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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.

For an explicit Runge-Kutta method, write its strictly lower-triangular stage matrix as $A$, weight vector as $b$, and $1=(1,\dots ,1{)}^T$. Applied to the scalar test equation, its stage vector satisfiesSince $A^s=0$, inversion is a finite Neumann series. Putting $z=\lambda h$ givesThis proves degree at most $s$. The printed statement of degree exactly $s$ needs a nondegeneracy assumption: a two-stage explicit scheme with $b=(1,0)$ is simply Euler's method with an unused second stage, and has polynomial $1+z$ of degree one.

Divide the two outer boundary equations respectively by $z$ and leave the derivative equation as written:Their difference gives $Cz+3D{z}^3=2A/z^2$. As $z\to \infty$, the inner equations require $A,B\to 1/4$. The two outer equations at leading order thus give $Cz+D{z}^3=-1/4$ and $Cz+3D{z}^3=0$, soThe inner equations also give $A+B=1/2-C-D$ and $B-A=-2C-4D$. Solving and substituting yieldsReplacing $z$ by $b/a$ gives all four requested approximations.

The $C$ term produces a nondecaying velocity at infinity and the $D$ term a growing one, so decay forces $C=D=0$. The first two boundary equations become $A+B=1/2$ and $-A+B=0$, givingThe remaining velocity terms decay respectively as $R^{-3}$ and $R^{-1}$.

For $\Psi =f(R){\sin }^2\theta$, the angular differentiation givesTrying $f=R^m$ in the fourth-order equation gives the indicial productwhose four distinct roots are $-1,1,2,4$. This proves that the stated four-term expression is the general separated solution of this angular type.

Put $z=b/a$ and $F(s)=A/s+Bs+C{s}^2+D{s}^4$. The velocity formulas become $u_R=2UF(s)\cos \theta /s^2$ and $u_{\theta }=-U{F}^{\prime }(s)\sin \theta /s$. The four boundary conditions therefore give

For occupation numbers zero or one, the energy and particle number are additive over modes. The grand canonical partition function factorizes:Dividing the product of Boltzmann weights by this product givesThe marginal mean occupation is thereforeThis is the Fermi-Dirac distribution of a noninteracting Fermi gas: the exclusion principle allows only zero or one fermion in each single-particle state, including its spin label.

The grand canonical partition function isThe probabilities in part (i) and their logarithms giveAt fixed volume and fixed $\mu$, differentiate the partition sum:Combining with the preceding identity yieldsThe variable held fixed is essential; this is the grand canonical derivative.

For independent weakly coupled systems, insert the product probabilities into the Gibbs entropy. Since each marginal sums to one and the logarithm of a product is a sum,At fixed total energy and particle number, $\delta E_2=-\delta E_1$ and $\delta N_2=-\delta N_1$. Part (i) givesIndependent exchange variations force both coefficients to vanish. Thus