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A prior distribution assigns uncertainty to a parameter before observing data. Multiplying its density by the likelihood and normalizing gives the posterior distribution. A Bayes estimator minimizes the posterior expected loss function for each observed sample, and therefore minimizes integrated Bayes risk when the expectations exist.

For quadratic loss $(d-\theta {)}^2$, the conditional risk is $\mathrm{Var}(\theta \mid x)+(d-E(\theta \mid x){)}^2$, so its minimizer is the posterior mean. For absolute loss $|d-\theta |$, the one-sided derivatives or subgradients show that a minimizer is any posterior median, characterized by $P(\theta <d\mid x)\le 1/2$ and $P(\theta >d\mid x)\le 1/2$.

For the stated uniform distribution sampling model, let $x_{\min }={\min }_i{x}_i$, $x_{\max }={\max }_i{x}_i$. The likelihood is $2^{-n}$ on $x_{\max }-1<\theta <x_{\min }+1$ and zero elsewhere. Intersecting this interval with the uniform prior givesWhen $A<B$, the bounded uniform-location posterior isThe posterior is symmetric, so its mean and unique median coincide. If $A\ge B$, the data have zero marginal likelihood under the stipulated prior/sampling model and this posterior density cannot be normalized. An empty or singleton intersection is not an ordinary uniform posterior of positive width.

Take an admissible variation $y+\epsilon \eta$ with $\eta ={\eta }^{\prime }=0$ at both regular endpoints. Its first variation is $\delta I=\int (f_y\eta +f_{y^{\prime }}{\eta }^{\prime }+f_{y^{\prime \prime }}{\eta }^{\prime \prime })dx$. Integrating the second term once and the third twice gives the boundary term $[f_{y^{\prime }}\eta +f_{y^{\prime \prime }}{\eta }^{\prime }-(d{f}_{y^{\prime \prime }}/dx)\eta {]}_a^b$, which vanishes. The fundamental lemma of the calculus of variations therefore yields the higher-order Euler-Lagrange equationHere $f_y=8y{y}^{\prime }$, $f_{y^{\prime }}=4y^2$, and $f_{y^{\prime \prime }}=2x^4{y}^{\prime \prime }$. The first two contributions cancel, leaving $(x^4{y}^{\prime \prime }{)}^{\prime \prime }=0$. For $x>0$, integrate twice to obtain $x^4{y}^{\prime \prime }=Ax+B$, and thenThe two singular-endpoint limits give $B=0$ and $A=2$. The conditions at one give $C=1$ and $D=0$. Hence the unique solution of the stated differential equation and limiting data isThere is a genuine distinction between solving this equation and stationarity of a finite improper action. Substitution gives $f=4x^2+4-4x^{-4}$, whose integral at zero diverges to $-\infty$. The displayed Euler-Lagrange equation remains valid for compactly supported variations on $(0,1)$, or on regular cut-off intervals, but the literal unregularized action is not a finite stationary functional on this solution. These singular endpoints in higher-order variational problems require a regularization or an admissible-space convention if a global variational claim is intended.

A nonzero residue class $a(\mathrm{mod}p)$ is a quadratic residue if $a\equiv x^2(\mathrm{mod}p)$ for some nonzero $x$. If it is a square, Fermat's little theorem gives $a^{(p-1)/2}=x^{p-1}\equiv 1$. Conversely, the map $x\mapsto x^2$ on the nonzero classes identifies exactly the pairs $\{x,-x\}$, giving $(p-1)/2$ distinct squares. They are all roots of $X^{(p-1)/2}-1$ over the field ${\mathbb{F}}_p$; a polynomial of this degree has at most that many roots. Its roots are therefore precisely the squares. This proves Euler's criterionApplying it to $-1$ gives $\boxed{(-1/p)=1⟺p\equiv 1(\mathrm{mod}4)}$. The exclusion of $a=0$ is the usual nonzero-residue convention needed for this equivalence.

For distinct odd primes $p,q$, quadratic reciprocity statesUsing the multiplicativity of the Legendre symbol, and the supplementary law $(2/p)=(-1{)}^{(p^2-1)/8}$,The first factor is $1$ because $73\equiv 1(\mathrm{mod}8)$; the second is $(73/3)=(1/3)=1$ because the reciprocity sign is positive. Thus $\boxed{73\text{ is a quadratic residue modulo }127}$.

Let $p(x)={\prod }_{i=0}^n(x-x_i)$ be the monic node polynomial. Suppose first that the quadrature is exact through degree $2n+1$. For any polynomial $r$ of degree at most $n$, $pr$ has degree at most $2n+1$, while every sampled value is zero. Exactness givesThus $p$ is an orthogonal polynomial of degree $n+1$, proving necessity in the orthogonal-node criterion for Gaussian quadrature.

Conversely, suppose the nodes are the zeros of such a polynomial. Rescale it to be monic without changing its zeros or orthogonality. For any $f$ of degree at most $2n+1$, polynomial division givesOrthogonality makes $\int pq=0$; the quadrature of $pq$ is zero because $p(x_i)=0$. The assumed degree-$n$ exactness applies to $r$, soHence degree-$2n+1$ exactness is equivalent to orthogonal nodes. The given distinct-zero property ensures the $n+1$ node values are independent interpolation data; the degree-$n$ exactness fixes the corresponding interpolatory weights. This is the construction of Gaussian quadrature.

Against the two pure column strategies, player one's expected payoffs are $L_1(p)=5-2p$ and $L_2(p)=2+(b-2)p$. A mixed strategy of the opponent cannot give a smaller expectation than their minimum. The zero-sum game optimization is thereforeThe first line decreases. If $0<b<2$, the second also decreases, and their minimum is maximized at $p=0$ with value $2$. If $b=2$, the second line is constantly $2$ and the first is at least $3$, so every $p$ is optimal. For $2<b\le 3$, the intersection $p=3/b$ lies at or beyond the right endpoint; the minimum increases throughout $[0,1]$ and is maximized at $p=1$ with value $b$. If $b>3$, the intersection lies inside the interval; the minimum increases before it and decreases afterwards. ThusFor the interior case the opponent can equalize row payoffs with first-column probability $(b-2)/b$, confirming the minimax theorem value.

For a connected cellular embedding on the sphere, the Euler formula for a connected planar graph isThe boundary graph of a convex polyhedron has such an embedding. Counting edge ends at vertices and edge sides at faces gives ${\sum }_{m\ge 3}m{V}_m=2E={\sum }_{n\ge 3}n{F}_n$. ThereforeFaces of size six contribute zero, and larger faces contribute negatively. The contribution of each triangle, quadrilateral or pentagon is at most three, soThus the small faces of a spherical polyhedral graph bound isThe tetrahedron shows that four is attainable. Connectedness and cellular faces are the standard hypotheses in this Euler formula; an arbitrary disconnected embedded graph has a different count.

Let the mass per unit length be $m$ and the constant tension be $T$. Small transverse displacement obeys the wave equationWith characteristic coordinates $\xi =x+ct$ and $\eta =x-ct$, the equation becomes $y_{\xi \eta }=0$. Integrating once in each variable gives D'Alembert's formulaFor a smooth solution with vanishing endpoint energy flux, multiplication by $y_t$ and integration by parts giveThe fixed endpoint has $y_t(0,t)=0$. Decay of displacement alone does not guarantee either a pointwise derivative-flux limit or finiteness of the energy integral; the conservation statement is understood in the usual finite-energy class. The profile formula below proves it directly there without requiring such a pointwise limit at infinity.

The endpoint condition gives $g(-ct)=-f(ct)$, allowing the reflected wave to be expressed through one profile:For forward time one may define the negative-argument part of this profile using the original $g$; no restriction on the independent initial waves is lost. Differentiation gives $y_t=c[f^{\prime }(x+ct)-f^{\prime }(ct-x)]$ and $y_x=f^{\prime }(x+ct)+f^{\prime }(ct-x)$. HenceThe two substitutions partition the real line at $ct$, making conservation of energy explicit.