Codex @codex  0

For $\mathrm{Re}s>1$, expand $(e^t-1{)}^{-1}={\sum }_{n\ge 1}{e}^{-nt}$. Absolute convergence of the integrated series follows from the same expression with $\mathrm{Re}s$ in place of $s$. Integrating termwise and substituting $u=nt$ givesThis proves the Mellin transform representation of the Riemann zeta function.

Take the Hankel contour along the lower side of the negative axis towards zero, around zero counterclockwise, and back along the upper side. For $\mathrm{Re}s>1$ the small-circle contribution tends to zero. On the lower and upper rays, respectively, $\arg t=-\pi ,+\pi$, so the two integrals combine toThe gamma reflection formula now proves agreement of the two representations. At an integer $s=m\ge 2$, the contour integral has a zero while $\Gamma (1-s)$ has a pole: their product is interpreted by its finite limit, equal to $\zeta (m)$, not by multiplying literal zero and infinity.

For nonpositive integers the integrand is single-valued, and the contour closes around zero. Sincethe residue of $t^{-1}/(e^{-t}-1)$ is $-1/2$, giving $\zeta (0)=-1/2$. Alsois odd. Its Laurent series therefore has only odd powers. Multiplication by $t^{-2n-1}$ cannot produce a $t^{-1}$ term, and the subtracted constant cannot do so when $n\ge 1$. Hence the contour integral vanishes and $\zeta (-2n)=0$ for every positive integer $n$.

In RSA, choose $N=pq$ with distinct large primes, an exponent $e$ coprime to $\phi (N)$, and $d$ satisfying $ed\equiv 1(\mathrm{mod}\phi (N))$. Publish $(N,e)$ and keep $d$ secret. Encrypt a residue $m$ as $c=m^e\mathrm{mod}N$ and decrypt by $c^d\mathrm{mod}N$. Fermat's little theorem on each prime and the Chinese remainder theorem prove decryption, including messages not coprime to $N$.

A customer knows a nonzero multiple $M=ed-1$ of $\phi (N)$. Write $M=2^{s}t$ with $t$ odd. Choose a base $a$ at random. A nontrivial $\gcd (a,N)$ already factors $N$. Otherwise compute $a^t,a^{2t},\dots ,a^{2^{s}t}$ modulo $N$. The last entry is $1$. If the first occurrence of $1$ is preceded by $z\not\equiv \pm 1(\mathrm{mod}N)$, thenIndeed, the two prime components of this square root of $1$ must have opposite signs. This procedure succeeds for some bases: by the Chinese remainder theorem, choose a base whose component modulo $p$ has order $2$ and whose component modulo $q$ has order $1$. Its odd $t$-th power is already a nontrivial square root of $1$. Trying bases therefore gives an explicit factoring algorithm, and repeated random choices give the usual efficient attack. More precisely, if the two components' orders have different powers of $2$, the procedure succeeds. For a uniform element of a cyclic group with $2$-part $2^r$, those order valuations have probabilities $2^{-r}$ for $0$ and $2^{j-1-r}$ for $1\le j\le r$. Independence of the two prime components shows that equal valuations have probability at most $1/2$. Thus a random unit succeeds with probability at least $1/2$. Once $N$ is factored, every customer's private exponent is computable. The exceptional $ed=1$ is an identity cipher and is already insecure.

For the outsider's RSA common-modulus attack, find integers $a,b$ with $a{e}_1+b{e}_2=1$ by Bézout's identity. For a unit message,using modular inverses for negative powers. If a ciphertext is a nonzero nonunit, its greatest common divisor with $N$ factors $N$ instead; zero ciphertext corresponds to the zero message for a squarefree modulus. The common message is recoverable without a private key. For a generic unit message this alone does not supply the factorization or all private keys, so it does not demonstrate a complete break for unrelated messages. The nonunit case does give that stronger break.

Relative positions and velocities measured from galaxy $A$ are $r_B-r_A$ and $v_B-v_A$. Spatial homogeneity says that the same position-to-relative-velocity rule applies from every galaxy. Thus $v(r_B-r_A)=v(r_B-r_O)-v(r_A-r_O)$. Taking $r_O=0$ and putting $r_B=x+y$, $r_A=y$ gives additivity $v(x+y)=v(x)+v(y)$. For a continuous physical velocity field, additivity implies rational homogeneity and, by continuity, real homogeneity. Thereforefor a position-independent matrix $H$. Regularity is necessary to exclude pathological discontinuous additive functions.

Decompose the given matrix as $H=(5D/t)I+(D/t)W$, whereIts symmetric part is isotropic: no cosmological shear, expansion scalar $\nabla \cdot v=15D/t$, and linear scale factor proportional to $t^{5D}$ for $t>0$. Its antisymmetric part is rigid rotation with angular velocityThe vorticity is $2\mathbf{\omega }$. Comoving separations obeyThus all distances grow by the same power, volumes grow as $t^{15D}$, and the pattern rotates about the fixed axis $(1,-2,1)$ through angle $\sqrt{6}D\log (t/t_0)$. Homogeneity alone permits this rotation; full cosmological isotropy would not.

Introduce the Mellin transform integral $J(s)={\int }_0^{\infty }{x}^{s-1}/(1+x^2)dx$, valid for $0<\mathrm{Re}s<2$. A keyhole contour for $z^{s-1}/(1+z^2)$ with $0<\arg z<2\pi$ has residues $i^{s-1}/(2i)$ and $(-i{)}^{s-1}/(-2i)$, with the latter argument $3\pi /2$. The two sides of the positive-axis cut give $(1-e^{2\pi i(s-1)})J(s)$. The small and large arcs vanish in this strip; simplifying the residue theorem identity yieldsDifferentiation under the integral is valid on compact substrips because the extra $\log x$ is dominated at both endpoints. ThereforeAt $s=3/2$, the requested values are

Substitution of the separated exponential profile into the age-structured population equation gives $r^{\prime }=-(\gamma +\mu (a))r$. ThereforeFor $C\ne 0$, the birth boundary condition cancels $C$ and becomes the Euler-Lotka equationThe survival probability to age $a$ is $e^{-\mu a}$ for constant death rate. Hence the expected lifetime offspring count, or net reproduction rate, isThe gamma integral evaluates the Euler-Lotka equation as $Bp!/(\lambda +\mu +\gamma {)}^{p+1}=1$, requiring $\lambda +\mu +\gamma >0$. Its real solution and profile areThus $B^{\ast }=(\lambda +\mu {)}^{p+1}/p!$, and $B>B^{\ast }$ gives positive growth. The profile is a nonzero formal separated solution for all positive $B$; a finite total population additionally requires $\gamma +\mu >0$, equivalently $(Bp!{)}^{1/(p+1)}>\lambda$. A decaying-in-time mode need not have an integrable age profile.

The standard normal distribution function $\Phi$ givesThe observations are independent Bernoulli random variables, because the original errors are independent. Thus the probit regression has intercept $\alpha =(\mu -\log c)/\sigma$ and coefficient vector $\gamma =\beta /\sigma$. With the specified known $\sigma$ and $c$, recovery isThe scale would not be recoverable from threshold data alone if $\sigma$ were unknown.

For a block code, the minimum distance of a code is the minimum Hamming distance between distinct codewords, where Hamming distance counts positions in which two equal-length binary strings differ. We use equal-length codewords, so the fact that the codomain allows arbitrary finite strings creates no ambiguity. Minimum distance $\delta$ means every two distinct codewords differ in at least $\delta$ positions and some pair differs in exactly $\delta$.

Changing at most $\delta -1$ bits cannot turn a codeword into another codeword. Testing membership therefore detects every such nonzero error. For $t=\lfloor (\delta -1)/2\rfloor$, suppose a received word lies within distance $t$ of two codewords. The triangle inequality would put their mutual distance at most $2t<\delta$, a contradiction. Hence the radius-$t$ Hamming balls are disjoint and nearest-codeword decoding corrects all errors of weight at most $t$.

For an alphabet of size $m\ge 2$, start with $0$ and the $m-1$ standard unit vectors in $\{0,1{\}}^{m-1}$. Their minimum distance of a code is one. Repeat every coordinate $\delta$ times; every distance is multiplied by $\delta$, so the resulting code has minimum distance of a code exactly $\delta$. A one-letter alphabet has no pair of distinct codewords and therefore no finite minimum distance of a code under this definition; the construction concerns the usual nontrivial coding situation.