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For the specified unit vector,For a cyclic matrix with first row $(a,b,c)$, its determinant is $a^3+b^3+c^3-3abc$. Here it givesTherefore $Ax=Cx$ has only the zero solution if $\alpha \ne 1$. At $\alpha =1$, $A=I$ and $A-C$ has kernel consisting of the equal-coordinate vectors. The complete answer isGeometrically, $C^{T}C=I$, $\det C=1$ and $Cn=n$, so $C$ is a rotation matrix about the $n$ axis. Its matrix trace is two, giving $\cos \theta =(\mathrm{tr}C-1)/2=1/2$; its skew part selects angle $-\pi /3$ about the oriented axis $n$. It has no fixed nonzero perpendicular vector. Meanwhile $A$ preserves perpendicular components and scales the axial component by $\alpha$. Comparing these components first forces $x_{\perp }=0$, and then $(\alpha -1)x_{\parallel }=0$, exactly matching the determinant calculation, including the noninvertible case $\alpha =0$.

Expanding the formula from part (a) in components gives $y_i=x_i+(\alpha -1)n_i{n}_j{x}_j$, with repeated indices summed. Hence the matrix of a linear map and its inverse, when $\alpha \ne 0$, have entriesHere ${\delta }_{ij}$ is the Kronecker delta. To check the matrix inverse, write $P=n{n}^T$; $P^2=P$ because $|n|=1$. Thus $A=(I-P)+\alpha P$ and $B=(I-P)+{\alpha }^{-1}P$, whose product is the identity matrix.

At $\alpha =0$, $\Phi$ is the orthogonal projection onto the plane perpendicular to $n$. Its image of a linear map and kernel of a linear map areEvery perpendicular vector is unchanged and every parallel vector is annihilated. The image plane is perpendicular to the kernel axis by the definition of $n^{\perp }$; their dimensions two and one also agree with the rank-nullity theorem.

The vector triple product identity gives $n\times (x\times n)=x-(n\cdot x)n$. Decompose $x=x_{\perp }+x_{\parallel }$, where $x_{\parallel }=(n\cdot x)n$ and $x_{\perp }\cdot n=0$. ThenThe linear map acts as the identity on $n^{\perp }$ and as multiplication by $\alpha$ along $n$. Consequently it is invertible exactly when $\alpha \ne 0$, withApplying $\Phi$ to this expression recovers the perpendicular and parallel components of $y$ separately.

The original PDF has coefficient $c_n=n^{n^{1/3}}$. Its $n$th root isThe Cauchy-Hadamard theorem therefore gives $\boxed{R=1}$. Again the power series converges absolutely inside the unit circle, while on that circle its terms do not tend to zero. The converted TeX loses the outer exponent and writes $n^{1/3}$; the calculation here uses the printed coefficient.

For the power series coefficient $c_n=n^2$, the Cauchy-Hadamard theorem givesThus the radius of convergence is $\boxed{R=1}$. The power series converges absolutely for $|z|<1$ and diverges for $|z|>1$. On $|z|=1$ its terms have modulus $n^2$, so they do not tend to zero and the series diverges there as well.

The floor function has even value on indices congruent to zero or one modulo four, and odd value on indices congruent to two or three. Accordingly the displayed sequence has terms $2+1/n$ in the first two residue classes and $-1/n$ in the other two. In particular,Every term is within $1/n$ of the set $\{0,2\}$. If a subsequence converged to $c\notin \{0,2\}$, its eventual distance from that set would be bounded below by a positive constant, contradicting this bound. The complete set of subsequential limits is $\{0,2\}$.

Define $a_{2k}=0$ and $a_{2k-1}=k$. The odd-indexed terms are unbounded, while the even-indexed subsequence is identically zero and hence converges to zero. This gives an unbounded sequence with a convergent subsequence.

Take $a_n=n$. Every subsequence has the form $a_{n_k}=n_k$, with $n_k\ge k$, so it tends to positive infinity and cannot converge to a real number. Thus this sequence has no convergent subsequence.

The Bolzano-Weierstrass theorem states that every bounded sequence in $\mathbb{R}$ has a convergent subsequence. More generally, a bounded sequence in ${\mathbb{R}}^d$ has a subsequence converging to a point of ${\mathbb{R}}^d$.

Away from the negative real axis, the complex conjugate has principal complex argument $-\theta$. ConsequentlyTaking the complex modulus gives $\cosh \theta =1$, hence $\theta =0$. The remaining equation is $e^{i\log r}=i$, soOn the negative real axis, both $z$ and $\overline{z}$ have principal complex argument $\pi$, so the left-hand side has modulus $2e^{-\pi }$, not two. There are no additional solutions there. The printed second term is the principal power of $\overline{z}$, not the complex conjugate of $z^i$.

The principal complex exponentiation gives $z^i=e^{-\theta }{e}^{i\log r}$. Equality to one requires its complex modulus to be one, hence $\theta =0$, and then its complex argument requires $\log r\in 2\pi \mathbb{Z}$. Thus all solutions are positive real, namelySubstitution verifies every value, and the modulus condition excludes every nonreal value.

Away from zero, the product and composition of the displayed smooth functions are continuous and differentiable. At zero,so $f$ remains continuous. Hence the continuity set is $\boxed{\mathbb{R}}$.

The difference quotient at zero is $f(h)/h=\sin (1/h)$. For $h_n=(\pi /2+2\pi n{)}^{-1}$ it equals one, and for $k_n=(3\pi /2+2\pi n{)}^{-1}$ it equals minus one. Both sequences tend to zero, so no derivative exists there. For $x\ne 0$, the product rule and chain rule giveThus the differentiability set is $\boxed{\mathbb{R}\setminus \{0\}}$.

At zero, $|f(x)-f(0)|=|x|\to 0$, so the function is continuous. At a nonzero $x_0$, approach along rational and irrational sequences; density of the rational numbers and density of the irrational numbers give limiting values $x_0$ and $-x_0$, which are different. Thus the continuity set is $\boxed{\{0\}}$.

At zero, the difference quotient is one for rational nonzero increments and minus one for irrational increments, so the derivative does not exist. Elsewhere differentiability implies continuity, and continuity already fails. Thus the differentiability set is $\boxed{\varnothing }$.

Yes: a tangential zero and a crossing zero can give exactly two fixed points. One explicit continuous piecewise linear example isAdjacent formulas agree at each junction. Their endpoint values lie in $[0,1]$, so linear interpolation proves that the whole image lies in $[0,1]$; also $f(0)=1$ and $f(1)=0$. On the four pieces, $f(x)-x$ is respectively $1-3x$, $3x/2-1/2$, $1-3x/2$, and $2-3x$. Their only zeros on their respective intervals are $1/3$ and $2/3$. Thus the complete fixed point set is $\boxed{\{1/3,2/3\}}$. Continuity forces at least one zero, but does not force every zero to change sign.

No. Define $f(x)=1$ for $0\le x<1$ and $f(1)=0$. This self-map of $[0,1]$ is discontinuous at one. Every point smaller than one is sent to a different point, and one is sent to zero, so it has no fixed point.

No. The continuous function $f(x)=x+1$ is a self-map of $\mathbb{R}$, but the fixed point equation would require $x+1=x$, which is impossible.

No. The continuous function $f(x)=(x+1)/2$ maps $(0,1)$ into $(0,1)$, but $f(x)-x=(1-x)/2>0$ throughout the domain. Its only possible fixed point is the excluded endpoint one.

An A-stable method must satisfy $|P_s(z)|\le 1$ for every $z$ with nonpositive real part. A consistent explicit Runge-Kutta method has $P_s^{\prime }(0)=1$, so its stability polynomial is nonconstant. Along the negative real axis the leading term of any nonconstant polynomial has magnitude tending to infinity. HenceThe positive order assumption in the question supplies the required consistency.

Order $p$ applied to this particular differential equation forces the local stability polynomial to agree with the exact exponential through degree $p$:As its degree is at most $s$, the nonzero coefficient $1/p!$ cannot be present for $p>s$. ThereforeThe implication is a necessary stability-polynomial condition; it does not assert that every such polynomial automatically gives the full nonlinear order conditions.