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Write $M(t)=\mathbb{E}{e}^{tX}$, with $M(0)=1$, and differentiate $\kappa (t)=\log M(t)$:Using $M^{(j)}(0)=\mathbb{E}{X}^j$ and $\mu =\mathbb{E}X$ givesThus the first three cumulants are the mean, variance and third central moment. A finite moment-generating function near zero justifies these differentiations. For positive claims with only three finite ordinary moments, the same calculations are valid as left derivatives at zero of the logarithm of the Laplace transform of a nonnegative random variable; positive exponential moments are not needed for the resulting moment identities.

For the independent Poisson distribution count, the random-sum transform identity givesConsequently the compound Poisson distribution hasThese are raw severity moments, not severity cumulants.

The printed gamma density uses rate $\nu$, not scale. Direct integration against $e^{ty}$ givesTherefore the required matching equations for the shifted gamma distribution areFor $\lambda >0$ and nonzero positive claims with finite $m_3$, these equations give positive $\alpha ,\nu$ and the unique solutionThis is the three-cumulant shifted gamma approximation. In particular, its skewness $2/\sqrt{\alpha }$ equals $m_3/(\sqrt{\lambda }{m}_2^{3/2})$, the skewness of $S$.

The competing normal approximation to a compound Poisson aggregate isTo justify its large-count limit, let ${\phi }_X$ be the claim characteristic function. For fixed severity law with $m_2<\infty$, its expansion is ${\phi }_X(u)=1+i{m}_{1}u-m_2{u}^2/2+o(u^2)$. The characteristic function of $(S-\lambda m_1)/\sqrt{\lambda m_2}$ isThus it converges in distribution to a standard normal distribution as $\lambda \to \infty$.

The normal approximation is simple, uses only $m_1,m_2$, and works well in the centre when the count is large and standardized skewness is small. It imposes symmetry, whereas positive compound-Poisson claims have positive third cumulant. The shifted-gamma approximation reproduces that third cumulant and can better describe moderate right skewness. It requires a finite and reliably known third moment, and matching three cumulants alone does not guarantee accurate extreme tails.

Neither continuous approximation reproduces the genuine atom $\mathbb{P}(S=0)=e^{-\lambda }$, which matters at small $\lambda$. The normal distribution also assigns probability $\Phi (-\sqrt{\lambda }{m}_1/\sqrt{m_2})$ to negative losses. The gamma approximation is bounded below by $k$, but its fitted shift need not be nonnegative: exponential claims of mean $\mu$ give $k=-\lambda \mu /3$, so it too can assign mass to negative losses. If $k>0$, it instead excludes some genuinely possible small losses. These support and atom discrepancies prevent either approximation from being a universal replacement for the exact aggregate law.

Use target signature $(-,+,\dots ,+)$ and a closed-string spatial coordinate $0\le \sigma \le 2\pi$. Write $Y$ for the compact coordinate and $X^A$ for the noncompact coordinates. In conformal gauge, the wave equation factorizes, so every coordinate is a sum of a function of $\tau +\sigma$ and a function of $\tau -\sigma$. The closed-string boundary conditions areThus the derivatives of each chiral part are periodic and have integer Fourier modes; only their worldsheet zero modes can produce the winding displacement. Integrating these Fourier series gives the closed-string mode expansionThe oscillator reality conditions are ${\alpha }_m^{†}={\alpha }_{-m}$ and similarly for the tilded modes. The coefficient of $\tau$ follows by integrating the canonical momentum density $\dot{Y}/(2\pi {\alpha }^{\prime })$. Single-valued wavefunctions of the center coordinate $y\sim y+2\pi R$ have momentum $n/R$, with $n\in \mathbb{Z}$. Equivalently, the compact chiral worldsheet zero modes are ${\alpha }^{\prime }{p}_L(\tau +\sigma )/2$ and ${\alpha }^{\prime }{p}_R(\tau -\sigma )/2$, whereHere tildes denote the left-moving sector. This convention will also fix the sign in compact-circle closed-string level matching.

Choose a noncompact light-cone pair and impose light-cone gauge in string theory $X^{+}=x^{+}+{\alpha }^{\prime }{p}^{+}\tau$, with $p^{+}\ne 0$. The Virasoro constraints $({\partial }_{\pm }X{)}^2=0$ then solve for ${\partial }_{\pm }{X}^{-}$ in terms of the transverse coordinates. For example, with ${\partial }_{\pm }={\partial }_{\tau }\pm {\partial }_{\sigma }$, one has ${\partial }_{\pm }{X}^{-}=({\partial }_{\pm }{X}^i{)}^2/(2{\alpha }^{\prime }{p}^{+})$. Thus there are two independent transverse oscillator Fock spaces, including $Y$ among the transverse coordinates. Quantization gives $[{\alpha }_m^i,{\alpha }_k^j]=m{\delta }^{ij}{\delta }_{m+k,0}$, and the same relation for the tilded oscillators, with the two sectors commuting. Lorentz consistency gives the critical dimension of the bosonic string $D=26$ and string intercept one in each sector. Accordinglycount 24 transverse oscillator species. The two zero-mode constraints give the lower-dimensional bosonic string mass spectrumTaking their difference and average yieldsEvery choice of transverse string oscillators and integer charges satisfying this closed-string level matching is a physical string state in light-cone gauge. The eliminated longitudinal oscillators do not supply additional polarizations.

To classify all massless states, positivity of the two compact terms implies $N+\tilde{N}\le 2$. When the sum is two, masslessness forces $n=w=0$, and closed-string level matching then forces $N=\tilde{N}=1$. The states ${\alpha }_{-1}^i{\tilde{\alpha }}_{-1}^j|0⟩$ have $24^2=576$ polarizations. In 25 noncompact dimensions, their decomposition gives a graviton, a Kalb–Ramond field, a dilaton, two Abelian gauge vectors from the metric and two-form with one compact index, and the radius scalar field. Their physical polarization counts areThe first two numbers are the symmetric traceless second-rank tensor and antisymmetric second-rank tensor of the massless little group $SO(23)$.

When $N+\tilde{N}=1$, closed-string level matching requires $|nw|=1$. The compact contribution obeysand equality occurs precisely at $R=\sqrt{{\alpha }^{\prime }}$. The charges $n=w=\pm 1$ have $(N,\tilde{N})=(1,0)$; the charges $n=-w=\pm 1$ have $(N,\tilde{N})=(0,1)$. Each charge has 24 polarizations supplied by its single excited sector. At a general radius these have

Finally, $N=\tilde{N}=0$ requires $nw=0$. The exceptional massless ground states of a bosonic circle occur atEach charge is a scalar field. These special radii must be included if “general radius” is to mean a complete classification, rather than only a generic radius. At a generic radius the massless spectrum is the neutral 576-state spectrum above; at exceptional radii one adds exactly the states just listed. The neutral ground state remains tachyonic.

The mapsends $p_L$ to $p_L$ and $p_R$ to $-p_R$. Reversing all right-moving compact string oscillators, ${\alpha }_m^Y\mapsto -{\alpha }_m^Y$, preserves their commutators and oscillator number; the left-moving oscillators are unchanged. Both the mass formula and compact-circle closed-string level matching are therefore unchanged, with a one-to-one map of the complete oscillator basis. This proves T-duality of the full spectrum, including winding, massive and tachyonic states.

At the self-dual circle the four single-oscillator charge families give $4\cdot 24=96$ new massless states. Each family decomposes into a 25-dimensional gauge vector with 23 polarizations and one scalar field. There are also four massless ground states, with charges $(\pm 2,0)$ and $(0,\pm 2)$. Thus the massless spectrum at the bosonic self-dual circle hasThe gauge interpretation follows directly from the chiral compact-boson operators. With $⟨Y_L(z)Y_L(0)⟩=-{\alpha }^{\prime }\log z/2$, the exponential $e^{ik{Y}_L}$ has conformal weight ${\alpha }^{\prime }{k}^2/4$. At the self-dual radius,have weight one and satisfy $J^3(z)J^3(0)\sim 1/(2z^2)$, $J^3(z)J^{\pm }(0)\sim \pm J^{\pm }(0)/z$, and $J^{+}(z)J^{-}(0)\sim z^{-2}+2J^3(0)/z$. These are the level-one $SU(2)$ current relations; the right sector supplies another copy. Multiplying these currents by a noncompact oscillator produces six gauge vectors in total. The nine scalar string vertex operators $J_L^a{J}_R^b$ form the $(3,3)$ representation, accounting for the radius scalar field and eight new scalar fields. The remaining neutral fields are the graviton, two-form and dilaton.

A radius deformation as a Higgs mechanism is the expectation value of the Cartan–Cartan scalar string vertex operator $J_L^3{J}_R^3\propto \partial Y_L\bar{\partial }{Y}_R$. Its stabilizer is $U(1{)}_L\times U(1{)}_R$, soThe four compact single-oscillator scalar field polarizations become longitudinal polarizations of the four massive vectors, exactly as in the Higgs mechanism. The other four extra scalar fields have masses $4/R^2-4/{\alpha }^{\prime }$ or $4R^2/{\alpha }^{\prime 2}-4/{\alpha }^{\prime }$ and need not remain massless; one pair becomes tachyonic on either side of the self-dual point. Together with the original bosonic tachyon, this means that the gauge-symmetry interpretation is formal and does not assert a stable bosonic-string vacuum.

Count oscillator polarizations at fixed center-of-mass momentum. In a critical closed bosonic string, each chiral sector has 24 transverse oscillator species. A species at frequency $m$ can be occupied $0,1,2,\dots$ times, so its contribution is $(1-q^m{)}^{-1}$. The full chiral generating function is thereforeThe two sectors are independent, while closed-string level matching restricts their oscillator numbers to the same $N$. Thus the number of states at the closed-string mass level is $g_N=d_N^2$, not $d_{2N}$.

Put $q=e^{-\beta }$, with $\beta >0$. The Dedekind eta function obeysThe transformed argument $2\pi i/\beta$ has large imaginary part, so its product approaches one and $\eta (2\pi i/\beta )\sim e^{-{\pi }^2/(6\beta )}$. HenceThis establishes the singular exponential as well as its power prefactor. The corrections to the transformed eta product are exponentially small in $1/\beta$.

The Cauchy coefficient formula, followed by $q=e^{-\beta }$, gives a steepest-descent integral with leading formLet $n=N-1$. The saddle of $S(\beta )=n\beta +4{\pi }^2/\beta$ isAlong the vertical steepest-descent direction $\beta ={\beta }_{\ast }+it$, the quadratic part is $-S^{\prime \prime }({\beta }_{\ast })t^2/2$. Thus the saddle-point approximation gives the Gaussian prefactorThe point $q=1$ supplies the largest exponential contribution. At another fixed root of unity of order $k>1$, the corresponding modular singularity gives a smaller exponential $e^{4\pi \sqrt{n}/k}$, so it does not change this leading asymptotic. Squaring the chiral result gives the large-level degeneracy of a closed bosonic stringReplacing $N-1$ by $N$ gives the equivalent usual leading asymptotic. The power $-27/2$ matters: the two-sector degeneracy is not just an unspecified exponential.

Since $M=2\sqrt{(N-1)/{\alpha }^{\prime }}$, the exponential is $e^{{\beta }_{H}M}$ withHere Boltzmann's constant is one. Coarse-graining the levels into a density per unit rest mass introduces the Jacobian $dN/dM={\alpha }^{\prime }M/2$. ConsequentlyThis is distinct from the degeneracy per level, whose power in $M$ is $-27$. The internal-state contribution to a canonical partition function behaves at large mass asIt converges for $\beta >{\beta }_H$ and diverges for $\beta <{\beta }_H$. At $\beta ={\beta }_H$, this particular internal rest-mass integral converges because of the power $M^{-26}$; one should not discard the prefactor and claim divergence there without specifying the full thermodynamic ensemble. For example, including continuous momenta in 25 spatial dimensions multiplies the high-mass integrand by a factor proportional to $M^{25/2}$, still leaving convergence of this ideal one-string integral at the endpoint, although sufficiently high derivatives are singular. Multi-string effects, volume and interactions affect the detailed limiting thermodynamics.

The leading microcanonical entropy grows as $S(E)\sim {\beta }_{H}E$, with logarithmic corrections, so increasingly large energies are stored in increasingly excited, long strings rather than in an arbitrarily hot ordinary gas. The Hagedorn temperature is the boundary of canonical convergence from above in inverse temperature; extrapolating the free-string canonical description to $T>T_H$ fails. This counting does not settle the interacting phase above that scale. Also, the bosonic ground-state tachyon is a separate vacuum instability; the high-level asymptotic is meaningful as a formal spectrum calculation without asserting a stable bosonic thermal vacuum.

Use natural units with $c=1$, write $H=\dot{a}/a$, and measure cosmic time from the Big Bang. The cosmological continuity equation follows by differentiating the Friedmann equation and using the Friedmann acceleration equation; for pressureless matter it gives $\dot{\rho }+3H\rho =0$, hence $\rho ={\rho }_0{a}^{-3}$ when the present scale factor is $a_0=1$. The present critical density and cosmological density parameter giveConsequently, with $C=H_0^2{\Omega }_0$, the Friedmann equation becomes ${\dot{a}}^2=C/a-k$.

Introduce conformal time through $dt=ad\tau$. A prime denotes a derivative with respect to $\tau$, so $a^{\prime }=a\dot{a}$ andSet $u=\sqrt{k}\tau$, with $u=0$ at the Big Bang. On the expanding branch the substitution $a=C(1-\cos u)/(2k)$ satisfies this equation. Integrating $dt=adu/\sqrt{k}$, with $t(0)=0$, gives the closed matter-dominated Friedmann solution:The parametrization continues smoothly through the maximum scale factor, reached at $u=\pi$, onto the contracting branch. Its curve in the $(t,a)$ plane is a rescaled cycloid.

For a presently expanding universe, $H_0>0$ selects $0<u_0<\pi$. The normalization $a(u_0)=1$ givesSubstitution yields the age of a closed dust universe:The first later zero of the scale factor occurs at $u=2\pi$, so the lifetime of a closed dust universe gives the Big Crunch timeThis is the total lifetime measured from the Big Bang; the remaining cosmic time is $t_{\mathrm{BC}}-t_0$. As a check, the limit ${\Omega }_0\downarrow 1$ gives $H_0{t}_0\to 2/3$, while the Big Crunch recedes to infinite time, as expected for a flat matter-dominated universe with zero cosmological constant.

Use a nonzero complex unital Banach algebra $B$ with submultiplicative norm, and write its unit as $e$. The complex scalar field is the usual convention in this spectral statement; for a real algebra one uses its complexification. We derive the needed analytic facts from norm-convergent series and scalar complex analysis.

If $\Vert a\Vert <1$, completeness makes the Neumann series ${\sum }_{j=0}^{\infty }{a}^j$ converge. Multiplying its finite partial sums by $e-a$ gives $e-a^{m+1}$, which tends to $e$. Hence its sum is the inverse of $e-a$. If $b$ is invertible, then $b+h=b(e+b^{-1}h)$ is invertible whenever $\Vert b^{-1}h\Vert <1$. This proves that the invertible elements form an open set and that inversion is locally norm-continuous.

Define the Banach algebra spectrum and the resolvent of an element byFor $|\lambda |>\Vert x\Vert$, the Neumann series givesThus $\sigma (x)$ is bounded by $\Vert x\Vert$ and is closed, because invertibility is open. Also $R(\lambda )\to 0$ in norm at infinity; for example its norm is at most $\Vert e\Vert /(|\lambda |-\Vert x\Vert )$, since $\Vert e\Vert \ge 1$ in a nonzero submultiplicative unital algebra. The standard normalization $\Vert e\Vert =1$ is not needed.

For points where the Banach algebra resolvent exists $\lambda ,\mu$, multiply out the inverses to get the resolvent identityFor ${\lambda }_0$ outside the Banach algebra spectrum, put $R_0=R({\lambda }_0)$. Locally,This series and its differentiated series converge uniformly on every smaller disk. It therefore proves $R$ is a Banach-space-valued holomorphic function, establishing complex differentiability in the Banach-space norm and gives $R^{\prime }=-R^2$, without assuming a Banach-algebra-valued analytic theorem. In particular every continuous complex linear functional $\varphi$ makes $\varphi (R(\lambda ))$ a scalar holomorphic function.

The Banach algebra spectrum is nonempty. Otherwise $R$ would be defined and analytic everywhere. Its local norm continuity and decay at infinity make it globally bounded. For every continuous linear functional $\varphi$, the scalar entire function $\varphi (R(\lambda ))$ is bounded and tends to zero at infinity, so it is identically zero: the scalar Cauchy estimate for its derivative on circles of radius $r$ is $M/r$, which tends to zero, and the limiting value then fixes its constant as zero. Such functionals separate points by the Hahn-Banach theorem. Explicitly a bounded real functional can be extended from the real span of any nonzero element using its norm, and complexifying it by $\varphi (v)=\ell (v)-i\ell (iv)$ gives a bounded complex linear functional nonzero on that element. Therefore $R$ itself would be zero, contradicting $(\lambda e-x)R=e$. We have proved the Banach algebra spectrum is a nonempty compact subset of the complex plane.

We next establish the spectral radius formula. The power norms are submultiplicative. If some $x^m=0$, their root norms are eventually zero. Otherwise, for any fixed $m$, write $n=qm+r$ with $0\le r<m$ to obtainTaking roots gives ${\mathrm{limsup}}_n\Vert x^n{\Vert }^{1/n}\le \Vert x^m{\Vert }^{1/m}$. Taking the infimum in $m$, and using that every root norm is at least that infimum, proves existence ofFor $|\lambda |>{\rho }_0$, the root test makes $\sum x^j/{\lambda }^{j+1}$ converge, even when $|\lambda |\le \Vert x\Vert$. The same telescoping multiplication proves it is a two-sided inverse. HenceFor the reverse bound fix $R>r_{\sigma }$. If $R_1>\max (R,\Vert x\Vert )$, integrate the uniformly convergent exterior series term by term on $|\lambda |=R_1$, obtainingThe same integral equals the one on $|\lambda |=R$: applying an arbitrary continuous linear functional reduces contour deformation through the Banach algebra spectrum-free annulus to the scalar Cauchy theorem, and separation of points restores equality in $B$. This proves the resolvent Cauchy coefficient formula rather than assuming vector-valued analytic calculus. Using the norm bound for the allowed vector integral givesTaking $n$th roots yields ${\rho }_0\le R$, and letting $R\downarrow r_{\sigma }$ finishes the proof:

For the first example take the commutative algebra of dual numbers $B=\mathbb{C}[\epsilon ]/({\epsilon }^2)$ with norm $\Vert a+b\epsilon \Vert =|a|+|b|$. Completeness follows from finite dimension, and the product estimate makes the norm submultiplicative. For $x=\epsilon$, $x\ne 0$ but $x^2=0$, so $\rho (x)=0$. Directly, $\lambda e-\epsilon$ has inverse ${\lambda }^{-1}e+{\lambda }^{-2}\epsilon$ for $\lambda \ne 0$, and its Banach algebra spectrum is $\{0\}$. For the second example take $B=\mathbb{C}$ with its absolute-value norm and $x=1$. Its Banach algebra spectrum is $\{1\}$, giving $\rho (x)=\Vert x\Vert =1$.

Let the first Bézier curve have degree $n$ and control points $P_0,\dots ,P_n$, and the next have degree $m$ and control points $Q_0,\dots ,Q_m$. Use one increasing parameter on the two adjacent intervals. For a continuous join, their endpoints must agree: $P_n=Q_0$. If each interval has length $h$, the endpoint derivatives are $n(P_n-P_{n-1})/h$ and $m(Q_1-Q_0)/h$. Therefore the parametric first-derivative join of Bézier curves is exactlyFor equal degrees, the two polygon edges incident on the common endpoint are equal vectors. Equivalently, $P_{n-1},P_n=Q_0,Q_1$ are collinear with the join as their midpoint. Agreement of edge directions without these lengths gives only geometric tangent continuity at a nonzero tangent, not equality of derivatives in the chosen parameter. The displayed condition also covers zero endpoint derivatives. With unequal interval lengths $h,k$, the condition is instead $n(P_n-P_{n-1})/h=m(Q_1-Q_0)/k$.

Applying the preceding derivative identity repeatedly yields the Bézier derivative control polygon for every order:where the forward finite difference is $\Delta P_j=P_{j+1}-P_j$ andThe induction step multiplies by the remaining degree $n-r$ and takes one more finite difference. Derivatives above order $n$ vanish. In particular,If the actual parameter interval has length $h$ and $t$ is its normalized coordinate, derivatives in the original parameter acquire the additional factor $h^{-r}$. These formulas allow derivative evaluation and join tests using only the control polygon.

Write $B_j^n(t)=\left(\genfrac{}{}{0ex}{}{n}{j}\right)t^j(1-t{)}^{n-j}$ for the Bernstein basis, and set $B_j^n=0$ when $j$ is outside its index range. Differentiating and using $j\left(\genfrac{}{}{0ex}{}{n}{j}\right)=n\left(\genfrac{}{}{0ex}{}{n-1}{j-1}\right)$ and $(n-j)\left(\genfrac{}{}{0ex}{}{n}{j}\right)=n\left(\genfrac{}{}{0ex}{}{n-1}{j}\right)$ givesHence, after shifting the first sum by one index,Thus the derivative is a degree-at-most-$(n-1)$ Bézier curve whose control points are $n(P_{j+1}-P_j)$. They are vectors rather than necessarily positions on the original Bézier curve. This is the Bézier derivative control polygon.

A cubic polynomial on either half is determined by its endpoint positions and endpoint derivatives. Use normalized parameters $s=2t$ on the first half and $s=2t-1$ on the second, so that derivatives with respect to $s$ are half the derivatives with respect to $t$. Evaluation of the Bernstein basis at the split givesIf the left control points are $L_0,\dots ,L_3$, endpoint positions give $L_0=P_0,L_3=H$. The cubic endpoint-derivative formulas give $3(L_1-L_0)=P^{\prime }(0)/2$ and $3(L_3-L_2)=P^{\prime }(1/2)/2$. On the right, $R_0=H,R_3=P_3$, $3(R_1-R_0)=P^{\prime }(1/2)/2$ and $3(R_3-R_2)=P^{\prime }(1)/2$. Substitution yieldsThese are exact restricted Bézier curves, since they match all four endpoint Hermite data of the respective cubics.

The construction is also the midpoint case of De Casteljau's algorithm. Form $A_j=(P_j+P_{j+1})/2$ for $j=0,1,2$, then $B_0=(A_0+A_1)/2$, $B_1=(A_1+A_2)/2$, and $H=(B_0+B_1)/2$. The two polygons are $(P_0,A_0,B_0,H)$ and $(H,B_1,A_2,P_3)$. Their common tangent edges satisfy $H-B_0=B_1-H$, as required for equal half-intervals. Thus the triangular midpoint construction simultaneously evaluates the split point and supplies the two exact representations.

A norm estimate controls the worst possible refined differences for every choice of the initial controls. Submultiplicativity turns a length-$p$ contraction into a bound for arbitrary long products, so it proves a certain number of continuous derivatives. Its absolute values can lose cancellations; failure of the estimate is inconclusive. It therefore gives a lower bound on smoothness.

Conversely, an observable eigenmode of a repeated local product exhibits a particular sequence of shrinking neighborhoods with inadequate decay. General-position controls have a nonzero coefficient in that mode, so greater derivative continuity is impossible. Special symmetric or polynomial data may suppress the bad mode; this is why the general-position qualification matters. The obstruction gives an upper bound on smoothness, while failure to find an obstruction proves nothing. A lower bound on spectral growth becomes an upper bound on smoothness because large surviving differences correspond to rough behavior.

For example, the binary Cardinal cubic B-spline subdivision mask has symbol $M(z)=(1+z{)}^4/8$, up to an irrelevant index shift. At derivative order $r=2$, the first-difference subdivision mask is $B_2=(1+z)/2$, whose maximum parity sum is $1/2$: the limit is at least $C^2$. At $r=3$, $B_3=1$, and the surviving difference mode does not decay. For generic controls the third derivative is piecewise constant with nonzero jumps, so the subdivision curve is not $C^3$. Here the lower and upper tests determine the exact integer continuity, but they need not coincide for an arbitrary subdivision mask. Stability, sum rules and borderline modes must be handled explicitly before interpreting any numerical estimate as a continuity level.