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For real $x$ at which the denominator is nonzero, mapping to the unit circle gives $|ax+b{|}^2=|cx+d{|}^2$. There are infinitely many such real $x$, so the resulting quadratic polynomial identity has all coefficients zero:If $c=0$, the first equality gives $a=0$, contradicting the nonzero determinant $ad-bc$. Thus $a,c\ne 0$. Put $\lambda =a/c$, $k=b/a$ and $\ell =d/c$. The equalities becomeTwo complex numbers with equal real parts and equal moduli have imaginary parts equal in magnitude. Consequently $\ell =k$ or $\ell =\overline{k}$. The first option would give $ad-bc=ac(\ell -k)=0$, so it is excluded. The second givesMoreover $k$ must be nonreal, since otherwise the two options coincide and the map would be constant. This proves the Möbius transformations from the real axis to the unit circle representation directly from coefficients.

Conversely, for real $x$, the numbers $x+k$ and $x+\overline{k}$ are complex conjugates, and their moduli agree. Nonreal $k$ ensures the denominator is nonzero on the real line; multiplication by $\lambda$ preserves modulus. At the point at infinity on the Riemann sphere, the map takes the value $\lambda$. Thus the representation also respects the extended real axis.

The quotient is defined only when $z+\overline{u}\ne 0$. Its unit-modulus condition then gives $|z+u{|}^2=|z+\overline{u}{|}^2$. On expanding with complex conjugates,Hence at least one imaginary part is zero, and either $z$ or $u$ is real. Conversely, either reality condition makes the two squared moduli equal; provided the denominator is nonzero, the quotient has modulus one. No conclusion is drawn from an undefined quotient.

Write a straight line in Cartesian form $Ax+By+C=0$, where $A,B,C$ are real and $(A,B)\ne (0,0)$. With $z=x+iy$, set $c=(A+iB)/2$ and $r=C$. Using the complex conjugate,Thus the real linear equation of a line in the complex plane isConversely, writing $c=u+iv$ gives $2ux+2vy+r=0$, an ordinary straight line with normal vector $(u,v)$. The nonzero condition on $c$ matters: $c=0$ would give either the whole complex plane or the empty set.

Set $s=x+y+z$ and $Q=x^2+y^2+z^2$. The expression in the equation is $4(Q+4)-(s+2{)}^2$. Apply the Cauchy-Schwarz inequality in ${\mathbb{R}}^4$ toTheir dot product is $s+2$, and $|\mathbf{e}{|}^2=4$. Thus the expression is nonnegative, and it is zero precisely when equality holds in Cauchy-Schwarz. That requires $\mathbf{v}=\lambda \mathbf{e}$. The fourth coordinate forces $\lambda =2$, and the other three coordinates then giveSubstitution gives $3(4+4+4+4)-2(4+4+4)-4(2+2+2)=48-24-24=0$, so this unique possible choice really is a solution.

Apply the Cauchy-Schwarz inequality to $(x,y,z)$ and $(1,1,1)$ in the Euclidean inner product on ${\mathbb{R}}^3$. It givesAfter expansion and division by two,The equality criterion from the introductory proof says $x=y=z$. One can also see both the inequality and equality case from $x^2+y^2+z^2-xy-yz-zx=\frac{1}{2}[(x-y{)}^2+(y-z{)}^2+(z-x{)}^2]$.

Let $P\mathbf{x}=(\mathbf{n}\cdot \mathbf{x})\mathbf{n}$ be the axial orthogonal projection and $N\mathbf{x}=\mathbf{n}\times \mathbf{x}$. Then $NP=PN=0$, while the vector triple-product identity gives $N^2=P-I$. The map is $aP+a(I-P)+bN$. Its inverse is $P/a$ on the axis. On the perpendicular plane,Thus the inverse of an axis-angle linear map isor, in the requested vector form,The denominators are nonzero because $a,b>0$. Multiplication on each invariant component verifies the matrix inverse without choosing a coordinate frame.

In suffix notation, $x_i^{\prime }=a{x}_i+b{ϵ}_{ikj}{n}_k{x}_j$, so the matrix entries areHere $\delta$ is the Kronecker delta and $ϵ$ the Levi-Civita symbol. The diagonal determines $a=2$. Comparing the off-diagonal entries withgives $b\mathbf{n}=(1,2,2)$. As $\mathbf{n}$ is a unit vector and $b>0$, $b=3$ and $\mathbf{n}=(1,2,2)/3$. ThusIn this example the transverse scaling is $\sqrt{13}$ and the axial scaling is two; the rotation angle is $\arctan (3/2)$.

Write $\mathbf{x}=s\mathbf{n}+\mathbf{v}$, with $\mathbf{v}\cdot \mathbf{n}=0$. The linear map sends $s\mathbf{n}$ to $as\mathbf{n}$ and sends $\mathbf{v}$ to $a\mathbf{v}+b(\mathbf{n}\times \mathbf{v})$. On the perpendicular plane, $\mathbf{v}$ and $\mathbf{n}\times \mathbf{v}$ are perpendicular vectors of equal length. ThusThe expression in parentheses is a rotation in three dimensions by $\vartheta$ about the oriented $\mathbf{n}$ axis, with the positive direction fixed by the cross product. Therefore the map scales the axial component by $a$, scales the perpendicular component by $\sqrt{a^2+b^2}$, and rotates the perpendicular plane by $\arctan (b/a)$. Since $a,b>0$, the angle lies between zero and $\pi /2$. This is an axis-angle decomposition, with different axial and transverse scale factors.

Decompose the position vector into its components parallel and perpendicular to the unit vector $\mathbf{u}$:This is an orthogonal projection, and the Pythagorean theorem gives $|\mathbf{r}{|}^2=s^2+|\mathbf{v}{|}^2$. The two surfaces are distinguished by fixing an angle in part (a), and fixing distance from an axis in part (b).

When $p=1$ but $t^2\ne 3t-2$, the dependent coefficient rows would require an impossible equality between their right-hand sides. Thus the augmented matrix has larger matrix rank than the coefficient matrix, and the linear system is inconsistent. For $p\ne 1$ part (i) already supplies a unique solution. Consequently there is no solution exactly when $p=1$ and $t\notin \{1,2\}$.

For $p=1$, the third coefficient row equals three times the second minus twice the first. The first two rows are independent, so the coefficient matrix rank is two. The same dependence must hold for the right-hand side, givingThus infinitely many solutions occur exactly for $p=1$ and $t\in \{1,2\}$. The null space has dimension one by the rank-nullity theorem, and an explicit parameterization, with $s\in \mathbb{R}$, isSubstitution verifies all three equations in each family.

The coefficient matrix has determinantA square linear system has a unique solution for every right-hand side precisely when its coefficient matrix is an invertible matrix. Here a unique solution exists exactly when $p\ne 1$, for every real $t$.

For an explicit Gaussian elimination, subtract twice the second equation from the third, then use the second equation, obtaining $x=2z+2t-t^2$ and $y=(t^2-t)/2-3z$. The first equation determinesand the preceding formulas determine $x,y$. When $p=1$ this last coefficient vanishes; the next parts determine compatibility rather than dividing by zero.

Use suffix notation, summing repeated indices, and the Levi-Civita symbol convention $(A\times B{)}_i={ϵ}_{ijk}{A}_j{B}_k$. Contracting the outer symbol with the symbol from the second cross product gives the epsilon-delta identitywhere $\delta$ is the Kronecker delta. ThusThe second term is zero because $A_p{A}_{\ell }$ is symmetric in $p,\ell$ and the Levi-Civita symbol is antisymmetric. The cyclic rearrangement of the remaining symbol has positive sign. Its contraction is the scalar triple product $A\cdot (B\times C)$. HenceThis cross products with a common vector identity requires no independence assumption, so it also covers zero or parallel vectors.

The three spherical basis vectors form an orthonormal basis: their displayed Cartesian components have unit lengths and pairwise zero dot products. For an orthonormal basis, a vector is the sum of its projections onto the basis vectors. Taking the dot product of each Cartesian unit vector with each spherical unit vector therefore gives the inverse change of basis:Equivalently, the change-of-basis matrix is orthogonal, so its matrix inverse is its transpose. These formulas use the stated angular convention for spherical polar coordinates; at a pole the tangential basis vectors still depend on the chosen azimuth.

The PDF defines this function using $\sin (1/x)$; the converted TeX contains a conflicting duplicate $\sin (1/2)$ before its later correct version. Use the PDF function. It is bounded by one and continuous at every positive $x$, although it is discontinuous at zero. Indeed the sequences $x_n=(\pi /2+2\pi n{)}^{-1}$ and $y_n=(3\pi /2+2\pi n{)}^{-1}$ tend to zero with function values $1$ and $-1$.

That single discontinuity does not prevent Riemann integrability. Given $\epsilon >0$, choose $0<\delta <\min \{1,\epsilon /4\}$. The cell $[0,\delta ]$ contributes at most $2\delta <\epsilon /2$ to the difference between the upper Darboux sum and the lower Darboux sum. On $[\delta ,1]$ the function is uniformly continuous, so choose a partition there whose oscillation sum is less than $\epsilon /2$, by the introductory proof. Joining that partition to the cell $[0,\delta ]$ givesThus the PDF's function is Riemann integrable despite its endpoint discontinuity. This is a direct application of bounded functions continuous away from one endpoint are Riemann integrable, proved here with the actual oscillation bound rather than an improper-integral argument.

For $x>0$, the function is a composition and product of continuous functions. At zero,Thus it is continuous on the whole closed interval, including its assigned value at zero. The continuous-function proof in the introductory solution applies, so this function is Riemann integrable. The vanishing factor $x$ controls the infinitely rapid oscillation at the endpoint.

Put $q=\lambda a+(1-\lambda )b$. Since $0<\lambda <1$ and $a<b$, $a<q<b$. Apply the mean value theorem separately to $[a,q]$ and $[q,b]$. There exist $x\in (a,q)$ and $y\in (q,b)$ such thatThese are the requested points, and $x<y$. Part (i) makes $f^{\prime }(x)\le f^{\prime }(y)$, so the corresponding secant slopes satisfyMultiply by the positive denominators and collect $f(q)$ to obtainThis is precisely the defining chord inequality for a convex function. The proof uses only existence and nonnegativity of $f^{\prime \prime }$, with no additional regularity hypothesis on that derivative.

Because $f^{\prime \prime }$ exists everywhere, $f^{\prime }$ is differentiable everywhere and therefore continuous everywhere; continuity of $f^{\prime \prime }$ is not required. For $x<y$, apply the mean value theorem to $f^{\prime }$ on $[x,y]$. It gives $\xi \in (x,y)$ withFor $x=y$ equality is immediate. Hence $f^{\prime }$ is nondecreasing: $x\le y$ implies $f^{\prime }(x)\le f^{\prime }(y)$.

For the similarity solution, differentiation givesChoose ${\lambda }^2\dot{\lambda }=e^{-\mu t}$. The remaining equation is $-(\eta f{)}^{\prime }=(f{f}^{\prime }{)}^{\prime }$. Its integration constant is zero for the required decaying or compactly supported profile, so $f(f^{\prime }+\eta )=0$. Where $f>0$, this gives $f^{\prime }=-\eta$. For a nonnegative population the complete profile is thereforeNormalization gives $Q=2b^3/3$, hence $b=(3Q/2{)}^{1/3}$. Integrating the scale equation with zero initial scale yieldsThusIts support satisfiesThe profile is a weak solution at its moving edges, where $n{n}_x$ vanishes. As $t\downarrow 0$ its support shrinks to the origin and its mass tends to $Q$, so its initial distribution is $Q{\delta }_0$. The bounded limiting range assumes $\mu >0$; at zero death rate the scale instead grows as $(3t{)}^{1/3}$.

Integrating the population equation over the line and using vanishing boundary flux gives $\dot{M}=-\mu M$, where $M(t)=\int n(x,t)dx$. Since $M(0)=Q$, one has $M(t)=Q{e}^{-\mu t}$. On the other hand the similarity substitution $x=\lambda (t)\eta$ givesConsequentlyThe decay condition is on the nonlinear diffusive flux $n{n}_x$; compact support also satisfies it.