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Using the Levi-Civita symbol and the Einstein summation convention in Cartesian coordinates, the product rule givesThe cyclic permutation in the first term preserves the sign, whereas exchanging the first two indices in the second reverses it. Recognizing the two curls proves

On a positively oriented contour encircling $z_0$ once inside an annulus of analyticity, the Laurent series coefficients areThis follows by inserting the convergent Laurent expansion and integrating term by term: only the term with exponent $-1$ has nonzero contour integral. Uniform convergence on compact circles in the annulus justifies that step.

For the function at hand,The geometric series converges for $|z|<1$, and the pole at zero excludes the centre. ThusThe coefficients are $a_n=-1$ for $n\ge -1$ and zero for $n<-1$. The coefficient formula gives the same result on $|z|=\rho <1$: $a_n=(2\pi i{)}^{-1}\oint z^{-n-2}/(z-1)dz$, and expansion of $(z-1{)}^{-1}$ selects exactly those coefficients.

The annulus is essential. For $|z|>1$, expand instead in inverse powers:These are different Laurent expansions about the same centre on the two annuli separated by the singularity at $z=1$. The series requested in the PDF is the inner one, not a global expansion.

For a counterclockwise simple contour containing $z_0$, and $f$ analytic on a neighbourhood of its interior and boundary, the Cauchy integral formula isFor the required rational integral, use partial fraction decomposition:Both points $1,-1$ lie inside $|z|=2$. Applying the formula with the constant analytic function one gives $2\pi i$ for each of the two terms. They cancel, soReversing the circle's orientation would leave this zero result unchanged.

Let $B=\left(\begin{array}{cc}0& b\\ c& 0\end{array}\right)$ be the lower block. ThenThe defining relation $a^4=I$ requires $T(a{)}^4=I_3$, so for real constants $bc=\pm 1$. Both possibilities ensure invertibility.

If $bc=-1$, the generator matrix has order four, because its square is not the identity while its fourth power is. Its four powers are distinct, giving a faithful representation. If $bc=1$, its square is the identity but it is not itself the identity; its lower diagonal entries are zero. The image then has order two and the kernel of a group homomorphism is $\{I,a^2\}$. Thus the faithfulness criterion for a block representation of the cyclic group of order four givesOther parameter values give no representation of $Z_4$, not additional unfaithful cases; $bc=0$ also makes the assigned matrix singular. No orthogonality of these matrices was required.

For the final construction takeMultiplication adds exponents, so $D(a^r)D(a^s)=D(a^{r+s})$ and $D(a{)}^4=I_2$. The representing matrix is the identity exactly for even exponents. Therefore this is a two-dimensional representation with kernel $\{I,a^2\}$, as required. It factors through the quotient cyclic group of order two.

The conjugacy class of $g\in G$ is $\{hg{h}^{-1}:h\in G\}$. An irreducible representation acts on a nonzero vector space with no nonzero proper subspace invariant under every group element.

In the quaternion group, $1$ and $-1$ commute with all elements, so each forms its own conjugacy class. The element $i$ commutes with $\pm 1,\pm i$, while conjugation by $j$ or $k$ changes it to $-i$. Thus its class is $\{i,-i\}$. The same argument for $j,k$ gives all five classes:Their sizes sum to eight. Equivalently each noncentral element has a four-element centralizer and hence a two-element class.

Use the following finite-group representation theorems over $\mathbb{C}$: the number of inequivalent irreducible representations equals the number of conjugacy classes, and their dimensions obey the sum of squares of irreducible representation dimensions, $\sum d_j^2=|G|$. Thus there are five irreducibles, with ${\sum }_{j=1}^5{d}_j^2=8$. Starting with five dimensions equal to one uses five of those eight units; increasing one dimension from one to two uses exactly the remaining three, while any larger increase would exceed eight. ThereforeThese are the dimensions underlying the character table of the quaternion group. The scalar field matters: the quaternionic irreducible becomes four-dimensional over the reals. The class-count and dimension-sum statements used here have their standard complex meaning.

By Fermat principle, a light ray makes optical travel time stationary. Multiplication by the constant vacuum speed of light does not change the stationary paths, so the functional can be taken to be the refractive index integrated over arc length. The constant denoted $c$ below is an integration constant, not that vacuum speed.

Take endpoint-preserving variations $y+\epsilon \eta$, with $\eta (a)=\eta (b)=0$. The first variation of the functional is $\delta I={\int }_a^b(f_y\eta +f_{y^{\prime }}{\eta }^{\prime })dx$; integration by parts supplies the differential equation and its conserved quantity below.

Since $P_0=1$, part (d) gives $Q_0^{\prime }=A_0/(1-x^2)$. A partial fraction decomposition and integration give, for $-1<x<1$,The condition at zero fixes $B=0$. Thus the full conclusion from the stated condition isThat condition alone does not fix the multiplicative constant. With the usual Legendre function of the second kind normalization $Q_0^{\prime }(0)=1$, one has $A_0=1$. On real exterior intervals replace the logarithm by $\log |(1+x)/(1-x)|$; complex continuation requires a choice of logarithm branch.

Let $W=P_{\ell }{Q}_{\ell }^{\prime }-P_{\ell }^{\prime }{Q}_{\ell }$. Differentiation cancels the cross terms, and the Legendre differential equation givesThus $[(1-x^2)W{]}^{\prime }=0$, sowith $A_{\ell }$ independent of $x$ on each interval avoiding the singular endpoints. This is the Abel identity. Linear independence forces $A_{\ell }\ne 0$, and its value depends on the chosen normalization of $Q_{\ell }$.

For $\ell =0$ the terminating even solution is constant; for $\ell =1$ the terminating odd solution is proportional to $x$. For $\ell =2$, the recurrence gives $a_2=-3a_0$; for $\ell =3$ it gives $a_3=-5a_1/3$. Imposing the Legendre polynomial normalization at $1$ yields

For a nonnegative integer $\ell$, the chain of the same parity as $\ell$ reaches $n=\ell$, where the numerator of the recurrence relation vanishes. The next coefficient and every later coefficient in that chain are zero, producing a polynomial of degree $\ell$. Normalizing it at $x=1$ gives the Legendre polynomial $P_{\ell }$.

The opposite-parity chain never reaches $n=\ell$. Its other possible zero, $n=-\ell -1$, is negative and cannot occur, so a nonzero initial coefficient produces a nonterminating power series. It is linearly independent of $P_{\ell }$ and can be scaled to the Legendre function of the second kind $Q_{\ell }$.

The degree statement requires $\ell \ge 0$, as the examples in part (c) indicate. For a negative integer $\ell$, the same equation has the nonnegative degree $-\ell -1$, because $\ell (\ell +1)=(-\ell -1)(-\ell )$. A polynomial cannot literally have negative degree.

Insert the power series and equate the coefficient of $x^n$. In the Legendre differential equation this givesHence the recurrence relation isThe free coefficients $a_0,a_1$ generate the independent even and odd series, respectively. They converge at least for $|x|<1$, up to the nearest singular points of the equation.

For $\mathrm{Re}z>0$, the defining Gamma integral converges at zero and at infinity and defines a holomorphic function. The Gamma function recurrence then supplies its meromorphic continuation to the rest of the complex plane, as used below.