For this tableau sign convention, a negative coefficient in the objective row means the corresponding nonbasic variable can increase profit. Initially $x_3$ has the most negative coefficient. The simplex ratio test gives $33/5$ in row one and $18/3=6$ in row two, so row two leaves to preserve feasibility. Next $x_1$ is the only column with a negative objective coefficient. The ratios are $3/(1/3)=9$ and $6/(1/3)=18$, so row one leaves. All resulting basic variables remain nonnegative.

Two dual faces of a plane graph share an edge exactly when their corresponding original vertices are adjacent. Their face-adjacency graph is therefore the original tree, which is a nontrivial bipartite graph. Colour the face for $r$ with one colour and alternate along the original tree. This gives a proper face colouring with two colours, and at least two are needed because there is an edge separating distinct faces. Hence the minimum is

A plane tree has one face of a plane graph, so its dual graph has one vertex. Every original edge is a bridge incident to that same face on both sides, so every dual edge is a loop. Thus the dual drawing in the figure is a bouquet of eight loops, arranged with the displayed nesting inherited from the tree. It has nine faces of a plane graph, matching $f^{\ast }=n=9$. The nesting matters for which faces meet; a bouquet drawn as eight unrelated petals would have a different face-adjacency graph.

If the starting length is $L=b-a$, then after $k$ interval halvings the current midpoint is at most $L/2^{k+1}$ from some root remaining in the bracket. Thus, with the midpoint returned after those halvings, it suffices to takeIf iterations instead count midpoint evaluations starting with the initial midpoint as evaluation one, the usual bound is $n\ge \lceil {\log }_2(L/ϵ)\rceil$, with at least one evaluation unless an endpoint root has already been found. Stating the counting convention avoids an apparent one-step discrepancy.

The bisection method begins with a finite interval $[a,b]$ on which the continuous function has opposite signs at its endpoints. The intermediate value theorem guarantees a root inside. Evaluate at the midpoint $m=(a+b)/2$. If $f(m)=0$, stop. Otherwise retain the half whose endpoints have opposite signs and repeat. An endpoint that is already a root is returned immediately. The nested intervals retain a root and their lengths halve, so their midpoints converge to a root.

Under the standard consistency and regularity hypotheses for the maximum-likelihood estimator,Equivalently, for large samples $\hat{\theta }$ is approximately $N(\theta ,I_n(\theta {)}^{-1})$. In a standard-error estimate, the unknown parameter in the information may be replaced by the consistent estimate $\hat{\theta }$.

For the sample likelihood function $L(\theta )={\prod }_{i}f(X_i;\theta )$ and log-likelihood $\ell =\log L$, the expected Fisher information isThe second equality uses the usual regularity assumptions permitting differentiation under the integral and a parameter-independent support. Independent identically distributed observations give $I_n(\theta )=n{I}_1(\theta )$.

From the preceding decomposition, $Y=(3/5)X+W$, where $W$ is independent of $X$ and has mean zero. Taking conditional expectation gives $\mathbb{E}[Y\mid X]=(3/5)X$. The corresponding continuous version at $X=5$ is

The identity $Y=aX+W$ forces $b=1-2a$ and $c=1-a$. Independence will follow from zero covariance, because $X,W$ are jointly Gaussian random variables. That covariance isThusThese values give $W=(-Z_1+2Z_2)/5$ with variance $1/5$ and zero covariance with $X$. Joint normality then proves the requested independence.

A linear combination of independent normal random variables is normal. Its mean is $2\cdot 0+0=0$, and independence removes the cross-covariance term in its variance:

The density is an even function on a symmetric interval, so the integrand $xf(x)$ is odd. Its integral vanishes:

Normalization of the probability density function requiresusing ${\cos }^{2}x=(1+\cos 2x)/2$. Therefore

The full-day count has Poisson distribution with mean and variance $2\cdot 450=900$. Its normal approximation is $N(900,900)$, with standard deviation $30$. With the usual continuity correction,Without the half-unit correction the central-limit approximation gives $1-\Phi (1)\approx 0.159$.

The one-minute increment has Poisson distribution with parameter two, independently of where that interval occurs. Hence

A rate-two Poisson process starts at $N_0=0$, has independent increments and stationary increments, and has $N_{t+h}-N_t\sim \mathrm{Poisson}(2h)$ for $h\ge 0$ when time is measured in minutes. Equivalently it is a counting process with independent stationary increments satisfying, uniformly in the start time,as $h\downarrow 0$. Thus an average rate alone is insufficient; the independence and increment assumptions are part of the model.

Every point of the indicated ray has the form $w_0=(r,2r,\sqrt{5}r)$ with $r>0$; the apex is excluded. Substitution into the preceding tangent space formula givesindependent of $r$. The affine tangent plane has the same equation, since the ray lies in this plane. Thus both interpretations of the requested tangent plane are constant along this generator of the cone.

The tangent space to a regular level set is the kernel of its derivative. Writing $z_0=\sqrt{x_0^2+y_0^2}$ givesIf an affine tangent plane through $v_0$ is intended, its equation is $x_0(x-x_0)+y_0(y-y_0)-z_0(z-z_0)=0$. Since $x_0^2+y_0^2=z_0^2$, this simplifies to the same plane $x_{0}x+y_{0}y-z_{0}z=0$, which passes through the apex.

Let $g(x,y,z)=x^2+y^2-z^2$. The upper cone away from the apex is the part of $g^{-1}(0)$ with $z>0$. ThereThe regular level set theorem says that a level set of a smooth scalar function with nonzero gradient is an embedded submanifold of codimension one. Intersecting it with the open half-space $z>0$ preserves this local property. Thus $M$ is a two-dimensional submanifold of ${\mathbb{R}}^3$.

A tangent vector to $M$ at $p$ is a velocity ${\gamma }^{\prime }(0)$ of a smooth curve $\gamma :(-\epsilon ,\epsilon )\to M$ with $\gamma (0)=p$. The tangent space is the set of all such velocities:It is a $k$-dimensional linear subspace of ${\mathbb{R}}^n$; for a local parametrization $\varphi$, it is the image of the injective derivative $D\varphi$ at the corresponding parameter point.

A $k$-dimensional embedded submanifold of ${\mathbb{R}}^n$ is a subset $M$ such that near each $p\in M$ there is a smooth diffeomorphism of ambient open neighbourhoods that sends $M$ to the coordinate plane ${\mathbb{R}}^k\times \{0{\}}^{n-k}$. More precisely, for some open $U\ni p$ and diffeomorphism $\Phi :U\to V\subseteq {\mathbb{R}}^n$,Thus locally the subset has $k$ freely varying smooth coordinates.