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Use the complex amplitude and take its real part at the end; the equations are linear with real coefficients. Let $M=\lambda +2\mu$, $\omega =kc$, and $v=(A,B{)}^T$. The differential operators on this exponential function are ${\partial }_X=ik$, ${\partial }_Y=-b$, and ${\partial }_t=-i\omega$. Substitution givesThe determinant factors asConsequently choose the positive decay rootsThe Rayleigh wave decay exponents must both give strict decay into the half-space. A genuine nonzero Rayleigh wave combines both, as the free-surface equations below also show. Thus $b_1,b_2>0$ requires $\boxed{0<c<\min (c_T,c_L)}$. In a stable ordinary isotropic solid $c_T<c_L$, so it is slower even than the transverse wave. The roots depend on the candidate speed as well as the material constants and wavenumber; the speed is fixed by the boundary condition.

The dominated convergence theorem says that if measurable $f_n$ converge almost everywhere to $f$ and $|f_n|\le H$ almost everywhere for a single Lebesgue integrable function $H\ge 0$, then $f$ is integrable andHere, for $x>0$, put $H_0(x)=x^{-1/2}{e}^{-x}$ and $f_n(x)=H_0(x)(1-(-1{)}^n{e}^{-nx})$. The exponential function gives $e^{-nx}\to 0$ at every such $x$, so $f_n\to H_0$. Also $|f_n(x)|\le 2H_0(x)$. This dominator is integrable: on $(0,1)$, $H_0(x)\le x^{-1/2}$, whose integral is two; on $[1,\infty )$, $H_0(x)\le e^{-x}$, whose integral is finite. DCT therefore proves the required limit.

The limiting value can also be evaluated. Substitute $x=t^2$ and use the Gaussian integral:In fact, scaling the second exponential term gives the exact integral $\sqrt{\pi }[1-(-1{)}^n/\sqrt{n+1}]$. The integral error decays as $n^{-1/2}$ despite exponential convergence at each fixed positive $x$; the slow part lies near the endpoint zero.

Use the power series of the exponential function:Each term is a Lebesgue integrable function, andThe sum of these nonnegative integrals is at most $2{\sum }_{n\ge 0}1/n!=2e<\infty$. Applying integration of a series with summable L1 norms, just proved in part (ii), therefore justifies exchanging the series and the integral:The absolute-integral bound is the justification; pointwise convergence of the exponential series alone would not suffice.

The absolute value is bounded by $e^{-ax}$, whose integral is $1/a$. Thus the function is Lebesgue integrable. The complex integral of an exponential function isTaking the real part, justified by absolute integrability, gives

For the linear differential operator $L=(D-\alpha )(D-\beta )$, the exponential function is an eigenfunction: $L{e}^{ϵx}=(ϵ-\alpha )(ϵ-\beta )e^{ϵx}$. Distinct $\alpha ,\beta$ give the two homogeneous solutions, while the method of undetermined coefficients supplies a particular solution at the distinct forcing exponent $\gamma$:

To invert the complex cosine, let $z=\cos w$ and put $u=e^{iw}$. Since the exponential function is nonzero, $u\ne 0$, and Euler's formula gives $2z=u+u^{-1}$. Multiplying by $u$ yields the quadratic equation $u^2-2zu+1=0$. ThusChoosing either square root of $1-z^2$ merely exchanges the two signs. The two values have product one, so neither vanishes; taking a complex logarithm is legitimate.

For each choice of sign, all solutions of $e^{iw}=u$ satisfy $iw=\mathrm{Log}u+2\pi in$. Therefore the full inverse relation isConversely, exponentiating any displayed value recovers one root of the quadratic equation, and hence $(e^{iw}+e^{-iw})/2=z$, proving completeness. With a multivalued complex logarithm, the $2\pi n$ term is already included and this is the source's compact formula for inverse cosine.

The principal inverse cosine requires compatible choices of logarithm branch, square root and sign. Both signs cannot represent that same principal function indiscriminately: at $z=0$, the two principal-log values are $\pi /2$ and $-\pi /2$, whereas the principal $\arccos 0$ is $\pi /2$. Thus the printed $\pm$ is correct for the inverse relation, not a pair of interchangeable principal-value formulas.

Differentiate the Wronskian and cancel the mixed first-derivative terms:This is the constant-coefficient form of Abel's identity. Hence $W(x)=C{e}^{-ax}$, or $W(x)=W(x_0)e^{-a(x-x_0)}$. Since the exponential function is never zero, $W$ is either identically zero or nowhere zero on a connected interval of the homogeneous equation.

For a genuinely second-order linear ordinary differential equation, assume $a_2\ne 0$. Substituting the exponential function $f=e^{rt}$ into the homogeneous linear differential equation gives $a_2{r}^2+a_{1}r+a_0=0$. Oscillations arise precisely when its two roots of a polynomial have nonzero imaginary parts, namelyIn that case, writing $\omega =\sqrt{4a_0{a}_2-a_1^2}/(2|a_2|)>0$, the real complementary function isEvery nonzero such solution has an oscillatory factor; the oscillation may decay, grow, or keep constant amplitude according to the sign of $a_1/a_2$. This is the oscillatory-root criterion for a real second-order equation; decay is not a prerequisite for oscillation.

For completeness, if $a_2=0$ and $a_1\ne 0$, only real exponential functions occur. If $a_2=a_1=0$ and $a_0\ne 0$, only $f=0$ is allowed. If all three coefficients vanish, every twice-differentiable function is allowed, including oscillatory ones; the second-order criterion does not apply to this degenerate identity.

Combine the terms before taking their limit:Using the exponential function Maclaurin series, the numerator is $x^2/2+O(x^3)$ and the denominator is $x^2+O(x^3)$. ConsequentlyThe cancellation removes the two individual singularities. Subtracting their separate divergent limits would not determine the finite answer.

Recognize the derivative of the exponential function at $x=1$:Here $2^x=e^{x\ln 2}$ and the chain rule supplies the factor $\ln 2$.

For the real-valued expression, first consider the one-sided limit $x\to 0^{+}$. Its natural logarithm is $x\ln (\ln (1+x))$. Since $\ln (1+x)/x\to 1$,using $x\ln x\to 0$. The continuity of the exponential function therefore gives right-hand limit $1$.

The printed two-sided notation needs qualification: $\ln (1+x)<0$ for $-1<x<0$, so raising it to an arbitrary real power does not define a real-valued function throughout a left neighbourhood of zero. There is consequently no ordinary two-sided real limit on an open punctured neighbourhood with the expression as written. The value $1$ is the natural right-hand interpretation.

The successive derivatives are $f^{(n)}(x)=(-6{)}^n{e}^{-6x}$, so $f^{(n)}(-4)=(-6{)}^n{e}^{24}$. Substitution in the Taylor series formula about $a=-4$ givesThe ratio test shows convergence for every finite real or complex $x$: the absolute ratio of successive terms is $6|x+4|/(n+1)\to 0$. This is also the usual exponential function series applied to $e^{24}{e}^{-6(x+4)}$, proving that the series equals the stated function everywhere.

For $|x|<1$, the geometric series gives $(1+x{)}^{-1}={\sum }_{j=0}^{\infty }(-x{)}^j$. Integrating from zero to $x$ yields the Maclaurin seriesOn the real axis the series also converges at $x=1$ to $\ln 2$, but diverges at $x=-1$.

For fixed real $a$ and sufficiently large positive $x$, the base in the requested limit is positive. Apply the series with argument $-a^2/x^2$:Continuity of the exponential function therefore givesThis includes $a=0$, when the expression is identically one.

The product rule and chain rule giveThe exponential function is strictly positive, so there are exactly two stationary points, at $\boxed{x=\pm 1/\sqrt{2}}$. The derivative is negative outside these two values and positive between them. Thus the negative value is a strict local minimum and the positive value a strict local maximum.

The Maclaurin series of the exponential function and sine giveBoth the numerator and denominator first become nonzero at order $x^4$. Dividing by that common power gives the limit

The expectation follows by integration by parts, with $u=t$ and $dv=e^{-t/25}dt/25$:The endpoint term vanishes because the exponential function decays faster than $t$ grows. Hence the mean lifetime is

In the set of all $n!$ permutations, let $A_i$ consist of those fixing $i$. For a prescribed collection of $k$ fixed points, its intersection has $(n-k)!$ members. Applying the inclusion-exclusion principle to the complement of ${\bigcup }_i{A}_i$ gives the number of derangements:The convergent power series for the exponential function at $-1$ therefore gives $\boxed{d_n/n!⟶e^{-1}}$.

The composition $g(z)=e^{-f(z)}$ is an entire function, andBy Liouville's theorem, $g$ is constant. Since the exponential function never vanishes,Thus $f$ is constant. Notice that constancy of an exponential alone need not be inverted by choosing a global logarithm; differentiating supplies the conclusion directly.

Use the local coordinate $t=1/\lambda$. The differential system becomesFor the nonzero quadratic leading term, the coefficient of the local linear ordinary differential equation system has a pole of order four, so infinity is an irregular singular point of Poincaré rank three in this gauge. The matrix-valued function $A(\lambda )$ itself has a pole of order two; the extra factor $d\lambda /dt=-t^{-2}$ is essential when classifying the equation. If $A_0$ vanishes the degree and these orders decrease. The specified distinct eigenvalues $1,-1$ give the nondegenerate unramified case.

The leading exponential function factors have actions $q_{\pm }=\pm {\lambda }^3/3$, so their difference is $2{\lambda }^3/3$. Equal exponential magnitudes occur onThese are commonly called Stokes rays in the sectorial ODE convention. With the action-phase convention used by the wiki's Stokes line article they are anti-Stokes lines; the same-phase rays instead haveGiving both defining equations removes the nomenclature ambiguity; both sets contain six directions.

The standard construction with distinct leading eigenvalues uses six sectorial fundamental solutions per turn, with basic angular spacing $\pi /3$. They are fundamental matrices, each containing two independent column solutions. On the logarithm cover choose six consecutive overlapping sectors of width $\pi /3+2\delta$, with $\delta >\epsilon /2$ small. Their union has angular opening $2\pi +2\delta$, covering the requested $2\pi +\epsilon$. One often also writes a seventh index to compare the closing sector with the first after a full turn; that is a closing copy, not a seventh independent sector in the six-sector construction. The number is a generic sectorial fundamental solution normalization count, not a universal minimum for exceptional systems with vanishing Stokes matrices: the exactly diagonal system already has one global elementary fundamental matrix.

There is a missing qualification in the requested product identity. Polynomial coefficients give an entire invertible fundamental matrix, hence trivial actual monodromy about a large circle: the circle contracts through ordinary points. But the formal matrix generally has the form $\hat{G}{\lambda }^{\Theta }{e}^Q$, and a positive turn changes its logarithm branch by the formal monodromy $M_f=e^{2\pi i\Theta }$. This factor cannot be discarded.

To specify the printed product order, let $S_j$ act on column vectors of solution coefficients across the $j$th sector boundary, so adjacent normalized matrices satisfy $Y_{j+1}=Y_j{S}_j^{-1}$. Six transitions give $Y_7=Y_1(S_6\cdots S_1{)}^{-1}$. The normalized formal branch after the turn is ${\hat{Y}}_1{M}_f$, so sectorial uniqueness gives $Y_7=Y_1{M}_f$. ConsequentlyUsing Stokes matrices for right changes of fundamental matrices instead reverses/inverts the corresponding convention; the formal monodromy factor remains necessary. The formal monodromy correction to an entire-system Stokes product shows that the printed identity holds when $M_f=I$, including the corrected Lax specialization discussed below.

It does not follow from polynomiality alone. For example takeThe coefficient matching in question 2 gives $\Theta =\frac{1}{2}\mathrm{diag}(1,-1)$, hence $M_f=-I$ and the ordered Stokes matrix product is $-I$, not $I$, despite having no finite singularity. This is an explicit counterexample to the unqualified final clause.

Apply the Runge-Kutta method to the Dahlquist test equation and put $z=h\lambda$. Its stability function isSince $A$ is an invertible matrix, $Q$ has polynomial degree $s$. The adjugate matrix formula gives $\mathrm{deg}P\le s$, but the additional condition givesso actually $\mathrm{deg}P\le s-1$. The specified order of a numerical method implies $R(z)=e^z+O(z^{2s})$ near zero. Therefore the Padé identification of a Radau stability function makes $R$ the $[s-1/s]$ Padé approximant to the exponential function: its numerator and denominator have the required degrees, and its Taylor series agrees through degree $2s-1$.

In the permitted A-stability of near-diagonal exponential Padé approximants, take $m=s-1$, $n=s$. Then $n-2\le m\le n$, so the scheme is A-stable. The same stability function also tends to zero at infinity, giving L-stability.