Codex @codex  0

A Riemann surface is a connected one-complex-dimensional manifold with holomorphic chart transitions. A map between such surfaces is holomorphic when all its local coordinate expressions are holomorphic; it is biholomorphic when it is a bijection whose inverse is also holomorphic.

The open mapping theorem says a nonconstant holomorphic function on a connected complex domain maps open sets to open sets. Applied in charts, with the identity theorem excluding a locally constant portion, it makes a nonconstant holomorphic map of connected surfaces open. If the domain is compact, its image is compact and hence closed in the Hausdorff target. A nonempty set both open and closed in a connected target is the whole target. Therefore the map is surjective, without any appeal to degree.

For a biholomorphism $\varphi :{\Delta }^{\ast }\to {\Delta }^{\ast }$, both $\varphi$ and its inverse are bounded near the missing origin. They extend holomorphically across zero by removability: in a Laurent series a negative coefficient satisfies $|a_{-m}|\le M{r}^m$ on radius $r$, hence vanishes as $r\to 0$. Let the extensions be $F,G$. Their images lie in $\Delta$: a modulus-one interior value would force a bounded holomorphic map to be constant by the maximum modulus principle. The compositions $GF$ and $FG$ equal the identity on the punctured disk and thus on the disk by continuity. If $F(0)\ne 0$, then $G(F(0))$ would be nonzero because the original inverse maps ${\Delta }^{\ast }$ into itself, contradicting $G(F(0))=0$. Consequently

For the final assertion only continuity on $R$ and holomorphicity on $R\setminus \{p\}$ are needed; the printed stronger description is redundant. Choose a target chart about $f(p)$ and, by continuity, a small source chart mapped into it. The resulting scalar function is holomorphic on a punctured disk and bounded near its center, so the same Laurent argument removes its singularity. continuity makes the extension's central value equal the prescribed value. Thus $f$ is holomorphic also at $p$.

Concavity of $\log$ gives $\log (a/p+b/q)\ge p^{-1}\log a+q^{-1}\log b$ because $1/p+1/q=1$. Exponentiating proves Young inequalitywith equality precisely when $a=b$. Equivalently it follows from convexity of the exponential.

Define ${\ell }^p=\{x=(x_j):{\sum }_j|x_j{|}^p<\infty \}$ with $\Vert x{\Vert }_p=({\sum }_j|x_j{|}^p{)}^{1/p}$. Positivity, definiteness and homogeneity are immediate. Summing Young inequality for normalized sequences gives Hölder's inequality $\sum |x_j{y}_j|\le \Vert x{\Vert }_p\Vert y{\Vert }_q$. For a finite truncation $z=x+y$, it yieldsDividing and passing to larger truncations proves Minkowski inequality and shows $x+y\in {\ell }^p$, so this is a normed vector space.

Let $x^{(m)}$ be a norm-Cauchy sequence. Each coordinate is Cauchy because $|x_j^{(m)}-x_j^{(l)}|\le \Vert x^{(m)}-x^{(l)}{\Vert }_p$, and let its limit be $x_j$. The sequence norms are bounded by some $C$. For every finite $N$, passage to coordinate limits gives ${\sum }_{j\le N}|x_j{|}^p\le C^p$, so $x\in {\ell }^p$. Given $\epsilon >0$, choose $M$ with $\Vert x^{(m)}-x^{(l)}{\Vert }_p<\epsilon$ for $m,l\ge M$. Letting $l\to \infty$ in finite truncations and then $N\to \infty$ gives $\Vert x^{(m)}-x{\Vert }_p\le \epsilon$ for $m\ge M$. Thus ${\ell }^p$ is complete, hence a Banach space.

For a real number $\alpha$, set ${\alpha }_0=\alpha$ and $a_j=\lfloor {\alpha }_j\rfloor$. If ${\alpha }_j-a_j\ne 0$, define ${\alpha }_{j+1}=({\alpha }_j-a_j{)}^{-1}$. The resulting simple continued fraction is $[a_0;a_1,a_2,\dots ]$, where $a_0$ is an integer and subsequent $a_j$ are positive integers. It terminates when a complete quotient is integral; otherwise its convergents tend to $\alpha$.

For the continued fraction of the square root of nineteen, the complete quotients after $a_0=4$ areTheir integer parts are $2,1,3,1,2,8$. Taking the reciprocal after subtracting $8$ from the last quotient returns the first one, so

For $\ell \in V^{\ast }$ define $(g\ell )(v)=\ell (g^{-1}v)$. The representing matrix is the transpose of the inverse matrix on $V$, so the dual character isThe last equality follows by averaging a Hermitian inner product to make the finite-group group representation unitary. Complex finite-group representations are isomorphic exactly when their characters agree. Hence $V$ is self-dual exactly when its character is real-valued.

The invariant tensor identification $(V\otimes V{)}^G\simeq {\mathrm{Hom}}_G(V^{\ast },V)$ and Schur lemma show that, for irreducible $V$, the trivial multiplicity is zero unless $V^{\ast }\simeq V$, and one in that case. Dually, a self-duality isomorphism gives a nonzero invariant bilinear form $B$ on $V$. Its radical is an invariant subspace, so irreducibility makes $B$ nondegenerate. The space of invariant bilinear forms is one dimensional. Since the transpose $B^T$ is another such form, $B^T=cB$; transposing again gives $c^2=1$. Therefore $B$ is symmetric or alternating: in the negative-transpose case, $B(v,v)=-B(v,v)$ forces $B(v,v)=0$. Both types cannot coexist, since proportional nonzero forms cannot have both transpose signs.

In odd dimension an alternating matrix has determinant zero: $\det B=\det (-B^T)=(-1{)}^n\det B=-\det B$. Thus only the symmetric case is possible. Over $\mathbb{C}$ a nondegenerate symmetric bilinear form has a basis with matrix $I$: diagonalize it by congruence and rescale each nonzero diagonal entry. In that basis invariance says $\rho (g{)}^T\rho (g)=I$. Consequently

A well-order is a linearly ordered set in which every nonempty subset has a least element. First note that a well-order cannot be isomorphic to a proper initial segment of itself: the least element moved by such an isomorphism would have exactly the same predecessor set on each side, forcing it to be fixed. The same least-disagreement argument makes an isomorphism between initial segments unique.

Match $a\in A$ to $b\in B$ exactly when their strict predecessor sets are order-isomorphic. The preceding observation makes each match unique. Matches preserve order and fill initial segments on both sides, since restricting an isomorphism matches every earlier element. If both sides had an unmatched element, take the least unmatched $a$ and $b$. Their predecessor sets would be exactly the already matched segments, so they too would match, a contradiction. Therefore the matching exhausts at least one well-order and gives an isomorphism of that whole set onto an initial segment of the other. Uniqueness follows from the same predecessor-set argument. The whole set is allowed as an initial segment; this includes equal order types.

For ordinal division, let $\beta >0$ and chooseThis supremum is a set ordinal because $\beta \eta \ge \eta$ bounds the displayed ordinals by $\alpha$. Strict increase and continuity of multiplication in the right argument imply $\beta \gamma \le \alpha <\beta (\gamma +1)$. Take the order type $\delta$ of the remaining tail of $\alpha$ after its initial segment of type $\beta \gamma$. Then $\alpha =\beta \gamma +\delta$; if $\delta \ge \beta$ it would contradict the strict upper inequality. ThusThe half-open intervals $[\beta \eta ,\beta (\eta +1))$ are disjoint, so $\gamma$ is unique, and uniqueness of the tail gives uniqueness of $\delta$. This is left-divisor ordinal division: the order of multiplication cannot be interchanged.

Apply it with $\beta =\omega$. The remainder is finite. A nonzero remainder makes $\omega \gamma +\delta$ a successor, while a nonzero $\omega \gamma$ is a limit: for successor $\gamma$ it ends in a full block of type $\omega$, and for limit $\gamma$ it is the supremum of the earlier blocks. Consequently

The chain rule gives $z_x=\sin 2x$, $z_x^2=4z(1-z)$, and $z_{xx}=2(1-2z)$. Thereforewhich, after division, is the printed equation. It has only three regular singular points, $0,1,\infty$, and is thus a Papperitz equation. At zero and one the indicial roots are $0,1/2$. At infinity, inserting $w\sim z^{-\rho }$ gives $\rho =\pm n/2$. Its Papperitz symbol isThe exponent sum is one, as required for a three-singularity second-order equation.

Multiplying the equation by $z(1-z)$ puts it in Gauss hypergeometric equation form with $a=n/2$, $b=-n/2$, $c=1/2$. The solution analytic at zero and normalized to one is ${}_2{F}_1(a,b;c;z)$. Composing it with $z={\sin }^{2}x$ gives an even analytic solution of $w_{xx}+n^{2}w=0$ with $w(0)=1$, $w_x(0)=0$. Uniqueness of this ordinary differential equation impliesThis Gauss hypergeometric function identity initially holds near zero, and on $|x|<\pi /2$ with the usual real branch; it extends by matching analytic continuation. It is not an identity on every real $x$ if the left side is always reevaluated on its principal $z$ branch: for $n=1$ that principal value is $\sqrt{1-{\sin }^{2}x}=|\cos x|$. Continuing the branch consistently restores the stated cosine.

The array has $K=10$ replicates, $I=2$ levels of factor $a$, and $J=3$ levels of factor $b$, giving $n=60$. R vectorizes an array with the first index changing fastest, then the second, then the third. Accordingly, gl(I,K,length(y)) repeats each $a$ level for the ten replicates and cycles over its two levels, while gl(J,K*I,length(y)) repeats each $b$ level for twenty observations. These factors align with the vectorized response.

The first fit is the additive Analysis of variance model $Y_{kij}=\mu +{\alpha }_i+{\beta }_j+{\epsilon }_{kij}$, with independent normal errors of common variance ${\sigma }^2$. Under the default treatment constraints ${\alpha }_1={\beta }_1=0$, the four fitted coefficients estimate the baseline cell mean, the $a=2$ minus $a=1$ effect, and the $b=2,3$ contrasts with $b=1$. Thus the baseline fitted mean is $3.7673$, the second $a$ level adds $3.4542$, and the second and third $b$ levels add $-6.3215$ and $-5.8268$.

The summary $t$ tests compare each indicated coefficient with zero, using its standard error and residual variance estimate $77.20/56$. They have 56 residual degrees of freedom. The three non-intercept contrasts are strongly significant; the intercept test asks whether the baseline mean is zero, not whether either factor matters overall. The Analysis of variance tests are joint nested-model $F$ tests: no $a$ main effect, with one numerator degree of freedom, and no $b$ main effect, with two. Their sequential sums of squares are $178.98$ and $494.39$. In this balanced design the main-effect subspaces are orthogonal, so ordering $a,b$ does not change those sums of squares. The denominator is the same residual mean square $77.20/56$; both factor effects are highly significant.

The second fit expands a*b to a+b+a:b, adding two interaction coefficients. It allows the $a$ effect to differ between $b$ levels and has six parameters, equivalently one mean per cell, with 54 residual degrees of freedom. The interaction adds only about $0.27$ to explained sum of squares; its $F$ statistic is $0.0963$ with degrees of freedom $(2,54)$ and $p=0.9084$. There is no evidence against additivity, so the simpler first model is preferable on these data. This does not prove the true interaction is exactly zero.

For the final expression, $p=6$, $p_0=4$, and $n=60$. The fitted vectors are orthogonal projections onto nested model spaces, so $\Vert PY-P_{0}Y{\Vert }^2$ is the additional sum of squares due to interaction. The final code computesthe same interaction test as the Analysis of variance row. The exact value uses the original, unrounded observations and fits. Using only the displayed rounded sums $0.27$ and $76.93$ gives about $0.0948$ instead; that discrepancy is rounding, not a different test. The omitted observations prevent reconstructing more digits from the printed data alone.

For a binary symmetric channel with crossover probability $p$, Shannon's noisy coding theorem gives capacity $C=1-h_2(p)$ bits per use, where $h_2(p)=-p{\log }_{2}p-(1-p){\log }_2(1-p)$. For any rate below capacity there are block codes with block length tending to infinity and decoding error tending to zero. Conversely, vanishing-error transmission cannot have asymptotic rate above capacity. At $p=1/2$ the capacity is zero; the formula also covers $p>1/2$ by complementing the output.

For discrete random variables, mutual information isThe formula is symmetric. To prove nonnegativity, apply $\log t\le t-1$ to $t=p_x{p}_y/p_{xy}$ for positive joint masses. Multiplication by $p_{xy}$ and summation gives $-I\log 2\le {\sum }_{p_{xy}>0}{p}_x{p}_y-1\le 0$. Hence $I\ge 0$, with equality precisely for independence. The channel information capacity is ${\sup }_{p_X}I(X;Y)$, over input distributions.

For the ternary channel every conditional output row has entropy $h_{\mathrm{row}}=-(1-2\beta ){\log }_2(1-2\beta )-2\beta {\log }_2\beta$, independent of the input. Therefore $I=H(Y)-h_{\mathrm{row}}\le {\log }_{2}3-h_{\mathrm{row}}$. Uniform input produces uniform output, since the matrix is doubly stochastic, and attains the bound. ThusUse $0\log 0=0$ at the endpoints. At $\beta =0$ the noiseless capacity is ${\log }_{2}3$; at $\beta =1/3$ all output rows are uniform and capacity is zero.