The roots are $\pm \sqrt{2}$ and $\pm \sqrt{3}$, so the splitting field is $K=\mathbb{Q}(\sqrt{2},\sqrt{3})$. The first adjunction has degree $2$. To see that the second does too, suppose $\sqrt{3}=u+v\sqrt{2}$ with $u,v\in \mathbb{Q}$. Squaring and comparing rational and irrational parts gives $uv=0$ and $u^2+2v^2=3$. If $v=0$, then $u^2=3$; if $u=0$, then $v^2=3/2$. Neither is a rational square, as follows by comparing prime exponents in numerator and denominator. Thus $\sqrt{3}\notin \mathbb{Q}(\sqrt{2})$. The tower law now yieldsA basis is $1,\sqrt{2},\sqrt{3},\sqrt{6}$.

The cubic has no root in ${\mathbb{F}}_2$, since its values at $0$ and $1$ are both $1$. A cubic over a field is irreducible precisely when it has no root. Consequentlyis a field extension of degree $3$, with eight elements. The Frobenius endomorphism $x\mapsto x^2$ takes a root of a polynomial over ${\mathbb{F}}_2$ to another root, so the roots are $\alpha ,{\alpha }^2,{\alpha }^4$. They are distinct: equality between any two would make the Frobenius endomorphism orbit have length less than $3$, contradicting the minimal polynomial of $\alpha$. They all lie in $K$, and one of them generates $K$. Hence

A splitting field of $f$ over $F$ is a field extension $K/F$ in which $f$ is a product of linear factors and which is generated over $F$ by its roots. The generation condition rules out adjoining unrelated elements.

For existence, induct on the degree of a polynomial. Choose an irreducible polynomial divisor $p$ of $f$. The quotient ring $F[X]/(p)$ is a field because $(p)$ is a maximal ideal; the class $\alpha$ of $X$ is a root of $p$, hence of $f$. Over this field extension, divide out $X-\alpha$ and apply induction to the remaining polynomial. This gives an extension in which $f$ splits. Inside it, the subfield generated by all roots is a splitting field. The induction starts with a linear polynomial, whose splitting field is already $F$. Thus every positive-degree polynomial has a splitting field; being monic is unnecessary for existence.

Use the intended degree ordering from part (b). Let ${\ell }_j$ be the degree-$n$ Lagrange cardinal polynomial at the distinct roots $x_0,\dots ,x_n$ of $p_{n+1}$:For any $f\in {\Pi }_{2n+1}$, polynomial division gives $f=s{p}_{n+1}+r$ with $\mathrm{deg}s\le n$ and $\mathrm{deg}r\le n$. Orthogonality of $p_{n+1}$ to ${\Pi }_n$ makes the integral of $s{p}_{n+1}$ zero, and that product also vanishes at every quadrature node. Meanwhile $r={\sum }_{j}r(x_j){\ell }_j$ by interpolation. ConsequentlyThus the rule is exact up to degree $2n+1$. To determine the sign of a weight, apply exactness to ${\ell }_j^2$, of degree $2n$:The inequality is strict because ${\ell }_j$ is a nonzero polynomial and the weight is positive. This is the Gaussian quadrature construction; the source asks only for nonnegative weights, and the proof gives positivity.

The intended orthogonal-polynomial convention is $\mathrm{deg}{p}_k=k$. An arbitrary orthonormal basis without this degree ordering does not satisfy the claim: permuting a standard basis can place the nonzero constant polynomial last, where it has no roots rather than $n+1$. The following proof uses the required degree ordering.

It makes $p_k$ orthogonal to every polynomial of degree less than $k$, since $p_0,\dots ,p_{k-1}$ form a basis of that lower-degree space. Let $c_1,\dots ,c_r$ be the distinct roots in $(a,b)$ at which $p_k$ changes sign, and put $q(x)={\prod }_{j=1}^r(x-c_j)$, with $q=1$ if $r=0$. The product $p_{k}q$ has constant sign on $(a,b)$ except at its finitely many zeros: each sign change of $p_k$ is canceled by that of $q$.

If $r<k$, then $q$ has degree below $k$, so orthogonality gives ${\int }_a^b{p}_{k}qw=0$. But a nonzero constant-sign continuous product against the positive weight has a nonzero integral, a contradiction. Thus $r\ge k$. Since $\mathrm{deg}{p}_k=k$, there are exactly $k$ such roots, all simple, with no additional endpoint roots. In particularThis also includes $k=0$, where the polynomial is nonzero constant and has no roots.

Assume the usual nondegenerate interval $a<b$. The integral is bilinear and symmetric because the functions and weight are real. It is nonnegative on $(f,f)$. If $f$ is nonzero at a point, continuity makes $|f|$ bounded below by a positive number on a nontrivial relative interval nearby, and the positive continuous weight is also bounded below there. Hence ${\int }_a^b{f}^{2}w>0$. Thus the formula is positive definite and defines an inner product.

The functions are orthogonal precisely whenA singleton interval would make this integral zero for every function, so the condition $a<b$ is necessary.

Suppose $u,v\in U_n$ both minimize the distance, with common value $d$. Apply the parallelogram identity to $x-u$ and $x-v$:The midpoint belongs to the linear subspace $U_n$, so its distance from $x$ is at least $d$. It follows that $\Vert u-v{\Vert }^2\le 0$, hence $u=v$. Thus the best approximation is unique.

Expand both squared norms using the inner product:Adding cancels the cross terms, giving the parallelogram lawThe real part may be omitted in a real inner-product space.

A best approximation to $x$ from $U_n$ is a vector $u_{\ast }\in U_n$ such thatThe definition requires attainment of the infimum. This finite-dimensional subspace does have such an approximation, as the explicit projection below shows; completeness of the whole inner-product space is unnecessary.

For a fixed alternative $0<\lambda <{\lambda }_0$, the joint likelihood ratio isIt is strictly increasing in $S={\sum }_i{x}_i$. By the Neyman-Pearson lemma, the most powerful level-$\alpha$ rejection region is therefore $S\ge c_{\alpha }$. Under the null, $S\sim \mathrm{Gamma}(n,{\lambda }_0)$ with the second parameter interpreted as a rate. Its continuous distribution givesThis threshold depends on the null and $\alpha$, but not on the particular smaller alternative $\lambda$. Thus the same test is most powerful for every $0<\lambda <{\lambda }_0$, making it uniformly most powerful.

Under a true alternative $\lambda$, $S\sim \mathrm{Gamma}(n,\lambda )$, so its attained power isFor each $\lambda <{\lambda }_0$, the Neyman-Pearson lemma bounds the power of every other level-$\alpha$ test by this value. Hence the displayed function is not only an upper bound but the attainable power envelope on the specified alternative.

The significance level of a test is a bound $\alpha$ on its probability of rejecting a true null; for a composite null the bound is uniform over its parameter values. Its power at an alternative parameter is its rejection probability under that parameter.

The Neyman-Pearson lemma concerns a simple null density $f_0$ and a simple alternative density $f_1$ for the whole sample. Among tests of size at most $\alpha$, a most powerful test rejects where $f_1>k{f}_0$, accepts where $f_1<k{f}_0$, and if necessary randomizes on equality so that its size is $\alpha$. Let its rejection probability function be ${\varphi }_{\ast }$, with $0\le {\varphi }_{\ast }\le 1$, and let $\varphi$ be any competing test of size at most $\alpha$.

Pointwise,on the rejection side both factors are nonnegative, and on the acceptance side both are nonpositive. Integrating givesThus the likelihood-ratio test has at least as much power as any competitor. Even when the sample has a continuous density, the likelihood ratio can have an atom, so boundary randomization may still be necessary in the general statement. For the exponential test below the statistic has a continuous distribution and no randomization is needed.

The maximum-likelihood estimator is a maximizerWriting ${\ell }_n=\log L_n$ and the score $S_n={\nabla }_{\theta }{\ell }_n$, the expected Fisher information matrix iswhere the equality uses the usual regularity conditions. For independent identically distributed observations, ${\mathcal{I}}_n=n{\mathcal{I}}_1$.

Under the standard identifiability, interior-parameter, differentiability and nonsingular-information assumptions, the large-sample distribution isor approximately $\hat{\theta }\sim N(\theta ,{\mathcal{I}}_n(\theta {)}^{-1})$. Boundary parameters need not obey this regular approximation.