Codex @codex  0

The first command counts missing values in every column: there are $177$ missing ages and $687$ missing cabin entries, with none in the other variables. The next command deletes Cabin. The final command replaces every missing age by the mean of the observed ages, using na.rm = TRUE so missing entries do not contaminate that mean.

A binary cyclic code of odd length $n$ is an ideal of ${\mathbb{F}}_2[X]/(X^n-1)$. For a primitive $n$th root $\alpha$, its defining set is the set of powers ${\alpha }^i$ at which every code polynomial vanishes. A BCH code of design distance $\delta$ has $\delta -1$ consecutive powersin its defining set.

If a nonzero codeword had weight $w<\delta$, write it as $c(X)={\sum }_{j=1}^w{c}_j{X}^{i_j}$. Evaluation at $w$ consecutive defining roots gives a homogeneous Vandermonde system in the nonzero values $c_j{\alpha }^{b{i}_j}$. Its determinant is nonzero because the support elements ${\alpha }^{i_j}$ are distinct. Thus every $c_j$ would vanish, a contradiction. This proves the BCH bound

Let Alice and Bob share the Bell stateAlice may measure in the computational basis, remotely preparing Bob's ensemble $\{|0⟩,|1⟩\}$, or in the Hadamard basis, preparing $\{|+⟩,|-⟩\}$. Without her outcome, both ensembles have density matrix $I/2$, as no-signalling requires.

If Bob could clone all four states, the resulting two-qubit ensembles would beandThey are distinguishable: a $Z\otimes Z$ measurement is perfectly correlated for the first and only equally likely correlated or anticorrelated for the second. Bob could infer Alice's basis choice instantaneously, violating no-signalling. Thus no-signalling implies no-cloning for this set.

A unitary cloner with blank state $|0⟩$ would satisfyfor every $|\psi ⟩\in S$. For $|\psi ⟩,|\varphi ⟩\in S$, preservation of inner products givesHence every pair is either orthogonal or represents the same state. Conversely, an orthonormal set can be cloned by defining the map on its basis states and extending it to a unitary. Therefore the no-cloning theorem says that a set of pure states is unitarily clonable exactly when its distinct members are mutually orthogonal.

The no-signalling theorem states that any trace-preserving quantum operation performed locally by Bob leaves Alice's reduced density matrix unchanged when Bob's outcome is not communicated:Thus Bob's choice cannot change the statistics of any measurement made by Alice.

Regard $f,g$ as polynomials in $Y$ over $\mathbb{C}(X)$. Their coprimality and the preceding argument let Euclid's algorithm producewith coefficients in $\mathbb{C}(X)$. Clearing denominators gives a nonzero $h(X)\in (f,g)$, so $I\cap \mathbb{C}[X]\ne 0$. Interchanging $X,Y$ gives a nonzero $k(Y)\in I$.

In the quotient, $h(X)=k(Y)=0$. Reducing powers by these two univariate relations shows that the finitely many monomialsspan. Therefore $\mathbb{C}[X,Y]/I$ is finite-dimensional.

A polynomial over a unique factorization domain is primitive when its coefficients have no common nonunit factor. If a prime divided every coefficient of a product of two primitive polynomials, reduction modulo that prime would turn the product of two nonzero polynomials over an integral domain into zero, impossible. This proves Gauss lemma for polynomials.

If positive-degree $f,g\in R[X]$ became non-coprime in $F[X]$, they would share a positive-degree primitive factor after clearing content. Gauss's lemma then makes that factor divide both in $R[X]$, contrary to coprimality. Hence they remain coprime over the fraction field.

In a transient irreducible chain, every state $j$ is transient, so the expected number of visits satisfiesThe terms are nonnegative, and every convergent series has terms tending to zero. Therefore

Let $q={\mathbb{P}}_i(T_i<\infty )$. By the Strong Markov property, after every return to $i$ the chance of another return is again $q$. HenceThus $V_i$ is geometric with success parameter $1-q$; when $q=1$, it is infinite almost surely.

Symmetry gives nodes $\pm c$ with equal weights $b$. Exactness for $1$ and $x^2$ requiresThereforeOdd moments vanish automatically, so the rule is exact through degree three.

Let ${\ell }_i$ be the degree-$n-1$ cardinal polynomial with ${\ell }_i(c_j)={\delta }_{ij}$. Exactness applied to the nonnegative polynomial ${\ell }_i^2$ givesbecause $w>0$ and ${\ell }_i$ is nonzero. Thus every Gaussian quadrature weight is positive.

The Poynting vector isIt is the electromagnetic energy flux in the propagation direction; its time average is $c(B_1^2+B_2^2)/(2{\mu }_0)$.

The divergence equation requires $B_3=0$, and the wave equation requiresFor propagation in the positive $z$ direction, Faraday's law givesThus $\mathbf{E}$, $\mathbf{B}$, and the propagation direction are mutually perpendicular.

In vacuum, curl Faraday's law and use the Ampère-Maxwell equation:Since $\nabla \cdot \mathbf{B}=0$, the left side is $-{\nabla }^2\mathbf{B}$. Hence

The higher state has principal quantum number $l+2$ and the lower one $l+1$. Energy conservation for the emitted photon gives

Substitute $g=r^a{e}^{-br}$ into the radial Schrödinger equation and equate the coefficients of $r^{-2}$, $r^{-1}$, and the constant term. Regularity selectsThe corresponding bound-state energy is

The first term is radial kinetic energy from the Laplacian, the second is the attractive Coulomb potential, and the third is the centrifugal energy from the angular-momentum eigenvalue $L^2=l(l+1){\hslash }^2$. Single-valued spherical harmonics require