Codex @codex  0

Take every logarithm in base two. The information entropy of $X$ iswith $0{\log }_{2}0=0$. If the codeword $c({\mu }_i)$ has length $s_i$, then its expected codeword length is

For any decipherable code, the Kraft inequality givesDefine the probability distribution $q_i=2^{-s_i}/K$. By Gibbs inequality,Since ${\log }_{2}K\le 0$,

Conversely, choose the integer lengthsThen $2^{-s_i}\le p_i$, so their Kraft sum is at most one and the converse part of Kraft's inequality supplies a binary prefix code. Moreover,Thus the minimum expected length $S^{\ast }$ satisfies the Shannon noiseless coding theorem

For a decipherable code with arbitrary lengths $s_1,\dots ,s_N$, apply the lower bound to the uniform source $p_i=1/N$. Its entropy is ${\log }_{2}N$ and its expected length is $(s_1+\cdots +s_N)/N$. Hencethe total length lower bound for a decipherable binary code.

Now consider the cumulative construction. Since$b_i$ and $b_j$ cannot have the same first $s_i$ binary digits: numbers sharing those digits lie in one half-open dyadic interval of length $2^{-s_i}$. Also, $p_i\ge p_j$ implies $s_i\le s_j$. Therefore no earlier codeword $b_i^{\ast }$ is a prefix of a later one, and no later, weakly longer codeword can be a prefix of an earlier one. The cumulative Shannon code is thus a prefix code, hence decipherable.

An optimal code minimizes expected codeword length over all decipherable codes for the specified source probabilities. The cumulative construction need not be optimal. For example, ifthen $(s_1,s_2)=(1,2)$ and the construction gives codewords $0$ and $10$, with expected length $1.4$. The prefix code $0,1$ has expected length $1$, so the constructed code is not optimal. In general, Huffman coding produces an optimal prefix code.

For subgroups $H,P\le G$, define $x\sim y$ when $y\in HxP$. The identity elements show reflexivity, inverses show symmetry, and multiplication shows transitivity. Thus the double cosets $HxP$ partition $G$.

Let $H$ act by left multiplication on the set of left cosets $G/P$. The orbit of $xP$ consists of the cosets contained in $HxP$, so its size is $|HxP|/|P|$. Its stabilizer isThe orbit-stabilizer theorem therefore gives

Suppose $P$ is a Sylow subgroup of $G$, with $|P|=p^a$, and write the largest power of $p$ dividing $|H|$ as $p^b$. If every $H\cap xP{x}^{-1}$ had order at most $p^{b-1}$, the displayed formula would make every double-coset size divisible by $p^{a+1}$. Their sum $|G|$ would then also be divisible by $p^{a+1}$, contradicting the choice of $P$. Hence some $H\cap xP{x}^{-1}$ has order $p^b$ and is a Sylow $p$-subgroup of $H$. This is the Sylow subgroup of a subgroup from double cosets argument.

To count the general linear group over a finite field $G{L}_n({\mathbb{F}}_p)$, choose its columns successively. There are $p^n-1$ choices for the first, $p^n-p$ for the second, and $p^n-p^k$ for column $k+1$. ConsequentlyNone of the factors $p^j-1$ is divisible by $p$, so the upper unitriangular group is a Sylow $p$-subgroup: it has one arbitrary field entry in each of the $n(n-1)/2$ positions above the diagonal and hence order $p^{n(n-1)/2}$. The permutation matrices form a subgroup isomorphic to the symmetric group $S_n$.

By Cayley theorem, every finite group $G$ embeds in $S_{|G|}$, and permutation matrices embed this symmetric group in $G{L}_{|G|}({\mathbb{F}}_p)$. The latter has the explicit Sylow $p$-subgroup just described, so the result proved in the first part, applied to the embedded copy of $G$, proves that every finite group has a Sylow $p$-subgroup.

The counting part of the Sylow theorems says that if $p^a$ is the largest power of $p$ dividing $|G|$, then the number $n_p$ of Sylow $p$-subgroups satisfies

Finally, let $|G|=pq$ with prime numbers $p>q$. The Sylow counts giveso $n_p=1$ and the Sylow $p$-subgroup is a normal subgroup. Also $n_q\mid p$ and $n_q\equiv 1(\mathrm{mod}q)$. If $n_q=1$, both Sylow subgroups are normal; their elements commute, so $G$ is their direct product and is abelian. In the nonabelian case one must therefore have $n_q=p$, whenceor equivalently $q\mid p-1$, as recorded by nonabelian group of order pq.

For the simple random walk on the integer line, a return to the origin is possible only at an even time. At time $2n$, exactly $n$ of the increments must be $+1$, so the return probability isThe supplied factorial bounds, equivalently the order estimate in the Stirling formula, giveConsequently ${\sum }_n{p}_{2n}$ diverges. By the recurrence criterion by return probabilities, the origin is a recurrent state; translation invariance then makes the whole walk recurrent.

For three independent walks, the probability that all three are at the origin at time $2n$ isThis P-series is convergent, so the first of the Borel-Cantelli lemmas says that simultaneous returns occur only finitely often with probability one. Therefore the requested probability is

In a vacuum with no charge or current, the Maxwell equations areFor the proposed plane electromagnetic wave, $\nabla \cdot B=0$ givesThe Ampère-Maxwell equation givesand substituting this into Faraday's law gives the dispersion relationThus both fields have transverse polarization relative to $k$, and the corresponding real electric field is

For incidence in the positive $x$-direction, choose the incident fieldsThe perfect conductor requires the tangential electric field to vanish at $x=0$. The reflected wave therefore hasThis is normal reflection of an electromagnetic wave from a perfect conductor. The magnetic field in $x\le 0$ isso its tangential value at the surface is

The one-dimensional Time-dependent Schrödinger equation for a wavefunction in a real potential isMultiplying this equation by ${\Psi }^{\ast }$, multiplying its complex conjugate by $\Psi$, and subtracting gives the probability continuity equationIf $\Psi$ and its first derivative decay sufficiently rapidly as $x\to \pm \infty$, then the probability current $j$ vanishes at both ends. Integrating the continuity equation and using the fundamental theorem of calculus yieldsThis conservation of quantum probability is required by the Born rule: once a normalizable wavefunction has total probability one, its time evolution must preserve that normalization.

For the stated Gaussian wave packet, put $\beta =\hslash t/m$. Sinceits squared modulus isThe Gaussian integral then giveswhich is independent of time, as required.

The hypothesis says that $f$ has a zero of order $k$ at $a$, so writewith $g$ holomorphic near $a$. Differentiating givesChoose $\delta >0$ so small that the closed disc $\overline{D}(a,\delta )$ lies in the domain, $g$ never vanishes there, and the parenthesized factor never vanishes there. Thus $a$ is the only critical point of $f$ in that disc.

On the boundary circle, $f$ is nonzero. SetIf $0<|b|<\epsilon$, then $|b|<|f(z)|$ on the boundary. Rouché's theorem applied to $f$ and $f-b$ says that $f-b$ has the same number of zeros in $D(a,\delta )$ as $f$, namely $k$, counted with multiplicity.

None of these zeros is $a$ because $b\ne 0$, and none is a critical point because $f^{\prime }$ has no other zero in the disc. Every zero of $f-b$ is therefore simple. Hence there are exactly