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inversion in a circle with center $c$ and radius $r>0$ is $I(z)=c+r^2/(\overline{z}-\overline{c})$, with $c$ and infinity interchanged. It fixes the circle pointwise and reverses complex orientation. Each inversion is fractional-linear in $\overline{z}$, so composing two eliminates the conjugation and gives $(az+b)/(cz+d)$ with nonzero determinant: a Möbius transformation.

Translate and rotate two distinct centers to $0,d$, where $d>0$, and let the radii be $r,s$. The composition of inversion in the first circle followed by the second isIts fixed points solve $d{z}^2-(d^2+r^2-s^2)z+d{r}^2=0$. The discriminant isDisjoint circle boundaries require either $d>r+s$ or $d<|r-s|$, so both factors have the same nonzero sign and $\Delta >0$. There are two distinct fixed points on the line of centers. If the circles are concentric, their radii differ and the composition is $z\mapsto (s^2/r^2)z$, with exactly the two fixed points $0,\infty$. Thus the composition has two distinct fixed points, also for nested disjoint circles. Reversing the order inverts the Möbius transformation and leaves its fixed points unchanged.

In the nonrelativistic dilute regime, $E(p)=m{c}^2+p^2/(2m)+\cdots$ and $(m{c}^2-\mu )/kT\gg 1$, so the Bose or Fermi correction in the denominator is negligible. The common Maxwell-Boltzmann distribution approximation givesDifferentiating the quoted Gaussian integral with respect to its squared scale gives ${\int }_0^{\infty }{p}^2{e}^{-a{p}^2}dp=\sqrt{\pi }/(4a^{3/2})$. Thereforechemical equilibrium of $p+n\leftrightarrow D$ requires ${\mu }_D={\mu }_p+{\mu }_n$. Apply the density formula to each species to obtainUsing $g_n=g_p=2$, the specified $g_D=4$, and $m_n\simeq m_p$, $m_D\simeq 2m_p$, givesFor $X_a=n_a/n_B$ and $n_B=\eta n_{\gamma }$, the corresponding deuterium equilibrium abundance isThe very small baryon-to-photon ratio means that even the high-energy tail contains many photons per baryon capable of destroying deuterium when $kT$ is of order $B_D$. Substantial deuterium survives only after that tail has been exponentially suppressed, so $B_D/kT$ must be large enough to offset the small prefactor. This is the deuterium bottleneck, not a threshold set just by equality of thermal energy and binding energy. The degeneracy $g_D=4$ is used as stipulated for this calculation.

With $x=\cos \theta$, the weighted inner product for the Chebyshev polynomials becomesThe identity $2\cos (n\theta )\cos (m\theta )=\cos ((n-m)\theta )+\cos ((n+m)\theta )$ and the vanishing of ${\int }_0^{\pi }\cos (k\theta )d\theta$ for positive integers $k$ give zero when $n\ne m$, $\pi /2$ when $n=m>0$, and $\pi$ when $n=m=0$. ThusFor Gaussian quadrature, choose the $n$ distinct nodes as the zeros of the degree-$n$ orthogonal polynomial for the weight, here $T_n$, and choose the weights as the weighted integrals of their Lagrange interpolation basis functions. If a polynomial $p$ has degree at most $2n-1$, divide it as $p=q{T}_n+r$, where both $q$ and $r$ have degree at most $n-1$. Orthogonality makes the integral of $q{T}_n$ zero, while its quadrature sum also vanishes; interpolation integrates $r$ exactly. This proves degree $2n-1$ exactness. It is optimal because the squared nodal polynomial, of degree $2n$, has zero quadrature sum and strictly positive integral.

For $n=3$, $T_3(x)=4x^3-3x$, so the nodes are $-\sqrt{3}/2,0,\sqrt{3}/2$. Their interpolation basis functions areThe weighted zeroth and second moments are $\pi$ and $\pi /2$, and the first moment is zero. Integrating these three basis functions gives equal weights $\pi /3$, hence the Chebyshev–Gauss quadrature rulewhich is exact for every polynomial of degree at most $5$.

For a time-independent Quantum Hamiltonian, a stationary state is an energy eigenstate. The Schrödinger equation separates its time dependence asIts probability density $|\psi (x,t){|}^2=|\chi (x){|}^2$ is time independent. Linearity of the Schrödinger equation givesPut $A_{jk}={\int }_{-\infty }^{\infty }{\chi }_j^{\ast }\hat{A}{\chi }_{k}dx$. The operator for a time-independent observable is a Hermitian operator, so $A_{21}=\overline{A_{12}}$. Substituting the evolved wavefunction into the expectation value givesand therefore, for a normalized state,The normalization is automatic for orthogonal ${\chi }_1,{\chi }_2$, in particular when $E_1\ne E_2$: self-adjointness of $\hat{H}$ gives $(E_1-E_2)⟨{\chi }_1,{\chi }_2⟩=0$. If the energies coincide, normalization of the individual eigenstates alone does not ensure normalization of their displayed sum; its squared norm is $1+\mathrm{Re}⟨{\chi }_1,{\chi }_2⟩$. For a nonzero sum not having unit norm, the physical expectation value is the displayed numerator divided by this squared norm. With orthonormal energy states, the stated formula holds in both the distinct and degenerate cases.

For the half-range Fourier cosine series on $(0,1)$, the coefficients areTwo integrations by parts, with $k=n\pi$, giveThusThe even, $2$-periodic extension is continuous and piecewise continuously differentiable, so the Fourier series converges to it, including at $x=1$. Equivalently, the summable coefficient bound gives uniform convergence and permits taking the limit from the interior. Evaluating at $x=1$, where $(-1{)}^n\cos (n\pi )=1$, yieldsThis is the Basel problem.

Set $h=f-g$, an entire function. By the Bolzano-Weierstrass theorem, a subsequence of the distinct zeros $z_i$ converges to a point $a$ in the closed unit disc. An analytic function has a convergent Taylor series in a neighbourhood of each point. If the first nonzero coefficient of $h$ at $a$ had index $m$, thenwith $k$ analytic and nonzero in a sufficiently small disc. The zero $a$ would then be isolated, contradicting the accumulating distinct zeros. All Taylor coefficients at $a$ vanish, so $h$ vanishes on a disc.

To justify continuation across the whole plane, let $Z$ be the set of points at which $h$ vanishes on a neighbourhood. This set is nonempty and open. It is also closed: if points of $Z$ tend to $b$, continuity of each derivative gives $h^{(m)}(b)=0$ for every $m$, and the Taylor series at $b$ gives a zero neighbourhood. Since $\mathbb{C}$ is a connected space, $Z=\mathbb{C}$. This proves the identity theorem here and gives $\boxed{f\equiv g}$.

Without boundedness, the distinct zeros need not have an accumulation point in the plane. For example,are distinct entire functions agreeing at all the indicated positive integers.

Both functionals are finite on $X$: a continuous function on the compact metric space $[0,1]$ is bounded and integrable. Homogeneity follows from $|af|=|a||f|$, and the triangle inequality follows by integrating, or taking the supremum of, $|f+g|\le |f|+|g|$. Each is nonnegative. If $\Vert f{\Vert }_{\infty }=0$, then $f$ vanishes everywhere. If a continuous function $f$ is nonzero at some point, continuity supplies an interval of positive length on which $|f|$ is bounded below by a positive number; hence $\Vert f{\Vert }_1>0$. Thus both are norms.

For the nonnegative $f_n$, direct integration gives the L1 normTherefore $\boxed{f_n⟶0\text{ in }\Vert \cdot {\Vert }_1}$. On the other hand, differentiation places the maximum of $t^n(1-t)$ at $t=n/(n+1)$, giving the supremum normFor $t<1$, exponential decay implies $n{t}^n(1-t)\to 0$, while $f_n(1)=0$. Thus the pointwise convergence is to zero throughout $[0,1]$. Any convergence in the supremum norm would imply uniform convergence, hence the same pointwise limit zero, contradicting the displayed positive limiting norm. Therefore there is no convergence in the supremum norm.

Write $q_n(X)=1+X+\cdots +X^{n-1}$, a monic primitive polynomial. If $n=ab$ with $a,b>1$, grouping terms gives a factorization in the polynomial ring $\mathbb{Z}[X]$:Both factors have positive degree, so $q_n$ is not an irreducible polynomial.

Conversely, if $n=p$ is a prime number, change variables by $X=Y+1$. The binomial theorem givesEvery coefficient except the leading one is divisible by $p$, because $p\mid \left(\genfrac{}{}{0ex}{}{p}{k}\right)$ for $1\le k<p$, whereas the constant coefficient $p$ is not divisible by $p^2$. The Eisenstein criterion makes this polynomial irreducible over $\mathbb{Q}$. The substitution $X\mapsto Y+1$ is an invertible ring homomorphism, so $q_p$ is also irreducible over $\mathbb{Q}$; the Gauss lemma for polynomials transfers this to $\mathbb{Z}[X]$. Hence

A Jordan normal form of a complex $n\times n$ matrix $A$ is a similarity transformation $P^{-1}AP=J$ in which $J$ is a direct sum of Jordan blocksFor each eigenvalue, the multiset of block sizes is uniquely determined by $A$; only the order of the blocks is arbitrary. The change-of-basis matrix $P$ is generally not unique.

For the given matrix, expansion of the characteristic polynomial givesThe upper-left $2\times 2$ minor of $A-I$ is $-5$, so its rank is $2$ and the eigenspace for $1$ has dimension $1$. Thus the two occurrences of the eigenvalue $1$ belong to a single size-$2$ Jordan block. More explicitly, the Jordan chain and the remaining eigenvector can be chosen asConsequently, with $P=(v_1{v}_2{v}_3)$, for which $\det P=1$, the answer is