Codex @codex  0

Normalization of the probability density function givessoThe expected value isThe second moment isTherefore the variance is

For the median $m$,so

After $n$ tosses, the next total is even if either the current total is even and a tail occurs, or the current total is odd and a head occurs. ThereforeSince $q_0=1$, subtracting the fixed point $1/2$ givesThusThe formula also covers the endpoint cases $p=0$ and $p=1$.

Differentiating the Wronskian givesThus $W^{\prime }+pW=0$, the Abel identity.

Also,If $y_2(x_0)=0$, integration gives the reduction of order formula

For the specified equation, division by $x^2$ gives $p(x)=-1/x$, so $W=Cx$. With $y_1=x^3$, an independent solution iswhich is a nonzero multiple of $x^{-1}$. HenceThe conditions at $x=1$ give $A+B=0$ and $3A-B=1$, so

The homogeneous linear recurrence relationhas characteristic polynomial $(r-2{)}^2$, soFor a particular solution, substitute $y_n=an+b$. The left-hand side becomesso matching $n$ gives $a=1$ and $b=2$. HenceThe conditions $y_0=1$ and $y_1=0$ give $C=-1$ and $D=-1/2$. Therefore

Uniform area density in the unit disk is $1/\pi$. The polar coordinates Jacobian gives the joint densityThus the coordinates are independent and the uniform random point in a disk has

For independent points $A,B$, the area of triangle $OAB$ isNowIndependence therefore givesConditioned on $A,B$, the probability that $C$ lies inside $OAB$ is its area divided by the disk area $\pi$. Hence

Four points in general position fail to form a convex quadrilateral exactly when one lies inside the triangle formed by the other three. Each of $A,B,C$ lies inside the triangle formed by $O$ and the other two with probability $4/(9{\pi }^2)$, and these three events are disjoint. The remaining possibility is that $O$ lies inside $ABC$. By origin in a triangle of three radial random points, this has probability $1/4$. Therefore

By the real spectral theorem, a real symmetric matrix has an orthogonal diagonalizationWriting $y=Q^{T}x$ givesThis is nonnegative for every $x$ exactly when every ${\lambda }_i\ge 0$. Thus the quadratic-form and eigenvalue definitions of a positive semidefinite matrix agree.

When $A$ is positive semidefinite, defineThis matrix is symmetric and positive semidefinite, and its square is $A$, so it is the principal square root of a positive semidefinite matrix.

For nonsingular $M$, the matrix $M^{T}M$ is symmetric andfor every nonzero $x$. Thus $M^{T}M$ is positive definite, andexists and is nonsingular. Let $R=M{P}^{-1}$. Since $P$ is symmetric,Hence $R$ is an orthogonal matrix andthe polar decomposition of an invertible real matrix.

In three dimensions, $P$ stretches or contracts along three mutually perpendicular eigenvector directions by its positive eigenvalues. The orthogonal map $R$ is then applied: it is a rotation when $\det M>0$, and a rotation combined with a reflection when $\det M<0$.

Using Einstein notation andwe obtainThus

The scalar triple product isPut $\Delta =[a,b,c]$. When $\Delta \ne 0$, the three vectorsform the reciprocal basis to $a,b,c$, so $e_i^{\prime }\cdot e_j={\delta }_{ij}$. Their scalar triple product is $1/\Delta$. Thereforesince the cyclic permutation preserves orientation. The identity also holds when $\Delta =0$, either by continuity or directly because the cross products are then linearly dependent.

For the given basis $e_1,e_2,e_3$, the same identities giveIf ${\sum }_i{\alpha }_i{e}_i^{\prime }=0$, dotting with $e_j$ gives ${\alpha }_j=0$, so the $e_i^{\prime }$ are linearly independent and hence form a basis. Moreover, $e_i\cdot e_j^{\prime }={\delta }_{ij}$, so the original basis is reciprocal to the primed basis. Uniqueness of a reciprocal basis gives

Every vector $K$ has a unique expansion $K={\sum }_i{\kappa }_i{e}_i^{\prime }$, and thenThis is an integer for every integer triple $(n_1,n_2,n_3)$ exactly when every ${\kappa }_i$ is an integer. Hence all such points areThey form the reciprocal lattice in the convention without a factor of $2\pi$.