At a point of complex differentiability, the complex derivative is independent of the direction of approach. Taking increments along the real axis gives $f^{\prime }=u_x+i{v}_x$; taking imaginary increments gives $f^{\prime }=v_y-i{u}_y$. Equality of their real parts and imaginary parts proves the Cauchy-Riemann equations $u_x=v_y$ and $u_y=-v_x$.

For the specified real part, observe that $z^2{e}^z$ has exactly that real part. Equivalently, integrating $v_y=u_x$ and checking $v_x=-u_y$ givesThusThe freedom is only a constant imaginary part: the difference of two such holomorphic functions has zero real part, so its Cauchy-Riemann equations make it constant on each connected domain.

For $s=|\mathbf{x}-{\mathbf{x}}_0|>0$, the radial Laplacian gives ${\nabla }^2(1/s)=s^{-2}{\partial }_s(s^2{\partial }_s(1/s))=0$. The outward flux of $\nabla [-1/(4\pi s)]$ through a small sphere centred at the source is $[1/(4\pi s^2)]4\pi s^2=1$. Applying Green's second identity to a smooth test function outside that small sphere and shrinking its radius therefore gives the distributional identityThe kernel tends to zero at infinity. Thus it is the fundamental solution with the sign convention $\Delta G=+\delta$. The sign will reverse when converting it into the electrostatic potential of a positive point charge through Poisson's equation.

A group has a closed binary operation, associativity, an identity, and an inverse for each element. For a finite group, Lagrange's theorem states $|G|=[G:H]|H|$ for each subgroup $H$, so $|H|$ divides $|G|$.

The order of a group element $g$ is the least positive $n$ with $g^n=e$, or infinity if none exists. In a finite group the powers form the cyclic subgroup $⟨g⟩$ of that order; Lagrange's theorem proves $\mathrm{ord}(g)\mid |G|$. If $|G|=p$ is prime and $g\ne e$, its order must be $p$. Thus $g$ generates the whole group, proving $G\cong C_p$.

If every nonidentity element has order two, every element is its own inverse. Then $ab=(ab{)}^{-1}=b^{-1}{a}^{-1}=ba$. This proves that a group of exponent two is abelian.

For the normalized second-order equation $y^{\prime \prime }+P(x)y^{\prime }+Q(x)y=0$, an ordinary point has analytic $P,Q$. A regular singular point is a singular point where $(x-x_0)P$ and $(x-x_0{)}^{2}Q$ extend analytically through $x_0$. The two finite singular points here are $0$ and $1/2$. At zero, the respective limiting coefficients are $1/2,0$; at $1/2$, they are $-1,0$. Thus both are regular singular points, with indicial roots $0,1/2$ and $0,2$, respectively.

For fixed endpoint values, stationarity requires the Euler-Lagrange equation $f_y-d(f_{y^{\prime }})/dx=0$. Differentiate $y^{\prime }{f}_{y^{\prime }}-f$ along an extremal: its derivative is $y^{\prime }[d(f_{y^{\prime }})/dx-f_y]-f_x$. Thus when $f_x=0$ the Beltrami identity makes that quantity constant.

Take $y$ as depth measured downward. Conservation of energy from rest gives $m{v}^2/2-mgy=0$, hence speed $v=\sqrt{2gy}$. With arc length $ds=\sqrt{1+y^{\prime 2}}dx$, the travel time is $T=(2g{)}^{-1/2}{\int }_0^{x_B}\sqrt{(1+y^{\prime 2})/y}dx$. For $f=\sqrt{(1+y^{\prime 2})/y}$, the Beltrami quantity is $-1/\sqrt{y(1+y^{\prime 2})}$. Thus $y(1+y^{\prime 2})=c>0$.

Put $y=c{\sin }^2\theta$ and take $y^{\prime }=\cot \theta$ along the first arch. Since $dy/d\theta =2c\sin \theta \cos \theta$, one obtains $dx/d\theta =2c{\sin }^2\theta$. Integrating from the initial cusp gives the cycloid solving the brachistochrone problem:For the equal-height target, the first complete arch has ${\theta }_B=\pi$, so $c=l/\pi$. Also $ds/d\theta =2c\sin \theta$ and $v=\sqrt{2gc}\sin \theta$ for $0<\theta <\pi$. ThereforeThe endpoint zeros of speed cancel the zeros of $ds/d\theta$, leaving finite travel time. Global minimality follows from the maximum-depth bound for a brachistochrone: any competing track with maximum depth $h$ satisfies $T\ge [l/\sqrt{h}+\pi \sqrt{h}]/\sqrt{2g}\ge \sqrt{2\pi l/g}$. The displayed cycloid attains both inequalities with $h=l/\pi$.

Under an orthogonal Cartesian change $x_i^{\prime }=R_{ij}{x}_j$, the chain rule gives ${\partial }_i^{\prime }=(\partial x_j/\partial x_i^{\prime }){\partial }_j=R_{ij}{\partial }_j$. Thus the derivative operator transforms as a rank-one Cartesian tensor. Under a general coordinate change it has the covector transformation law when applied to a scalar; derivatives of vector components then need the usual connection terms. The calculations below use Cartesian coordinates.

Let $r$ be a quarter-turn and $s$ a reflection of the square. The dihedral group is generated by $r,s$ with $r^4=s^2=1$ and $srs=r^{-1}$, and its elements are $1,r,r^2,r^3,s,sr,s{r}^2,s{r}^3$. Two generators are minimal because this group is not cyclic. Its subgroups of order two are $⟨r^2⟩$ and $⟨s{r}^j⟩$ for $j=0,1,2,3$. Its subgroups of order four are $⟨r⟩$, $⟨r^2,s⟩$ and $⟨r^2,sr⟩$.

A group homomorphism obeys $\varphi (gh)=\varphi (g)\varphi (h)$ for all elements; a group isomorphism is a bijective group homomorphism. Take the rotation subgroup $⟨r⟩$ and defineThe quaternion element $i$ has order four, so the map is well defined modulo four, preserves products and is injective onto $\{1,i,-1,-i\}$. Surjectivity onto all eight quaternion elements is not required of a homomorphism.

A cyclic group is generated by the powers of one element. The final claim needs the usual qualification no nontrivial proper subgroup, and $|G|>1$: the trivial subgroup is proper in every nontrivial group, even one of prime order. With that interpretation, if $G=⟨g⟩$ has prime order $p$, Lagrange's theorem permits only subgroup orders one and $p$. Conversely, if its order $n>1$ is composite, choose a divisor $d$ with $1<d<n$. The subgroup $⟨g^d⟩$ has order $n/d$, strictly between one and $n$. This proves the assertion described by groups with no proper nontrivial subgroups.

A group representation is a homomorphism $D:G\to \mathrm{GL}(V)$ into invertible linear maps. It is a faithful representation when it is injective, equivalently when its kernel contains only the identity. A faithful real two-dimensional representation of the cyclic group of order four isThese are rotations of the plane through $0,\pi /2,\pi ,3\pi /2$. Since $J^2=-I$ and $J^4=I$, their multiplication agrees with the group, and all four matrices are distinct.

For any invertible $S$, set $\tilde{D}(g)=SD(g)S^{-1}$. Thenso this also is a group representation. The character of a representation is ${\chi }_D(g)=\mathrm{tr}D(g)$. For the displayed quarter-turn representation, its values on $1,i,-1,-i$ are $2,0,-2,0$.

The homogeneous Cauchy-Euler equation has powers $x^r$ with $r(r-1)+2r-2=(r-1)(r+2)=0$. Boundedness at zero selects $x$, and boundedness at infinity selects $x^{-2}$. Put $G=Ax$ for $x<\zeta$ and $G=B{x}^{-2}$ for $x>\zeta$. Continuity gives $B=A{\zeta }^3$. Since the operator is $L=(x^{2}d/dx{)}^{\prime }-2$, the required delta-function normalization is ${\zeta }^2[G_x{]}_{{\zeta }^{-}}^{{\zeta }^{+}}=1$. Thus $-3A{\zeta }^2=1$, giving the bounded Green kernel for a radial Euler operator:In the requested notation, $\boxed{a=1,b=-2,c=1,d=-2,k_{-}=k_{+}=-1/3}$. The two expressions agree at $x=\zeta$.

With ordinary measure $d\zeta$, the solution isFor bounded forcing these integrals converge and give a bounded solution. It is unique because no nonzero combination of $x,x^{-2}$ is bounded at both ends. For a more general forcing, convergence and boundedness of the displayed integral must be checked.

The Wronskian is $W=y_1{y}_2^{\prime }-y_1^{\prime }{y}_2$. If constants $A,B$ obey $A{y}_1+B{y}_2=0$, differentiating gives $A{y}_1^{\prime }+B{y}_2^{\prime }=0$. At any point where $W\ne 0$, these form an invertible two-by-two system, forcing $A=B=0$. Thus the two solutions are linearly independent.

For the given equation, the normalized coefficients are $p=(1-x)/(2x)$ and $q=-1/(4x)$, so zero is a regular singular point. Substitute a Frobenius series $y=x^r{\sum }_{n=0}^{\infty }{a}_n{x}^n$, $a_0\ne 0$. The lowest power gives the indicial equationThe coefficient of $x^{n+r}$ givesFor either root the common factor is nonzero for $n\ge 0$, soWith leading coefficient one, the two independent solutions areTheir analytic series factors both have infinite radius of convergence, because the ratio of consecutive terms tends to zero for every fixed $x$. The factor $x^{1/2}$ still gives the second solution a branch point at zero; it is not an entire function or an ordinary Taylor series there. On $x>0$, the general real solution is $C_1{y}_1+C_2{y}_2$. This is the $b=1/2$, $z=x/2$ case of the exponential reduction of a confluent hypergeometric equation with equal parameters.