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Separate the variables in this first-order ordinary differential equation:Integration gives $\frac{1}{2}\ln (y^2+2)=\frac{1}{2}\ln |x^2-3|+C$, so $y^2+2=K|x^2-3|$. On the interval containing the prescribed point $x=3$, we have $x^2-3>0$. The initial condition $y(3)=4$ therefore gives $18=6K$, hence $K=3$. Choosing the positive square root to match that initial condition yieldsThe maximal interval containing $3$ on which this solves the explicit ordinary differential equation is $\boxed{x>\sqrt{11/3}}$. At the left endpoint $y=0$, the original denominator vanishes and the derivative becomes unbounded. Directly, $y^{\prime }=3x/y$ and $x(2+y^2)/[y(x^2-3)]=3x/y$, verifying the equation throughout this interval.

The right-hand side has modulus $4$ and argument $\pi /3$. Its square roots have modulus $2$ and arguments $\pi /6$ and $7\pi /6$, giving

On an interval avoiding zero put $v=y/x$. A real solution requires $|v|\le 1$, and the principal square root is $\sqrt{x^2-y^2}=|x|\sqrt{1-v^2}$. The homogeneous first-order differential equation reduces toWhere $|v|<1$, separation of variables gives $\arcsin v=\mathrm{sgn}(x)\log |x|+C$. Thus the nonconstant pieces arerestricted to intervals where the cosine of the phase is nonnegative. This sign condition is essential when substituting back into the original square root.

There are also the solutions $y=x$ and $y=-x$. They can join the nonconstant pieces with a continuous first derivative, because the transformed rate vanishes but is not Lipschitz continuous at $v=\pm 1$. For example, on the positive axis every $A>0$ givesThe sticking solutions of a bounded square-root differential equation describe the same joining mechanism on either half-axis after using the increasing variable $t=\mathrm{sgn}(x)\log |x|$.

Isolate the square root: $\sqrt{x+16}=x-4$. A real solution must have $x\ge 4$, because a real principal square root is nonnegative. Squaring gives $x+16=(x-4{)}^2$, hence $x(x-9)=0$. The candidate zero violates $x\ge 4$; substitution verifies the other candidate. ThusThe sign restriction prevents the extraneous solution introduced by squaring.

Isolate the square root: $\sqrt{2x^2+17}=3-x$. Its right side must be nonnegative, so retain the condition $x\le 3$ before squaring. Squaring then givesBoth candidates pass the original equation: $\sqrt{25}-2=3$ and $\sqrt{49}-4=3$. ThusThe substitution check excludes the possibility of extraneous roots introduced by squaring.

Use the integration by substitution $x=\sin \theta$, with $-\pi /6\le \theta \le \pi /6$. On this interval $\cos \theta >0$, so it cancels the positive denominator square root. The transformed integral isIts integrand is even. Therefore the answer in the PDF's modulus-amplitude convention for the elliptic integral of the second kind isThe modulus is $2/3$, not the squared parameter $4/9$.

Write a square root as $u+iv$, with $u,v$ real. Equating the real part and imaginary part of its square gives $u^2-v^2=-1$ and $2uv=1$. Taking complex modulus also gives $u^2+v^2=\sqrt{2}$. ThereforeSince $uv>0$, the signs agree. The two roots areThe $\pm$ changes both signs together, rather than independently.

Independence makes the variances of the attendance indicators additive. The binomial distribution therefore gives $\mathrm{Var}(N)=np(1-p)=24(3/4)(1/4)=9/2$. Its positive square root is the standard deviation:

A constant sequence must satisfy the fixed point equation $y_{\ast }=\sqrt{2+y_{\ast }}$. The square root forces $y_{\ast }\ge 0$. Squaring gives $(y_{\ast }-2)(y_{\ast }+1)=0$, so the negative algebraic root is inadmissible. HenceStarting at $y_0=2$ produces the constant sequence $2,2,\dots$.

Solving for $y$ gives $y=\pm (b/a)\sqrt{x^2-a^2}$ with $x\ge a$. For large $x$, the square root is $x-a^2/(2x)+O(x^{-3})$, so the two asymptotes areThe curve lies between these lines on the right side and has its nearest point, with vertical tangent, at $(a,0)$. As $t$ increases it travels from the lower unbounded end through that vertex to the upper unbounded end. The left hyperbola branch is not part of the given curve.

For the composite square root, $y>1$ throughout the restricted $x$ interval. Differentiate with respect to $y$:The second expression comes from differentiating both $y^{-1}$ and $(\log y{)}^{-1/2}$. The endpoints where $y=1$ are excluded here.

In $A^{\mathsf{T}}A$, every row sums to one, so $(1,1,1)$ is an eigenvector with eigenvalue $1$. The antisymmetric direction $(1,0,-1)$ gives eigenvalue $2$. A direction orthogonal to both is $(1,-2,1)$, and direct multiplication shows that it lies in the kernel of a linear map. Normalizing all three gives the complete systemThey form an orthonormal basis of ${\mathbb{R}}^3$. The extra zero eigenvalue reflects the one-dimensional kernel of a linear map for this rank-two rectangular matrix. The question's symbols ${\sigma }_m$ denote the eigenvalues of $A^{\mathsf{T}}A$; the nonzero singular values of $A$ are their square roots, namely $1$ and $\sqrt{2}$.

The Newton root-finding iteration follows the current tangent line to its horizontal-axis intersection. Since $f^{\prime }(x)=2x$, it givesA positive initial guess is appropriate for the positive square root; negative starting values instead target the negative root.

Put $\theta =\pi /12$. The double-angle trigonometric identities imply ${\cos }^2(2\theta )=1-{\sin }^2(2\theta )$. Since $2\theta =\pi /6$ is in the first quadrant, the positive square root is required. The half-angle identities then give the requested expression in terms of $\sin (\pi /6)$:

Branches and cuts. For the square root of z squared minus one, a branch point is a point around which analytic continuation along an arbitrarily small closed loop can change the value of a chosen branch of a multivalued function. Near $z=1$, the square root behaves as $\sqrt{2}(z-1{)}^{1/2}$; near $z=-1$ it is a nonzero analytic factor times $(z+1{)}^{1/2}$. One circuit around either point reverses its sign. Thus the finite branch points are $\boxed{z=\pm 1}$. Infinity is not a branch point: in $w=1/z$ the expression is $w^{-1}\sqrt{1-w^2}$, which has two single-valued meromorphic choices near zero.

A single branch cut can join $-1$ to $1$; its shortest choice is the real segment $[-1,1]$. Any surviving loop then surrounds either both branch points or neither, so it produces no sign change. Alternatively, use two branch cuts, $(-\infty ,-1]$ and $[1,\infty )$, each joining a branch point to infinity. On the resulting slit region one can choose a consistent branch of a multivalued function.

For the finite cut, fix $f(z)\sim z$ at infinity, equivalently $f(x)>0$ for $x>1$. To give its actual values on both sides, putFor $y\ne 0$, the selected square root isHere $\mathrm{sgn}(0)=0$; when $x=0$, $U=0$. Squaring verifies the formula because $UV=|xy|$. Its real part has the sign of $x$, its imaginary part has the sign of $y$, and it matches the chosen value at infinity. In particular, the two boundary values on the branch cut areOutside the cut, $f(x)=\mathrm{sgn}(x)\sqrt{x^2-1}$. Reversing the global branch of a multivalued function reverses every displayed value.

Differential equation. Write $f(\zeta )=\sqrt{{\zeta }^2-1}$ with this finite cut, and orient $C$ counterclockwise with one circuit around it. Since $C$ is fixed and avoids the branch points, differentiation under the integral sign givesOn the contour $f^{\prime }=\zeta /f$. ThereforeThe last contour integral vanishes because the expression in brackets returns to its original value after circling both branch points. This proves $\boxed{z{u}^{\prime \prime }+u^{\prime }-zu=0}$, including at $z=0$; division by $z$ is unnecessary.

Value at zero. Near either endpoint ${\zeta }_0=\pm 1$, the integrand at zero is $O(|\zeta -{\zeta }_0{|}^{-1/2})$. A small arc of radius $\epsilon$ has length $O(\epsilon )$ and hence contributes $O(\sqrt{\epsilon })\to 0$. The upper lip runs from $1$ to $-1$, and the lower lip from $-1$ to $1$. ThusUsing $x=\sin t$ evaluates the final integral as $\pi$, and soThe same deformation gives $u(z)=2i{\int }_{-1}^1{e}^{zx}(1-x^2{)}^{-1/2}dx$. The sign of $u(0)$ reverses if either the global square-root choice or the contour orientation is reversed; the differential equation is unaffected.

Choose the square root of one minus z squared with $f(0)=1$: explicitly $f(z)=\sqrt{1-z}\sqrt{1+z}$ with principal square roots. It is single valued and holomorphic away from the two specified exterior real branch cuts. If $q(x)=\sqrt{x^2-1}>0$ for $|x|>1$, its boundary values areFor example, above the right cut $1-z$ approaches the negative real axis from below, selecting $-i\sqrt{x-1}$; above the left cut $1+z$ approaches it from above, selecting $+i\sqrt{-x-1}$. Continuing once around either branch point changes the sign, which explains the need for cuts.

Apply the residue theorem to $F(z)=1/[zf(z)]$ on a large disk slit along both cuts, with small arcs excluding $\pm 1$. The only enclosed pole is at zero, and its residue is $1/f(0)=1$. On the positively oriented boundary of the slit domain, the upper right lip runs from one to $R$, and the lower right lip returns from $R$ to one. The jump in $F$ gives their combined integral as $2i{I}_R$, where $I_R={\int }_1^{R}dx/[x\sqrt{x^2-1}]$. The left lips similarly give $2i{I}_R$ after the substitution $x=-t$.

The small arcs contribute $O(\sqrt{\epsilon })$ because the branch singularities have order $-1/2$. The large arc contributes $O(R^{-1})$ because $F(z)=O(|z{|}^{-2})$. Taking limits therefore yields $4iI=2\pi i$, and henceThe real substitution $x=\sec \theta$ independently reduces the integral to ${\int }_0^{\pi /2}d\theta$, confirming its value and sign.

Use the same principal square root $S(z)=\sqrt{(z-a)/(z-b)}$, which is holomorphic on $A$ because its argument avoids the nonpositive real axis. Thenis a holomorphic function satisfyingThus it is a holomorphic branch of the requested square root. This construction does not require a global holomorphic logarithm of $z-a$ or $z-b$ separately. Its sign is fixed by $S(z)\to 1$ at infinity, so $g(z)/z\to 1$. The negative of this function is the other holomorphic branch; any two branches have a continuous quotient taking values in $\{1,-1\}$, which is constant because $A$ is connected.

There is no such map for the domain printed in the paper. The planar domain $A=\mathbb{C}\setminus [a,b]$ is not simply connected. For example, a sufficiently large positively oriented circle $\gamma$ surrounds $[a,b]$ and lies in $A$, whileThe integrand is holomorphic on $A$. If $A$ were simply connected, the Cauchy integral theorem would force this integral to be zero. A conformal bijection is a homeomorphism and preserves simple connectivity, whereas $H$ is simply connected. Thus a conformal bijection $A\to H$ is impossible.

The explicit construction also pinpoints the missing point. DefineThis Möbius transformation sends the removed line segment to the nonpositive real axis, together with infinity. More precisely, its image on the stated planar domain issince $T(z)=1$ has no finite solution when $a\ne b$. Conversely, its inverse $z=(bw-a)/(w-1)$ is defined for every $w$ in this set and cannot lie on $[a,b]$, proving this image description exactly.

Take the principal square root of $T(z)$, characterized by positive real part, and then multiply by $i$. The result is the conformal mapEach stage is injective and onto the indicated image; the square root maps the slit plane to the right half-plane, with the point $1$ omitted here. Hence this is a conformal bijection onto the punctured upper half-plane. Its inverse isIf the source is instead $(\mathbb{C}\cup \{\infty \})\setminus [a,b]$ on the Riemann sphere, the same map extends with $F(\infty )=i$ and becomes a conformal bijection onto all of $H$. Indeed, in the local coordinate $u=1/z$ it has expansion $F=i+\frac{i}{2}(b-a)u+O(u^2)$, whose derivative at $u=0$ is nonzero. This gives the sphere version of the requested construction without silently changing the printed planar domain.

A real square matrix $Q$ is an orthogonal matrix when $Q^{T}Q=I$, equivalently its columns form an orthonormal basis. Here $Q^T$ is the transpose and $I$ the identity matrix. For every vector $x$,Taking nonnegative square roots provesThus an orthogonal matrix preserves the Euclidean norm.

Since every $u_{jj}>0$, the diagonal matrix $D$ is invertible. SetThe division is by the diagonal entry of row $i$. Thus $\hat{U}$ is an upper triangular matrix with diagonal entries one and $A=LD\hat{U}$.

To use symmetry without assuming the conclusion, multiply by the matrix inverse of $L$ and its transpose:The matrix $M$ is symmetric because $A$ is a symmetric matrix. It is also an upper triangular matrix, since every factor on the right is an upper triangular matrix. A symmetric upper triangular matrix is diagonal. Its diagonal is exactly $D$, because $\hat{U}$ and $L^{-T}$ have diagonal entries one. Hence $M=D$, givingThis is the LDL decomposition.

Take the positive square roots of the diagonal entries:The matrix $\hat{L}$ is a lower triangular matrix with positive diagonal, andThis is a Cholesky decomposition. For a nonzero vector $x$, the vector ${\hat{L}}^{T}x$ is nonzero because ${\hat{L}}^T$ is invertible. ThereforeHere $\Vert \cdot {\Vert }_2$ is the Euclidean norm. Thus $A$ is a positive-definite matrix.