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Write $y={\sum }_{k\ge 0}{c}_k{x}^k$ with $c_0=1$. Substitution into the ordinary differential equation givesUnless the series terminates, its coefficient ratio is $O(1/k)$, so it converges for every finite $x$. A coefficient first vanishes exactly when $\lambda$ is a nonnegative integer $n$, and then $c_n=(-1{)}^n/n!\ne 0$, $c_{n+1}=0$. Thus $y$ is a polynomial precisely for $\lambda =n\in \{0,1,2,\dots \}$, in which caseis the normalized Laguerre polynomial.

Multiplying the differential equation by $e^{-x}$ gives $(x{e}^{-x}{y}_n^{\prime }{)}^{\prime }+n{e}^{-x}{y}_n=0$. Multiply by $y_m$, subtract the equation with the indices exchanged, and integrate over $[0,\infty )$. The boundary term $x{e}^{-x}(y_m{y}_n^{\prime }-y_n{y}_m^{\prime })$ vanishes both at zero and infinity. ThereforeThis is the weighted orthogonality of the Laguerre polynomials.

Their leading coefficients show $\mathrm{deg}(y_m{y}_n)=m+n$. Since $y_0,\dots ,y_{m+n}$ have distinct degrees, they form a basis of that polynomial space. Hence $y_m{y}_n={\sum }_{p=0}^{m+n}{a}_p{y}_p$, and comparison of leading coefficients gives $a_{m+n}=(m+n)!/(m!n!)\ne 0$. Likewise the degrees of $y_m$ and $y_m{y}_n$ for $0\le m<n$ run through $0,\dots ,n-1$ and $n,\dots ,2n-1$, respectively. A nontrivial linear combination cannot cancel its highest-degree term, so these $2n$ functions are linearly independent.

For $n=2$, this gives the basis $y_0,y_1,y_2,y_1{y}_2$ of the cubic polynomial space, establishing the specified expansion. Its weighted integral is $a_0$: $y_0=1$ integrates to one, $y_1$ and $y_2$ are orthogonal to $y_0$, and $y_1{y}_2$ integrates to zero. At the two roots ${\alpha }_1,{\alpha }_2$ of $y_2$, the expansion reduces to $f({\alpha }_i)=a_0+a_1{y}_1({\alpha }_i)$. Solving these two equations for $a_0$ givesFor explicit verification, the recurrence gives $y_1=1-x$, $y_2=1-2x+x^2/2$. Thus ${\alpha }_1=2-\sqrt{2}$, ${\alpha }_2=2+\sqrt{2}$ are distinct positive roots and $w_1=(2+\sqrt{2})/4$, $w_2=(2-\sqrt{2})/4$. This is the two-node Gauss-Laguerre quadrature rule, exact for every cubic.

Let $r=\dim \mathrm{im}T$. Since the kernel of a linear map and its image agree, the rank-nullity theorem gives $\dim V=r+r$. Thus $\dim V=2r$ is even. Every $Tv$ lies in $\ker T$, so $T^2=0$. The minimal polynomial therefore divides $X^2$. It cannot be $1$, and it cannot be $X$ since $T\ne 0$; hence the minimal polynomial is $X^2$.

Subtract the exact solution from the Jacobi method to get $e_n=H{e}_{n-1}$, hence $e_n=H^n{e}_0$. Convergence from every initial vector is equivalent to $H^n\to 0$. If every eigenvalue has modulus below one, a Jordan block $\lambda I+N$ has powers that are finite sums $\left(\genfrac{}{}{0ex}{}{n}{k}\right){\lambda }^{n-k}{N}^k$. The polynomial factor in $n$ is dominated by geometric decay; for $\lambda =0$, the block is nilpotent. Consequently $H^n\to 0$.

Conversely, for a complex eigenvector $v$ with eigenvalue $\lambda$, $H^{n}v={\lambda }^{n}v$. Convergence for real errors also implies convergence for their complexification, so this tends to zero only if $|\lambda |<1$. Therefore the iteration converges from every starting vector if and only if $\rho (H)<1$, where $\rho$ is the spectral radius.

For the specified matrix,Its eigenvalues are $0,\mu ,-\mu$, so $\rho (H)=|\mu |$. Hence $-1<\mu <1$ is the convergence range. The original coefficient matrix has determinant $12(1-{\mu }^2)$, so the endpoints also violate its assumed nonsingularity.

The dual space is $V^{\ast }={\mathrm{Hom}}_F(V,F)$, the vector space of all linear functionals with pointwise operations. For finite $d=\dim V$, choose a basis $v_1,\dots ,v_d$ and define its dual basis by ${\epsilon }_j(v_i)={\delta }_{ij}$. Every linear functional is $\ell ={\sum }_j\ell (v_j){\epsilon }_j$, and evaluation on each $v_i$ proves that the ${\epsilon }_j$ are linearly independent. Thus $\dim V^{\ast }=\dim V=d$. This dimension formula is a finite-dimensional assertion.

On the space of polynomials of degree at most $n$, define the evaluation map $E(p)=(p(a_0),\dots ,p(a_n))$. Its kernel is zero: a nonzero degree-at-most-$n$ polynomial cannot have $n+1$ distinct roots. Since domain and codomain both have dimension $n+1$, $E$ is an isomorphism. The evaluations therefore form a basis of the dual space. The linear functional $p\mapsto p^{\prime }(0)$ has a unique expansion in that basis, proving the existence and uniqueness of the requested weights.

Explicitly, use the Lagrange interpolation polynomial basisThe differentiation weights from nodal evaluations are ${\lambda }_j={\ell }_j^{\prime }(0)$. If a node is zero, no singular reciprocal formula is needed: differentiate the finite product directly. For $n=0$ the sole weight is zero.

Assume $L$ is linear, annihilates all polynomials of degree at most $k$, is defined on the truncated-power kernels and permits interchange with the remainder integral. For $k\ge 1$, a convenient sufficient condition is that $L$ be a bounded linear functional on $C[a,b]$; more general approximation functionals need the corresponding continuity/interchange property on their domain. Taylor's integral remainder givesApplying $L$, the polynomial terms vanish, and the Peano kernel theorem givesHere $t_{+}=\max (t,0)$. This is the printed raw-kernel convention; the normalized Peano kernel includes $1/k!$ inside its definition instead, and then has no factorial outside the integral. The interchange is justified by boundedness applied to the limit of Riemann sums of the continuous kernel-valued integrand. For a kernel with $k=0$, its discontinuity requires a function space/interchange assumption accommodating the step functions as well.

For the specified Simpson's rule error, direct evaluation on $1,x,x^2,x^3$ gives zero, so $k=3$. For $0\le \theta \le 2$,Simplifying separately on the two halves givesBoth pieces are strictly negative in the open interval; they agree at the midpoint. Their integral is $-1/15$, as direct polynomial integration shows. ThusThis is optimal, not merely an upper bound. For $f(x)=x^4/24$, the fourth derivative is identically one and $L(f)=-1/90$, so the minimum constant is $\rho =1/90$.

The polynomial ring $F[X]$ is an integral domain, since the product of two nonzero polynomials has nonzero leading coefficient. Let $I$ be an ideal. If $I=0$, it is generated by zero. Otherwise choose a nonzero polynomial $g\in I$ of minimum degree. For any $f\in I$, the polynomial division algorithm over the field gives $f=qg+r$ with $r=0$ or $\mathrm{deg}r<\mathrm{deg}g$. Since $r=f-qg\in I$, minimality forces $r=0$. Hence every element of $I$ is a multiple of $g$, and $I=(g)$. Thus $F[X]$ is a principal ideal domain.

The nonprincipal coordinate ideal in two variables provides the contrast:If it were $(g)$, then $g$ would divide $X$ and $Y$. Total degrees add when nonzero polynomials over a field are multiplied. A nonconstant divisor of $X$ consequently has degree one, its cofactor is constant, and it is a nonzero scalar multiple of $X$. Such a polynomial cannot divide $Y$, as setting $X=0$ shows. Thus $g$ would be a nonzero constant and the ideal would contain one. But all elements of $(X,Y)$ vanish at $(0,0)$, so this is impossible.

If $\dim V=n$, then $\dim \mathrm{End}(V)=n^2$. The $n^2+1$ endomorphisms $I,T,\dots ,T^{n^2}$ are therefore linearly dependent, so some nonzero polynomial $p$ satisfies $p(T)=0$. The minimal polynomial is the monic polynomial of least degree annihilating $T$. Uniqueness follows because the difference of two such monic polynomials would be a nonzero annihilator of smaller degree.

A Jordan block $J_r(\lambda )$ has minimal polynomial $(x-\lambda {)}^r$, and the minimal polynomial of a direct sum is the least common multiple of those of its blocks. Thus an eight-dimensional example is

The two algebras have the same identity, so an inverse in $A$ is also an inverse in $B$. Therefore $G(A)\subseteq G_B(A)$.

For strictness, take $B=C_{\mathbb{C}}(\mathbb{T})$, where $\mathbb{T}$ is the unit circle, and let $A$ be the uniform closure of the polynomials in the coordinate function $z$. This is a closed unital subalgebra of the Banach algebra $B$. The coordinate $z$ has inverse $\overline{z}$ in $B$. But $\overline{z}$ cannot belong to $A$: for every polynomial $p$,and the same identity holds for every element of $A$ by uniform convergence. It fails for $\overline{z}$, since $z\overline{z}=1$. Thus $z\in G_B(A)\setminus G(A)$. This algebra is the boundary realization of the disk algebra.

Take complex-valued functions. Pointwise multiplication is associative and commutative, with identity the constant function $1$, whose norm is one. The product rule givesTo prove completeness, let $(f_n)$ be a Cauchy sequence in this norm. The functions and their derivatives converge uniformly to continuous functions $f,g$. Passing to the limit in $f_n(t)-f_n(0)={\int }_0^t{f}_n^{\prime }(s)ds$ gives $f(t)-f(0)={\int }_0^{t}g(s)ds$. The fundamental theorem of calculus yields $f^{\prime }=g$, including the one-sided endpoint derivatives. Thus $f\in C^1[0,1]$ and $f_n\to f$ in the stated norm. Consequently $A$ is a unital commutative Banach algebra.

Let $u(t)=t$. If $\lambda \notin [0,1]$, then $(u-\lambda 1{)}^{-1}(t)=1/(t-\lambda )$ belongs to $C^1[0,1]$. If $\lambda \in [0,1]$, the function $u-\lambda 1$ vanishes somewhere and cannot have a pointwise inverse. Thus ${\sigma }_A(u)=[0,1]$. Every character of an algebra satisfies $\varphi (u)=t_0$ for some $t_0\in [0,1]$, by part (d), and therefore $\varphi (p(u))=p(t_0)$ for every polynomial $p$.

Polynomials are dense in this stronger norm, not just the supremum norm. By the Weierstrass approximation theorem, approximate the real and imaginary parts of $f^{\prime }$ uniformly by polynomials, combining them into $q_n\to f^{\prime }$. Put $p_n(t)=f(0)+{\int }_0^t{q}_n(s)ds$. Then $p_n$ is a polynomial andContinuity of the algebra character now gives $\varphi (f)=\lim p_n(t_0)=f(t_0)$. Conversely each point evaluation is a continuous algebra character. Hence the character space of C1 on a compact interval consists exactly of the evaluation functionals, and part (c) givesDistinct points give distinct maximal ideals, since the coordinate function separates them.