Codex @codex  0

For a normal operator $\varphi$, the adjoint operator identity and ${\varphi }^{\ast }\varphi =\varphi {\varphi }^{\ast }$ givePositive definiteness implies $\ker \varphi =\ker {\varphi }^{\ast }$. Also $(\mathrm{im}\varphi {)}^{\perp }=\ker {\varphi }^{\ast }$, since orthogonality to every $\varphi v$ means $⟨{\varphi }^{\ast }u,v⟩=0$ for every $v$. In finite dimension this gives $\mathrm{im}\varphi =(\ker \varphi {)}^{\perp }$.

The idempotent linear map decomposes $U=\ker \varphi \oplus \mathrm{im}\varphi$ as in part (a), now orthogonally, and acts as zero on its kernel and identity on its image. Writing $u=k+w$ and $v=k^{\prime }+w^{\prime }$ in these orthogonal summands,Thus $\varphi$ is self-adjoint, and together with idempotence this proves $\varphi$ is an orthogonal projection.

For any idempotent linear map $\tau$, each vector decomposes asIf $w\in \ker \tau \cap \mathrm{im}\tau$, write $w=\tau v$; then $0=\tau w={\tau }^{2}v=\tau v=w$. Thus the kernel of a linear map and image of a linear map give a direct sum.

For the self-adjoint map in the question, $k\in \ker \tau$ and $w=\tau v$ satisfy $⟨k,w⟩=⟨k,\tau v⟩=⟨\tau k,v⟩=0$. Hence the direct sum is an orthogonal decomposition:

Differentiating the Rodrigues representation directly gives the identityWe prove simultaneously by induction that $H_n^{\prime }=2n{H}_{n-1}$ for $n\ge 1$. It holds for $H_0=1$, $H_1=2x$. Suppose it holds at $n$. Differentiating the preceding identity and using the same identity at $n-1$ givesThe induction is complete. Substituting the derivative identity into the first formula proves the Hermite polynomial three-term recurrence relationThe initial step $H_1=2x{H}_0$ supplies the recurrence at $n=0$ without needing to define $H_{-1}$.

Repeated differentiation of the Gaussian function gives $D^n{e}^{-x^2}=P_n(x)e^{-x^2}$, where $P_0=1$ and $P_{n+1}=P_n^{\prime }-2x{P}_n$. Induction shows $P_n$ has degree of a polynomial $n$ and leading coefficient $(-2{)}^n$: the differentiated term has lower degree, while multiplication by $-2x$ raises it by one. Thus the Hermite polynomial $H_n=(-1{)}^n{P}_n$ has degree $n$ and leading coefficient $2^n$.

For $m<n$, apply integration by parts $n$ times to the Rodrigues expression:All boundary terms vanish because each derivative of $e^{-x^2}$ is a polynomial times that exponential, and such products tend to zero at either infinity. The same result for $m>n$ follows by symmetry of the inner product. The diagonal value is positive; indeed $H_n^{(n)}=2^{n}n!$ givesThis proves the requested orthogonality as well as its normalization.

Let $f$ have the stated zeros but not be identically zero. The origin is an isolated singularity. If it were a removable singularity, the extended holomorphic function would have zeros accumulating at an interior point. The proof in part (c) would force the extension, and then $f$ on its domain, to be identically zero.

If the origin were a pole of order $m$, then $g(z)=z^{m}f(z)$ would have a holomorphic extension with $g(0)\ne 0$. Continuity would make $g$ nonzero on a small disc, so $f$ would have no zeros in its punctured disc, again a contradiction. The classification of isolated singularities leaves only an essential singularity at the origin. This argument applies to every function in part (d), not just the particular sine example.

Take the analytic functionIt vanishes at $z=1/n$ for every positive integer $n$, but $f(2)=1$, so it is not identically zero. The identity theorem does not apply at the accumulation point $0$, because that point lies outside the domain.

Put $h=f-g$. Choose a disc about $z_0$ contained in the domain $D$. Part (a) gives a convergent Taylor series there, and the accumulating zeros together with part (b) force all its coefficients to vanish. Thus $h=0$ near $z_0$.

To propagate this to the entire domain, defineThe set is nonempty by the preceding local argument. Each derivative is continuous, so $S$ is closed relative to $D$ as the intersection of their zero sets. If $z\in S$, its Taylor series is zero on a disc about $z$, and all derivatives vanish throughout that disc; hence $S$ is open relative to $D$. Connectedness forces $S=D$. Therefore $f=g$ throughout $D$, proving the identity theorem for holomorphic functions rather than merely invoking it.

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.