The two products are conjugate:Conjugation by $a^{-1}$ is a group automorphism, so part (a) provesDirectly, cancellation gives $(ba{)}^r=a^{-1}(ab{)}^{r}a$ for every positive integer $r$. Thus a power of one product is the identity exactly when the same power of the other is the identity. This proves that conjugation preserves element order, even without the finiteness assumption.

Let $n=\mathrm{ord}(a)$, which is finite. A group homomorphism preserves powers and the identity, soIf $r=\mathrm{ord}(f(a))$, divide $n$ by $r$: $n=qr+s$ with $0\le s<r$. Then $e^{\prime }=f(a{)}^n=f(a{)}^s$, and minimality of $r$ forces $s=0$. ThusIf $f$ is a group isomorphism, apply the same argument to its inverse: $\mathrm{ord}(a)\le \mathrm{ord}(f(a))$. The two inequalities yield

Choose generalized coordinates $q_i$ measured from a static equilibrium point, and assume a conservative system with no velocity-dependent forces. Keeping terms of degree two in displacements and velocities gives the small oscillation LagrangianHere $A$ is the positive-definite symmetric mass matrix evaluated at equilibrium, and $B$ is the symmetric Hessian matrix of the potential energy there. The linear potential term is zero by equilibrium, and an additive constant has been dropped. Coordinate-dependent corrections to the kinetic matrix contribute only at higher order.

The Euler-Lagrange equations are $A\ddot{\mathbf{q}}+B\mathbf{q}=0$. For a normal mode $\mathbf{q}=\mathbf{v}{e}^{i\omega t}$, they give the generalized eigenvalue problem $(B-{\omega }^{2}A)\mathbf{v}=0$. A nonzero mode shape exists exactly whenThis is a degree-$n$ polynomial in ${\omega }^2$. At a stable equilibrium its roots are nonnegative; zero roots correspond to zero modes.

Split the Cartesian second-rank tensor into symmetric and antisymmetric parts:Define $u_k=\frac{1}{2}{ϵ}_{kij}{T}_{ij}$. The symmetric part contributes zero to this contraction. The epsilon-delta identity then givesThereforeThe symmetric part and the three independent antisymmetric components determine these quantities uniquely. Under proper rotations $u$ is a vector; under reflections it is an axial vector.

For the critically damped case, causality reduces the convolution toAlthough the forcing is not integrable on the full real line, this integral converges absolutely because the Green's function is integrable. Thus the calculation is valid directly; the corresponding Fourier transform can be interpreted as a tempered distribution.

Use the complex exponential and ${\int }_0^{\infty }s{e}^{-as}ds=a^{-2}$ for $\mathrm{Re}a>0$:Since $(1+i{)}^2=2i$, the convolution response isThis is the bounded steady response to the stated forcing acting for all time. Without an initial condition or the convolution selection, the general solution additionally contains $(C_1+C_{2}t)e^{-\kappa t}$. No switch-on at $t=0$ was specified, so zero initial displacement and velocity should not be imposed on this all-time forcing.

Let $\beta =\sqrt{{\Omega }^2-{\kappa }^2}$, interpreted as $i\sqrt{{\kappa }^2-{\Omega }^2}$ in the overdamped case. Factoring the denominator givesFourier inversion and the convolution theorem givewhereThus the requested numerator isFor all positive $\kappa ,\Omega$, both poles lie in the upper half-plane. In the overdamped case their imaginary parts are $\kappa \pm \sqrt{{\kappa }^2-{\Omega }^2}$, both positive. At equality they merge into a double pole.

Apply the Fourier transform of a derivative twice to the damped harmonic oscillator equation. Under the corresponding convergence assumptions,ThusSince $\kappa ,\Omega >0$, this denominator has no real zero: its imaginary part could vanish only at $\omega =0$, where its real part is positive. The transform therefore defines a regular response on the real frequency axis.

A sufficient assumption is that $f$ is absolutely continuous, $f,f^{\prime }\in L^1(\mathbb{R})$, and $f(t)\to 0$ at both infinities. Integration by parts in the Fourier transform givesThe decay assumption removes the boundary term, and integrability makes both transforms ordinary convergent integrals. The same derivative identity extends to tempered distributions.

For a positively oriented simple closed contour, the residue theorem states that the integral of a function holomorphic on and inside the contour except at finitely many isolated interior singularities is $2\pi i$ times the sum of its residues. There must be no singularity on the contour.

Factor the denominator as $(z^2+z+1)(z^2-z+1)$. For $R>1$, the upper semicircle encloses the two simple polesThe other two poles are in the lower half-plane. Since $D^{\prime }(z)=4z^3+2z=2z(2z^2+1)$,Writing $h=\sqrt{3}/2$ and using $1/a=1/2-ih$, $1/b=-1/2-ih$, their numerator difference simplifies toThus for every $R>1$ the closed-contour integral isTo remove the arc, on $C_R$ we have $|e^{iz}|=e^{-\mathrm{Im}z}\le 1$ and $|z^4+z^2+1|\ge R^4-R^2-1$. For sufficiently large $R$, the complex line integral estimate givesThe real-line integral is absolutely convergent. Its imaginary part is zero by oddness of $\sin x/(x^4+x^2+1)$. Therefore

The only denominator zero is $z_0=-1$, which lies strictly inside the given circle. The numerator is an entire function. Apply the derivative form of the Cauchy integral formula with $n=6$:The sixth derivative is $-\sin z$, so its value at $-1$ is $\sin 1$. Consequently

A sufficient contour setting for both results is a positively oriented, simple, closed, piecewise smooth contour $C$, with the function holomorphic on an open neighbourhood of $C$ and its entire interior. The Cauchy integral theorem then statesIf $z_0$ is strictly inside $C$, the Cauchy integral formula statesThe contour must not pass through $z_0$. Positive orientation means anticlockwise; reversing it reverses the sign of the contour integral. The holomorphy assumption throughout the interior excludes hidden singularities there.