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Let $h=f-g$, whose power series coefficients are $c_n=a_n-b_n$. Suppose some coefficient is nonzero, and let $N$ be the least such index. Inside the common convergence disc,The parenthesized analytic function tends to $c_N\ne 0$ as $z\to z_0$, so it does not vanish in some neighborhood. Hence $h$ has no zeros sufficiently near $z_0$ except possibly $z_0$ itself. This contradicts $h(z_k)=0$ with $z_k\ne z_0$ and $z_k\to z_0$. Thus all coefficients agree:This also proves the local isolation of zeros of a nonzero analytic function.

The complex Taylor series theorem states that if $f$ is holomorphic in $|z-z_0|<R$, thenThe power series converges absolutely there and uniformly on each smaller closed disc; its coefficients are unique. For $|z-z_0|<r<R$, the Cauchy integral formula also gives $f^{(n)}(z_0)/n!=(2\pi i{)}^{-1}{\int }_{|\zeta -z_0|=r}f(\zeta )(\zeta -z_0{)}^{-n-1}d\zeta$. This is the analytic, rather than finite real-variable, form of Taylor theorem relevant here.

A mode has dimensionless damping ratio $k/{\alpha }_n$, and characteristic rootsIf $k/{\alpha }_n\ll 1$, the normal mode oscillates at ${\omega }_n={\alpha }_n+O(k^2/{\alpha }_n)$ with exponentially decaying envelope $e^{-kt}$. As $k\to 0$, the solution approaches the undamped wave equation response.

If $k/{\alpha }_n\gg 1$, the roots are real and negative, withThe mode has a fast transient and a slow nonoscillatory relaxation. After the fast transient, neglecting $y_{tt}$ gives the diffusion equation $y_t\simeq c^2{y}_{xx}/(2k)$ for these long-wavelength modes. At $k={\alpha }_n$, the repeated root produces $t{e}^{-kt}$: this is critical damping.

Thus the critical parameter is $k/{\alpha }_n=1$ for each mode, and the fundamental threshold isWhen $k<\pi c/l$ every mode is oscillatory. For any fixed, finite $k$, sufficiently large $n$ still gives oscillatory modes, so large damping of the lowest modes does not mean that every mode of the full string is overdamped.

Assume the impulse point is in the interior, $0<a<l$. Expand the displacement in the fixed-end Fourier sine seriesOrthogonality gives the initial velocity coefficient by applying the Dirac delta to the sine function:Substitution in the wave equation yields $T_n^{\prime \prime }+2k{T}_n^{\prime }+{\alpha }_n^2{T}_n=0$. For $k<{\alpha }_n$, its roots are $-k\pm i{\omega }_n$, where ${\omega }_n=\sqrt{{\alpha }_n^2-k^2}$. The two initial conditions determineHence the normalized velocity-impulse Green function for a damped string isFor a mode with critical damping, use the limit $e^{-kt}t$ in place of the time factor. For $k>{\alpha }_n$, put ${\eta }_n=\sqrt{k^2-{\alpha }_n^2}$ and use $e^{-kt}\sinh ({\eta }_{n}t)/{\eta }_n$. These choices solve the same modal ordinary differential equation and initial conditions, so the formula covers arbitrary $k>0$. The impulse initial velocity is a distribution, so initial differentiation of the series is interpreted distributionally.

The printed formula requires correction: the factor depending on $n$ must lie inside the sum, and sine-mode normalization requires $2/l$, rather than $2$. The printed slash after the final sine is also stray. Direct differentiation at $t=0$ verifies the corrected normalization: the initial sine coefficients are exactly those of $\delta (x-a)$.

Parametrize the same contour by $z=e^{it}$, $0\le t\le 2\pi$. Then $dz/z=idt$ and $z-z^{-1}=2i\sin t$. Part a becomesCanceling the nonzero common factors gives

Take the unit circle counterclockwise and $n\ge 0$. The integrand is a Laurent polynomial with its only possible singularity at zero. Its expansion isThe $z^{-1}$ term is the term $j=n$, so the residue is $(-1{)}^n(\genfrac{}{}{0ex}{}{2n}{n})$. The residue theorem gives

Suppose $f(n)\le N$ for all $n$ and $f(n)=N$ eventually. If $k>N$, then $f^{-1}(I_k)=\varnothing$. If $k\le N$, the preimage contains all sufficiently large indices and is cofinite. Every open set therefore has open preimage, proving $f$ is continuous. Together with the introductory argument this proves the claimed if-and-only-if classification.

If $f(n)\to \infty$, then for each $k$ the complement of $f^{-1}(I_k)$ contains only finitely many indices. Thus every nonempty open tail has a cofinite preimage, and the empty set has empty preimage. Hence $f$ is continuous from the cofinite space to the tail-topology space.

Let $r=\mathrm{tr}C$, $s=\mathrm{tr}D$, and $\beta =(1-i)/2$. Their traces are real by part c, and $\Phi (C)=\beta rI+iC$. Thus $\mathrm{tr}\Phi (C)=(2\beta +i)r=r$. Expanding the product and using $2{\beta }^2+2i\beta =1$ givesThereforeIn particular the trace expression is real and symmetric on $Q$. Under the coordinates $C=aI+i\mathbf{v}\cdot \mathbf{A}$, $D=dI+i\mathbf{w}\cdot \mathbf{A}$, it is $-ad+\mathbf{v}\cdot \mathbf{w}$. Thus the trace form on scalar-plus-skew-Hermitian two-by-two matrices has rank $4$, inertia $(3,1)$, or numerical signature $2$, agreeing with $b$ under the real isomorphism.

Taking traces in the defining equation gives $\mathrm{tr}C+\overline{\mathrm{tr}C}=2\mathrm{tr}C$, so the trace is real. Put $a=\mathrm{tr}C/2$. Then $K=C-aI$ is traceless and satisfies $K^{\ast }=-K$, so $-iK$ is a traceless Hermitian matrix. Hence every element has a unique representationConversely each such matrix satisfies the defining equation. This proves $Q$ is a real four-dimensional vector space, with basis $I,i{A}_1,i{A}_2,i{A}_3$.

Substitution into the given map givesIts values are Hermitian, and it acts on real coordinates by $(a,v_1,v_2,v_3)\mapsto (a,-v_1,-v_2,-v_3)$. This invertible real linear coordinate transformation proves the required isomorphism. The inverse sends $aI+{\sum }_j{u}_j{A}_j$ to $aI-i{\sum }_j{u}_j{A}_j$.

The displayed matrices are the three Pauli matrices, in the order ${\sigma }_3,{\sigma }_1,{\sigma }_2$. They satisfy $\mathrm{tr}{A}_i=0$ and $\mathrm{tr}(A_i{A}_j)=2{\delta }_{ij}$. Every Hermitian matrix has a unique expression $aI+{\sum }_i{v}_i{A}_i$ with all four coefficients real. Therefore they form a basis together with $I$.

The preceding trace identities give $b(I,I)=-1$, $b(I,A_i)=0$, and $b(A_i,A_j)={\delta }_{ij}$. The Gram matrix is thusConsequently rank $=4$ and inertia signature $(3,1)$. If signature is defined as the number of positive squares minus the number of negative squares, its value is $2$. This is the Hermitian matrix representation of Minkowski four-vectors, with the negative-determinant sign convention.

A Hermitian two-by-two matrix has the formAddition and real scalar multiplication preserve this condition, and the two diagonal entries plus the real and imaginary parts of $z$ give four independent real coordinates. Thus $S$ is a real four-dimensional vector space. The trace product is real because $\overline{\mathrm{tr}(AB)}=\mathrm{tr}(BA)=\mathrm{tr}(AB)$; this also proves symmetry of $b$.

Directly, $\mathrm{tr}(A^2)={\alpha }^2+{\delta }^2+2|z{|}^2$ and $(\mathrm{tr}A{)}^2={\alpha }^2+2\alpha \delta +{\delta }^2$. HenceSince $\mathrm{tr}I=2$ and $\mathrm{tr}(AI)=\mathrm{tr}A$,

By part a the transformed function solves the same equation with ${\mu }_n$, and by part c its polynomial factor has degree $n$. The uniqueness proved in part b therefore gives $q_n=c_n{p}_n$. Comparing their leading coefficients givesFor the unitary convention used above, the Gaussian integral gives $c=1$, so $c_n=(-i{)}^n$. With the unnormalized phase-$e^{-i\lambda x}$ convention, $c=\sqrt{2\pi }$ instead. This proves that Hermite functions are Fourier eigenfunctions, with the normalization-dependent scalar fully specified.

Multiplication by $x$ transforms to $id/d\lambda$. Write $p_n(x)={\sum }_{j=0}^n{a}_j{x}^j$ and use $\hat{g}=cg$ for the Gaussian $g=e^{-x^2/2}$. ThenSuccessive differentiation of the Gaussian gives a polynomial of degree $j$ with leading term $(-1{)}^j{\lambda }^j$ multiplying that same Gaussian. Thus $q_n$ has leading coefficient $c{a}_n(-i{)}^n\ne 0$. This proves $q_n$ is a polynomial of degree exactly $n$, rather than merely at most $n$.

Set $f=p(x)e^{-x^2/2}$. Differentiating givesThus the polynomial must satisfy $p^{\prime \prime }-2x{p}^{\prime }-(\mu +1)p=0$. If it has degree $n$ and nonzero leading coefficient, comparison of highest powers gives $\mu =-(2n+1)$. For this value the equation becomes the Hermite differential equationWrite $p={\sum }_{j=0}^n{a}_j{x}^j$, with coefficients beyond $n$ zero. Comparing powers yieldsThe equation at $j=n-1$ forces $a_{n-1}=0$, and downward recurrence removes all terms of the opposite parity and determines all same-parity terms from $a_n$. Taking $a_n=1$ produces the explicit polynomialIt satisfies every coefficient equation, so existence as well as uniqueness up to a nonzero scalar is proved. HenceThe monic $p_n$ is $2^{-n}$ times the usual physicists' Hermite polynomial.

Use $\hat{f}(\lambda )=(2\pi {)}^{-1/2}{\int }_{\mathbb{R}}{e}^{-i\lambda x}f(x)dx$. The transform statement requires sufficient decay and regularity for the transform and its derivatives; functions in the Schwartz space suffice, and all the functions constructed below belong to that space. Integration by parts twice and differentiation under the integral giveTransforming the differential equation therefore givesThe differential equation by itself does not ensure that an ordinary Fourier transform exists: $f=e^{x^2/2}$ solves it with $\mu =1$ but grows too rapidly. Thus the usual transformability assumptions are implicit in this part.

Choose a fixed reference basis and let $V,W$ have columns $v_i,w_i$. Both matrices are invertible. The vectors $v_i+\lambda w_i$ are dependent exactly when $\det (V+\lambda W)=0$. This is a polynomial in $\lambda$ of degree $m$, whose leading coefficient is $\det W\ne 0$. A nonzero degree-$m$ polynomial has at most $m$ distinct real roots. Thus there are at most $m$ exceptional parameters. This is the determinant argument for the matrix pencil $V+\lambda W$.

Assume for contradiction that $n>m$. Apply part b to the matrix of the given relation coefficients, whose columns are nonzero. For its first dependent prefix choose $b_1,\dots ,b_k$ withMultiply the $j$th given vector relation by $b_j$ and sum. Interchanging the finite sums givesThe $w_i$ are a basis, so all $z_i$ must be zero, contradicting part b. Therefore $n\le m$.

Choose any nonzero vector $b$ in the one-dimensional kernel of $A_k$. It gives ${\sum }_{j=1}^k{a}_{ij}{b}_j=0$ for every row. Because no column is zero, $b$ must have at least two nonzero entries: a relation supported at a single index would make that column zero.

Suppose, contrary to the claim, that $A_k\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_k)b=0$. Its kernel is spanned by $b$, so there is a scalar $c$ with ${\lambda }_j{b}_j=c{b}_j$ for every $j$. At the two nonzero entries this gives two equal weights ${\lambda }_j=c$, contradicting their distinctness. Thus the weighted sum is nonzero in at least one row. This proves that distinct diagonal weights break a first column dependence. Some entries of $b$ may be zero; only two nonzero entries are needed.

By the rank-nullity theorem, $\mathrm{nullity}A=n-\mathrm{rank}A\ge n-m>0$. Thus a first dependent column prefix exists. Minimality of $k$ makes the first $k-1$ columns independent, so $\mathrm{rank}{A}_{k-1}=k-1$. Adding one column cannot reduce this rank, while nonzero nullity of $A_k$ implies rank at most $k-1$. HenceFor $k=1$ the earlier prefix is the zero-dimensional empty matrix, and the same argument applies.