With fixed endpoint values, a smooth extremal satisfies the Euler-Lagrange equation $F_y-(d/dx)F_{y^{\prime }}=0$. If $F$ has no explicit $x$ dependence, differentiation givesThus $F-y^{\prime }{F}_{y^{\prime }}$ is constant, the Beltrami identity.

For this integrand, $F_y=2y$ and $F_{y^{\prime }}=1+2y^{\prime }$, so the equation is $y^{\prime \prime }=y$. ConsequentlyThe two constants are fixed uniquely by the prescribed values at distinct endpoints. Equivalently, writing these values as $y_1,y_2$,Its first integral is $y^2-y^{\prime 2}$, which is constant. Using only that first integral without the Euler-Lagrange equation could admit spurious constant functions, so the differential equation was used directly.

For any nonzero endpoint-vanishing perturbation $\eta$, the linear terms in $J[y+\eta ]-J[y]$ vanish by the Euler-Lagrange equation and integration by parts; also $\int {\eta }^{\prime }=0$. The remainder isHence the extremal is the unique strict global minimum, and cannot be a maximum.

The quadratic reciprocity law states that distinct odd primes $a,b$ satisfyFirst suppose $p\ne q$. If the odd prime $r$ equalled one of them, divisibility of $p-q$ by $r$ would force the other to equal $r$ too, a contradiction. Thus reciprocity applies to $r,p$ and $r,q$. The modular congruence modulo $4r$ gives both $p\equiv q(\mathrm{mod}r)$ and $p\equiv q(\mathrm{mod}4)$. Therefore $(p/r)=(q/r)$, and the reciprocity signs are identical. Rearranging the two reciprocity identities proves $\boxed{(r/p)=(r/q)}$. If $p=q$, the equality is immediate, including a zero Legendre symbol when $p=r$.

For general odd $n>1$, factor $n={\prod }_j{r}_j^{e_j}$. If $p=q$ there is nothing to prove. Otherwise neither $p$ nor $q$ divides $n$, since any such divisor, together with $4n\mid p-q$, would force $p=q$. Each $4r_j$ divides $p-q$, so the first result and multiplicativity of the Legendre symbol giveFor $n=1$, both symbols equal one.

Eisenstein criterion says that a primitive integer polynomial is irreducible over $\mathbb{Q}$ if some prime divides all coefficients except the leading one, does not divide the leading coefficient, and its square does not divide the constant coefficient. The passage from primitive integer factorization to rational factorization is Gauss lemma for polynomials.

Translate the variable:This is Eisenstein at two. Translation is an automorphism of $\mathbb{Q}[x]$, so $x^8+1$ itself is irreducible.

Set $\zeta =e^{i\pi /8}$ and $c=\cos (\pi /8)$. Since $\zeta$ is a root of that irreducible degree-eight polynomial, $[\mathbb{Q}(\zeta ):\mathbb{Q}]=8$. Also $c=(\zeta +{\zeta }^{-1})/2$, so $\mathbb{Q}(c)\subseteq \mathbb{Q}(\zeta )$ and $\zeta$ solves $T^2-2cT+1=0$. The extension degree is at most two. It is not one, since every element of $\mathbb{Q}(c)$ is real and $\zeta$ is not. The tower law for field extensions now gives

Fix ${\omega }_1$. For any $u,v\ne {\omega }_1$, apply the given two-point transitivity property to $({\omega }_1,u)$ and $({\omega }_1,v)$. The resulting element fixes ${\omega }_1$ and sends $u$ to $v$, so the stabilizer is transitive on the complement.

The whole group is transitive on $\Omega$: extend any desired first-coordinate pair by choosing a different second point on each side. By the orbit-stabilizer theorem, $|G|=n|{\mathrm{Stab}}_G({\omega }_1)|$. Applying the same theorem inside the stabilizer to its orbit of size $n-1$ shows that $n-1$ divides the stabilizer order. HenceIn the action on ordered pairs there are exactly two orbits. One is the diagonal, transitive because the action on $\Omega$ is transitive. The other comprises distinct-coordinate pairs and is transitive by hypothesis. Equality of the two coordinates is preserved by every group element, so the two sets cannot merge into one orbit. Faithfulness of the action is not required.

Richardson extrapolation cancels a leading discretization error by combining two mesh sizes. If $A(h)=I+c{h}^p+O(h^q)$ with $q>p$, thenUse a uniform grid $x_i=a+ih$, $0\le i\le n$, with even $n$. The composite trapezium rule isThe composite Simpson rule, formed by integrating a quadratic interpolant on each pair of successive panels, isAssume an even-power trapezium error expansion $T_h=I+c_2{h}^2+O(h^4)$; continuous fourth derivative is a standard sufficient smoothness hypothesis. The coarse grid has step $2h$, so one extrapolation step is $(4T_h-T_{2h})/3$. Its endpoint coefficients are $h/3$, its odd-index coefficients are $4h/3$, and its interior even-index coefficients are $2h/3$. ThereforeThe algebraic identity between the two rules holds for any sampled values; the smoothness and error assumptions are used to infer the improved accuracy.

A Givens rotation equals the identity except in coordinates $i,j$, where its block can be taken asThe two block columns are orthonormal because $c^2+s^2=1$; other coordinates are unchanged, so the full matrix is orthogonal. If $(v_i,v_j)\ne (0,0)$, take $r=\sqrt{v_i^2+v_j^2}$, $c=v_i/r$ and $s=v_j/r$. The transformed pair is $(r,0)$. If both entries are zero, the identity already suffices.

For the specified vector one possible matrix isIts rows are orthonormal. It is not unique: premultiply by any orthogonal transformation fixing the second coordinate axis pointwise. This rotates or reflects the other two output coordinates without changing the required image. Equivalently its second row must be $v^T/\Vert v\Vert$, but any orthonormal completion of that row works.

The power of a test at a parameter value is its probability of rejecting the null there. Its size is the supremum of this probability over the null parameter set. The Neyman-Pearson lemma states that for simple null and alternative densities $f_0,f_1$, a test rejecting where $f_1>c{f}_0$, accepting where $f_1<c{f}_0$, and randomizing as needed on equality to attain size $\alpha$, has greatest power among tests of size at most $\alpha$.

The specified function is nonnegative. Its integral isFor independent observations write $S={\sum }_i{X}_i$. The likelihood ratio of the smaller alternative parameter to the null isIt is strictly increasing in $S$. Thus the most powerful level-$\alpha$ test rejects forThe continuous distribution permits this size without randomization for $0<\alpha <1$. Its threshold depends only on the null and the desired size, not on which ${\theta }_1<{\theta }_0$ was chosen. The same test is therefore uniformly most powerful against the entire one-sided alternative $\theta <{\theta }_0$.

A frequentist confidence interval is a pair of data-dependent endpoints whose random interval covers the fixed parameter with probability at least $1-\alpha$ under every allowed parameter value. In repeated uses of the same sampling procedure, at least the advertised fraction of intervals cover the true value. It is not a posterior probability assigned to the parameter after observing the data.

For a sample of size $n$ from a Poisson law with mean $\mu$, the sum $S$ is Poisson with mean $n\mu$. At observed $s>0$, define the lower and upper limits byAt $s=0$ set $L=0$ and use the same upper equation. This inverts the two exact one-sided tail tests. Discreteness means the resulting central interval is generally conservative: its coverage is at least the nominal value, not identically equal to it.

Here one observes a four-year total of 39, so the lower endpoint satisfies ${\varphi }_L(38)=0.975$ and the upper satisfies ${\varphi }_U(39)=0.025$. Selecting the corresponding entries supplied in the source givesThis interval concerns the expected four-year total, not an annual rate. The other two supplied numbers correspond to the wrong inclusive tail for one of these endpoints.