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The conclusions of (b) and (c) remain true. Indeed, the unit circle is a path-connected space, and the upper semicircle supplies a path from $-1$ to $1$; restricting the proposed closed cover or two-valued continuous function to that path reduces either claim to the corresponding result on $I$.

The fixed-point conclusion (d) fails. For example, any nontrivial spatial rotation $f(z)=e^{i\theta }z$ of the unit circle has no fixed point when $e^{i\theta }\ne 1$; the antipodal map $f(z)=-z$ is the simplest example.

Suppose $I=A\cup B$, where $A$ and $B$ are disjoint closed sets with $-1\in A$ and $1\in B$. Since each is the complement of the other, both are also open in $I$. Define $f=-1$ on $A$ and $f=1$ on $B$. This locally constant function is continuous and contradicts part (c). Therefore $A\cap B\ne \varnothing$.

Conversely, suppose a continuous $f:I\to \{-1,1\}$ satisfies $f(-1)=-1$ and $f(1)=1$. The function $g=-f:I\to I$ is continuous, so part (d) gives $x=g(x)=-f(x)$. If $f(x)=1$, this forces $x=-1$, contradicting $f(-1)=-1$; if $f(x)=-1$, it forces $x=1$, contradicting $f(1)=1$. Hence no such $f$ exists.

Suppose that the continuous function $f:I\to I$ has no fixed point. Then $g(x)=f(x)-x$ is continuous and never zero, sois a continuous function from $I$ to $\{-1,1\}$. Since $f(-1)\ge -1$ and equality would be a fixed point, $h(-1)=1$. Similarly $h(1)=-1$. Multiplying $h$ by $-1$ contradicts part (c). Thus $f$ has a fixed point.

If such a continuous function existed, the inverse imageswould be closed, would cover $I$, and would contain $-1$ and $1$ respectively. Part (b) gives a point in $A\cap B$, which is impossible.

Suppose for a contradiction that $A\cap B=\varnothing$. Each grid point then belongs to exactly one of $A$ and $B$; colour it red in $A$ and green in $B$. By part (a), there are adjacent points $a_n\in A$ and $b_n\in B$, in either order, withThe interval $I$ is a compact set, so a subsequence of $(a_n)$ converges to some $x\in I$. Along the same subsequence $b_n\to x$. Since $A$ and $B$ are closed sets, $x\in A\cap B$, a contradiction. Hence $A\cap B\ne \varnothing$.

Read the $2n+1$ colours from left to right. If every neighbouring pair had the same colour, mathematical induction would make every point the same colour as $-1$, contradicting the different colour of $1$. Therefore at least one adjacent pair has different colours.

Condition on a position retained after basis reconciliation, so Bob used Alice's basis. If Eve does not intercept, no error occurs. If she intercepts, her uniformly chosen basis agrees with Alice's with probability $1/2$, again causing no error. With probability $1/2$ she uses the conjugate basis; her measurement and resent state then make Bob's bit uniformly random, giving an error with probability $1/2$.

Therefore the bit error rate in the retained key is

Alice sends the $m$ conjugate-coded qubits through the noiseless quantum channel. Bob independently chooses a uniformly random basis bit $y_i^{\prime }$ for each qubit and performs a projective measurement in the computational basis when $y_i^{\prime }=0$ or the Hadamard basis when $y_i^{\prime }=1$.

After all measurements, Alice and Bob publicly reveal $Y$ and $Y^{\prime }$ but keep the data bits secret. They discard every position with $y_i\ne y_i^{\prime }$ and retain the measured bits at positions where the bases agree. In the ideal channel those retained bits equal Alice's $x_i$ with certainty. Each pair of bases agrees with probability $1/2$, so the shared secret key has expected length $m/2$.

For each position $i$, Alice uses $y_i$ to choose a basis and $x_i$ to choose a vector in that basis. Explicitly, she sends the qubitThus $y_i=0$ uses the computational basis $\{|0⟩,|1⟩\}$, while $y_i=1$ uses the Hadamard basis $\{|+⟩,|-⟩\}$.

In cylindrical coordinates,The screw transformationleaves $r$ and $\varphi -kz$ fixed, so it is the helical symmetry of a potential. Its infinitesimal generator has ${\xi }_z=1$ and ${\xi }_{\varphi }=k$. Noether's theorem therefore gives the conserved quantityThis is a linear combination of axial linear momentum and axial angular momentum.

Letbe the infinitesimal generator of the continuous symmetry, and letbe the canonical momentum. Noether theorem states that along every trajectory satisfying the Euler-Lagrange equations,is conserved.

Indeed, differentiate the invariance of $L$ with respect to $s$ at zero:On an Euler-Lagrange trajectory, $\partial L/\partial q_i=d{p}_i/dt$, soThis proves the Noether conserved quantity for a mechanical point symmetry.

Use the suggested substitutionIt maps $[0,1]$ to itself and satisfiesHencewhere $K$ is the complete elliptic integral of the first kind. Part (a) and the Gamma function recurrence $\Gamma (5/4)=\Gamma (1/4)/4$ now give

With $u=x^4$, the change of variables formula givesThe beta-gamma identity, $\Gamma (1/2)=\sqrt{\pi }$, and the Gamma reflection formula giveTherefore

Take context-free grammars $G_L,G_M$ for $L,M$ and rename their nonterminals so that they are disjoint. If their start symbols are $S_L,S_M$, add a new start symbol $S$ and the productionAll productions remain context-free. A derivation from $S$ produces a word from $L$ followed by a word from $M$, and every such concatenation has a derivation. Thus the new grammar generates $LM$, proving closure of context-free languages under concatenation.

The pumping lemma for context-free languages states that for every context-free language $L$ there is a pumping length $p$ such that every $w\in L$ with $|w|\ge p$ can be writtenwith

The Householder-John theorem states: if $A=M-N$ is Hermitian positive definite and $M^{\ast }+N$ is also Hermitian positive definite, then $M$ is invertible and

Now let $A=D+L+L^T$ be real symmetric positive definite and tridiagonal. For the Jacobi splitting, $M=D$ and $N=-(L+L^T)$. Positive definiteness of $A$ implies every diagonal entry is positive, so $D$ is positive definite. Introduce the alternating-sign diagonal matrixTridiagonality givesSince $S$ is orthogonal, $SAS$ is positive definite whenever $A$ is. Both hypotheses of the Householder-John theorem hold, and hence the Jacobi iteration matrix satisfies $\rho (M^{-1}N)<1$. Therefore the Jacobi convergence for a symmetric positive-definite tridiagonal matrix follows.