The Dirichlet series for the Riemann zeta function converges for $\Re z>1$. In the Hankel integral representation the branch is a branch of $t^{z-1}$, so the argument restriction concerns $t$, not the parameter $z$. Use the principal branch cut along the negative real axis and a positive Hankel loop with a fixed small indentation around zero. Its integral $I(z)$ is an entire function of $z$, since the tails have exponential decay and the fixed contour avoids zero. Multiplication by $\Gamma (1-z)/(2\pi i)$ gives the meromorphic continuation of $\zeta$: it is directly defined away from positive integers, has a pole at $1$, and the apparent singularities at $2,3,\dots$ are interpreted by limits.

For a check in the initial half-plane, collapsing the indentation for $\Re z>1$ givesExpansion of $(e^r-1{)}^{-1}$ into exponentials justifies the last equality there, and the Gamma reflection formula recovers the printed integral representation.

Take the upper semicircle contour with the straight segment oriented from left to right, making the total orientation positive. Its enclosed poles are $t=2\pi ij$, $j=1,\dots ,N$; each has residue $-(2\pi ij{)}^{z-1}$. ThusThe reverse orientation changes the sign.

Initially take $\Re z<0$, so the series on the right converges as $N\to \infty$ and the large arcs can be discarded under the given assumption. Denote by $L_{+}$ and $L_{-}$ the integrals along horizontal lines just above and below the real axis, both from left to right. The residue computation and its lower-half-plane clockwise counterpart giveContour deformation about the negative-axis branch cut gives $I=L_{-}-L_{+}$. HenceMultiplying by $\Gamma (1-z)/(2\pi i)$ yieldsIt extends by meromorphic continuation to all $z\ne 1$, with removable products evaluated by limits where individual factors diverge. The derivation does not assert ordinary convergence of the original positive-term series outside $\Re z>1$.

For $s\ge 0$ and $\delta >0$, define the scale-restricted Hausdorff content at scale $\delta$ byusing countable covers. The Hausdorff measure is ${\mathcal{H}}^s(F)={\lim }_{\delta \downarrow 0}{\mathcal{H}}_{\delta }^s(F)$; the limit exists by monotonicity. A conventional dimension-dependent normalizing constant does not change the Hausdorff dimensionAt $s=0$, a nonempty singleton has contribution one, so the measure counts points.

Let $\mathcal{T}(K)={\bigcup }_{i=1}^k{S}_i(K)$ and $c={\max }_i{c}_i<1$. For compact nonempty $K,L$, matching a point of $S_i(K)$ with the image of a nearby point of $L$ gives $d_H(\mathcal{T}K,\mathcal{T}L)\le c{d}_H(K,L)$. Choose a sufficiently large closed ball $B$ with every $S_i(B)\subset B$. The nonempty compact sets ${\mathcal{T}}^m(B)$ are nested, so their intersection $I$ is nonempty and compact. The finite union of similarity images commutes with this decreasing compact intersection: a point lying in all unions has some repeated branch index, and compactness supplies a preimage in all the nested sets. Consequently $\mathcal{T}(I)=I$. If $J$ is another invariant compact set, the contraction estimate gives $d_H(I,J)\le c{d}_H(I,J)$, hence $I=J$. This proves existence and uniqueness of the invariant set of a finite similarity iterated function system.

Under the open set condition, there is a nonempty bounded open set $O$ such that the $S_i(O)$ lie inside $O$ and are pairwise disjoint. The dimension formula isOverlaps can invalidate this formula if that assumption is omitted.

For the printed template, subdivide the central straight run of length two into its two collinear unit steps. The eight directed steps, before dividing all coordinates by four, have successive verticesEach step defines a similarity with ratio $1/4$, with rotation matching its direction. The open diamond $O=\{|x-1/2|+|y|<1/2\}$ is an explicit witness for the open set condition. Its eight image diamonds have radius $1/8$ in the ${\ell }^1$ metric, lie in $O$, and their centres have pairwise ${\ell }^1$ distance at least $1/4$; their interiors are therefore disjoint. The Minkowski sausage dimension is consequentlyThis makes the subdivision convention explicit. If the length-two run were instead treated as one unsplit edge replaced at scale $1/2$, there would be six maps of ratio $1/4$ and one of ratio $1/2$. The same diamond separation works, but the equation becomes $64^{-s}+2^{-s}=1$, giving $s={\log }_{2}3$, a different fractal. The requested $3/2$ corresponds to the eight-unit-step interpretation.

The Poisson bracket is $\{f,g\}={\sum }_a(f_{q_a}{g}_{p_a}-f_{p_a}{g}_{q_a})$. For a function with no explicit time dependence, the chain rule and Hamilton equations giveWith $L_b={ϵ}_{bcd}{x}_c{p}_d$, direct differentiation givesApply the product rule to $L_a={ϵ}_{acd}{x}_c{p}_d$ and substitute these two brackets. Contracting the epsilon symbols yields $\boxed{\{L_a,L_b\}={ϵ}_{abc}{L}_c}$. Equivalently, the two product terms are precisely the infinitesimal rotation of the vector $x\times p$ about the $b$ axis, with the same bracket convention as the position and momentum vectors.

Assume the Laplace integrals converge absolutely in a common right half-plane. The Fubini theorem and the substitution $t=u-v$ on $u\ge v\ge 0$ give the convolution theorem for Laplace transforms:For the specified inverse-square-root function, the supplied identity says $f\ast f=\pi$ on $t>0$. Consequently $\hat{f}(p{)}^2=\pi /p$. For real $p>0$ the defining integral is positive, fixing the positive square root; analytic continuation into $\mathrm{Re}p>0$ giveswith the branch positive on the positive real axis. The square identity alone would not select the sign.

At a hyperbolic saddle, the stable invariant subspace of $Df(0)$ is the sum of its generalized eigenspaces with negative real parts; the unstable subspace uses positive real parts. The stable manifold theorem gives invariant local manifolds tangent to these subspaces at the equilibrium, with the corresponding dimensions and the appropriate forward or backward convergence to the equilibrium.

Here the Jacobian is $\mathrm{diag}(1,-1)$, so the unstable tangent is the $x$ axis and the stable tangent the $y$ axis. For the stable graph write $x=h(y)=A{y}^2+B{y}^3+O(y^4)$. Invariance requires $h^{\prime }(y)(-y+3h(y{)}^2)=h+h^2+2hy+3y^2$. Comparing quadratic and cubic terms gives $-2A=A+3$ and $-3B=B+2A$, so $A=-1,B=1/2$.

For the unstable graph $y=g(x)=C{x}^2+D{x}^3+O(x^4)$, invariance is $g^{\prime }(x)(x+x^2+2xg+3g^2)=-g+3x^2$. Coefficients give $2C=3-C$ and $2C+3D=-D$, so $C=1,D=-1/2$. Thus the cubic graph expansion of a saddle invariant manifold isThese describe the local manifolds to cubic order, not exact global polynomial curves.

A digital signature lets a holder of a private key authenticate a message while anyone with the public key verifies it. Signing a message digest ties the signature to the message and detects alterations; it does not conceal the message. Authentication also requires a trustworthy association of the public key with its owner.

For the ElGamal signature scheme, choose a large prime $p$ and a generator $g$ of ${\mathbb{F}}_p^{\times }$. The secret is $x$, and the public key is $(p,g,y)$ with $y=g^x\mathrm{mod}p$. To sign a message whose hash is represented by $m\mathrm{mod}(p-1)$, choose a fresh secret nonce $k$ coprime to $p-1$ and setTransmit $(r,s)$ with the message. Verification checks the permitted ranges and $\boxed{g^m\equiv y^r{r}^s(\mathrm{mod}p)}$. Correctness follows from $xr+ks\equiv m(\mathrm{mod}p-1)$. The secret exponent prevents an outsider from directly constructing this relation for an arbitrary hashed message; nonce reuse can expose the exponent through the repeated linear congruences. Hashing is part of the signature scheme, rather than signing arbitrary raw exponents with no message encoding.

A nonidentity Möbius transformation has either two distinct fixed points on the Riemann sphere or one double fixed point. With two fixed points, conjugate them to zero and infinity. A Möbius map fixing both is $z\mapsto kz$ with $k\ne 0$. With one fixed point, conjugate it to infinity; the resulting map is affine, $z\mapsto az+b$. If $a\ne 1$ it would have another fixed point, so $a=1$ and $b\ne 0$. Scaling the coordinate makes it $z\mapsto z+1$. The identity is already the case $k=1$.

Define ${\mathrm{tr}}^2(g)$ using a determinant-one representing matrix; the two possible lifts differ by sign, so its square is well defined and invariant under conjugation. For $z\mapsto kz$ it is $k+2+k^{-1}$, while for a nontrivial translation it is four. If two nonidentity dilations have equal trace-square, their multipliers satisfy the same quadratic and are either equal or reciprocal. Inversion $z\mapsto 1/z$ conjugates one reciprocal multiplier to the other. Trace-square four forces the dilation multiplier to be one, which is the excluded identity; thus every nonidentity map with that invariant is in the translation class. This proves $\boxed{g\sim h⟺{\mathrm{tr}}^2(g)={\mathrm{tr}}^2(h)}$ for the stated nonidentity maps.

Not every such map fixes an interior point of hyperbolic three-space. In its upper-half-space extension, $z\mapsto z+1$ translates the horizontal coordinate and has no fixed point at positive height. A dilation extends as $(z,t)\mapsto (kz,|k|t)$, so it fixes the vertical axis only when $|k|=1$; if $|k|\ne 1$, the height changes. Boundary fixed points therefore do not imply an interior fixed point.