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The inverse is nonconstant; locally at zero,If the assumed meromorphic function $\mathcal{H}$ had no poles, it would be entire. It is bounded on the compact closure of a fundamental parallelogram, and its two periods then make it bounded throughout the complex plane. The Liouville theorem would force it to be constant, a contradiction. Therefore $\mathcal{H}$ has at least one pole, as stated by the general result that a nonconstant elliptic function has a pole.

Part (b) gives paths with the same endpoint $z$ but integral valuesApplying the local inverse and then continuing it shows thatSince $K,L>0$, the first period is real and the second is purely imaginary, so they are linearly independent over $\mathbb{R}$. Hence $\mathcal{H}$ is an elliptic function with two periodsIndeed, it is the Jacobi elliptic sine, and these generators give its period lattice.

Write the intercepted ciphertext, known plaintext, and desired plaintext without spaces:Because the pad is additive, Eve applies the malleability of an additive one-time pad and sendsLetter-by-letter, using $A=1,\dots ,Z=26$, this givesThus Eve should deliverIndeed, if $C=P+K$, then $C^{\prime }=C-P+P^{\prime }=P^{\prime }+K$, so Ollie decrypts the desired message.

A one-time pad over the finite additive group ${\mathbb{Z}}_n$ uses a key $K$ that is uniform on ${\mathbb{Z}}_n$, independent of the plaintext $X$, as long as the plaintext, and never reused. Encryption and decryption are

For independent random variables $X,Y$, conditioning reduces entropy and translation by a known element of the finite additive group preserves entropy, soInterchanging $X$ and $Y$ gives $H(X+Y)\ge H(Y)$, henceThis is the entropy of a sum of independent finite-group variables.

The result explains why adding independent pad symbols cannot reduce uncertainty. For a uniform pad, $X+K$ is itself uniform and is independent of $X$, which is the perfect-secrecy property. Independence is necessary: if $X$ is nonconstant and $Y=-X(\mathrm{mod}n)$, then $X+Y=0$ is constant, so

For $f\notin E$ and rational $q\in (0,1)$, every $M$ admits $m\ge M$ withConsequentlyBy the Cauchy-Hadamard theorem, the Taylor series of $f$ at $q$ has radius of convergence zero.

Suppose some Taylor series represented $f$ throughout a neighborhood of a point $x\in [0,1]$. That neighborhood contains a rational $q\in (0,1)$. A function represented locally by a convergent power series is a real analytic function, and re-expanding that series about $q$ gives the Taylor series of $f$ at $q$ a positive radius of convergence. This contradicts the preceding conclusion. Thus $f$ agrees with no Taylor series on a neighborhood of any point; it is a nowhere-analytic generic smooth function.

For rational $q\in (0,1)$ and positive integer $M$, defineEvaluation of the $m$th derivative at $q$ is continuous in the smooth-function metric, so $U_{q,M}$ is open.

It is also dense. Given $f$ and a metric tolerance $\epsilon >0$, choose $R\ge 1$ so that the contribution of all derivatives of order at least $R$ is below $\epsilon /2$, and choose $m\ge \max \{M,R\}$. For large $K$, perturb $f$ byFor every $r<R$, $\Vert h_K^{(r)}{\Vert }_{\infty }\le \epsilon /(4R)$, so $d(f+h_K,f)<\epsilon$. On the other hand,For sufficiently large $K$, $|f^{(m)}(q)+h_K^{(m)}(q)|>m!m^m$, proving density.

The space is complete by part (b). The Baire category theorem therefore makesdense. Put $E=C^{\infty }([0,1])\setminus G$. Each complement $U_{q,M}^c$ is closed and nowhere dense, so $E$ is a meagre set, or a set of first category. Every $f\notin E$ has the required derivative growth. This is the generic superfactorial derivative growth at rational points.

Every summand is at most $2^{-r}$, and ${\sum }_{r\ge 0}{2}^{-r}=2$, so $d(f,g)$ is finite and well defined. Nonnegativity and symmetry follow from the uniform norm. If $d(f,g)=0$, its $r=0$ summand gives $\Vert f-g{\Vert }_{\infty }=0$, hence $f=g$. Finally,and the triangle inequality for each uniform norm give the triangle inequality for $d$. Thus $d$ is a metric.

Let $(f_n)$ be $d$-Cauchy. For fixed $r$ and $0<\epsilon <1$, sufficiently large $m,n$ satisfywhich forces $\Vert f_m^{(r)}-f_n^{(r)}{\Vert }_{\infty }<\epsilon$. Hence, for every $r$, the continuous functions $f_n^{(r)}$ converge uniformly to some continuous $g_r$. The fundamental theorem of calculus givesUniform convergence permits passage to the limit, yieldingTherefore $g_r^{\prime }=g_{r+1}$ for every $r$, so $g_0\in C^{\infty }([0,1])$ and $g_0^{(r)}=g_r$.

To prove convergence in $d$, first choose $R$ so that ${\sum }_{r\ge R}{2}^{-r}$ is small, then use uniform convergence for the finitely many derivatives $r<R$. Thus $d(f_n,g_0)\to 0$. This proves the completeness of the smooth-function metric.

The Baire category theorem states that every countable intersection of open dense subsets of a complete metric space is dense. Equivalently, no nonempty open subset of a complete metric space is a countable union of nowhere dense sets.

To prove the first form, let $G_1,G_2,\dots$ be open dense subsets of a complete metric space $X$, and let $V\subseteq X$ be nonempty and open. Choose a closed ballInductively, density and openness of $G_{n+1}$ allow a closed ballThe balls are nested and $d(x_m,x_n)<2^{-n}$ for $m>n$, so $(x_n)$ is Cauchy. Completeness gives $x_n\to x\in X$. For every $n$, all later centres lie in $\overline{B}(x_n,r_n)$; closedness gives $x\in \overline{B}(x_n,r_n)\subseteq G_n$. The first ball also lies in $V$, henceSince every nonempty open $V$ meets the intersection, that intersection is dense.

No. In quantum dense coding, Alice's four encodings produce four different joint Bell states, but the qubit she transmits has reduced statein every case. Charlie intercepts only that qubit, so every measurement he can perform has the same probability distribution for messages $00,01,10,11$. He gains no information about the two-bit string.

Expanding either Bell state and using $⟨0|1{⟩}_B=0$ removes the cross terms:Thus the two expectation values agree. Equivalently, both states have the same reduced density matrix $I/2$ on qubit $A$; the positivity of $Q$ makes it a possible measurement effect but is not needed for the algebraic equality.