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For any linear system $Ax=c$, a solution exists exactly when $c$ belongs to the image of a linear map represented by $A$. If $x_0$ is one solution, all others are exactlysince $A(x-x_0)=0$, and conversely adding a kernel vector preserves the equation. Thus there are no solutions if $c\notin \mathrm{im}A$, one if $c\in \mathrm{im}A$ and $\ker A=\{0\}$, and infinitely many if $c\in \mathrm{im}A$ and the kernel is nontrivial: every real multiple of a nonzero kernel vector gives a distinct solution. Any two solutions differ by a kernel vector. Equivalently, consistency means the augmented matrix and $A$ have equal rank of a matrix; a consistent square linear system is unique exactly when its determinant is nonzero.

For the given matrix, set $d=b-a$ and $s=2a+b$. Perform the column operation $C_1\leftarrow C_1+C_2+C_3$, then the row operations $R_2\leftarrow R_2-R_1$ and $R_3\leftarrow R_3-R_1$. Factoring the first column givesThese operations are determinant-preserving, and the displayed polynomial identity remains valid even when a factored expression vanishes.

To find all kernels and images of a linear map, putIf $J$ is the all-ones matrix, then $A=aJ+dP$. The direct sum ${\mathbb{R}}^3=\mathbb{R}u\oplus H$ is invariant under $A$: on $\mathbb{R}u$ it acts by multiplication by $s$, and on $H$ by $dP$. Since $P^3=I$, its restriction to $H$ is invertible.

For $d\ne 0$ and $s\ne 0$, the kernel is $\{0\}$ and the image of a linear map is all of ${\mathbb{R}}^3$. Decompose the right-hand side asApplying the inverse on each invariant summand gives the unique solutionFor $d\ne 0$ and $s=0$, the kernel is $\mathbb{R}u$ and the image of a linear map is $H$. A solution exists exactly when $c+2=0$. For $c=-2$, all solutions areFor $d=0$ and $a\ne 0$, so $a=b$, the kernel is $H$ and the image of a linear map is $\mathbb{R}u$. A solution exists exactly when $c=1$. All such solutions areFinally, for $a=b=0$, the kernel is ${\mathbb{R}}^3$ and the image of a linear map is $\{0\}$. The right-hand side has first entry one, so there is no solution for any $c$. These four cases exhaust all parameter values.

Use suffix notation, summing repeated indices from one to three. The contraction identity for the Levi-Civita symbol givesHere $a\cdot a=1$ because $a$ is a unit vector. For the second cross product identity, use ${ϵ}_{ijk}{ϵ}_{j\ell m}={\delta }_{im}{\delta }_{k\ell }-{\delta }_{i\ell }{\delta }_{km}$:The first term vanishes because $a_{\ell }{a}_n$ is symmetric while the Levi-Civita symbol is antisymmetric in those indices. The last equality uses cyclicity and antisymmetry of the scalar triple product.

On the unit sphere, a great circle has radius one, so the spherical distance is its shorter arc length, equal to its central angle in $[0,\pi ]$. The dot product of its endpoint unit vectors is the cosine of that angle. HenceFor the angular formulas, take the ordinary nondegenerate convex minor spherical triangle: no pair of vertices is antipodal and the three vertices do not lie on one great circle. These are needed for the displayed denominators and interior angles to be defined. Merely requiring three distinct points, as the PDF does, does not exclude degeneracy.

Put $p=\delta (A,B)$, $q=\delta (A,C)$ and $r=\delta (B,C)$. The outgoing unit tangent vectors at $A$ areTheir dot product is $\cos \alpha$. Equivalently, they are obtained from the normalized plane normals $a\times b$ and $a\times c$ by the same right-angle rotation in the tangent plane: $u=(a\times b)\times a/|a\times b|$, and similarly for $v$. Rotation preserves the dot product, givingSince $|a\times b|=\sin p$ and $|a\times c|=\sin q$, the first proved identity becomes the spherical law of cosines:For the spherical law of sines, take norms in the second proved identity. Writing $\tau =a\cdot (b\times c)$, and using $0<\alpha <\pi$, givesCyclic permutation gives the corresponding formulas at $B$ and $C$. Dividing each by the positive product $\sin p\sin q\sin r$ provesFinally, if all sides equal $s$, the spherical law of cosines givesbecause $0<s<\pi$ makes $1+\cos s>0$ and $\cos s<1$. Cosine is strictly decreasing on $(0,\pi )$, so every angle of a nondegenerate equilateral spherical triangle is greater than $\pi /3$.

The key fact is that convergence of a power series at one nonzero point forces absolute convergence at every smaller modulus. Indeed, if $\sum a_n{w}^n$ converges and $w\ne 0$, its terms form a bounded sequence, say $|a_n{w}^n|\le M$. For $|z|<|w|$,by convergence of a geometric series.

Let $S$ be the set of moduli of points at which the power series converges, and put $R=\sup S$, allowing $R=\infty$. The set contains zero, so its supremum is defined in $[0,\infty ]$. Whenever $|z|<R$, the definition of supremum provides a convergence point $w$ with $|w|>|z|$. The preceding estimate gives absolute convergence at $z$. Whenever $|z|>R$, convergence at $z$ would put $|z|$ in $S$, contradicting its upper bound. This proves the asserted radius of convergence property, including $R=0$ and $R=\infty$. Nothing in the argument decides convergence on $|z|=R$.

For the requested boundary behavior, take the power seriesIts coefficients are $a_{2k}=1/k$ for $k\ge 1$, $a_{2k+1}=0$ and $a_0=0$. At either $z=1$ or $z=-1$ it is the divergent harmonic series. At either $z=i$ or $z=-i$ it is ${\sum }_{k\ge 1}(-1{)}^k/k$, which converges by the alternating series test. In fact its radius of convergence is one, since inside the unit circle the terms are bounded by a geometric series, and outside it they do not tend to zero.

Rolle's theorem states that a continuous function $f:[a,b]\to \mathbb{R}$, differentiable on $(a,b)$ with $a<b$ and $f(a)=f(b)$, has $f^{\prime }(c)=0$ for some $c\in (a,b)$. If $f$ is constant this is immediate. Otherwise the extreme value theorem supplies a maximum or minimum with value different from the common endpoint value, so one such extremum occurs in $(a,b)$. At an interior maximum, the difference quotients for positive increments are nonpositive and those for negative increments are nonnegative. Their common limit, the derivative, must be zero. At a minimum the inequalities reverse, with the same conclusion. This proves Rolle's theorem.

For the two-function identity, write $\Delta f=f(b)-f(a)$ and $\Delta g=g(b)-g(a)$, and set $\varphi (x)=g(x)\Delta f-f(x)\Delta g$. This is a continuous function on the closed interval and is differentiable inside it. Its endpoint values agree:They need not be zero, but Rolle's theorem only requires equality. At the resulting point $c$,which proves the Cauchy mean value theorem identityFor the final endpoint limit, first state the denominator convention explicitly. In the usual zero-over-zero version of L'Hopital rule, $g^{\prime }(x)\ne 0$ on some whole interval $(0,\delta )$; this is understood if the displayed derivative quotient is to be a function defined throughout a deleted right neighborhood of zero. Define $f(0)=g(0)=0$, making both functions continuous at zero. For $0<x<\delta$, Rolle's theorem implies $g(x)\ne 0$, since otherwise $g$ would have equal zero values at $0$ and $x$ and a zero derivative between them. Apply the proved Cauchy mean value theorem on $[0,x]$ to obtain $c_x\in (0,x)$ withGiven $\epsilon >0$, the hypothesis on the derivative quotient supplies $\eta >0$ such that its distance from $\ell$ is less than $\epsilon$ whenever $0<t<\eta$. If $0<x<\min (\eta ,\delta )$ then $0<c_x<x<\eta$, so the same bound holds for $f(x)/g(x)$. ThereforeNo continuity of the derivatives is needed.

The PDF does not explicitly state the nonvanishing condition on $g^{\prime }$. If “the limit exists” permits a limit only along the natural domain where $g^{\prime }\ne 0$, the assertion is false, even with $g>0$ everywhere. The distinction concerns derivative quotient limits on partial domains. Here is a concrete counterexample to that weaker reading. Put $x_n=2^{-n}$ and $h_n=4^{-n}$ for $n\ge 0$. On each interval $[x_n/2,x_n]$, split off the plateau $[3x_n/4,x_n]$. On the plateau setOn the transition interval $[x_n/2,3x_n/4]$ setExtend $g=1,f=0$ for $x\ge 1$. The endpoints fit because $S(0)=0$, $S(1)=1$, while both $S^{\prime }$ and $q^{\prime }$ vanish at zero and one and $q$ also vanishes there. Thus $f,g$ are differentiable for all $x>0$ and tend to zero at zero. The derivative $g^{\prime }$ is nonzero exactly inside the transition intervals, where $f^{\prime }=0$, so the derivative quotient has natural-domain limit zero. But $f(x_n)/g(x_n)=0$, whereas $f(7x_n/8)/g(7x_n/8)=1$. Hence $f/g$ has no limit. The preceding qualified proof is the intended valid form of L'Hopital rule.

The intermediate value theorem says that if $f:[a,b]\to \mathbb{R}$ is a continuous function, every value between $f(a)$ and $f(b)$ is attained at some point of $[a,b]$.

For a proof, endpoint values are immediate, and replacing $f$ by $-f$ reduces the remaining case to $f(a)<y<f(b)$. The set $S=\{x\in [a,b]:f(x)\le y\}$ is nonempty and bounded above; let $c=\sup S$ by the least-upper-bound property. Continuity and the strict endpoint inequalities ensure $a<c<b$: near $a$ there are points in $S$, and near $b$ there are none. If $f(c)>y$, continuity gives a neighborhood of $c$ containing no points of $S$, contradicting the defining approximation property of the supremum. If $f(c)<y$, continuity gives points to the right of $c$ still in $S$, contradicting its being an upper bound. Hence $f(c)=y$, proving the intermediate value theorem.

For the distance problem, define the continuous functionIt satisfies $d(x)\ge |x-a|$, so it is a coercive function: $d(x)\to \infty$ as $|x|\to \infty$. Choose $R>|a|+d(0)+1$. Outside $[-R,R]$, the distance is greater than $d(0)$. On $[-R,R]$, the extreme value theorem gives a minimizer $x_0$, with $A=d(x_0)\le d(0)$. It is therefore a global minimizer and $A\ge 0$.

For any $y>A$, choose $X>x_0$ so large that $d(X)>y$. Apply the intermediate value theorem to $d$ on $[x_0,X]$ to obtain a point whose distance is $y$. The distance $A$ itself is attained at $x_0$, and no smaller value is possible. This is the range of a continuous coercive real function. Thus the distance set is exactlyIn particular $A=0$ precisely when the fixed point lies on the graph.