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Set $r=k-1$, with $k\ge 2$ for the final linear-reproduction assertion, and use the normalized knot polynomial ${\psi }_i={\omega }_i/r!$. Differentiate the Marsden identity $r-j$ times with respect to the auxiliary variable $x$, for $0\le j\le r$. This givesThe Taylor polynomial expansion of an arbitrary polynomial $p$ of degree at most $r$ about $x$ is exact:Substitute the differentiated identity into each term and collect the B-spline coefficients. The result isThis is the normalized Marsden dual functional. There is no extra factorial outside the sum, because it is already included in ${\psi }_i$.

To prove independence of $x$ directly, differentiate the finite sum. The contribution from differentiating ${\psi }_i$ in term $j$ cancels the contribution from differentiating $p$ in term $j-1$. Explicitly,The two uncancelled endpoint terms vanish because $p^{(r+1)}={\psi }_i^{(r+1)}=0$. Thus the coefficient functional is constant in the auxiliary variable, as required.

The leading two terms of the knot polynomial giveFor a linear polynomial $p(t)=A+Bt$, only $j=0,1$ survive, whenceThe coefficient points $t_i^{\ast }$ are the Greville abscissae. Inserting these coefficients proves exact linear reproduction:All identities hold on the basic knot interval $[t_k,t_{n+1}]$, including the endpoints with the standard continuous endpoint convention. Repeated knots cause no difficulty in these polynomial differentiations.

Use the standard nonnegative partition-normalized B-splines $N_j$; their sum is at most one, and is exactly one for the complete clamped basis on $[0,1]$. The printed question introduces $M_i$ only in its last matrix formula without defining it. The intended normalization is the unit-integral normalization of a B-spline:For supports wholly inside the interval, $c_i=(t_{i+k}-t_i)/k$. The positivity and partition convention, sometimes called $L_{\infty }$ normalization, should not be confused with individually rescaling every spline to have maximum exactly one.

The characterization in part 1(a) gives the normal equations for the orthogonal projection:Dividing each equation by $c_i$ yieldsThis is the mixed-normalization spline Gram matrix. It need not be symmetric, but $G=D^{-1}H$ with $D=\mathrm{diag}(c_i)$ and $H_{ij}=(N_i,N_j{)}_{L_2}$. The ordinary Gram matrix $H$ is positive definite because the splines are linearly independent, so $G$ is invertible and $a=G^{-1}d$.

Since each $M_i$ is nonnegative and has integral one,The induced matrix operator norm is the maximum absolute row sum, soFinally, nonnegativity and the subpartition of unity for B-splines giveTaking the supremum over $t$ and then over $\Vert f{\Vert }_{\infty }\le 1$ proves the maximum-norm bound for spline projection:The formulas also establish linearity and boundedness of the projection on continuous inputs. Its range is exactly the finite-dimensional spline space $\mathcal{S}$, since it fixes every spline in that space. Thus, when the splines are continuous, it is an operator $C[0,1]\to C[0,1]$ with range $\mathcal{S}$; it is not onto all of $C[0,1]$ as the printed wording says. For example, $e^t$ cannot belong to a fixed finite-degree spline space because it is not a polynomial on any nonempty knot interval. The missing $M_i$ normalization and this range correction are needed for a precise reading of the result.

A practical family consists of constant-distance parallel surfaces and smooth variable normal offsetsConstants cover wall thickness, cutter radius and clearance. A scalar B-spline or piecewise polynomial displacement field permits thickness variation, local bumps and smooth blending; compact support makes local edits economical. Use the base parameterization or compatible data on its subdivision hierarchy so that evaluating the displacement and its derivatives requires only a small local stencil. Choose displacement size and gradients to avoid unintended folds, and trim global self-intersections where the physical application requires an envelope. A constant displacement smaller than the local curvature radii prevents the corresponding local focal singularities, although global clearance still needs a separate check.

When remaining within a polynomial or rational representation is more useful than prescribing an exact distance, a related practical form is $P+\lambda (u,v)(P_u\times P_v)$ with a low-degree scalar field $\lambda$. Its displacement is normal but has magnitude $|\lambda |\Vert P_u\times P_v\Vert$, so it is not a constant-distance parallel surface. For polynomial $P$ and $\lambda$, it stays polynomial; for rational input it stays rational. More general low-degree vector displacement fields similarly preserve efficient evaluation, while relinquishing the normal-offset condition. These forms make explicit the tradeoff between an exact metric offset, which involves normalization, and an easily represented deformation. Simple scalar normal-displacement fields are useful and directly evaluable; algebraic displacement fields are useful when representation closure is required.

Interpret polynomial reproduction as a guarantee for every polynomial of degree at most $d$, with the planar grid coordinates fixed by the affine-exact refinement. Constants are unchanged because every stencil sums to one. Linear height functions are unchanged because each stencil's weighted position is its refined grid location. Therefore the Loop subdivision limit reproduces all affine polynomials exactly.

Quadratic data already disprove the next degree. Use extruded data $g_j=j^2$ on a unit-spaced transverse row coordinate $y$. At spacing $h$, both even and odd rules from the induced subdivision mask give the value of $y^2$ at the refined location plus $h^2/4$. For the even stencil this follows fromand for the odd stencil from the average of the values at $y-h/2$ and $y+h/2$. Constants are reproduced, so the accumulated vertical displacement over all refinement levels isThe limit is $y^2+1/3$, not the original quadratic graph. HenceThe word “every” matters: some special higher-degree bivariate polynomials have cancelling second moments and can be reproduced. The conclusion concerns the entire polynomial space, which already fails at degree two.

Let the three grid-edge vectors be $v_1=a$, $v_2=b$, $v_3=b-a$, and define the constant-coefficient differential operator $D={\sum }_{r=1}^3(v_r\cdot \nabla {)}^2$. For a stencil at grid spacing $h$, its displacement vectors $\delta$ about the new vertex have vanishing first and third moments by central symmetry. All four stencil types have the same second momentFor the old-vertex stencil, each pair $\pm h{v}_r$ contributes $h^2{v}_r{v}_r^T/8$. For an edge parallel to $a$, the endpoint displacements are $\pm ha/2$ with weights $3/8$, and the opposite displacements are $\pm h(b-a/2)$ with weights $1/8$. Their moment is $h^2[a{a}^T/4+b{b}^T/4-(a{b}^T+b{a}^T)/8]$, precisely the same expression; the other edge orientations follow by symmetry.

If $f$ is a multivariate polynomial of total degree at most three, its Taylor expansion is exact through the third-order term. Thus every refined vertex samples the single polynomialThe polynomial $Df$ has degree at most one, so subsequent refinement reproduces it exactly. Starting at unit spacing and successively halving $h$, the limit therefore isThe correction lowers degree by two, leaving the leading homogeneous term unchanged. Hence every sampled polynomial of degree at most three generates a polynomial of that same degree, even though quadratics and cubics need not be reproduced identically. For the extruded cases this gives $y^2\mapsto y^2+1/3$ and $y^3\mapsto y^3+y$.

Degree four is not guaranteed. Extruded data $g_j=j^4$ reduce to the Cardinal cubic B-spline $G(y)={\sum }_j{j}^4{M}_3(y-j)$, which is piecewise cubic and hence cannot be a single polynomial of degree four. It is not a lower-degree polynomial either: at integer rows, $G(k)=[(k-1{)}^4+4k^4+(k+1{)}^4]/6=k^4+2k^2+1/3$, which cannot agree with a cubic at all integers. This counterexample excludes every guaranteed degree $d\ge 4$. ConsequentlyThis distinguishes polynomial generation from exact reproduction and again refers to all polynomials through the stated degree.

The limiting polynomial has a double root at $x=-1$, so two roots require a Puiseux series. For positive $ϵ$, put $\delta =\sqrt{ϵ}$ and $x=-1+a\delta +b{\delta }^2+\cdots$. Substitution and dominant balance giveThus $a=\pm 1$ and $b=-3/2$. The two finite roots have the three-term asymptotic expansionsThis is square-root splitting of a double polynomial root. For negative $ϵ$, the same formulas apply with a chosen complex square root and describe a complex conjugate pair.

The third root is a divergent algebraic root in a singular perturbation. Set $x=X/ϵ$: the equation for $X$ is $X^3+X^2+2ϵX+{ϵ}^2=0$. Its nonzero limiting root is $X=-1$. Substituting $X=-1+Aϵ+B{ϵ}^2+\cdots$ yields $A=2$ and $B=3$, henceAs an independent check, Vieta formulas give the sum of roots as $-1/ϵ$: the two finite-root expansions sum to $-2-3ϵ+O({ϵ}^2)$, giving the same large-root expansion.

Over a finite field ${\mathbb{F}}_q$, the splitting field is a finite extension ${\mathbb{F}}_{q^m}$, where $m$ is the least common multiple of the degrees of the distinct irreducible factors. Its Galois group is cyclic of order $m$, generated by $x\mapsto x^q$.

Over ${\mathbb{F}}_2$, the polynomial has no linear factor: its values at zero and one are both one. The only irreducible quadratic is $x^2+x+1$; reduction modulo it gives $x$, not zero. A reducible quartic without linear factors would split into quadratics, so $f$ is irreducible. Thus its splitting field is ${\mathbb{F}}_{16}$ and

The nonzero leading coefficients and distinct degrees make $P_0,\dots ,P_{n-1}$ a basis of the polynomials of degree at most $n-1$. Consequently the given orthogonality implies ${\int }_a^{b}f(x)Q(x)dx=0$ for every such polynomial $Q$.

If $f$ has infinitely many distinct zeros in $(a,b)$, there is nothing to prove. Otherwise suppose it has fewer than $n$. Among those zeros, list the sign-changing ones as $t_1<\cdots <t_m$, with $m<n$, and put $Q(x)={\prod }_{j=1}^m(x-t_j)$, using $Q=1$ when $m=0$. On each interval between zeros, $f$ has a fixed sign by continuity. Crossing a sign-changing zero flips both $f$ and $Q$; crossing any other zero flips neither sign. Thus $fQ$ has one constant sign wherever nonzero. Since $f$ is not identically zero, continuity gives an open interval on which $fQ$ is strictly of that sign. Hence its integral is nonzero, contrary to orthogonality to $Q$. Therefore $f$ has at least $n$ distinct zeros in $(a,b)$. This is the reusable orthogonality forces many interior zeros argument.

The polynomial $T^4-2$ is irreducible by Eisenstein criterion at two, so $1,\alpha ,{\alpha }^2,{\alpha }^3$ is a rational basis and each of these elements is integral. The nontrivial automorphism of $K/k$ sends $\alpha$ to $-\alpha$. For an algebraic integer $\theta$, its relative trace and the trace of the algebraic integer $\alpha \theta$ areBoth are in ${\mathcal{O}}_k$, proving $2a,2b,2c,4d\in \mathbb{Z}$. Multiplying $\theta$ by its conjugate gives the field normAgain this is in ${\mathcal{O}}_k$, so both displayed coefficients are integers.

To finish the denominator argument, put $A=2a$, $B=2b$, $C=2c$, $D=4d$, all integers. The two norm coefficients implyThe second congruence modulo two first gives $D$ even. Modulo four it then gives $B$ even. Therefore $2BD$ vanishes modulo four, and the first congruence reads $A^2+2C^2\equiv 0(\mathrm{mod}4)$. The possible squares modulo four force $A$ even and then $C$ even. Returning to the second congruence now gives $D^2\equiv 0(\mathrm{mod}8)$, so $D$ is divisible by four. Thus $a,b,c,d$ are all integers.

Conversely, every integer linear combination of these powers is integral. We have proved the integral basis of the fourth-root-of-two field:

The polynomial Runge theorem states that, for a compact set $K\subset \mathbb{C}$ with connected complement and a function $f$ holomorphic on a neighborhood of $K$, for every $\epsilon >0$ there is a polynomial $P$ such thatThe connected-complement hypothesis permits polynomial approximation rather than approximation by rational functions with poles in bounded complementary components.

A binary cyclic code is a linear subspace of ${\mathbb{F}}_2^N$ invariant under cyclic coordinate shift. Associate a word $(c_0,\dots ,c_{N-1})$ to $\sum c_j{X}^j$ in ${\mathbb{F}}_2[X]/(X^N-1)$. Cyclic shift is multiplication by $X$, so linearity and shift invariance make the code an ideal of this quotient.

Its inverse image in ${\mathbb{F}}_2[X]$ is an ideal containing $(X^N-1)$. Since a polynomial ring over a field is a principal ideal domain, that inverse image is $(g)$ for a unique monic polynomial $g$. Containment gives $X^N-1=gh$ for some polynomial $h$. Thus the generator polynomial of a cyclic code satisfiesFor a nonzero code $g$ is also its monic word polynomial of smallest degree. For the zero code use $g=X^N-1$, whose residue is zero. Conversely every monic divisor defines a cyclic code, proving the full correspondence.

Over ${\mathbb{F}}_2$,The quartic factor has no root in ${\mathbb{F}}_2$. The only monic irreducible quadratic over this field is $X^2+X+1$, and division leaves remainder $X+1$, so the quartic is not divisible by it. A reducible quartic with no linear factor would have two irreducible quadratic factors; hence this quartic is irreducible. There are exactly four binary cyclic codes of length five, given by the generator polynomials of a cyclic codeThey are respectively the full space, the even-weight code, the repetition code and the zero code, with dimensions $5,4,1,0$.

If $A$ is nonempty, take $a\in A$. The stated implication repeatedly gives $a+jt\in A$ for $j=0,1,\dots ,p-1$. These residues are all distinct: equality of two would give $(j-\ell )t=0$ in the prime field ${\mathbb{Z}}_p$, and the nonzero $t$ can be cancelled. They therefore exhaust the field. ThusEquivalently, a nonzero element generates the additive cyclic group of prime order.

Now assume $0<k<p$. If $A+s=A+t$ with $s\ne t$, translating by $-s$ gives $A=A+(t-s)$, contradicting the preceding result. Hence all $p$ translates are distinct. Translation defines equivalence classes on the collection of all $k$-element subsets: two subsets are equivalent if one is a translate of the other, and each class contains exactly $p$ subsets. Since there are $\left(\genfrac{}{}{0ex}{}{p}{k}\right)$ such subsets, their partition into classes givesThis is the free translation action on nontrivial subsets of a prime cyclic group.

The binomial theorem now gives, in the prime field,Starting with $0^p=0$ and applying this identity successively to the representatives $0,1,\dots ,p-1$ proves $a^p=a$ for every residue $a$. For nonzero $a$, cancellation giveswhich is Fermat's little theorem, or $a^p\equiv a(\mathrm{mod}p)$ for every integer $a$ and $a^{p-1}\equiv 1$ when $p\nmid a$.

The same vanishing of intermediate binomial coefficients proves $(R+S{)}^p=R^p+S^p$ for any commuting elements in a ring whose ring characteristic is $p$, in particular for polynomials over ${\mathbb{Z}}_p$. Applying this repeatedly to the terms of $Q$, and applying Fermat's little theorem to each coefficient, gives the Frobenius endomorphism identityThis is an equality of formal polynomials. It does not assert $x^p=x$ as a polynomial; that latter equality holds only after evaluating at elements of ${\mathbb{Z}}_p$.

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.