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Adding the three equations for $s,t$ gives the necessary solvability condition $a+b+c=3$. Conversely, subtraction of the first two and use of the third forceIf $a+b+c=3$, these values also satisfy the first two equations: for example,The analogous calculation gives $1-s+t=b$. Thus existence and uniqueness hold exactly under the stated compatibility condition:

For the three-variable system the matrix and data vector arePerform Gaussian elimination on the augmented matrix. The operations $R_2\leftarrow 5R_2-2R_1$ and $R_3\leftarrow 5R_3+R_1$ giveEach operation is an invertible row scaling followed by a row addition, so the solution set is preserved. Next $R_3\leftarrow 7R_3+R_2$ givesBack substitution now yieldsand thenAll three pivots are nonzero. Thus the solution exists uniquely for every $s,t$, the matrix rank is $3$, and the kernel of a linear map represented by $A$ is $\{0\}$. The general rank-nullity theorem for a matrix with $m$ columns ishere $m=3$.

Finally, reverse the viewpoint and regard $x,y,z$ as prescribed. The necessary and sufficient condition already proved applies with $a=5x+2y-z$, $b=2x+5y-z$, $c=-x-y+8z$. Their sum is $6(x+y+z)$. Consequently $s,t$ can be recovered exactly whenWhen this holds, their unique values are $s=\frac{3}{2}(x-y)$ and $t=\frac{1}{2}(1+x+y-8z)$. The allowable triples form an affine plane, not all of ${\mathbb{R}}^3$.

The chosen unit normal points toward the plane from the origin, so the plane equation isTo locate the circle centre, project $\mathbf{p}$ onto this plane. PutThen $\mathbf{n}\cdot \mathbf{h}=d$, so $\mathbf{h}$ lies in the plane. For any other point $\mathbf{r}$ of the plane, $\mathbf{r}-\mathbf{h}$ is perpendicular to $\mathbf{n}$. The Pythagorean theorem therefore givesRestricting the sphere equation to the plane consequently gives a circle centred at $\mathbf{h}$. Thus $\mathbf{m}=\mathbf{h}$, proving that the displacement of its centre is purely normal:The same Pythagorean theorem gives $q^2={\rho }^2+{\gamma }^2$, and henceThe sign of $\gamma$ is fixed by which side of the plane contains $\mathbf{p}$; $q$ and $\rho$ alone determine only its magnitude. Substituting the signed offset yields the sphere-plane intersection radiusA genuine circle requires $|d-\mathbf{n}\cdot \mathbf{p}|<q$; equality gives tangency and $\rho =0$, while a larger distance gives no intersection. The signed distance from a point to a plane in the direction $\mathbf{n}$ is $\mathbf{n}\cdot \mathbf{p}-d=-\gamma$; the ordinary distance from a point to a plane is its absolute value.

The two displacement vectors $\mathbf{b}-\mathbf{a}$ and $\mathbf{c}-\mathbf{a}$ lie in the required plane. Because the three points are non-collinear, their cross productis nonzero and is a normal vector to that plane. A point with position vector $\mathbf{r}$ lies in the plane exactly whenExpanding the cross product givesThe terms $\mathbf{a}\cdot (\mathbf{a}\times \mathbf{b})$ and $\mathbf{a}\cdot (\mathbf{c}\times \mathbf{a})$ vanish, so the equation of a plane through three points becomesNon-collinearity is important: it ensures this is a genuine plane equation with nonzero normal, rather than the vacuous equality $0=0$.

For the specified coordinates,Their sum is $(3,3,6)$, with Euclidean norm $3\sqrt{6}$. Consequently either orientation of the unit normal is valid:As a check, the scalar triple product is $12$, so the plane equation simplifies to $x+y+2z=4$, which all three supplied points satisfy. The normal direction is $(1,1,2)$.

Use suffix notation with indices ranging from $1$ to $3$, and sum over repeated indices. By the component definitions of the dot product and cross product,The Levi-Civita symbol is unchanged by a cyclic permutation of its three indices: the cycle is two transpositions and hence has positive sign. Thus ${ϵ}_{ijk}={ϵ}_{kij}$. Commuting the scalar components givesTherefore the scalar triple product obeysCyclically moving the vectors preserves the scalar triple product; interchanging just two reverses its sign.

Using the exponential definition of the hyperbolic cosine and the complex exponential function,ThusFor real $x$, the hyperbolic sine vanishes only at $x=0$. Also $|y|\le 1<\pi$, so the only zero of sine available to $y$ is $y=0$. The product therefore vanishes exactly when $b=0$ or $4-b^2=0$:There are three possible parameter values. At $b=0$ the roots are $\pm i$ and $\cosh (\pm i)=\cos 1$ is real; at $b=\pm 2$ the roots themselves are real.

Write the complex number as $z=x+iy$. Since $iz=-y+ix$, the modulus of the complex exponential givesThis equals one precisely when $y=0$. The quadratic formula shows that the imaginary part of either root is $\pm \sqrt{4-b^2}/2$. Consequently the roots are real exactly at the two endpoints:Both endpoints satisfy the condition. The repeated roots are respectively $1$ and $-1$; their multiples by $i$ have exponentials on the unit circle.

The quadratic formula givesso both roots of a polynomial have real part $-b/2$. Writing $z=x+iy$, the complex exponential function satisfies $e^z=e^x(\cos y+i\sin y)$, and hence its modulus is $e^x$. ThereforeIntersecting this condition with the allowed parameter interval gives the answer for either root:The endpoint $b=2$ is included: the repeated root of a polynomial is $z=-1$, whose exponential has modulus $e^{-1}<1$.

Rationalize the numerator of the summand:For every $n\ge 1$ we have $\sqrt{1+1/n}\le \sqrt{2}$, and thereforeThe harmonic series diverges, so the comparison test for series makes the given positive-term series diverge as well:For completeness, divergence of the harmonic series follows by grouping the terms with $2^k<n\le 2^{k+1}$: each such block contributes at least $2^k/2^{k+1}=1/2$. The same rationalization also shows that $n$ times the summand tends to $1/2$. The summands tend to zero, but their harmonic-size tail prevents convergence.

Use rationalization of a square-root difference to avoid subtracting two large, nearly equal quantities:Since $1/n\to 0$ and the square root is continuous at $1$, the denominator tends to $2$. The reciprocal limit law established above is applicable because this limit is nonzero. HenceThe limit is one half; rationalization exposes the finite difference hidden by cancellation.

True. First suppose $p^e\mid N$ with $e\ge 2$ maximal. By the Chinese remainder theorem, choose $a\equiv 1+p^{e-1}(\mathrm{mod}{p}^e)$ and $a\equiv 1$ modulo all other prime-power factors. This is a unit, with Jacobi symbol one, because it is one modulo every prime. For $m=(N-1)/2$, the binomial theorem givesbecause $p\nmid m$ and all higher terms contain $p^e$. It is therefore a witness to the desired failure.

If $N$ is square-free and composite, it has at least two prime factors. Choose $a$ to be a quadratic nonresidue at one prime $p$ and one at every other prime, again using the Chinese remainder theorem. Then $(a/N)=-1$. At a different odd prime $q\mid N$, however, $a^m\equiv 1\not\equiv -1(\mathrm{mod}q)$. These two cases prove that every odd composite has an Euler-Jacobi witness, unlike the situation for the ordinary Fermat primality test.

False. The Carmichael number $561=3\cdot 11\cdot 17$ supplies a counterexample. For every $a$ coprime to $561$, the Fermat little theorem gives $a^{560}\equiv 1$ modulo each of these primes, since $2,10,16$ all divide $560$. The Chinese remainder theorem then gives $a^{560}\equiv 1(\mathrm{mod}561)$. Thus the proposed Fermat primality test can fail to detect a composite for every coprime base.

True. Since $N$ is not a square, some exponent $e_i$ is odd. Choose a quadratic nonresidue modulo that $p_i$ and choose residue one at every other prime. The Chinese remainder theorem produces a unit $b$ modulo $N$ with $(b/N)=-1$. Multiplication by $b$ permutes the residue classes modulo $N$. Thus, by multiplicativity of the Jacobi symbol,The sum is therefore $\boxed{0}$. Nonunits contribute zero, so the permutation argument covers every term.

False. The Jacobi symbol $(2/15)=(2/3)(2/5)=(-1)(-1)=1$, yet $2$ is not a quadratic residue modulo three. A solution modulo fifteen would give one modulo three, which is impossible. A positive Jacobi symbol need not indicate a square modulo a composite modulus.

For a normal mode write $x_j=A_j{e}^{i\omega t}$. The generalized eigenvalue problem isThe three independent eigenvectors and corresponding squared frequencies areThe last two have zero mass-weighted displacement and are orthogonal in the mass inner product. The lowest frequency is zero: it represents a rigid translation with both springs unchanged. Its general motion includes uniform centre-of-mass velocity, not just a constant displacement. Superpositions of the two oscillatory normal modes and this translation give the general solution.

Let $x_1,x_2,x_3$ be longitudinal displacements from equilibrium. The original diagram has masses $m,2m,m$, and no attachments to fixed walls. The Lagrangian isThe Euler-Lagrange equations giveThe forces sum to zero, consistent with momentum conservation for an isolated molecule.

The delayed positive term models recruitment depending on the population a time $T$ earlier; the quadratic negative term represents instantaneous density-dependent losses. Thus this delay differential equation delays recruitment rather than the strength of crowding acting on the present population.

With $x=1+y$, the linearized equation is $\dot{y}=\alpha y(t-T)-2\alpha y(t)$. Its characteristic equation of a delay differential equation is $\lambda +2\alpha =\alpha e^{-\lambda T}$. A root with nonnegative real part would satisfy $|\lambda +2\alpha |\ge 2\alpha$, whereas $|\alpha e^{-\lambda T}|\le \alpha$. This is impossible. Consequently the equilibrium is locally exponentially asymptotically stable for every $\alpha >0$ and $T>0$. There is no upper bound on $\alpha T$ in this model.

Put $x=1+y$ and retain linear terms. The delay differential equation is $\dot{y}(t)=-\alpha y(t-T)$, with characteristic equation of a delay differential equation $\lambda +\alpha e^{-\lambda T}=0$. Write $z=\lambda T$ and $q=\alpha T$. If a root has $\mathrm{Re}z\ge 0$, then $|z|=q{e}^{-\mathrm{Re}z}\le q$. Writing $z=a+ib$ gives $a=-q{e}^{-a}\cos b$. For $q<\pi /2$, $|b|<\pi /2$ and the right side is negative, contradicting $a\ge 0$. All characteristic roots therefore lie in the left half-plane.

An imaginary root $z=ib$ requires $\cos b=0$ and $b=q\sin b$. The first crossing is $q=\pi /2$, $z=\pm i\pi /2$. Implicit differentiation gives $dz/dq=-e^{-z}/(1+z)$, whose real part at the positive crossing is $(\pi /2)/(1+{\pi }^2/4)>0$. The later imaginary crossings occur at $q=\pi /2+2\pi k$ and also enter the right half-plane. Thus the equilibrium is locally exponentially asymptotically stable whenand unstable when $\alpha T>\pi /2$. At equality the linearized equation has neutral oscillatory modes; it is the Hopf bifurcation threshold. If stability is meant for the nonlinear equation itself, the borderline can also be resolved. In dimensionless time $s=t/T$, let $q_0=\pi /2$. A small center oscillation has the form $y=R\cos \theta +R^2(\cos 2\theta /5+\sin 2\theta /10)+O(R^3)$. Substitution at quadratic order gives these second-harmonic coefficients. At cubic order the fundamental forcing is $q_0{R}^3(\cos \theta /20+3\sin \theta /20)$. Projection onto the center mode divides the complex forcing by the characteristic derivative $1+i{q}_0$, yielding the Hopf normal form amplitude equationIts cubic coefficient is negative, so the equilibrium at equality is locally asymptotically stable, with algebraic rather than exponential decay. Thus the usual strictly decaying linear condition is $\alpha T<\pi /2$, while nonlinear local asymptotic stability includes $\alpha T=\pi /2$. The purely imaginary linear modes alone would not justify that borderline conclusion.

The linear-system residual at an approximate solution $x_m$ is $r_m=b-A{x}_m$. For symmetric positive definite $A$, the conjugate gradient method has $x_m\in x_0+K_m(A,r_0)$ and minimizes the squared $A$-norm of the error over that affine space; its residual is orthogonal to $K_m(A,r_0)$.

If $A$ has $t$ distinct eigenvalues, interpolate $q(\lambda )=1/\lambda$ on them with a polynomial of degree at most $t-1$. The finite-dimensional spectral theorem gives $A^{-1}=q(A)$, so the exact correction $x_{\ast }-x_0=A^{-1}{r}_0$ lies in $K_t(A,r_0)$. The minimum possible error at step $t$ is therefore zero. Thus $\boxed{\text{CG terminates in at most }t\text{ steps}}$ in exact arithmetic, possibly earlier if the initial residual has fewer eigencomponents. If $r_0=0$, it terminates immediately.

A symmetric positive definite matrix preconditioner $M$ changes the system to $M^{-1/2}A{M}^{-1/2}z=M^{-1/2}b$, with $x=M^{-1/2}z$; practical preconditioned CG applies solves with $M$. It can accelerate convergence by reducing the spectral condition number of a positive-definite matrix, which improves the usual bound involving $(\sqrt{\kappa }-1)/(\sqrt{\kappa }+1)$, and by clustering eigenvalues, which allows low-degree residual polynomials small on the spectrum even when the extreme-eigenvalue bound is pessimistic.

Choosing $\boxed{M=A}$ makes the transformed matrix $I$, so one step suffices. But applying $M^{-1}$ already requires solving the original problem, and constructing an exact factorization generally incurs its original computational cost. It is therefore not a useful shortcut for a single unsolved system, though an available factorization can be reused for multiple right-hand sides. Effective matrix preconditioning balances spectral improvement against inexpensive application.