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The original PDF has coefficient $c_n=n^{n^{1/3}}$. Its $n$th root isThe Cauchy-Hadamard theorem therefore gives $\boxed{R=1}$. Again the power series converges absolutely inside the unit circle, while on that circle its terms do not tend to zero. The converted TeX loses the outer exponent and writes $n^{1/3}$; the calculation here uses the printed coefficient.

For the power series coefficient $c_n=n^2$, the Cauchy-Hadamard theorem givesThus the radius of convergence is $\boxed{R=1}$. The power series converges absolutely for $|z|<1$ and diverges for $|z|>1$. On $|z|=1$ its terms have modulus $n^2$, so they do not tend to zero and the series diverges there as well.

Let $R_f,R_g$ be the two radii of convergence. First take $|x|<t<R_f$, with $t>0$. Absolute convergence at $t$ bounds the terms: $|a_n{t}^n|\le M$ for all $n$. ThusAt $x=0$ the right-hand series is finite. Otherwise it converges by the ratio test, because its limiting ratio is $|x|/t<1$. So the differentiated power series converges absolutely at every interior point of $f$, proving $R_g\ge R_f$. This proof also covers $R_f=\infty$ by choosing a finite $t$ above any given $|x|$; if $R_f=0$, that inequality is automatic.

Conversely, if $0<|x|<R_g$, absolute convergence givesThe constant term does not affect convergence, and both series converge at zero. Hence $R_f\ge R_g$, including the zero and infinite cases. Therefore $\boxed{R_f=R_g}$. This proof uses coefficient estimates, rather than assuming the termwise differentiation of a power series result whose radius assertion is being established.

The radius of convergence of a power series is the extended real number $R\in [0,\infty ]$ characterized by absolute convergence for $|x|<R$ and divergence for $|x|>R$. Equivalently,For $R=0$, only the central point is guaranteed; for $R=\infty$, every real point is inside. At finite endpoints $x=\pm R$, convergence must be investigated separately and is not determined by $R$.

Since the common radius of convergence is infinite, the termwise differentiation of a power series theorem applies on the entire real line. For the even series, differentiation and the index change $m=n-1$ giveThe odd series similarly gives $g^{\prime }(x)={\sum }_{n=0}^{\infty }(-1{)}^n{x}^{2n}/(2n)!=f(x)$. Thus $\boxed{f^{\prime }=-g,g^{\prime }=f}$ everywhere. By the product rule,The mean value theorem implies that a differentiable function with zero derivative is constant on the real line. At zero, the series give $f(0)=1$ and $g(0)=0$. Consequently $\boxed{f(x{)}^2+g(x{)}^2=1\text{ for every }x\in \mathbb{R}}$. This derives the identity from the power series without assuming a trigonometric identity in advance.

For any fixed real $x\ne 0$, the ratios between consecutive nonzero terms in the absolute-value series areandThe ratio test proves absolute convergence of both power series at every such $x$; convergence at zero is immediate. Both radii of convergence are therefore $\boxed{\infty }$. Applying the test to the nonzero terms avoids the undefined coefficient ratios caused by alternate zero coefficients when these are viewed as full power series in $x$.

The bipartite graph property means that a walk starting in $V_0$ ends there precisely when its length is even. Thus the even-length walk generating function on a bipartite graph is $Z_G^0(x)={\sum }_{k\ge 0}{\sigma }_{2k}{x}^{2k}$. Deleting the last edge of an odd-length self-avoiding walk leaves an even-length one; each such prefix has at most $\Delta$ possible last-edge choices. Hence ${\sigma }_{2k+1}\le \Delta {\sigma }_{2k}$, and for $x\ge 0$,These inequalities also hold with infinite values. Their finite positive prefactor shows that the two power series have the same radius of convergence $1/\mu$; alternatively apply the root formula to the even subsequence.

The equal rooted counts imply that the self-avoiding walk generating function is $Z_G(x)={\sum }_{n\ge 0}{\sigma }_n{x}^n$. The Cauchy-Hadamard theorem identifies the radius of a power series $\sum a_n{x}^n$ as the reciprocal of $\mathrm{limsup}|a_n{|}^{1/n}$. ConsequentlyEquivalently, the root test gives convergence when $x\mu <1$ and divergence when $x\mu >1$. At the positive boundary $x={\mu }^{-1}$, the preceding infimum formula gives ${\sigma }_n{\mu }^{-n}\ge 1$, so this particular series diverges there as well.

A Lie group is a finite-dimensional smooth manifold with a group structure for which multiplication and inversion are smooth. The group defined here is the compact symplectic group $Sp(n)=U(2n)\cap Sp(2n,\mathbb{C})$, rather than the full complex symplectic group. Also, the displayed matrix expression requires $B$ to be $2n\times 2n$; the printed $n\times n$ size is incompatible with $J$.

Since $J^2=-I$, we have $J^{-1}=-J$. The matrix exponential commutes with conjugation, as follows term by term from its absolutely convergent power series. ConsequentlyTo construct logarithm charts for the compact symplectic group, consider the real vector spaceDifferentiating the defining identities at $I$ gives precisely these conditions. Conversely if $X\in \mathfrak{k}$, $e^X$ is unitary andso $e^X\in Sp(n)$. These are the infinitesimal conditions of the compact symplectic Lie algebra.

Near $I$, the convergent matrix logarithm series $\log (I+C)={\sum }_{m\ge 1}(-1{)}^{m+1}{C}^m/m$ is smooth and inverse to the matrix exponential near zero. These local inverses respect transpose, conjugate transpose, and conjugation; also $\log (A^{-1})=-\log A$ when both matrices are sufficiently close to $I$. Shrink their neighborhoods accordingly. If $A\in Sp(n)$ there, unitarity givesThe symplectic identity is equivalent to $A^t=J^{-1}{A}^{-1}J$, so, putting $X=\log A$, it gives $X^t=-J^{-1}XJ=JXJ$, equivalently $XJ+J{X}^t=0$. Thus this local logarithm restricts to a bijection between a neighborhood of $I$ in $Sp(n)$ and an open neighborhood of zero in $\mathfrak{k}$. Its inverse is the restricted matrix exponential. These restrictions are manifold charts; left multiplication translates them to every $A_0\in Sp(n)$ via $A\mapsto \log (A_0^{-1}A)$. The chart overlaps are smooth compositions of multiplication, exponential and logarithm. The subspace topology is Hausdorff and second countable, inherited from the finite-dimensional matrix space, so these charts give a smooth manifold.

Closure under products and inverses follows from $AJ{A}^t=J$ and unitarity. Matrix multiplication is polynomial in real and imaginary entries, and inversion on the unitary group is $A\mapsto A^{\ast }$, a real linear operation. Their restrictions are smooth in the charts just constructed. Hence $Sp(n)$ is a Lie group without needing a closed-subgroup theorem.

Write $X$ in $n\times n$ blocks. The two infinitesimal conditions giveThe skew-Hermitian matrix $P$ has $n^2$ real parameters; the complex symmetric matrix $Q$ has $n(n+1)/2$ complex parameters, hence $n(n+1)$ real parameters. ThereforeFor $n=1$, any two-by-two matrix satisfies $AJ{A}^t=(\det A)J$, so the group is $SU(2)$. Explicitly,The rows are orthonormal and the determinant is one; conversely unitarity and determinant one force this form. This is the SU(2) as the three-sphere parametrization. The map and its inverse, extraction of the first row, are smooth. Thus $Sp(1)$ is diffeomorphic to $S^3$.

The Taylor series definition says that $f\in C^{\infty }(\mathbb{R})$ and, for every $a$, there is $r>0$ such thatThe equivalent factorial derivative criterion for real analyticity says that, for every $a$, there are a neighborhood $I$ of $a$ and constants $C,A>0$ such thatThe uniformity over $I$ matters: bounds only at $a$ do not exclude a flat function.

Assume the factorial derivative criterion for real analyticity. The Taylor theorem with Lagrange remainder giveswhen the segment from $a$ to $a+h$ is contained in $I$. For sufficiently small $|h|<A^{-1}$ the Taylor remainder tends to zero, proving the Taylor series definition.

Conversely, write the convergent power series at $a$ as $\sum b_k{h}^k$. Choose $\rho$ strictly inside its radius of convergence; then $|b_k|\le M{\rho }^{-k}$ for some $M$. Termwise differentiation on $|h|\le \rho /2$ givesHere the sum is $(1-1/2{)}^{-n-1}$, obtained by differentiating the geometric series. This is the required locally uniform bound. The two definitions of a real analytic function are equivalent.

The total differential order of ${\partial }_t-{\partial }_x^2$ is two, so its principal symbol is $p(\tau ,\xi )=-{\xi }^2$. The conormal to $t=0$ is $(1,0)$, on which $p$ vanishes. The initial line is a characteristic hypersurface for this total-order symbol; the first-order time derivative does not enter it.

Suppose a real analytic solution existed near $(0,0)$. Repeated use of the heat equation gives ${\partial }_t^{k}u={\partial }_x^{2k}u$. The initial power series is ${\sum }_{j\ge 0}(-1{)}^j{x}^{2j}$ near zero, henceThe time Taylor series at $x=0$ would therefore have coefficients $(-1{)}^k(2k)!/k!$. The ratio of successive absolute coefficients is $2(2k+1)\to \infty$, giving radius of convergence zero. This contradicts the assumed real analytic regularity. No such analytic local solution exists, although the initial function itself is real analytic.

For $x>0$, the exponent $x/(1+x)$ lies in $(0,1)$, and $0\le {\cos }^{4}x\le 1$. ConsequentlyThe last inequality uses the positive-term power series $e^x={\sum }_{k\ge 0}{x}^k/k!\ge x^2/2$ for $x>0$. The final bound tends to zero, so the squeeze theorem gives limit $0$.

A complex power series with radius of convergence $R$ converges absolutely at every $z$ with $|z|<R$ and diverges at every $z$ with $|z|>R$. This makes no general assertion at $|z|=R$: boundary points must be examined separately. The series always converges at $z=0$. If $R=0$, no nonzero point lies in the convergence disk; if $R=\infty$, the series converges absolutely throughout the complex plane.

If $p$ is the zero polynomial, every coefficient vanishes, so $R=\infty$. Otherwise let $d$ be its degree and $c\ne 0$ its leading coefficient. Since $p(n)=c{n}^d(1+o(1))$,The ratio test therefore proves absolute convergence when $|z|<1$. For $|z|>1$, the terms $p(n)z^n$ do not tend to zero, so the series diverges. Thus the polynomial-coefficient power series has $R=1$ for every nonzero polynomial. In fact, it also diverges at every $|z|=1$, because $|p(n)z^n|=|p(n)|$ does not tend to zero, whether $d=0$ or $d>0$.

Introduce a parameter $t$ to keep track of total degree. The power series for the matrix exponential givesSet $S=X+Y$ and $C=\frac{1}{2}[X,Y]$, where the commutator is $[X,Y]=XY-YX$. Expanding a second matrix exponential givesSincethe two expressions agree through total degree two. More explicitly, applying the local matrix logarithm to the first expansion gives $\log (e^{tX}{e}^{tY})=t(X+Y)+\frac{1}{2}{t}^2[X,Y]+O(t^3)$. This proves the displayed order of the Baker--Campbell--Hausdorff formula. The Lie algebra is closed under the commutator, so its displayed exponent lies in the same Lie algebra. The expansion is a formal power series identity, or a convergent identity sufficiently near the identity matrix; this argument does not assert unrestricted global convergence.

The next two terms of the Baker--Campbell--Hausdorff formula are the two cubic commutators:Here $O(4)$ means total degree at least four in $X,Y$. If one continues by one more degree, the quartic contribution is $-\frac{1}{24}[Y,[X,[X,Y]]]$.

For the Pauli matrices, use the Pauli matrix multiplication lawPut $N=\mathbf{n}\cdot \mathbf{\tau }$. The antisymmetric term disappears when contracted with $n_a{n}_b$, so $N^2=|\mathbf{n}{|}^{2}I=I$. Consequently every even power of $N$ equals $I$, and every odd power equals $N$. Splitting the matrix exponential into even and odd terms of its power series yieldsThe Pauli matrices are Hermitian matrices, and the components of $\mathbf{n}$ are real, so $N^{†}=N$. For real $\alpha$, the resulting matrix $U$ satisfiesAlso $\mathrm{tr}N=0$ and $N^2=I$, so its two eigenvalues are $1,-1$. The eigenvalues of $U$ are therefore $e^{i\alpha },e^{-i\alpha }$, whose product is one. Thus $U$ is a unitary matrix with determinant one: it belongs to the SU(2) group.

Multiplying the two exact Pauli matrix exponentials, and using ${\tau }_1{\tau }_2=i{\tau }_3$, givesIn particular, the ${\tau }_3$ term has a minus sign. Its quadratic Taylor expansion isTaking $X=i\alpha {\tau }_1$, $Y=i\beta {\tau }_2$, the Pauli matrix commutator identity gives $[X,Y]=-2i\alpha \beta {\tau }_3$. The quadratic exponent in the Baker--Campbell--Hausdorff formula is therefore $Z=i\alpha {\tau }_1+i\beta {\tau }_2-i\alpha \beta {\tau }_3$. In $e^Z=I+Z+\frac{1}{2}{Z}^2+O(3)$, only the linear part of $Z$ contributes to $Z^2$ through degree two. The anticommutator ${\tau }_1{\tau }_2+{\tau }_2{\tau }_1=0$ gives $Z^2=-({\alpha }^2+{\beta }^2)I+O(3)$, reproducing the exact product's quadratic expansion.

Use the same method of multiple scales and its solvability condition in the method of multiple scales. The position-dependent damping function is odd, so the lemma that odd position-dependent damping has zero first-order amplitude drift applies. In its wave amplitude average, $\psi \mapsto \psi +\pi$ changes $\sin (A\cos \psi )$ to its negative while preserving ${\sin }^2\psi$, so the average is zero. Hence $A^{\prime }=0$; the wave phase average again gives ${\theta }^{\prime }=0$.

Equivalently, expand the position-dependent damping in its uniformly convergent power series for bounded $A$. Each term ${\cos }^{2m+1}\psi \sin \psi$ has only even sine Fourier harmonics, so none resonates with the unit-frequency oscillator. With the initial wave amplitude and wave phase unchanged, the leading uniform approximation isThere is no first-order slow drift. Effects at the next order accumulate only an $O(ϵ)$ correction on this time scale.

Analytic continuation extends a holomorphic function by holomorphic elements agreeing on overlaps; along a path, a chain of overlapping neighbourhoods transports the initial element. The identity theorem makes compatible continuation unique on connected overlaps, although different paths can lead to different branches.

Choose the principal complex logarithm and the slit domain $D=\mathbb{C}\setminus [1,\infty )$. On $D$, the functionis holomorphic, with the removable value $h(0)=1$. The domain $D$ is simply connected, so $h$ has a primitive and its integral from $0$ to $z$ is independent of the path in $D$. For $|z|<1$, the uniformly convergent power series on compact subdiscs givesTherefore $f$ is a single-valued analytic continuation of the dilogarithm to $\mathbb{C}\setminus [1,\infty )$. The apparent problem at $0$ is removable; $1$ is a genuine branch point, even though the function has a finite one-sided limit there, because its derivative has a logarithmic singularity. Other cuts from $1$ to infinity give corresponding branches.

Every counted point has integer coordinates and lies in the square $[-n,n{]}^2$, while the origin is always counted. HenceNo precise circle-area estimate is needed; this elementary bound controls the growth of the power series coefficients.

The comparison test for series states that if $0\le u_n\le v_n$, convergence of $\sum v_n$ implies convergence of $\sum u_n$. Equivalently, divergence of the smaller nonnegative series forces divergence of the larger one.

For $|z|=r<1$,For $0<r<1$, the ratio of successive terms on the right tends to $r<1$, so the ratio test and the comparison test for series show absolute convergence. The case $r=0$ is immediate.

If $|z|\ge 1$, then $|a_n{z}^n|\ge 1$, so the terms do not tend to zero. The term test for divergence therefore rules out convergence, including every point on the boundary $|z|=1$. Thus the radius of convergence isThe series converges exactly in the open unit disc. This is an example of polynomial coefficient growth with unit power-series radius: the lattice-count bound is polynomial rather than exponential.

For the exponent $\sigma =m$, the linear recurrence relation reduces to $(m+n)a_n=-a_{n-1}$. The excluded nonpositive integers ensure that no denominator is zero. Normalize $a_0=1$ and iterate:The ratio test gives an infinite radius of convergence for the power series factor. Hence a solution isFor nonintegral $m$, choose instead the exponent $\sigma =0$. Since $n-m\ne 0$ for every positive integer $n$, its linear recurrence relation gives $n{a}_n=-a_{n-1}$. ThusThe distinct nonintegral leading exponents establish linear independence. More explicitly, $W(e^{-x},y_m)=m{e}^{-x}{x}^{m-1}$ by the Abel identity and the leading coefficient at zero, so the Wronskian is nonzero. The factor $x^m$ uses the real branch on $x>0$.

Use the physical Coulomb bound state boundary condition: the spherically symmetric wavefunction is regular at the origin and square integrable with measure $4\pi r^{2}dr$. For $E<0$, setAt large $r$, the decaying branch behaves exponentially as $e^{-Kr}$. Substituting $\psi =f{e}^{-Kr}$ into the radial Schrödinger equation givesThe regular solution has a power series $f(r)={\sum }_{j\ge 0}{a}_j{r}^j$, with recurrenceA nonzero regular solution has $a_0\ne 0$. If the series does not terminate, its coefficients eventually have one sign after choosing an overall real phase. For any fixed $0<\delta <1/2$ and sufficiently large $j$, the ratio of consecutive magnitudes is at least $2K(1-\delta )/(j+2)$. The tail therefore grows at least as a positive constant times $e^{2K(1-\delta )r}$ divided by a power of $r$; a finite polynomial of earlier terms cannot cancel it. Multiplication by $e^{-Kr}$ leaves exponential growth, contradicting normalizability.

Thus the series must terminate at degree $N-1$, where $KN=\lambda$ and $N$ is a positive integer. Conversely, termination gives a regular polynomial times $e^{-\lambda r/N}$, which is normalizable. ThereforeThese are the $\ell =0$ members of the hydrogen spectrum. Regularity at the origin is essential: square integrability alone on the punctured radial interval would also admit a singular $1/r$ branch. Such a branch is not a physical eigenfunction of the ordinary three-dimensional Coulomb Hamiltonian; its singularity introduces an inadmissible point-source contribution. In the reduced radial variable $u=r\psi$, the physical condition is $u(0)=0$.

Non-characteristic. The first-order principal symbol is ${\nu }_t-i{\nu }_x$, which equals one on $(0,1)$.

Set $z=x+it$. The equation is equivalent to ${\partial }_{\bar{z}}\varphi =({\varphi }_x+i{\varphi }_t)/2=0$. Because $\varphi$ is continuously differentiable, the Cauchy-Riemann equations imply that it is a holomorphic function of $z$. Its convergent complex Taylor series restricts to a real power series along $t=0$, so $g$ is a real analytic function.

For smooth-data instability of the Cauchy-Riemann Cauchy problem, chooseThese are entire functions and satisfy the equation. Their initial derivatives obey ${\sup }_x|{\partial }_x^j{g}_n|\le n^j{e}^{-\sqrt{n}}$, so every finite sum tends to zero. ButThus analytic solutions can exist while continuous dependence on initial data fails for this smooth-data topology.

The Cauchy estimate follows directly from the differentiated Cauchy integral formula:The circle has length $2\pi r$, giving the bound for every integer $j\ge 0$.

Put $w=\varphi -\psi$ and integrate on the straight segment $z=\tau t/|t|$. For fixed $|x|<sR$, a disc centred at $x$ with radius approaching $R(\sigma (\tau )-s)$ stays inside $|\xi |\le \sigma (\tau )R$. Applying the first-derivative Cauchy estimate there givesIntegrating its modulus and taking the supremum over $|x|<sR$ is exactly the asserted integral estimate. The source cancels from $T\varphi -T\psi$.

To obtain a contraction mapping, use the weighted holomorphic norm on a shrinking time domain. Its finite-norm space consists of holomorphic functions on ${\mathcal{C}}_{\alpha }$ with zero initial value; the apparent quotient at $t=0$ is interpreted by a limit. Completeness follows because convergence in this norm implies locally uniform convergence of holomorphic functions, including near $t=0$, and the limiting pointwise bounds give convergence in the norm.

Here is the explicit contraction estimate on a shrinking holomorphic domain. Write $N=\Vert w{\Vert }_{\alpha }$, $r=|t|$, $A=\alpha (1-s)$, and chooseThen $\sigma -s=(A-\tau )/(2\alpha )$ and $\alpha (1-\sigma )-\tau =(A-\tau )/2$. The whole auxiliary polydisc lies inside ${\mathcal{C}}_{\alpha }$, and its supremum of $|w|$ is at most $2N\tau /(A-\tau )$. ConsequentlyThus $\Vert T\varphi -T\psi {\Vert }_{\alpha }\le (4\alpha /R)\Vert \varphi -\psi {\Vert }_{\alpha }$. Choose $0<\alpha <\min (\eta ,R/4)$. The source is bounded on the closed polydisc, say by $M_f$, and $\Vert T0{\Vert }_{\alpha }\le \alpha M_f$. Therefore $T$ maps the Banach space into itself and is a strict contraction mapping. The Banach fixed-point theorem gives a holomorphic fixed point with $\varphi (x,0)=0$. Differentiating the integral identity yields ${\varphi }_t-i{\varphi }_x=f$, establishing this case of the Cauchy-Kovalevskaya theorem.

Finally, near each real initial point, a real analytic function $g$ has a holomorphic extension $G$. Apply the same construction to $\varphi =G(x)+v$ with $v(x,0)=0$ and source $i{G}^{\prime }(x)$, restricting the discs if necessary. This gives a local real analytic function of $(x,t)$ with the prescribed initial value. Equivalently the solution is $\boxed{\varphi (x,t)=G(x+it)}$ wherever the extension is defined.