-x$ the value is $-x^3$, which takes both signs arbitrarily near zero. ThereforeIn particular a vanishing Hessian determinant does not exclude a saddle point.

The gradient and constant Hessian matrix areAt ${\mathbf{x}}_0=(1,0{)}^T$, $\nabla F=(10,-8{)}^T$ and $H_F^{-1}\nabla F=(1,-1{)}^T$. Hence one multivariable Newton iteration givesAt this point the gradient vanishes, so every subsequent iterate is $(0,1)$. The one-step result follows because the gradient of a quadratic polynomial is affine and its linearization is exact. Since the leading principal minors of $H_F$ are $6>0$ and $8>0$, it is a positive-definite matrix. Alternatively,directly proves that $(0,1)$ is the unique global minimum, with value $-1$.

For a twice continuously differentiable function $F$, its gradient and Hessian matrix areA critical point solves $\nabla F=0$. The Jacobian matrix of this vector equation is $H_F$, so the multivariable Newton iteration isEquivalently ${\mathbf{x}}_{k+1}={\mathbf{x}}_k-H_F({\mathbf{x}}_k{)}^{-1}\nabla F({\mathbf{x}}_k)$ when the Hessian matrix is invertible. Solving the linear system is the direct computational form. A critical point found in this way need not be a local minimum; its type requires a separate check.

Write $A=a^2>0$ and $E=e^{-(x^2+y^2)/A}>0$. The partial derivatives factor asA critical point has both partial derivatives zero. If $x,y\ne 0$, this would require $x^2-2y^2=A$ and $x^2-2y^2=-2A$, which is impossible. On the axes the remaining conditions giveAt either critical point with $x\ne 0$, the Hessian matrix isIt is negative definite, so the second-derivative test shows that both $(\pm |a|,0)$ are strict local maxima, of value $a^2/e$. They also attain the global maximum: $f\le x^2{e}^{-x^2/A}\le A/e$. For completeness, the Hessian matrix at $(0,\pm |a|)$ is $\mathrm{diag}(6/e,8/e)$, giving strict local minima, while at the origin it is $\mathrm{diag}(2,-4)$, giving a saddle point.

Differentiate the ordinary least squares objective in a. Its normal equations for linear least squares areUsing ${\sum }_i{x}_i=0$ and $S_{xx}={\sum }_i{x}_i^2>0$ givesThe Hessian matrix of the objective in $(a,b)$ is $2\mathrm{diag}(n,S_{xx})$, which is positive-definite, so this is its unique global minimum. For the observed estimates, replace $Y_i$ by $y_i$.

Near a nondegenerate stationary point, the quadratic part of the Taylor expansion determines the local contour plot. For a strict local maximum, the model $f=f_0-X^2-Y^2$ gives nested closed level sets, with values decreasing as one moves outward. For a strict local minimum, the model $f=f_0+X^2+Y^2$ gives the same nested shape with values increasing outward. Unequal nonzero quadratic coefficients stretch the circles into ellipses after a rotation to the Hessian matrix's eigenvectors.

For a nondegenerate saddle point of a scalar function, the model $f=f_0+X^2-Y^2$ gives hyperbolas. Levels above $f_0$ open along the positive-curvature direction; levels below $f_0$ open along the negative-curvature direction. The critical level set is the two crossing lines $Y=\pm X$. Degenerate stationary points can have different local geometry, so these are the standard nondegenerate sketches.

Set both first partial derivatives equal to zero:Thus $x=0,\pm 1/\sqrt{2}$ and $y=0,\pm \sqrt{2}$, independently, giving exactly nine stationary points. The Hessian matrix is diagonal everywhere:Its diagonal entries are its eigenvalues, so the second-derivative test immediately classifies the points:

In particular,All nine stationary points are nondegenerate, so no higher-order test is needed. Neither type of local extremum is global: $f(x,0)=x^4-x^2\to +\infty$ as $|x|\to \infty$, whereas $f(0,y)=-y^4+4y^2\to -\infty$ as $|y|\to \infty$.

To organize the contour plot, complete the squares:The level set through all four outer saddle points of a scalar function factors exactly:It is therefore the union of the circle $x^2+y^2=5/2$ and the hyperbola $y^2-x^2=3/2$, which cross at those four saddle points of a scalar function. Near the origin, $f=-x^2+4y^2+O((x^2+y^2{)}^2)$, so the zero contours have tangents $y=\pm x/2$. These two exact structural features, together with the nested contours around the four local extrema, determine the requested sketch. They are an example of a separable quartic landscape with nine stationary points.

Write $c=\cos \theta$, $s=\sin \theta$. Since $\theta$ is constant, the multivariable chain rule gives ${\partial }_x=c{\partial }_u-s{\partial }_v$ and ${\partial }_y=s{\partial }_u+c{\partial }_v$. Applying these operators twice to a twice continuously differentiable function givesThe mixed terms cancel and $c^2+s^2=1$, soThis is rotational invariance of the Laplacian: an orthogonal change of coordinates preserves the trace of the Hessian matrix. A nonconstant angle would introduce additional derivatives and is not covered by this constant rotation calculation.

The stationary point equations are $x^2+2y-21/4=0$ and $y^2+2x-21/4=0$. Subtracting them factors as $(x-y)(x+y-2)=0$, so all solutions lie either on $x=y$ or on $x+y=2$. The first case gives $x=y=3/2$ or $x=y=-7/2$; the second gives $(x,y)=(-1/2,5/2)$ or $(5/2,-1/2)$. This also verifies the two stated points and exhausts the possible solutions.

The Hessian matrix is $H=\left(\begin{array}{cc}2x& 2\\ 2& 2y\end{array}\right)$, with determinant $4(xy-1)$. The second-derivative test gives:

These extrema are local. Along $x=y=t$ the cubic term makes the function unbounded above and below, so it has no global maximum or minimum on the whole plane.

At $(t,t)$ with $t^2=1/2$, differentiating the gradient yields the Hessian matrix $H=-tI$. At the positive point it is negative definite, and at the negative point it is positive-definite; in both cases $f_{xy}=0$ and $\det H=1/2$. Therefore the positive point is a maximum, the negative point is a minimum, and there are no saddle points.

They are global extrema: with $r=\sqrt{x^2+y^2}$, the Cauchy-Schwarz inequality gives $|f|\le \sqrt{2}r/(1+r^2)\le 1/\sqrt{2}$, with equality at exactly those two points. This is the extremum bound underlying the level sets of a linear numerator over a radial quadratic.

Put $S={\theta }_1+{\theta }_2$ and $D={\theta }_1-{\theta }_2$. The gradient has components $3\sin S+\sin D$ and $3\sin S-\sin D$, so a stationary point requires $\sin S=\sin D=0$. In the closed square, $S\in [0,2\pi ]$ and $D\in [-\pi ,\pi ]$, and the compatible possibilities are precisely the four corners and $(\pi /2,\pi /2)$.

The Hessian matrix isIts eigenvectors $(1,1)$ and $(1,-1)$ have eigenvalues $6\cos S$ and $2\cos D$. The second-derivative test gives the following classification:

The corner classifications also hold for the smooth extension beyond the square, and hence for the restricted region. The global bounds $-4\le U\le 4$ show that the stated minima and maxima are global within it. The central point has opposite curvature along the two diagonal directions.

Inside the triangle, both $x$ and $y$ are nonzero. Thus $\nabla f=0$ reduces to $1-2x+y=0$ and $1-x+2y=0$, whose unique solution is $(1/3,-1/3)$. At that pointThe Hessian matrix has positive eigenvalues $1$ and $1/3$, so this is a strict local minimum. It is also the minimum on the closed triangle: $f=0$ on the boundary, $f<0$ inside, and there is no other interior stationary point. Therefore

The gradient is $(-y^2-2x-3x^2,2y(1-x))$. Its second component vanishes only if $y=0$ or $x=1$. The latter gives the impossible first-component equation $-y^2-5=0$. With $y=0$, the first component gives $x=0$ or $x=-2/3$. The Hessian matrix isAt $(0,0)$ it has eigenvalues $-2,2$, giving a saddle point. At $(-2/3,0)$ its eigenvalues are $2,10/3$, giving a strict local minimum, of value $-4/27$. These are all real stationary points. The minimum is local: for example $g(x,0)\to -\infty$ as $x\to +\infty$.

At $(X,y)=(0,0)$ the Hessian matrix is $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, with eigenvalues $1$ and $-1$, so $(x,y)=(1,0)$ is a saddle point. At the other four points the mixed second derivative vanishes and the two diagonal entries both equal $-2Xy/e$. HenceThe four extrema are global: $|z|=(|X|e^{-X^2/2})(|y|e^{-y^2/2})\le e^{-1}$ because each factor has maximum $e^{-1/2}$ at absolute argument one.

As printed, the function is undefined at the origin. For $r^2=x^2+y^2>0$, its gradient isThe scalar prefactor is positive, and $(x,y)\ne (0,0)$, so this gradient never vanishes. Thus there are no stationary points on the stated domain. Its infimum is three, approached as $r\to 0$, but that value is not attained there.

If the intended definition includes the natural extension $f(0,0)=3$, that extension has a unique strict global minimum at the origin, with value three. Indeed, $e^{-1/r^2}>0$ off zero, and $e^{-1/r^2}/r\to 0$ proves differentiability at zero with zero gradient. More strongly, subtracting three gives the radial Gaussian flat function: every derivative of that difference vanishes at the origin. The vanishing Hessian matrix does not turn this minimum into an inflection point. These are two different domain interpretations; the extension must be stated rather than silently inserted.

Write $h(x,y)=y(r-4)$, where $r=\sqrt{x^2+y^2}$. Because this hyperbolic function has strictly positive derivative $\cosh$, stationary points under an increasing outer function have the same positions and types for $g=\sinh h$ and $h$. For $r>0$,The first equation requires $x=0$ or $y=0$. If $y=0$, the second gives $|x|=4$. If $x=0$, it gives $2|y|=4$. These are all the nonzero stationary points. At the origin, $h=-4y+O(r^2)$, so its gradient is $(0,-4)$; the origin is not stationary.

At $(0,2)$ the Hessian matrix of $h$ is $\mathrm{diag}(1,2)$, and at $(0,-2)$ it is $\mathrm{diag}(-1,-2)$. At $(4,0)$ and $(-4,0)$ its Hessian matrices are respectively $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ and $\left(\begin{array}{cc}0& -1\\ -1& 0\end{array}\right)$. ThereforeThe extrema are local: $g$ is unbounded above and below on the full plane.

For the level curve geometry, a level $g=c$ is $y(r-4)=\mathrm{arsinh}c$. In particular, the zero level curve is the union of the entire $x$ axis and the circle $r=4$, crossing at the two saddle points. Inside the circle, $g<0$ above the axis and $g>0$ below it; outside, these signs reverse. Levels between $-\sinh 4$ and zero have closed loops around the upper minimum as well as unbounded lower-half-plane components. Reflection across the $x$ axis reverses the sign and gives the loops about the lower maximum. At the two extremal heights, the corresponding closed loop shrinks to its extremum. These features determine the sketch.

Since $f>0$, a level $f=c$ can occur only for $c>0$. Taking its natural logarithm gives $xy=\ln c$. Thus $c>1$ produces rectangular hyperbolas in the first and third quadrants, $0<c<1$ produces them in the second and fourth quadrants, and $c=1$ consists of both coordinate axes.

The gradient $e^{xy}(y,x)$ vanishes only at $(0,0)$, where $f=1$. The Hessian matrix there is $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, with negative determinant. Equivalently, $f(x,x)>1$ while $f(x,-x)<1$ for nonzero small $x$. Hence the only stationary point is a saddle point at $(0,0,1)$ on the graph; its location in the contour plane is the origin.

At the origin the Hessian matrix isThus the second-derivative test gives a saddle point. Its opposite signs can also be checked without a test formula: $g(x,0)=x^2{e}^{-x^2/2}>0$ for $x\ne 0$, whereas $g(0,y)=-y^2{e}^{-y^2/2}<0$ for $y\ne 0$. Both types of values occur arbitrarily close to the zero value at the origin, excluding either a local maximum or a local minimum.