Codex @codex  0

For the given Möbius transformation, the fixed point equation is $z(z+3)=3z+1$, hence

For the general Möbius transformation, its finite fixed points satisfyWith $c\ne 0$ and nonzero discriminant, the quadratic formula gives the two distinct fixed points of a Möbius transformation:The fixed-point equation for $\alpha$ also gives $b-d\alpha =-\alpha (a-c\alpha )$. ThereforeDividing yields the conjugation of a Möbius transformation with two distinct fixed points:Neither factor in this quotient vanishes, since $(a-c\alpha )(a-c\beta )=ad-bc\ne 0$. The formula extends over the Riemann sphere, including the transformation's pole. For the special transformation and the ordering $\alpha =1,\beta =-1$, $k=1/2$.

For every integer $m$, De Moivre's theorem states $(\cos \theta +i\sin \theta {)}^m=\cos (m\theta )+i\sin (m\theta )$. Taking the real part for $m=5$ givesReplace ${\sin }^2\theta$ by $1-{\cos }^2\theta$ and expand:The resulting polynomial is the fifth Chebyshev polynomial.

A trace-zero Hermitian matrix has the unique formSet $x_3=h$, $x_1=\mathrm{Re}w$ and $x_2=-\mathrm{Im}w$. Then it is exactly $R(x)$; these assignments also prove uniqueness. This is the real Pauli matrix representation $R(x)=x_1{\sigma }_1+x_2{\sigma }_2+x_3{\sigma }_3$.

For a unitary matrix, $U^{-1}=U^{†}$. Thusand invariance of the matrix trace under similarity transformations gives $\mathrm{tr}(U^{-1}AU)=0$. Therefore conjugation maps the space to itself.

Finally,The determinant is invariant under similarity transformations, so $\det R(y)=\det R(x)$ implies

Let $[v{]}_{\times }$ denote the cross product matrix, with $[v{]}_{\times }x=v\times x$. ThenThe two eigenvalue calculations use an orthonormal basis adapted to $u,v$.

Differentiating givesThereforeFor real coordinates take $q>0$ and a local branch of the angle. The transformation is locally canonical transformation because its determinant is one, but periodicity in $p$ prevents a globally one-to-one map if all real $p$ are retained.

The Poisson bracket is $\{Q,P\}=\alpha q^{\beta -2}$. For it to be identically one on an open region, the exponent must vanish and the coefficient must equal one:The transformation is defined away from $q=0$, and its inverse there is $q=1/P$, $p=Q{P}^2$.

Unmodified Hebbian learning reinforces correlated input and output but has no mechanism limiting synaptic strength. The second rule adds activity-dependent negative feedback and stabilizes the norm at an input-scale-independent value, provided the input keeps exciting the output. The third has a cubic weight penalty independent of the instantaneous output; its equilibrium norm instead depends on the strength of the input second moment.

For the third averaged rule, an equilibrium along eigenvector $i$ has radial linearized eigenvalue $-2{\lambda }_i/\tau$ and transverse eigenvalues $({\lambda }_j-{\lambda }_i)/\tau$. Thus a positive simple largest eigenvalue gives stable equilibria in its two opposite directions, whereas lower eigen-directions are unstable to larger components. With generic initial weights, it learns the dominant correlation direction, linking the rule to principal component analysis. These are idealized normalization and feature-extraction mechanisms, not claims that biological synapses implement these exact equations.

Let the active block have size $m$. Using a stored skew triangle, the product $Dv$ costs $m(m-1)$ scalar multiplications: two for each unordered off-diagonal pair. Updating its stored triangle with $v{w}^T-w{v}^T$ costs another $m(m-1)$ products. Reflector construction, scalings and the active column take only $O(m)$ additional products. Summing over $m=n-1,\dots ,2$ gives

For a symmetric block, an optimized matrix-vector calculation costs $m^2$ products, and the symmetric rank-two update of one triangle costs $m(m+1)$ products. The additional scalar $v^{T}Dv$ and its correction to the update vector cost only $O(m)$ per step. ThusThe leading terms are the same under the requested product-count convention. Skew symmetry removes the diagonal work and the scalar correction, giving lower-order savings, not a factor-of-two leading advantage over an already symmetry-aware implementation. Counting additions as well would change the reported flop constants; it is not the convention requested here.

At step $k=1,\dots ,n-2$, isolate the active column tail $a=A_{k+1:n,k}$. If its subtail is already zero, no reflection is needed. Otherwise choose a Householder reflection $H=I-2v{v}^T/(v^{T}v)$ on the active coordinates such that $Ha$ has only its first component nonzero. A cancellation-resistant choice is $v=a+\mathrm{sgn}(a_1)\Vert a\Vert e_1$, with an arbitrary positive sign if $a_1=0$. Embed $H$ as $P=\mathrm{diag}(I_k,H)$ and replace $A$ by $P^{T}AP$. Previous reduced columns are untouched, and column $k$ now has no entries below the first subdiagonal.

Similarity by an orthogonal matrix preserves skew symmetry. Consequently an upper-Hessenberg result also has zeros above the first superdiagonal, so it is tridiagonal with zero diagonal and opposite-sign corresponding off-diagonal entries.

For the active skew-symmetric block $D$, compute $w=2Dv/(v^{T}v)$. Because $v^{T}Dv=0$, expanding the two-sided reflection simplifies toStore and update just one triangle, obtain the other by negation, and keep the diagonal exactly zero. Matrix-vector multiplication can likewise use each stored entry once to contribute to two output coordinates. This is skew-symmetric Householder tridiagonalization; there is no need for two dense matrix-matrix multiplications.

An elementary geometric-series construction suffices, without assuming a general polynomial-approximation theorem. Rotate the square by a multiple of $\pi /2$ so that $\Re w>1$. Put $c=-R$ with $R>0$. Every point of the square then satisfies $|z-c{|}^2\le (R+1{)}^2+1$, whereasfor sufficiently large $R$. Thus the whole square lies inside a disk centred at $c$ whose radius is less than $|w-c|$. The other three ways $w$ can lie outside the square are handled by the indicated rotations, after which rotating back preserves polynomial form.

Now expandLet $\rho ={\max }_{z\in K}|z-c|/|w-c|<1$. The tail after degree $m$ is uniformly bounded by ${\rho }^{m+1}/[|w-c|(1-\rho )]$, tending to zero. Hence the displayed partial sums are the required uniform polynomial approximations. This is the Taylor series about a carefully chosen distant centre, rather than a potentially inadequate series about $0$.

The complex Taylor theorem says that an analytic function has the expansionthroughout every disk centred at $z_0$ contained in $D$. In particular it converges for $|z-z_0|<\mathrm{dist}(z_0,\mathbb{C}\setminus D)$, and uniformly on every smaller closed disk. The coefficients are uniquely its derivatives divided by factorials. The convergence radius can be larger if the function extends beyond the boundary of $D$; the theorem does not say that the given domain determines its exact radius.

Assume a positive barrier $B\le K<S_0$. Under the zero-rate risk-neutral Black-Scholes model, $X_t=\log (S_t/B)$ starts at $x=\log (S_0/B)>0$ and is Brownian motion with variance rate ${\sigma }^2$ and drift $\mu =-{\sigma }^2/2$. The drifted transition density killed at zero isTo see the factor, multiply the zero-drift reflected Gaussian difference by $\exp [\mu (y-x)/{\sigma }^2-{\mu }^{2}t/(2{\sigma }^2)]$; its second term is the ordinary drifted Gaussian from starting point $-x$ times $e^{-2\mu x/{\sigma }^2}$.

Integrate the payoff $(B{e}^y-K{)}_{+}$ against this density. The first term is $C(S_0,K)$. Since $K\ge B$, the payoff is zero for $y\le 0$, so the second integral may be extended to all real $y$ and equals the ordinary call price from reflected spot $B{e}^{-x}=B^2/S_0$. The reflection weight is $e^x=S_0/B$. Thus the down-and-out European call value isThis assumes continuous monitoring and knock-out on first hitting, with no rebate.

At zero interest, interchanging spot and strike gives $d_1(K,S_0)=-d_2(S_0,K)$ and $d_2(K,S_0)=-d_1(S_0,K)$. Substitution into the put formula therefore givesScaling both positive arguments by $\alpha$ leaves $\log (S_0/K)$ and both $d$ values unchanged, while multiplying every price prefactor by $\alpha$. ThusThese are Put-call symmetry in the Black-Scholes model and degree-one homogeneity.

For positive spot, strike, volatility and maturity, the Black-Scholes formula iswhere $d_1=[\log (S_0/K)+(r+{\sigma }^2/2)T]/(\sigma \sqrt{T})$ and $d_2=d_1-\sigma \sqrt{T}$. Degenerate zero-volatility or zero-maturity cases are obtained by limits.