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Using the Taylor series of the exponential and series division,Thus the first three nonzero Laurent terms areThe singularities occur at $z=k\pi i$. The nearest nonzero ones are $\pm \pi i$, so this Laurent series is valid on $0<|z|<\pi$.

On an annulus $r<|z-z_0|<R$, the Laurent theorem represents an analytic function uniquely asFor any positively oriented simple closed contour $C$ in the annulus winding once around $z_0$, the coefficients in the Laurent series areThe formula follows from the Cauchy integral formula applied term by term, and it is independent of the chosen contour by the Cauchy integral theorem.

Let $P_n$ be the finite-dimensional vector space of real polynomials of degree at most $n$. For fixed $a>0$,is a norm: if it vanishes, the polynomial vanishes on an interval and hence is the zero polynomial. Alsois a norm on $P_n$. By finite-dimensional equivalence of norms, there is $C>0$, depending only on $n$ and $a$, such that $M(p)\le CN(p)$. ThereforeThe extreme value theorem supplies $y\in [0,a]$ at which the last supremum is attained, and hence

Two norms $N_1,N_2$ are equivalent norms when constants $c,C>0$ satisfyfor every $v$.

For $x,y\in {\mathbb{R}}^n$, the reverse triangle inequality givesby the Cauchy-Schwarz inequality. Hence $g$ is Lipschitz continuous, and therefore continuous.

On the compact Euclidean unit sphere, $g$ is positive and continuous, so the extreme value theorem gives $0<m\le g\le M<\infty$. Homogeneity then yieldsThus every norm on a finite-dimensional vector space is equivalent to the Euclidean norm; comparing two such bounds proves any two norms on $V$ are Lipschitz equivalent.

A map $f:A\to \mathbb{R}$ is Lipschitz continuous with Lipschitz constant $L$ whenfor all $y,z\in A$.

Fix $y_0\in A$. The Lipschitz inequality and the triangle inequality implywhile choosing $y=y_0$ gives a finite upper bound. Hence the infimum defining $F(x)$ is a real number.

If $x\in A$, then $f(x)\le f(y)+Ld(x,y)$ for every $y\in A$, while $y=x$ gives the reverse inequality after taking the infimum. Thus $F(x)=f(x)$. For arbitrary $x,z\in X$,Taking infima gives $F(x)\le F(z)+Ld(x,z)$; interchanging $x,z$ givesThis is the McShane extension theorem: $F$ extends $f$ without increasing its Lipschitz constant.

First, every nonzero nonunit in a principal ideal domain factors into irreducible elements. Otherwise repeatedly choose a proper nonunit factor $a_{j+1}$ of $a_j$; then the principal idealsform a strictly increasing chain, contradicting the ascending chain condition, since every principal ideal domain is a Noetherian ring.

Next every irreducible $p$ is a prime element. If $p\mid ab$ but $p\nmid a$, then $\gcd (p,a)=1$. The Bezout identity gives $up+va=1$, so multiplying by $b$ shows $p\mid b$. Thus irreducibles are prime.

Existence of an irreducible factorization and primality give uniqueness: an irreducible on one side divides the product on the other, hence is associate to one factor; cancel and continue. Therefore every principal ideal domain is a unique factorization domain.

If six triangles met at every vertex, counting vertex-face incidences would give $3F=6V$, hence $F=2V$. But part b gives $F=2V-4$, a contradiction. Therefore no such triangulation exists.

For a geodesic triangulation with $V,E,F$ vertices, edges, and faces, summing the spherical excess gives $4\pi =2\pi V-\pi F$, because all angles around each vertex sum to $2\pi$. Also $3F=2E$. ThereforeThus every geodesic triangulation of the sphere has Euler number $2$.

For a geodesic triangle on the unit sphere with angles $\alpha ,\beta ,\gamma$, the Gauss-Bonnet theorem gives

Composing the maps givesa conformal bijection from $S$ to the unit disc.

For $\mathrm{Re}z>0$,so $w=(z-1)/(z+1)$ lies in the unit disc. Its inverse is $z=(1+w)/(1-w)$, whose real part is $(1-|w{|}^2)/|1-w{|}^2>0$. Both derivatives are nonzero, proving this Möbius transformation is conformal.

ChooseIt is holomorphic on the right half-plane with derivative $1/z\ne 0$, and its inverse $z=e^w$ has positive real part precisely for $|\mathrm{Im}w|<\pi /2$. Thus this branch of the complex logarithm is a conformal bijection onto $S$.

Minimizing $\Vert Y-X\theta {\Vert }^2$ gives the normal equations $X^{T}X\hat{\theta }=X^{T}Y$. Full column rank makes $X^{T}X$ invertible, soWith the hat matrix $H=X(X^{T}X{)}^{-1}{X}^T$, the residual is $\hat{\epsilon }=(I-H)\epsilon$. The symmetric projection $I-H$ has rank $n-p$, henceand the latter estimator is unbiased.

The volume flux per unit width isThereforeThe first is the stress-free film; the second has an effectively immobilized free surface.