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There is a genuine indexing defect in the printed PDF: its new value is at $m+1$, while every old-time coefficient is centered at $m$. On a fixed grid the literal residual containsFor a general solution the leading term is $u_x/(\mu \delta )$, so this is not a consistent diffusion discretization. It also leaves one new interior value unspecified while imposing a generally incompatible update on a prescribed boundary value. The calculation that follows is for the intended same-node update $U_m^{n+1}=U_m^n+k{L}_{\delta }{U}_m^n$; that correction is necessary, not an unnoticed transcription change.

For the midpoint flux stencil for one-dimensional diffusion, defineAssume sufficient smoothness of $a$ and the solution for the displayed remainders. Multiplying the centered Taylor expansions of $a$ and $u$ givesThe first time difference is $(u(t+k)-u(t))/k=u_t+k{u}_{tt}/2+O(k^2)$. After using the diffusion equation, the normalized local truncation error isTherefore, for the corrected update,The first is the residual divided by the time step; the second is the unnormalized one-step defect. At fixed $\mu =k/{\delta }^2$ they are respectively $O({\delta }^2)$ and $O({\delta }^4)$. A rate based on derivatives is not justified for a merely bounded nonsmooth $a$.

Divide the recurrence by $2k$ and substitute a smooth exact solution. Centered Taylor expansions in time and space giveThus the leapfrog advection scheme is consistent with normalized residual $O(k^2+{\delta }^2)$. Its unnormalized defect is $O(k^3+k{\delta }^2)$. For a fixed $0<\nu <1/2$, the uniform two-level bound proved above and accumulation over $O(1/k)$ steps yield, for a sufficiently smooth solution,Both starting levels must be supplied consistently; errors of order $k^2+{\delta }^2$ give second-order convergence. For general square-integrable functions as initial data, point sampling is not defined on equivalence classes. Use cell averages or another bounded grid projection, together with a bounded consistent startup. Approximate the initial data by smooth functions, apply the smooth-data convergence estimate, and use the uniform numerical and exact translation bounds to pass to the L2 space limit. This is the content of Lax equivalence theorem in the present setting.

At $\nu >1/2$ the growing Fourier mode prevents general convergence. At the endpoint, initial data supported in increasingly narrow neighborhoods of the double root can have a second-level error of norm $O(k)$ but a final error of order one after $O(1/k)$ steps. These are vanishing starting errors, so the endpoint fails general two-level convergence as well. Consequentlyis also the general convergence range, with a consistent stable initialization. Claims for a particular specially chosen endpoint startup would be a different, restricted assertion.

Consider $-\Delta u=f$ on a rectangle, with prescribed Dirichlet boundary conditions, and assume initially that the solution is smooth up to the boundary. A finite difference method chooses grid values as unknowns, approximates the differential operator by a local stencil, incorporates the boundary data, and solves the resulting sparse linear system. Accuracy of a stencil by itself is insufficient: one also needs a well-posed discrete problem, a stability estimate uniform under refinement, and a sufficiently accurate algebraic solve.

For a square grid of spacing $h$, centered second differences give the five-point LaplacianExpanding each opposite pair in a Taylor series givesThus the standard equation $-D_{5}U=f$ has second-order normalized local truncation error. Known boundary values move to the right-hand side; they are not iterated as independent unknowns. Nonrectangular geometry requires a consistent boundary approximation rather than silently retaining a full interior stencil outside the domain.

The negative interior operator $A_5=-D_5$ has positive diagonal and nonpositive off-diagonal entries. For a grid vector extended by zero on the boundary,The edge sum includes links to the fixed boundary. Connectivity makes the matrix positive definite, proving uniqueness and providing an energy structure. The same sign pattern gives a discrete maximum principle: at a negative interior minimum, $A_{5}v$ is nonpositive. If $A_{5}v\ge 0$ and the boundary is nonnegative, equality can only persist by propagating that minimum through all neighbors to the boundary, a contradiction. Therefore $v\ge 0$.

On the unit square, the barrier $b(x,y)=[x(1-x)+y(1-y)]/4$ satisfies $A_{5}b=1$ exactly. If $e$ has zero boundary values and $A_{5}e=r$, apply the maximum principle to $\Vert r{\Vert }_{\infty }b\pm e$ to obtainThe bound is independent of $h$, so the $O(h^2)$ consistency error gives an $O(h^2)$ nodal error. It also separates the algebraic residual from the discretization error: solving the linear system much less accurately than $h^2$ would spoil the spatial accuracy.

A direct higher-order design is possible. In one coordinate,has fourth-order accuracy. Adding the corresponding formula in the other coordinate gives a wider stencil. Its advantages must be balanced against boundary closures, more distant couplings, and the possible loss of monotonicity. Higher order does not automatically imply a stable or geometrically convenient scheme.

The Mehrstellen method takes a different route: it uses the Poisson equation to eliminate leading truncation derivatives while keeping a compact unknown stencil. Define the nine-point finite-difference stencilHere each sum has four terms. Equivalently $D_9=D_5+(h^2/6){\delta }_{xx}{\delta }_{yy}$. Expanding givesThe uncorrected nine-point operator is only second order for a general function. But $-\Delta u=f$ implies ${\Delta }^{2}u=-\Delta f$. Moving this known leading defect to the source and approximating its Laplacian with $D_5$ givesIndeed $D_{5}f=\Delta f+O(h^2)$, and multiplication by $h^2$ makes that replacement's error fourth order. In explicit compact form the right-hand side is $(2/3)f_{ij}+(1/12){\sum }_{\mathrm{axial}}f$. Only known source samples, not additional distant solution unknowns, enter this correction. The corresponding unscaled stencil residual is $O(h^6)$. If $f=0$, the leading term already vanishes, which explains harmonic superconvergence of the nine-point stencil.

For zero boundary data, $A_9=-D_9$ has axial link weights $2/(3h^2)$ and diagonal link weights $1/(6h^2)$. Their energy sum proves positive definiteness, and their sign pattern again proves the discrete maximum principle. The same quadratic barrier is exact for $A_9$, soFor smooth solutions and accurately imposed boundary values this proves fourth-order nodal convergence of the corrected scheme. Singular corners, rough sources or low-order boundary treatment can reduce the observed rate; a formal interior expansion is not a global regularity theorem.

More generally, compact high-order schemes choose a solution stencil and a source stencil together, match their Taylor expansions, and use differentiated forms of the PDE to remove derivatives that would otherwise need a wide solution stencil. Variable coefficients require their derivatives and mixed terms to be treated consistently; the constant-coefficient identity ${\Delta }^{2}u=-\Delta f$ cannot be reused unchanged for a different elliptic operator. On irregular meshes, conservation and consistent fluxes may be more useful starting points than a symmetric Cartesian formula.

Finally, discretization and solution cost are coupled. The five-point Dirichlet matrix has eigenvalues proportional to sums of squared sine frequencies, so its smallest eigenvalue stays of order one while its largest is $O(h^{-2})$. Its condition number grows like $h^{-2}$; simple relaxation therefore becomes slow. Sparse direct elimination, Conjugate gradient method with appropriate matrix preconditioning, and the multigrid method are alternatives. A sound design combines local accuracy, boundary fidelity, mesh-independent stability, and an algebraic solver whose residual is small enough for the intended error tolerance.

Finite second moment implies $\mathbb{E}X<\infty$. For $\lambda$ near any positive ${\lambda }_0$, the derivative of $e^{-\lambda X}$ is bounded in absolute value by $X$, provided the neighbourhood stays in $(0,\infty )$. The dominated convergence theorem therefore permits differentiation under the expectation:For the expansion, put $r(z)=e^{-z}-1+z-z^2/2$. Taylor expansion gives $r(z)/z^2\to 0$ as $z\downarrow 0$. Also, for $z\ge 0$,The inequalities follow by integrating $0\le 1-e^{-u}\le u$ from zero to $z$. Consequently $r(\lambda X)/{\lambda }^2\to 0$ pointwise and its absolute value is bounded by $X^2/2$. Another application of the dominated convergence theorem gives the second-moment expansion of a Laplace transform:No third moment is needed for the remainder estimate.

The force from one location is defined almost surely, since the uniform distribution assigns probability zero to the origin. Symmetry of that uniform distribution makes its characteristic function real:Near zero the integrand is bounded by $2$. For large $x$ it is at most $m^2{t}^2/(2x^4)$. The monotone convergence theorem therefore gives the finite integralIndependence of the locations gives the characteristic function of the total force as $q_n(t{)}^n$. For fixed $t$, $I_n(t)$ is bounded, so Taylor expansion of the logarithm givesThe logarithm is legitimate for all sufficiently large $n$.

For $t\ne 0$, substitute $u=m|t|/x^2$. The cosine integral for a symmetric stable exponent, evaluated directly in part (b), givesAt $t=0$ both sides vanish. The limiting characteristic function is continuous at zero, so the Lévy continuity theorem proves weak convergence of random variables to the symmetric stable distribution of index $1/2$ withThe coefficient reflects the spatial density $n/(2n)=1/2$. Symmetry was used in the characteristic function, without assuming that the singular force has a finite mean.

Indifference means equality of the risky expected utility and the utility of the sure reduced wealth. A second-order Taylor expansion, using $\mathbb{E}\varphi =0$, givesFor the sure alternative,Equating these and retaining the leading terms produces the small multiplicative risk premiumFor increasing concave utility functions this is positive. The approximation assumes genuinely small shocks with controlled Taylor remainders and positive final wealth; small variance alone would not justify it for rare large shocks.

The working score function and its derivative areAt an interior working maximizer, $U(\hat{\beta })=0$. A Taylor expansion at ${\beta }_0$ yieldsfor an intermediate value ${\beta }^{\ast }$. When the last term is negligible relative to the linear one, this givesThis is the Poisson score linearization under proportional variance. Correct specification of the mean alone implies $EU({\beta }_0)=0$. The stipulated independence and variance relation giveTaking the mean and variance of the linear approximation therefore provesThese are first-order conclusions, not exact finite-sample identities. The mean and variance assumptions alone do not guarantee that the approximation is accurate: an interior consistent solution, increasing information, control of leverage, and a suitable central limit condition are needed for normal inference.

The quasi-likelihood score for variance $\varphi {\mu }_i$ is $U/\varphi$, so its zero is unchanged by the unknown common dispersion parameter. Thus the working Poisson estimating equation remains appropriate for the mean parameter, but its unadjusted Poisson variance estimate is wrong when $\varphi \ne 1$. A usual Pearson dispersion estimator for this one-parameter model isIn particular overdispersion multiplies coefficient standard errors by approximately $\sqrt{\varphi }$, even though it leaves the coefficient estimating equation unchanged.

Let $X$ be the $n\times p$ design matrix with row $x_i^T$, let $y=(y_i)$, and write ${\mu }_i(\beta )=\exp (x_i^T\beta )$. Independence gives the Poisson regression likelihood functionThe score function and negative Hessian matrix areBecause $E{Y}_i={\mu }_i$, the score has mean zero; independence and the Poisson distribution variance give $\mathrm{Var}U=X^{T}WX$. Thus the observed and expected Fisher information matrices are the same here.

Assume $X$ has full column matrix rank $p$. For a nonzero vector $a$, $a^{T}Ia={\sum }_i{\mu }_i(x_i^{T}a{)}^2>0$. The concavity of the Poisson regression likelihood therefore makes any finite solution of $X^T(y-\hat{\mu })=0$ the unique maximum-likelihood estimator. If the design is rank deficient, parameters differing by a vector in $\ker X$ give the same distribution, so only identifiable combinations can be estimated. Full rank alone does not ensure a finite maximizer: for an intercept-only model with all observations zero, $\ell =-n{e}^{\beta }$ approaches its supremum as $\beta \to -\infty$. More generally, directions that decrease only fitted means attached to zero counts can produce boundary fits. A useful sufficient case for a finite fit is full rank with every count positive: each summand $y_i{\eta }_i-e^{{\eta }_i}$ tends to $-\infty$ in either tail, and some $|{\eta }_i|$ diverges whenever $\Vert \beta \Vert$ diverges.

For an interior fit, Newton's method and Fisher scoring give the same update:This is also iteratively reweighted least squares. Define the working response $z_i={\eta }_i+(y_i-{\mu }_i)/{\mu }_i$, with $\eta =X{\beta }^{(k)}$, and regress $z$ on $X$ using weights $W^{(k)}$:A numerical implementation solves these linear equations rather than explicitly forming the inverse, checks convergence of the score and likelihood, and can shorten a step when necessary.

Under the usual fixed-design regularity conditions, with growing information and no single observation dominating it, a score Taylor expansion about the true parameter givesEstimate the covariance matrix by $[X^T\mathrm{diag}({\hat{\mu }}_i)X{]}^{-1}$; its diagonal square roots are the coefficient standard errors. This justifies large-sample Wald confidence intervals and Wald tests for specified coefficients or contrasts. Finite-sample unbiasedness is not being asserted. A likelihood-ratio test compares nested fits through twice their maximized log-likelihood difference. The Poisson deviance against the saturated fit iswith $0\log 0=0$. Residual diagnostics and a check of overdispersion assess whether the assumed Poisson variance is adequate; the variance correction required when it is not is derived in the next part.

A kernel for density estimation is an integrable function K of unit integral, ordinarily nonnegative and centered. Square integrability is needed for integrated variance; a symmetric second-order kernel for density estimation also has $\int uK(u)du=0$ and finite nonzero second moment ${\mu }_2(K)$. Its scaled form is $K_h(u)=h^{-1}K(u/h)$, and the kernel density estimator isHigher-order signed kernels are possible, but need not give a nonnegative probability density function estimate.

The mean integrated squared error is $\mathbb{E}\int ({\hat{f}}_h(x)-f(x){)}^{2}dx$. Put $g_h=K_h\ast f$. Independence gives the pointwise mean $g_h$ and variance $n^{-1}[(K_h^2\ast f)-g_h^2]$. Integrating, use $\int f=1$ and Fubini to obtain $\int (K_h^2\ast f)=R(K_h)$. The bias-variance decomposition proves the exact formulaEquivalently this is $R(f)-2\int f(K_h\ast f)+(1-n^{-1})R(K_h\ast f)+R(K)/(nh)$. These identities assume the displayed integrals are finite; they are exact, rather than a small-smoothing bandwidth expansion.

For a symmetric second-order kernel for density estimation, Taylor expansion gives leading bias $h^2{\mu }_2(K)f^{\prime \prime }/2$, and the leading integrated variance is $R(K)/(nh)$. Thus the asymptotic mean integrated squared error balances inverse smoothing bandwidth against fourth-power bias, with optimal smoothing bandwidth $[R(K)/(n{\mu }_2(K{)}^{2}R(f^{\prime \prime })){]}^{1/5}$. Its appearance of the integrated squared density curvature motivates the scale comparison.

For $f_a(x)=af(ax)$, $a>0$, two differentiations give $f_a^{\prime \prime }(x)=a^3{f}^{\prime \prime }(ax)$. A change of variables therefore gives $R(f_a^{\prime \prime })=a^{5}R(f^{\prime \prime })$, which tends to zero as $a\downarrow 0$ if the original curvature integral is finite. The distribution with probability density function $f_a$ is that of X divided by a, so $\sigma (f_a)=\sigma (f)/a$. ConsequentlyThis scale-invariant density curvature measures shape without dependence on the units used to measure X.

For the final minimization, write $c=35/32$ and use the triweight kernel $f_0=c(1-x^2{)}^3$ on $(-1,1)$, zero outside. Integration gives total mass one, mean zero and variance $1/9$. Its value and its first two derivatives vanish at both endpoints, so its extension is twice continuously differentiable. On the interior,In particular $f_0^{\prime \prime \prime }(1)=-48c$ and $f_0^{\prime \prime \prime }(-1)=48c$.

Put $e=h-f_0$. The mass and moment constraints give $\int e=\int x^{2}e=0$, since both means vanish and their variances match. If $R(h^{\prime \prime })=\infty$ the claim is immediate; otherwise expand the squared norm and integrate its cross term twice by parts over $[-1,1]$. The first endpoint term vanishes because $f_0^{\prime \prime }(\pm 1)=0$, but the next one must be retained:For the second equality use the zero mass and second moment of e and $e=h\ge 0$ outside the support. All terms in the last expression are nonnegative, including the endpoint values of a probability density function. It follows thatThis proves that the triweight density minimizes integrated squared curvature under the specified constraints. Equality forces $e^{\prime \prime }=0$; a globally affine integrable difference of probability density functions is zero, so equality occurs only at $h=f_0$.

Assume a symmetric second-order probability density function kernel for density estimation and enough smoothness for the displayed derivative functionals. The benchmark smoothing bandwidth is $h_0=[R(K)/(n{\mu }_2(K{)}^{2}R(f^{\prime \prime })){]}^{1/5}$, balancing $R(K)/(nh)$ against $h^4{\mu }_2(K{)}^{2}R(f^{\prime \prime })/4$. Three widely used selection methods are as follows.

For least-squares density cross-validation, expand integrated squared error as $R({\hat{f}}_h)-2\int {\hat{f}}_{h}f+R(f)$. Estimate the middle integral by leaving out the observation at which the estimate is evaluated, and minimizeThe left-out observation is independent of its estimator, so the second term has expectation $-2\int (K_h\ast f)f$. Thus $\mathbb{E}\mathrm{CV}(h)=\mathrm{MISE}(h)-R(f)$ exactly. The subtraction of the self-contribution is essential. Computation uses $R({\hat{f}}_h)=n^{-2}{\sum }_{i,j}(K_h\ast {\tilde{K}}_h)(X_i-X_j)$, where ${\tilde{K}}_h(u)=K_h(-u)$. This method directly targets integrated error without estimating a curvature functional first. Under standard smoothness and a suitable search interval its selected smoothing bandwidth is ratio-consistent with the MISE optimum, and hence with $h_0$. Relative fluctuations are ordinarily $O_p(n^{-1/10})$, so the selected values can be quite variable in finite samples.

For biased density cross-validation, estimate curvature from the second-derivative probability density function estimator and subtract its diagonal variance term:Differentiating the kernel for density estimation estimator and applying the integrated-variance identity gives the exact expectation $\mathbb{E}R({\hat{f}}_h^{\prime \prime })=R(K^{\prime \prime })/(n{h}^5)+(1-n^{-1})R(K_h\ast f^{\prime \prime })$. Under the derivative regularity assumptions, $R(K_h\ast f^{\prime \prime })=R(f^{\prime \prime })+O(h^2)$, so $\mathbb{E}{\hat{A}}_h=R(f^{\prime \prime })+O(h^2)+O(n^{-1})$. Minimize BCV over a reasonable smoothing bandwidth range. It is a smoothed criterion directed at AMISE; the name reflects its finite-sample bias rather than an exact unbiased-risk identity. Standard conditions again give ratio consistency and relative fluctuation order $n^{-1/10}$. It often has a smoother objective than ordinary least-squares cross-validation, though numerical behavior and relative variance depend on the probability density function and kernel for density estimation.

For plug-in bandwidth selection, obtain an independent pilot scale g, estimate $A=R(f^{\prime \prime })$, then setThe pilot should be positive and stable; consistent positive estimates make this expression ratio-consistent by continuity. A simple normal-reference initialization uses $R(f^{\prime \prime })=3/(8\sqrt{\pi }{\sigma }^5)$ under a normal reference model, substituting a sample scale for sigma. For a Gaussian density kernel this gives $h\approx 1.06\hat{\sigma }{n}^{-1/5}$. A pure normal-reference rule is quick but is asymptotically correctly calibrated only when its assumed curvature matches the actual probability density function; it is useful as the first stage of a data-driven plug-in procedure.

A more adaptive fourth-derivative pilot estimate of density curvature uses $R(f^{\prime \prime })=\int f{f}^{(4)}$, obtained by two integrations by parts, and estimates it byThe diagonal expectation contributes $K^{(4)}(0)/(n{g}^5)$; the leading smoothing bias is $-g^2{\mu }_2(K)R(f^{\prime \prime \prime })/2$. Balancing these terms produces the stated pilot scale $g\asymp R(f^{\prime \prime \prime }{)}^{-1/7}{n}^{-1/7}$, with another pilot or reference model supplying $R(f^{\prime \prime \prime })$. Under sufficient derivatives and tail control, its error is bounded byat that pilot scale. Taylor expansion of the fifth-root formula gives ${\hat{h}}_{\mathrm{PI}}/h_0-1=-(\hat{A}-A)/(5A)+o_p(|\hat{A}-A|)$, so it inherits that bound. A calibrated cancellation of leading pilot bias can give a faster rate. This illustrates why an appropriately constructed plug-in selector can converge more quickly than cross-validation, while requiring more smoothness and pilot choices. For each ratio-consistent method, its attained AMISE divided by the minimal AMISE tends to one.

Write $h=\Delta x$, $k=\Delta t$, and assume $a,u$ are sufficiently smooth for the following Taylor expansions. The symmetric half-grid diffusion consistency calculation giveswhereThe normalized local truncation error of the full update is consequentlyThe PDF prints an advection equation, $u_t=a(x)u_x$, whereas this stencil approximates conservative diffusion, $u_t=(a{u}_x{)}_x$. For the literal printed equation the leading defect is $(a-a^{\prime })u_x-a{u}_{xx}$, which generally does not vanish: the method is inconsistent, with an $O(1)$ normalized defect.

For the intended variable-coefficient conservative diffusion equation, the first two terms cancel and ${\mathcal{T}}_{h,k}=O(k+h^2)$. Under the parabolic step scaling $k=O(h^2)$, this is $O(h^2)$, while the unnormalized one-step local truncation error is $k{\mathcal{T}}_{h,k}=O(h^4)$. These two orders use different normalizations. With numerical stability and compatible smooth initial and boundary conditions, the intended diffusion problem has global error $O(k+h^2)$ on a fixed time interval. Positivity and boundedness of $a$ alone do not justify the displayed Taylor expansion; coefficient regularity is also needed for that error formula.

Use the Gaussian pole transition for a large gamma parameter. With $t=1+s/\sqrt{n}$ and $\lambda =n+i\mu \sqrt{n}+\nu +o(1)$, Taylor expansion givesThe upper semicircle is traversed from left to right, so $\int ds/s=-i\pi$ on the indentation. The minus sign in $dt/(1-t)$ therefore makes its contribution $+i\pi$. The remaining leading integral isAt $\mu =0$ its Cauchy principal value vanishes by oddness. Differentiating with respect to $\mu$ removes the pole, leaving the Fourier transform of a Gaussian:Integrating from zero givesThis is uniform for bounded real $\mu ,\nu$ (and extends analytically over bounded complex values where the local contour is continued consistently). Contributions away from the saddle are exponentially small after localization. The bounded shift $\nu$ enters at the next order, not the leading error function. For example, retaining the displayed cubic correction gives $\sqrt{2\pi /n}{e}^{-{\mu }^2/2}[\nu +(2+{\mu }^2)/3]$ as the next term. This also checks the constant and contour sign.

At a critical point $p$ of a twice continuously differentiable real function, $\nabla f(p)=0$ and the Taylor expansion isThe real symmetric Hessian matrix has an orthonormal eigenbasis, so writing $h={\sum }_i{a}_i{u}_i$ gives $h^{T}Hh={\sum }_i{\lambda }_i{a}_i^2$. If every eigenvalue is positive, this quadratic form is at least ${\lambda }_{\min }|h{|}^2$, and the remainder cannot change its sign for sufficiently small nonzero $h$. Thus $p$ is a strict local minimum. If every eigenvalue is negative, the analogous argument gives a strict local maximum. If positive and negative eigenvalues both occur, displacement along the corresponding eigenvectors gives values above and below $f(p)$, so $p$ is a saddle point of a scalar function.

The second-derivative test can be inconclusive when the Hessian matrix is singular and semidefinite: higher-order terms along directions in the kernel of a linear map matter. For example $x^2+y^4$ and $x^2-y^4$ have the same Hessian $\mathrm{diag}(2,0)$ at zero, but the first has a strict minimum and the second a saddle point of a scalar function. A zero eigenvalue does not prevent a saddle conclusion if other eigenvalues already have both signs.

For the given function write $r^2=x^2+y^2$. Its first derivatives areThe origin is a critical point for every real $\alpha$, withIts eigenvalues are $1,-1$, so it is always a saddle point of a scalar function. If one coordinate at a critical point is zero, the other must also be zero. If both are nonzero, the derivative equations force $x^2=y^2=1/(2\alpha )$, possible only for $\alpha >0$. Hence for $\alpha \le 0$ the origin is the only critical point.

For $\alpha >0$ there are four additional critical points,At such a point,Thus the two equal-sign points are strict maxima, and the two opposite-sign points are strict minima. They are global: $2|xy|\le r^2$ implieswhere the last maximum occurs at $r^2=1/\alpha$ and equality also requires $x^2=y^2$.

As $\alpha \downarrow 0$, the four nonzero critical points have distance ${\alpha }^{-1/2}$ from the origin and escape to infinity, rather than merging with the origin. Their values diverge as $\pm 1/(2e\alpha )$. As $\alpha \uparrow 0$ from below there are no extra critical points, while the origin stays a saddle throughout. This is critical points escaping to infinity in a Gaussian-weighted bilinear function.

Use Minkowski spacetime coordinates $x^{\mu }=(ct,x,y,z)$ and the Minkowski metric of signature $(+,-,-,-)$. For a massive particle with three-velocity $\mathbf{v}=d\mathbf{x}/dt$ and $|\mathbf{v}|<c$, proper time satisfiesThus the four-velocity components areHere upper indices indicate components in the chosen coordinates; lowering with the Minkowski metric changes the signs of the spatial components. Its invariant product defined by the Minkowski metric is