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For fixed real $x$, define $u_0=x$ and $u_n=\sin (0.99u_{n-1})$. This makes $u_n$ exactly the summand containing $n$ nested sines. The elementary inequality $|\sin t|\le |t|$ gives inductivelyThe fact that an iterated contraction gives a summable orbit applies here. Comparison with a geometric series therefore givesThus the nested-sine series converges absolutely for every real $x$; there are no divergent or merely conditionally convergent cases.

Let $r=|\sin x|\in [0,1]$. For $r<1$, the absolute summand $r^n/\sqrt{n}$ is at most $r^n$, so a geometric series comparison proves absolute convergence. For $r=1$, the series is $\sum (-1{)}^n/\sqrt{n}$: its positive term sizes decrease to zero, so the alternating series test proves convergence. Its absolute series is a divergent p-series with exponent $1/2$. HenceIt converges for every real $x$ and never diverges.

Put $s=\sin x$, so $-1\le s\le 1$. The printed summand is $(3s^n+s^{2n})/n$. If $|s|<1$, its absolute value is at most $3|s{|}^n+|s{|}^{2n}$, and the two geometric series converge. Thus there is absolute convergence.

For $s=1$ the series becomes $\sum 4/n$, a divergent multiple of the harmonic series. For $s=-1$ it becomesThe first series converges by the alternating series test, while the second diverges to $+\infty$, so their partial sums tend to $+\infty$. In particular, alternation of some terms does not imply convergence here. The complete classification isThere are no conditionally convergent cases.

A member of both groups satisfies $A^{T}A=I$ and $A^{T}JA=J$. Multiplication of the second identity by $A$ on the left, using $A{A}^T=I$, gives $JA=AJ$. A real two-by-two matrix commuting with $J=\mathrm{diag}(1,-1)$ must be diagonal, since the two off-diagonal entries change sign in this equation. Its diagonal entries are $\pm 1$ by the orthogonal matrix identity, and the restriction $a>0$ forces the first to be $1$. Therefore the intersection is the two-element group

For the Minkowski metric $J=\mathrm{diag}(1,-1)$, preservation means $(Ax{)}^{T}J(Ay)=x^{T}Jy$ for all real vectors $x,y$. This is equivalent to $A^{T}JA=J$, or, entrywise,The restriction $a>0$ puts the first column on the right branch of the unit hyperbola, so uniquely $a=\cosh t$, $c=\sinh t$ for a real parameter $t$. The third equation makes the second column a scalar multiple of $(\sinh t,\cosh t{)}^T$: this also covers $t=0$, when $b=0$. The second equation makes that multiple $\epsilon =\pm 1$. Thus all the required matrices areThe hyperbolic function addition formulas give $H(t)H(s)=H(t+s)$, while $D_{\epsilon }H(s)=H(\epsilon s)D_{\epsilon }$. ConsequentlyThe product remains of the required form and its upper-left entry is positive. The identity is $H(0)D_1$, and the inverse is $H(-\epsilon t)D_{\epsilon }$. Together with associativity of matrix multiplication, these identities establish the group property. This is the time-orientation-preserving Lorentz group in one spatial dimension.

Preserving the Euclidean scalar product means $(Ax{)}^T(Ay)=x^{T}y$ for every pair of real vectors $x,y$. Since these bilinear expressions agree on all pairs of basis vectors, this is equivalent to $A^{T}A=I$: the columns of $A$ form an orthonormal basis.

The first column is a unit vector, so it is $(\cos \theta ,\sin \theta {)}^T$ for some real $\theta$. A perpendicular unit vector is one of the two choices $\epsilon (-\sin \theta ,\cos \theta {)}^T$, where $\epsilon \in \{1,-1\}$. Hence all the required orthogonal matrices areHere $R(\theta )$ is a planar rotation, and $\theta$ is determined modulo $2\pi$. To check the group laws explicitly, $D_{\epsilon }R(\varphi )=R(\epsilon \varphi )D_{\epsilon }$, soThis proves closure. The identity is $R(0)D_1=I$, the inverse of $R(\theta )D_{\epsilon }$ is $R(-\epsilon \theta )D_{\epsilon }$, and associativity is inherited from matrix multiplication. Thus the set is the orthogonal group $O(2)$.

There is no such universal constant. For a real parameter $M$, consider the real matrixIts characteristic polynomial is $(t-1{)}^2$, so both eigenvalues are real and ${\lambda }_{\max }(B_M)=1$. ButFor any proposed finite $C$, choosing $M>2(C-1)$ violates the proposed bound. This is an explicit illustration that nonnormal eigenvalues do not bound a quadratic quotient.

Expand $x={\sum }_j{c}_j{v}_j$ in the orthonormal basis from part (a). ThenFor $x\ne 0$, the weights $|c_j{|}^2/{\sum }_k|c_k{|}^2$ are nonnegative and sum to one. The Rayleigh quotient therefore lies between the smallest and largest eigenvalues. Conversely, normalized extreme eigenvectors $v_{\min },v_{\max }$ give, for $0\le t\le 1$,Thus the exact Hermitian Rayleigh quotient range isWhen $n=1$, the range is the singleton containing the sole eigenvalue.

Use the conventionfor the characteristic polynomial. Choosing $\det (A-tI)$ instead multiplies it by $(-1{)}^n$ and leaves the requested zero identity unchanged. For each eigenvector $v_j$, induction gives $A^r{v}_j={\lambda }_j^r{v}_j$, including $r=0$. Thusbecause ${\lambda }_{j}I-A$ has the nonzero kernel vector $v_j$ and therefore has zero determinant. Part (a) supplies a basis consisting of these $n$ eigenvectors. A linear map vanishing on every basis vector vanishes everywhere. ConsequentlyThis proves the Cayley-Hamilton theorem identity directly in the distinct-eigenvalue case assumed here.

Take eigenvectors $v_i,v_j$ for distinct eigenvalues ${\lambda }_i,{\lambda }_j$. Since the eigenvalues are real and $A$ is a Hermitian matrix,Hence $({\lambda }_j-{\lambda }_i)v_i^{†}{v}_j=0$, so eigenvectors for distinct eigenvalues are orthogonal vectors. They are therefore linearly independent; after normalization the given $n$ eigenvectors form an orthonormal basis of ${\mathbb{C}}^n$.

Separate off the integer part of the phase. Set $d_n=\sqrt{n^2+n}-n$. Rationalization givesSince squaring removes the sign in $\cos (\pi n+u)=(-1{)}^n\cos u$, the given sequence satisfies $b_n={\cos }^2(\pi d_n)$. The continuity of the cosine now gives

Rationalizing avoids cancellation between the two large terms of the sequence:By continuity of the square root and the reciprocal at the limiting values,

The shifted matrix $N=D-\lambda I$ is still a normal matrix, sinceand $N{N}^{†}$ has the same expansion except that its first term is $D{D}^{†}$, which equals $D^{†}D$. For the given eigenvector, $Ne=0$. Part (b), applied to $N$, therefore gives $N^{†}e=0$. Since $N^{†}=D^{†}-\overline{\lambda }I$,The corresponding eigenvalue of the Hermitian conjugate is the complex conjugate of the original eigenvalue.

Using the complex Euclidean norm and the normal matrix identity,Both sides are nonnegative. Therefore $\Vert Cx\Vert =0$ if and only if $\Vert C^{†}x\Vert =0$. This proves the stronger reusable fact that normal matrices have equal adjoint norms.

Since $A$ is an invertible matrix, so is its Hermitian conjugate, andIf $A$ is a normal matrix, substitute $A^{†}A=A{A}^{†}$ into the second identity to obtain $B{B}^{†}=I$, so $B$ is a unitary matrix. Conversely, if $B$ is a unitary matrix, multiply $B{B}^{†}=I$ on the left by $A$ and on the right by $A^{†}$. This gives $A^{†}A=A{A}^{†}$, the defining normal matrix identity. Thus $B$ is a unitary matrix exactly when $A$ is a normal matrix.

A branch of the complex logarithm must be specified to obtain a unique locus. First use the principal complex logarithm. Away from the negative real axis, write $z=r{e}^{i\theta }$ with $r>0$ and $-\pi <\theta <\pi$. Then the complex logarithms of $z$ and its complex conjugate are $\log r+i\theta$ and $\log r-i\theta$. Equating real and imaginary parts gives $\log r=\theta$. Consequently the principal-branch locus is the open segment of a logarithmic spiralThe origin is excluded because its complex logarithm does not exist. If the principal value is assigned argument $\pi$ on the negative real axis, a point there equals its own complex conjugate, so its principal complex logarithm would have to equal $i$ times itself. That forces the complex logarithm to be zero and cannot hold there; the endpoints remain excluded.

If the printed equation instead means that some values of the multivalued complex logarithms satisfy it, write those values as $\log r+it$ and $\log r+iu$, with $e^{it}=z/r$ and $e^{iu}=\overline{z}/r$. Equality gives $\log r=-u=t$, hence $u=-t$, which is compatible with conjugation. Every real $t$ is allowed. Under this interpretation the locus is the entire logarithmic spiral $z=e^{(1+i)t}$, $t\in \mathbb{R}$. The PDF does not choose between these branch conventions.

Write $z=x+iy$. Dividing by two givesThe left side is nonnegative, so squaring is reversible only when $x+a\ge 0$. Squaring yields $y^2=4ax$.

If $a>0$, this equation implies $x\ge 0$, so the sign condition holds automatically. The locus is the parabolawith vertex at the origin, focus of a conic $(a,0)$ and directrix of a conic $x=-a$. Indeed the original equation equates distance from the focus of a conic with distance from the directrix of a conic on its right side.

If $a<0$, the squared equation forces $x\le 0$, making $x+a<0$. This violates the sign condition. Thus the locus is empty for $a<0$; squaring without checking that condition would introduce a spurious parabola.

Write $z=x+iy$ and $\alpha =p+iq$. The complex conjugates in the PDF giveSince $(p,q)\ne (0,0)$, the locus is the straight line through the origin perpendicular to the position vector of $\alpha$ in the complex plane. Equivalently,

Let $K$ be the set where the indicator function $f$ equals one. Fix a positive integer $N$. Each permitted length-$N$ decimal prefix $w=(w_1,\dots ,w_N)\in \{2,7{\}}^N$ gives the intervalThere are $2^N$ such intervals, each of length $10^{-N}$, and they cover $K$. Every point of $K$ is strictly inside the interval for its prefix, since its remaining tail lies between $2/(9\cdot 10^N)$ and $7/(9\cdot 10^N)$. Thus none of the interval endpoints belongs to $K$. The intervals are disjoint and separated: their distinct prefix integers cannot be consecutive because their final digits are only $2$ and $7$.

The complement consists of finitely many closed intervals on which $f=0$, of total length $1-2^{N}10^{-N}=1-5^{-N}$. Part (b) therefore already proves Riemann integrability, since this length tends to one.

To compute the Riemann integral directly, make a partition of an interval using $0,1$ and all the endpoints of the $I_w$. On complementary cells the upper Darboux sum contribution is zero, and on the $I_w$ it is at most their total length. Moreover every nonempty cell contains a terminating decimal, which is not in $K$: neither its terminating expansion with a zero tail nor its alternative expansion with a nine tail consists entirely of twos and sevens. Such decimals are dense, so every cell has infimum zero. ThereforeLetting $N\to \infty$ gives the exact answerThis is an example of an indicator of a restricted-digit decimal set: infinitely many, indeed uncountably many, points can support an indicator function whose Riemann integral is zero.

Let $|f|\le M$. If $M=0$ the result is immediate, so take $M>0$. Given $\epsilon >0$, choose $0<\delta <\epsilon /(4M)$ and the finitely many intervals supplied by the hypothesis. On each $[a_i,b_i]$, choose a partition of an interval whose Darboux sum gap is less than $\epsilon /(2n)$, using Riemann integrability there.

Combine all these partitions with $0,1$ and all interval endpoints to obtain a global partition of an interval $P$. On the chosen intervals the gaps add to less than $\epsilon /2$. The complementary intervals have total lengthand on each the difference between the supremum and infimum is at most $2M$. ConsequentlyThe Riemann integrability criterion proves $f$ is Riemann integrable on $[0,1]$. Shared endpoints cause no problem: the Darboux sums use closed intervals, and the complementary interval estimate includes every endpoint value. This proves Riemann integrability from almost full interval coverage.