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Local errors of the Euler schemes. Let $h=\Delta t>0$, and evaluate $f$ and its partial derivatives at an exact starting point $(y(t),t)$. DefineThe exact Taylor expansion is $y(t+h)=y+hf+\frac{1}{2}{h}^{2}A+O(h^3)$. The forward Euler method retains only $y+hf$, so its one-step defect, exact minus computed value, isFor the Backward Euler method, substitution of the exact solution into the scheme gives the residualbecause $f(y(t+h),t+h)=f+hA+O(h^2)$. Under local smoothness, solving the implicit step changes this residual by a factor $1+O(h)$, so the one-step defect has the same leading term:These are unnormalized local truncation errors, as in the question. A convention dividing by $h$ instead gives $O(h)$. Both schemes have first-order global error under the usual smoothness and stability hypotheses.

Euler stability. For the Dahlquist test equation $y^{\prime }=-\Lambda y$, put $z=h\Lambda$, with $\mathrm{Re}z>0$. The numerical amplification factors areStrict decay requires $|R|<1$; the boundary $|R|=1$ gives bounded but generally nondecaying values. For Forward Euler method,For positive real $\Lambda$, this reduces to $0<h\Lambda <2$. The absolute stability region in the $z$ plane is the disk $|z-1|\le 1$. For Backward Euler method, $|1+z|>1$ whenever $\mathrm{Re}z>0$, so it decays for every step size. It is A-stable and also L-stable, since $R_{BE}(z)\to 0$ for large $z$ in this half-plane. Here $z$ has the opposite sign from the alternative convention $z=h\lambda$ for $y^{\prime }=\lambda y$.

Predictor-corrector accuracy. The printed two-stage construction is Heun's method, equivalentlyAlthough the second stage resembles a Backward Euler method evaluation, it uses the predicted value, so the combined method is explicit. Expand that stage around the exact starting point, puttingThenand the numerical step is $y+hf+\frac{1}{2}{h}^{2}A+\frac{1}{4}{h}^{3}B+O(h^4)$. Along the exact solution, $y^{\prime \prime \prime }=B+f_{y}A$, so comparison with the exact Taylor expansion givesThus the local defect improves by one power of $h$, giving second-order global error under the usual hypotheses. For the linear test equation the defect reduces to $-h^3{\Lambda }^{3}y/6+O(h^4)$.

Predictor-corrector stability. Applied to the test equation, the predictor is $y^{(1)}=(1-z)y_n$ and the second value is $y^{(2)}=(1-z+z^2)y_n$. Their average givesFor real positive $z$, $R_H(z)=\frac{1}{2}[(z-1{)}^2+1]$, so strict decay holds exactly for $0<z<2$, the same real-axis step bound as Forward Euler method. For complex $z=x+iy$ the precise condition isThe region is bounded and does not contain the full right half-plane: for example $z=3$ gives $R_H=5/2$. It is therefore not A-stable. Its imaginary-axis values satisfy $|R_H(iy){|}^2=1+y^4/4>1$ for $y\ne 0$. Better accuracy has not inherited the unconditional stability of an actual implicit Backward Euler method step.

Fourth-order spatial formula. For a sufficiently smooth function, pair the Taylor expansions at opposite shifts:Multiply the first sum by sixteen, subtract the second, and subtract $30u(x)$. The constant and fourth-derivative terms cancel, leavingHence the fourth-order centered second derivative satisfiesRearranging proves the requested $u^{\prime \prime }=D_{4,h}u+O(h^4)$, including its leading error coefficient.

Explicit diffusion scheme. Take $x_j=jh$, $h=1/N$, $t_n=n\tau$, and let $U_j^n$ approximate $\psi (x_j,t_n)$. Combining the forward Euler method in time with $D_{4,h}$ in space givesTo check the error, insert the exact solution. The heat equation gives ${\psi }_t={\psi }_{xx}$, while the time Taylor expansion and the spatial formula give the one-step defectThus its unnormalized local truncation error isas requested. Dividing this defect by $\tau$ gives the normalized residual $O(\tau )+O(h^4)$; these are distinct error conventions.

Boundary implementation. The centered stencil requires two ghost values at each end. An even-reflection closure for Neumann diffusion imposesThe reflected first centered difference at either endpoint is zero. Apply the same evolution formula at the endpoints after substituting these ghost values; for example,The ghost values also close the stencil at $j=1$ and $j=N-1$. They are determined auxiliary values, not new physical data. For a smooth solution compatible with the Neumann boundary conditions, differentiating those conditions in time and using the heat equation gives ${\psi }_{xxx}={\psi }_{xxxxx}=\cdots =0$ at the endpoints. Thus even reflection preserves the smoothness needed for fourth-order spatial accuracy. If the initial data fail the required higher compatibility conditions, a uniform smooth error expansion through $t=0$ cannot be assumed.

For completeness, the reflected spatial operator has cosine eigenvectors $\cos (\ell \pi j/N)$ and eigenvaluesThe most negative one is $-16/(3h^2)$. Requiring $|1+\tau {\lambda }_{\ell }|\le 1$ gives the forward Euler method stability restriction $\boxed{\tau /h^2\le 3/8}$ in the natural trapezoid-weighted discrete norm. A fourth-order spatial stencil does not remove the explicit diffusion time-step restriction.

Implicit alternative. The Crank-Nicolson diffusion scheme using the same spatial operator and boundary closure isIt has second-order time accuracy and fourth-order spatial accuracy for smooth compatible solutions; its one-step defect is $O({\tau }^3)+O(\tau h^4)$. Each spatial eigenvalue $\lambda \le 0$ has amplification factor $(1+\tau \lambda /2)/(1-\tau \lambda /2)$ of magnitude at most one, so it removes the explicit restriction for this boundary closure.

The price is solving a banded linear system at every step; the same matrix factorization can be reused for fixed $h$ and $\tau$. Large time steps still reduce accuracy. Moreover, the amplification factor tends to $-1$ as $\tau |\lambda |\to \infty$, so short spatial wavelengths can alternate sign with little numerical damping. Thus unconditional linear stability does not guarantee a monotone or well-resolved transient. Boundary accuracy and initial compatibility remain necessary with either time discretization.

Substitution of $y=x^p$ into the homogeneous Euler-Cauchy equation gives $(p(p-1)+p-1)x^{p-2}=(p^2-1)x^{p-2}$, so $\boxed{p=1,-1}$. Use the solution $u(x)=x$ that vanishes at zero and $v(x)=1/x$ that decays at infinity. Their Wronskian is $u{v}^{\prime }-u^{\prime }v=-2/x$. The half-line Green function for the first cylindrical harmonic is thereforeIt is continuous at $x=s$, its first derivative has jump $1/2-(-1/2)=1$, and it solves the homogeneous equation elsewhere. Thus $L_{x}G=\delta (x-s)$ for integration with ordinary measure $ds$. The boundary conditions hold for each fixed $s>0$.

Integrating against the forcing givesSince the two integrals are $2-(x^2+2x+2)e^{-x}$ and $e^{-x}$, respectively,A Taylor expansion gives $y=-x/2+O(x^2)$ near zero, while $y\sim -1/x$ at infinity. This verifies the two endpoint limits; direct differentiation verifies $Ly=e^{-x}$ on $x>0$. The differential expression itself is singular at zero, so the endpoint condition is imposed by continuity. No nonzero combination of $x$ and $1/x$ satisfies both homogeneous endpoint conditions, proving uniqueness.

Write $A=Q/(2\pi {ϵ}_0{a}^2)$. The Taylor expansion $\sqrt{a^2+z^2}=a+z^2/(2a)+O(z^4/a^3)$ gives the near-axis resultTo extend off the axis, work separately above and below the charged plane. In either source-free half-space the regular axisymmetric solid harmonics are $r^n{P}_n(\cos \theta )$. Matching the positive-axis coefficients, using $P_n(1)=1$, and then reflecting in the disk givesHere $P_2(t)=(3t^2-1)/2$ is a Legendre polynomial. The quadratic term is $A(z^2-{\rho }^2/2)/(2a)$. The linear term is $-Az$ above and $+Az$ below; it represents the discontinuity of the normal electric field. A single regular spherical expansion across the entire charged sheet would incorrectly remove that jump.

For a harmonic function, put $a=u_{xx}$ and $b=u_{xy}=u_{yx}$. Since $u_{yy}=-u_{xx}$, its Hessian matrix isA nonzero determinant is therefore negative, and the two real Hessian eigenvalues are $\pm \sqrt{a^2+b^2}$. At a stationary point the quadratic term of the Taylor expansion has both signs. Consequently every such nondegenerate stationary point is a saddle point. This is the nondegenerate stationary points of a planar harmonic function criterion. The same argument applies to $v$, whose stationary points coincide with those of $u$. A zero Hessian determinant is outside this second-order classification.

In Cartesian coordinates a rank-$r$ tensor is an array whose components transform under a change of orthonormal axes by one rotation matrix factor per index:An isotropic tensor has unchanged components under every proper rotation. The Levi-Civita symbol obeysfor $Q\in SO(3)$, proving rotation invariance of the Levi-Civita symbol. Under an orthogonal transformation of determinant $-1$ its sign reverses; including such transformations requires the pseudotensor distinction. Here isotropy is with respect to proper rotations.

Contract the determinant identity with ${ϵ}_{ijk}$. Since ${ϵ}_{ijk}{ϵ}_{ijk}=3!=6$, one obtainsEvery index on the right is contracted, so the tensor component transformation law leaves no free rotation factor: the determinant is a scalar. Equivalently, $A^{\prime }=QA{Q}^T$ gives $\det A^{\prime }=(\det Q{)}^2\det A=\det A$.

For the matrix inverse, defineThis is the cofactor matrix. To verify the inverse formula rather than just recognize it, contract with $A_{ib}$ and use the determinant identity again:Here the contraction of two Levi-Civita symbols gives ${ϵ}_{ajk}{ϵ}_{ijk}=2{\delta }_{ai}$. Therefore $A{C}^T=(\det A)I$ and, when $\det A\ne 0$,These are the Levi-Civita formulas for a three-dimensional determinant and inverse.

Now differentiate the cubic determinant formula with respect to the individual entry $A_{ab}$. The three differentiated factors give equal contributions: exchanging two paired indices changes the sign of both epsilon symbols, leaving their product unchanged. ThusSubstituting the inverse expression proves the entrywise derivative of the determinant:The cofactor version remains valid for singular matrices; only the inverse version requires invertibility.

For the time-dependent Jacobian matrix, the stated smooth expansion impliesApplying the chain rule and the entrywise determinant derivative gives Jacobi's formula near $t=0$:At zero, $C=I$, so ${{\partial }_t\det A|}_0={\sum }_a{\partial }_a{u}_a=\nabla \cdot u$. A Taylor expansion therefore yieldsThe Jacobian determinant measures local volume change, so this establishes volume preservation to first order. An arbitrary family specified only through its first-order expansion need not preserve volume exactly at higher orders.

The Lagrangian is $\mathcal{L}=T-\mathcal{V}$, and the Euler-Lagrange equations areAt a rest equilibrium $q=0$, ${\partial }_i\mathcal{V}(0)=0$ for all three coordinates. The Taylor expansion therefore starts asThe symmetric Hessian matrix is the matrix $\mathsf{V}$ of the quadratic potential energy. If a quadratic coefficient vanishes along a direction, higher-order terms can be the first nonconstant terms there; equilibrium by itself does not imply strict stability.

Write ${\mathsf{T}}_0=\mathsf{T}(0)$. Terms involving the variation of $\mathsf{T}(q)$ are cubic in small displacements and velocities, so the quadratic small oscillation Lagrangian isA normal mode has fixed displacement ratios, $q(t)=v\cos (\omega t+\varphi )$. Its angular frequency $\omega$ is an eigenfrequency, or normal frequency, and satisfies the generalized eigenvalue problem for small oscillationsHere ${\mathsf{T}}_0$ is positive definite for independent physical coordinates and positive masses. A stable quadratic minimum makes ${\omega }^2\ge 0$; a negative value would describe exponential instability rather than oscillation.

For the suspended bar, let ${\kappa }_0$ and ${\kappa }_1$ denote the actual incremental spring constants of the upper and each lower spring. The physical displacement of the bar is $q_1$, while the lower spring extensions change by $q_2-q_1$ and $q_3-q_1$. ConsequentlyThe linear gravitational and elastic terms cancel at the loaded equilibrium. Expanding the remaining squares gives the small-oscillation mass and stiffness matricesThusSource convention: the printed matrix is obtained by setting ${\kappa }_0=\lambda /L$ and ${\kappa }_1=\mu /l$. Although the PDF prose calls $\lambda ,\mu$ spring constants, that matrix treats them as elastic parameters whose ratios to the stated lengths are the incremental stiffnesses. If they are read literally as Hooke-law spring constants, the divisions by $L,l$ should instead be omitted. The numerical case has $L=l=1$, so this discrepancy does not affect its results.

With the numerical parameters, ${\mathsf{T}}_0=\mathrm{diag}(2,1,1)$ andUse the exchange symmetry of the lower masses to find the vertical modes of a horizontally constrained suspended bar. The antisymmetric vector $v_a=(0,1,-1{)}^T$ obeys $\mathsf{V}{v}_a={\mathsf{T}}_0{v}_a$, giving ${\omega }_a=1$. In the symmetric sector put $v=(A,B,B{)}^T$ and $s={\omega }^2$. The two independent equations areTheir determinant gives $s^2-4s+2=0$, so $s=2\pm \sqrt{2}$. Taking $A=1$, the full set isIn the lowest normal mode, all three masses move in phase, with the lower masses moving about $2.414$ times as far as the bar. In the antisymmetric normal mode, the bar stays still and the lower masses move equally in opposite directions; their changing forces on the bar cancel. In the highest normal mode, the lower masses move together against the bar, with magnitude about $0.414$ of its displacement. Any small motion is a superposition of these three modes.

Let $L=-(1-x^2)D^2+xD$, so $L{y}_n=n^2{y}_n$ and the forced equation is $({\gamma }^2-L)Z=e^{\epsilon x}$. The small-parameter Taylor expansion of the forcing isEach Chebyshev polynomial can be inverted separately under ${\gamma }^2-L$. None of the required denominators vanishes because $\gamma$ is a real noninteger. Thus the regular solution of a forced Chebyshev equation hasHere we select the same regular endpoint domain as in part (a), with finite function values and finite endpoint derivatives. This qualification is needed: the printed forced equation alone does not state endpoint conditions. Without this selection one can add $c_1\cos (\gamma \arccos x)+c_2\sin (\gamma \arccos x)$, which remain finite but generally have singular endpoint derivatives. With $x=\cos \theta$, regularity gives $Z_{\theta }(0)=Z_{\theta }(\pi )=0$; the corresponding homogeneous equation has only the zero solution for noninteger $\gamma$. This makes the displayed expansion unique.

The Taylor expansions giveand hence $f(z)=b/z+iab+O(z)$. Since $b>0$, zero is a simple pole with residue $b$. It is not a removable singularity, despite the vanishing numerator, because the denominator has a zero of order two.

The zeros of the hyperbolic sine are exactly $z=k\pi i$, $k\in \mathbb{Z}$, since $\sinh z=0$ is equivalent to $e^{2z}=1$. Each zero is simple because $\cosh (k\pi i)=(-1{)}^k\ne 0$. The quotient is therefore holomorphic precisely onAt zero, the Taylor expansion $\sinh z=z+O(z^3)$ gives $z/{\sinh }^{2}z=1/z+O(z)$, so zero is a simple pole. At $k\pi i\ne 0$, the denominator has a double zero and the numerator is nonzero, giving a pole of order two. Thus none of the omitted points admits a holomorphic extension.

For $g=\nabla f(x)\ne 0$, the unit steepest descent direction is $-g/\Vert g{\Vert }_2$ by the Cauchy-Schwarz inequality. The usual unnormalized direction is $\boxed{p=-g}$; choosing its scale separately as a line-search step gives the same ray. It satisfies $\boxed{g^{T}p=-\Vert g{\Vert }_2^2<0}$, so it is a descent direction. The first-order Taylor expansion then guarantees decrease for sufficiently small positive steps. The subsequent bounds use this unnormalized negative-gradient convention.

A descent direction $s$ satisfies $g^{T}s<0$, where $g=\nabla f(x)$; the first-order Taylor expansion then gives $f(x+\alpha s)<f(x)$ for sufficiently small $\alpha >0$. To compare directions one must fix their scale. Among Euclidean unit vectors, the Cauchy-Schwarz inequality givesThis is the unit steepest descent direction. The conventional unnormalized direction used in the algorithm is the same ray, $\boxed{s=-g}$. Equivalently it is the unique minimizer of $g^{T}s+\frac{1}{2}\Vert s{\Vert }_2^2=\frac{1}{2}\Vert s+g{\Vert }_2^2-\frac{1}{2}\Vert g{\Vert }_2^2$. If $g=0$ there is no strict first-order descent direction, and the conventional update is zero.

A local minimizer of a differentiable function on ${\mathbb{R}}^N$ necessarily satisfies $\boxed{\nabla f(x_{\ast })=0}$. A sufficient condition for a strict local minimum is $\boxed{\nabla f(x_{\ast })=0,{\nabla }^{2}f(x_{\ast })\text{ positive definite}}$. Indeed, the second-order Taylor expansion has a strictly positive quadratic term that dominates its $o(\Vert x-x_{\ast }{\Vert }^2)$ remainder near $x_{\ast }$.

For Newton's method, solve $H_n{s}_n=-g_n$ with $H_n={\nabla }^{2}f(x_n)$, $g_n=\nabla f(x_n)$, and set $x_{n+1}=x_n+s_n$. A positive-definite Hessian makes this a descent direction; general Hessians may need modification or a separate line search for a globally useful method.

The sequence is q-quadratically convergent if there is $C<\infty$ such that $\Vert x_{n+1}-x_{\ast }\Vert \le C\Vert x_n-x_{\ast }{\Vert }^2$ eventually. Sufficient local hypotheses are $\nabla f(x_{\ast })=0$, a nonsingular $H_{\ast }={\nabla }^{2}f(x_{\ast })$, a Lipschitz continuous Hessian near $x_{\ast }$, and $x_0$ sufficiently close. For convergence to a local minimum take $H_{\ast }$ positive definite. Near $x_{\ast }$, inverse Hessians are uniformly bounded, and integration along the error segment givesA Hessian Lipschitz constant $L_H$ yields $\Vert x_{n+1}-x_{\ast }\Vert \le \frac{1}{2}{L}_H\Vert H_n^{-1}\Vert \Vert x_n-x_{\ast }{\Vert }^2$. Choosing a sufficiently small invariant neighborhood proves quadratic convergence. Continuous second derivatives alone do not guarantee this quadratic rate; the Lipschitz hypothesis supplies the needed remainder bound.

At the current exact path point $v(\mu )$, stationarity is already zero and $Cy=\mu e$. Therefore $F_{\bar{\mu }}(v(\mu ))=(0,(\mu -\bar{\mu })e{)}^T$. Its Newton correction solvesThe matrix has no explicit dependence on the barrier parameter, so comparison with part (iv) giveswhen $J$ is nonsingular. If it is singular, the displayed vector is still a solution of the correction equations, but uniqueness of the correction cannot be inferred from differentiability of the path alone.

This step is exactly the tangent predictor in the stated first-order Taylor expansion. For nearby positive $\bar{\mu }$, it approximates the new path point with error $o(|\bar{\mu }-\mu |)$; a bounded second path derivative improves this to $O((\bar{\mu }-\mu {)}^2)$. Continuity also preserves strict positivity for sufficiently small parameter changes. Larger changes may require damping to keep both the constraint values and multiplier estimates positive.

For each fixed $\theta$, the Taylor expansion of the exponential yieldsThe limiting moment-generating function $e^{{\theta }^2/2}$ is that of the standard normal distribution and is finite around zero. The moment-generating function continuity theorem therefore gives

The right-sign denominator has its own small mixing contribution. In the same CP symmetry convention its decay amplitude is ${\mathcal{A}}_{\mathrm{RS}}=g_{+}+r{e}^{-i\delta }{g}_{-}$. After cancelling $e^{-\tau }$, the exact numerator and denominator areDefine $u=y\cos \delta$ and $v=x\sin \delta$. Their Taylor expansions are $N=r^2+r(u-v)\tau +\{x^2+y^2+r^2(y^2-x^2)\}{\tau }^2/4+O({\tau }^3)$ and $D=1+r(u+v)\tau +\{y^2-x^2+r^2(x^2+y^2)\}{\tau }^2/4+O({\tau }^3)$. Dividing the series gives the strict second-order answerThe usually quoted approximation additionally drops terms suppressed by powers of $r$ in the right-sign denominator:For $r=0.1$ and $x=y=0.01$, this is $0.01+0.001\tau +0.00005{\tau }^2$ for $\delta =0$, and $0.01-0.001\tau +0.00005{\tau }^2$ for $\delta =\pi /2$. The first increases; the second initially decreases, reaches approximately $0.005$ at $\tau =10$, then rises. The strict series instead has $(C_1,C_2)=(0.00099,0.000049005)$ and $(-0.00101,0.000051005)$. At $\tau =20$ the exact ratios are approximately $0.04913$ and $0.009934$; the leading approximations give $0.05$ and $0.01$.

With no neutral meson mixing, $x=y=0$ and $g_{-}=0$, so $\boxed{R=r^2=0.01}$ at all decay times. The wrong-sign signal then remains through direct doubly Cabibbo suppressed decay.

Away from the boundary, derivatives of a smoothly varying solution are much smaller than $k^{2}f$ unless the outer solution vanishes. Nonzero boundary data therefore require a rapidly varying layer. If its thickness is $\delta$, the radial second derivative is of order $f/{\delta }^2$; balancing it with $k^{2}f$ gives $\delta =k^{-1}$.

Put $R=k(1-r)$ and write $f(r,\theta )=F(R,\theta )$. The equation becomesFor fixed $R$ and boundary data with angular variation independent of $k$, the leading equation is $F_{RR}=F$. Matching to the zero outer solution removes the growing exponential, and the boundary value at $R=0$ fixes its coefficient:This approximation describes the layer; it is not a uniform expansion in polar coordinates at the origin.

Laplace's method localizes an exponential integral near maxima of its real phase. At an isolated smooth interior maximum $t_0$ with $G^{\prime \prime }(t_0)<0$ and $F(t_0)\ne 0$, Taylor expansion and Gaussian integral giveAn endpoint maximum with nonzero inward phase slope instead gives an inverse-$k$ factor. Degenerate maxima require their first nonzero local phase term; continuous differentiability alone does not guarantee the quadratic formula.

Here $\cos (ix\sin \varphi )=\cosh (x\sin \varphi )$. In its two exponentials, the positive exponential dominates at the interior maximum $\varphi =\pi /2$ of $\sin \varphi$. Since $\sin (\pi /2+s)=1-s^2/2+O(s^4)$,The negative exponential has no comparable positive maximum and is negligible. Consequently, where $kr\to \infty$,In the boundary layer $r=1-R/k$, $r^{-1/2}=1+O(k^{-1})$, so this reduces to $e^{-R}$, agreeing with the formal result when $g=1$.

A fixed small-angle approximation alone is insufficient because its error is multiplied by $x$. Introduce the shrinking cutoff $\eta =x^{-1/3}$, which satisfies $\eta \ll ϵ$ but $\sqrt{x}\eta \to \infty$.

On $[0,\eta ]$, the Taylor expansion ${\sin }^{2}t=t^2+O(t^4)$ gives $x({\sin }^{2}t-t^2)=O(x{\eta }^4)=O(x^{-1/3})$ uniformly. Thusuniformly there, and the two central integrals are asymptotically equal. On $[\eta ,ϵ]$, choose a fixed $c>0$ with $\sin t\ge ct$ for $0\le t\le ϵ$. The sine-integral tail is at most $ϵe^{-c^{2}x{\eta }^2}=ϵe^{-c^2{x}^{1/3}}$; the Gaussian tail has the same exponential bound with $c=1$. Both are $o(x^{-1/2})$, while the central Gaussian integral is asymptotic to $\sqrt{\pi }/(2\sqrt{x})$. HenceThis justifies the localized Laplace method rather than interchanging a nonuniform approximation with integration.