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Differentiate the polynomial and factor:The stationary points therefore have horizontal coordinatesFor a check, $y^{\prime \prime }=6x-4$ is negative at the first and positive at the second, giving a local maximum and a local minimum respectively.

The second derivative is $y^{\prime \prime }=6x$, so $y^{\prime \prime }(1)=6>0$. By the second-derivative test, the stationary point is a strict local minimum, at $(1,5)$. It is not a global minimum, since this cubic is unbounded below toward negative infinity.

The partial derivatives are $g_x=\sin x$ and $g_y=y$, so all stationary points are $(n\pi ,0)$ for integers $n$. Their Hessian matrix is $\mathrm{diag}(\cos x,1)$. At even multiples of $\pi$ it is a positive-definite matrix, whereas at odd multiples it has one positive and one negative eigenvalue. ConsequentlyThere are no local maxima. Since both $1-\cos x$ and $y^2/2$ are nonnegative, these are also global minima. In the requested open plotting window, the stationary points are $(0,0)$ and $(\pm \pi ,0)$; the local minima at $(\pm 2\pi ,0)$ lie on excluded edges.

For a level set $g=c$, solving for the vertical coordinate givesFor $0<c<2$ these level sets are closed ovals around each local minimum. At $c=0$ the level sets shrink to the local minima. The level set through the saddle points iswhere the two signed branches together describe the same set as $y=\pm 2|\cos (x/2)|$. They cross at $(\pm \pi ,0)$ and reach $(0,\pm 2)$. The crossing slopes at a saddle point are $\pm 1$, as also follows from the local quadratic form $(-\delta x^2+\delta y^2)/2$. For $c>2$, the upper and lower level sets remain separate and extend across the plotting window without meeting the horizontal axis.

An interior stationary point has vanishing gradient:For a differentiable function, this is the necessary first-order condition for an interior local minimum or local maximum.

For $a=-1$, the local minimum at the origin has $f=1$. Its full level set isThe connected component containing the minimum is just the isolated origin. The same level also has the outer branches $y=\pm \sqrt{x^4/2-x^2}$ for $|x|\ge \sqrt{2}$; these are different connected components, not curves approaching the origin.

Both saddle points $(\pm 1,0)$ have $f=e^{1/4}$. Their common level set factors asIts explicit equations are thereforeThe two parabolas cross at each saddle, explaining the change of contour connectivity there. Each displayed equation can be restricted to the square of part (ii) if only that sketch window is wanted.

Because the exponential function is strictly increasing, the level sets $f=L>0$ are exactly those of $G=\ln L$. For $a=1$ they satisfyThere are no contours for $L<1$, the level $L=1$ is the origin, and $L>1$ gives nested closed ovals symmetric about both axes. Their stronger fourth-power growth in $x$ makes them relatively narrower horizontally as the level rises.

For $a=-1$, the contour equation becomesThe central local minimum is surrounded by closed ovals for $1<L<e^{1/4}$, accompanied by outer open branches. At $L=e^{1/4}$, the level passes through the two saddle points $(\pm 1,0)$ and changes its connectivity. Levels $L<1$ have only outer branches, while levels above the saddle value have an upper and a lower branch. The sketch is restricted to the requested square; labels give $G=\ln f$, so they identify the corresponding contours of $f$ without exponentially large labels.

The function graph is a horizontally shifted and scaled fourth power. It is nonnegative and touches the horizontal axis at its fourfold polynomial root. The axis intersections areThe derivative is $8(1+2x{)}^3$, so the graph decreases to its flat local minimum at $x=-1/2$ and then increases. Its endpoint values on the requested interval are $f(-1)=1$ and $f(1)=81$. The original sketch includes a closer view of the minimum so that the small values are visible beside the much larger right endpoint.

Eliminate $z$ using the constraint: $z=x-x^2-y^2$. The restricted objective is the quadratic polynomial $F(x,y)=axy+x-x^2-y^2$, with constant Hessian matrixIts eigenvalues are $-2+a$ and $-2-a$. HenceFor $|a|<2$, both eigenvalues are negative and the quadratic polynomial is strictly concave, so the maximum is global on the entire constraint surface. For $|a|>2$, the opposite signs give increasing and decreasing directions. At $a=\pm 2$ there is no stationary point, as already found, rather than a degenerate extremum to classify. No parameter value produces a constrained local minimum.

The Hessian matrix isAt the origin it is the identity, so the origin is a strict local minimum. At $(0,1)$ it is $\mathrm{diag}(3,-1)$, so this point is a saddle point. At either remaining stationary point,which again gives a saddle point. All three saddle points have eigenvalues $3$ and $-1$; there are no local maxima.

The most useful guide to the contour plot is the common saddle value. Direct polynomial factorization givesThus the level set $f=1/6$ is exactly the three complete linesTheir intersections are the three saddle points, and they enclose the origin in an equilateral triangle. Near the local minimum, the level sets are nearly circles because the leading term is $(x^2+y^2)/2$. For $0<c<1/6$, a closed contour surrounds the origin, while three separate open branches lie outside the triangular barrier. At the saddle level these components meet at the saddle points; above it the central closed contour has opened into the three outer channels.

The identity in polar coordinatesshows the rotation symmetry through $2\pi /3$. It identifies the function as the Hénon-Heiles potential and helps organize all three channels in the original sketch below. The dashed straight lines are the saddle level; labels on the other contours give $f$.

At each regular point, the gradientis normal to the level set and points towards increasing $f$. The arrows in the contour plot above show this direction; their lengths have been normalized to keep the outer arrows legible. Near the origin they point outwards, as expected for a local minimum. Near $(0,1)$, write $\eta =y-1$; the linear approximation is $\nabla f\simeq (3x,-\eta )$. Thus the arrows point away along the horizontal direction and towards the point along the vertical direction. The other two saddle points have the same behaviour after rotations through $2\pi /3$. All four stationary points have zero gradient, so no direction arrow is assigned at the points themselves.

Near each local minimum, the gradient points outward; near each local maximum, it points inward. The five remaining points have saddle directions, determined by their indefinite Hessian matrices.

On the $x$-axis the exact gradient is $(0,x(x^2-1){)}^T$, and on the $y$-axis it is $(y(y^2-1),0{)}^T$. Thus arrows are perpendicular to the axes and reverse direction at the origin and the unit-axis stationary points.

On the unit circle, substitute $x^2+y^2=1$ into the gradient to obtainIt is radial outward in quadrants I and III, radial inward in quadrants II and IV, and zero at the four axis intersections. These arrows, together with the nearby saddle and extremum patterns, are shown in the right panel of the contour figure. The gradient is normal to every regular level curve and points toward increasing values of $f$.

The chain rule and the quotient rule giveThe exponential function and denominator are positive, so the stationary points occur exactly at $x=\pm 1$. Their values areThe derivative is negative for $|x|>1$ and positive for $|x|<1$, so $x=-1$ is a local minimum and $x=1$ a local maximum. They are also global extrema, since $-1/2\le x/(1+x^2)\le 1/2$.

The Hessian matrix is diagonal:Its eigenvalues are $c$ and $d$. Hence the second-derivative test gives a strict local minimum if $c,d>0$, a strict local maximum if $c,d<0$, and a saddle point of a scalar function if $cd<0$. No degenerate case occurs because both are nonzero.

The local minimum is the unique global minimum, since both separated summands are strictly convex functions and tend to $+\infty$ at either end of the real line. Similarly, when both signs are negative there is a unique global maximum.

First take $a\ne 0$, put $A=|a|$, and suppose $k>0$. The force is the negative derivative of the potential energy:The potential energy is an odd function, passes through $U(0)=0$, and approaches zero from above as $x\to -\infty$ and from below as $x\to +\infty$. Its stationary points are $x=\pm A$:Thus the sole stable equilibrium is the local minimum at $x=A$; the local maximum at $x=-A$ is an unstable equilibrium. The requested values at the stable equilibrium areThe force is an even function, positive for $|x|<A$, negative for $|x|>A$, and zero at $x=\pm A$. Its central maximum is $F(0)=k/A^2$. Sinceits two minima occur at $x=\pm \sqrt{3}A$, with $F=-k/(8A^2)$, and $F\to 0^{-}$ at both ends. These values fix the shapes of both requested sketches.

In one dimension, attraction towards the origin means $xF(x)<0$, while repulsion from the origin means $xF(x)>0$. With that explicitly stated convention,A positive force alone means rightward, not necessarily outward. At the origin $F(0)\ne 0$, so it is not an equilibrium; the inward/outward distinction is undefined exactly there.

The sign assumptions matter because the paper calls $a,k$ merely constants. For $k<0$ both curves reverse sign, the stable equilibrium is $x=-A$, its potential energy is $-|k|/(2A)$, and the attraction/repulsion intervals interchange. For $k=0$ the force vanishes everywhere on the nonsingular domain and there is no isolated potential energy minimum. If $a=0$ and $k\ne 0$, then $U=-k/x$ for $x\ne 0$ and $F=-k/x^2$, with no finite equilibrium.

At a regular constrained stationary curve, choose a constant multiplier $\lambda$ and make the augmented functional $I-\lambda J$ stationary under endpoint-fixed variations. Its integrand is $H=F-\lambda G$ and the equation is $d{H}_{y^{\prime }}/dx-H_y=0$. This normal multiplier formulation requires the first variation of the constraint not to vanish identically; the usual abnormal case would require two multipliers. When $H_x=0$,so $\boxed{y^{\prime }{H}_{y^{\prime }}-H=c_0}$ is constant, proving the requested first integral.

For the chain, $ds=\sqrt{1+(y^{\prime }{)}^2}dx$, so the fixed physical length is exactly $J[y]$. Its gravitational potential, apart from the constant factor $\rho g$, is $I[y]={\int }_{-a}^{a}y\sqrt{1+(y^{\prime }{)}^2}dx$. Thus $H=(y-\lambda )\sqrt{1+(y^{\prime }{)}^2}$, and the Euler–Lagrange equation isThe first integral becomes $(y-\lambda )/\sqrt{1+(y^{\prime }{)}^2}=c$, after renaming the constant with the opposite sign. The differential equation also gives $(y-\lambda )y^{\prime \prime }=1+(y^{\prime }{)}^2$, so the nondegenerate solution integrates to $y=\lambda +c\cosh ((x-x_0)/c)$. Equal endpoint heights force $x_0=0$: the two cosh values agree only when their arguments are equal or opposite, and $a>0$ excludes the former. HenceFor the hanging energy minimum take $c>0$. Since $\sqrt{1+(y^{\prime }{)}^2}=\cosh (x/c)$,For $L>2a$ the length equation has exactly one positive solution: its right side decreases from infinity to $2a$ as $c$ increases from zero to infinity. For $L=2a$ the only feasible shape is the straight segment, obtained as a limiting case; for $L<2a$ no chain can span the endpoints. A negative $c$ gives the inverted stationary catenary, rather than the physically hanging minimum.

Minimizing $\Vert Y-X\theta {\Vert }^2$ gives the normal equations $X^{T}X\hat{\theta }=X^{T}Y$. Since $X$ has full column rank, $X^{T}X$ is positive definite and invertible, and the objective has a unique minimum. Set $A=(X^{T}X{)}^{-1}{X}^T$; then $\hat{\theta }=AY$ and $AX=I_p$. The model gives $\mathbb{E}[Y]=X\theta$ and $\mathrm{Var}(Y)={\sigma }^2{I}_n$, so $\mathbb{E}[\hat{\theta }]=\theta$ and $\mathrm{Var}(\hat{\theta })={\sigma }^{2}A{A}^T={\sigma }^2(X^{T}X{)}^{-1}$. For the alternative estimator the corresponding general formulas are $\mathbb{E}[\tilde{\theta }]=DX\theta$ and $\mathrm{Var}(\tilde{\theta })={\sigma }^{2}D{D}^T$.

The multinomial likelihood is a parameter-independent factor times ${\prod }_i{\pi }_i^{n_i}$. Maximize its logarithm subject to $\sum {\pi }_i=1$: the multiplier equations give $n_i/{\pi }_i=c$, and summation makes $c=n$. Thus ${\hat{\pi }}_i=n_i/n$. Positivity of all counts makes this the unique interior maximum by strict concavity.

The likelihood ratio compares the fitted null likelihood with this saturated likelihood. Taking minus twice its logarithm gives $\Lambda (\hat{\theta })=2{\sum }_i{n}_i\log [n_i/(n{\pi }_i(\hat{\theta }))]$. Under regularity and large expected counts, compare it to ${\chi }_{k-2}^2$ for one fitted parameter: the saturated probabilities have $k-1$ free coordinates and the null has one. If the parameter is fixed at ${\theta }_0$, use $\Lambda ({\theta }_0)$ and ${\chi }_{k-1}^2$. Reject for a value above the chosen upper quantile.

For the ungrouped logarithmic-series observations write $n=1188$ and $S={\sum }_r{X}_r=6968$. The log likelihood isSetting its derivative to zero givesBisection of this monotone mean equation, or Newton's method with a valid bracket, obtains ${\hat{\theta }}_{\mathrm{full}}\approx 0.9442280$.

The supplied table instead uses ten grouped cells, with the last probability ${\pi }_{\ge 10}(\theta )=1-{\sum }_{i=1}^9{\pi }_i(\theta )$. Its likelihood is determined by the cell counts and is not the ungrouped likelihood just differentiated. From its listed values, the minimum deviance occurs near $\boxed{{\hat{\theta }}_{\mathrm{grouped}}\approx 0.93}$; direct minimization of this grouped deviance gives about $0.9333714$. The supplied two forms of the estimate therefore concern different likelihoods. They must not be identified: the full-data score equation gives about $0.94423$, while the grouped table points near $0.93$. The following requested tests use the table's grouped likelihood, with ten cells. The source values and totals are preserved unchanged in TeX.

With one parameter fitted to the grouped cells, the reference degrees of freedom are $10-1-1=8$. The table already gives a deviance of $8.29$ at $0.93$, so the minimized deviance is no larger than that; its direct grouped minimum is about $7.76$. Either is below the supplied critical value $15.51$. Therefore do not reject the fitted grouped log-series model at level five percent. The estimate being profiled in this test is the grouped-likelihood estimate described above, not the different full-data score estimate.

A best approximation to $x$ from $U_n$ is an element $u_{\ast }\in U_n$ satisfying $\boxed{\Vert x-u_{\ast }\Vert =\underset{u\in U_n}{\inf }\Vert x-u\Vert }$. This is a minimum over the specified approximation subspace, not over all of $H$.