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Let $f$ be a real function with $\Vert f{\Vert }_w^2={\int }_a^{b}f(x{)}^{2}w(x)dx<\infty$, and let $V_m$ be the vector space of real polynomials of degree at most $m$. Its orthogonal projection is the least-squares polynomial in an orthogonal-polynomial basisEach coefficient depends on just one weighted inner product with $f$ and one squared norm. In particular, the coefficients $c_0,\dots ,c_m$ stay the same when the allowed polynomial degree increases. This statement assumes a fixed weight, interval and normalization of the orthogonal polynomials.

To see why the least-squares normal equations decouple, first write a trial polynomial as $q={\sum }_{j=0}^m{a}_j{\varphi }_j$ in any basis of $V_m$. The residual must be orthogonal to each permitted direction, givingThe Gram matrix usually couples the unknown coefficients. With ${\varphi }_j=p_j$, it becomes $G_{ij}=h_i{\delta }_{ij}$, leaving the separate equations $h_i{a}_i=b_i$. For the monomial basis on $[-1,1]$ with $w=1$, by contrast, $G_{ij}$ is zero when $i+j$ is odd and $2/(i+j+1)$ when $i+j$ is even. Distinct even powers, or distinct odd powers, remain coupled.

There is also a proof of the minimum that uses no differentiation. The defining coefficients give $⟨f-q_m,p_n{⟩}_w=0$ for $0\le n\le m$. Thus $f-q_m$ is orthogonal to all of $V_m$. For every $v\in V_m$, the Pythagorean theorem in an inner-product space givesThe second term is nonnegative and vanishes exactly when $v=q_m$, proving that the least-squares solution is unique. This is an integrated squared-error optimum; it does not assert a smallest error at every point or a smallest uniform norm.

For $N>m$, the nested orthogonal projections satisfy the exact residual budgetIndeed, $q_N-q_m$ is a linear combination of mutually orthogonal polynomials, and $f-q_N$ is orthogonal to it. Every newly retained component removes exactly $c_n^2{h}_n$ from the squared error. No infinite expansion or completeness assumption is needed for this finite identity. The Legendre least-squares projections of a cubic make every term explicit.

An orthogonal polynomial family chooses polynomials that do not overlap under a specified inner product. For real polynomials on an interval $(a,b)$, use the weighted inner productAssume $a<b$, the weight $w$ is measurable and positive almost everywhere, and its absolute moments ${\int }_a^b|x{|}^{k}w(x)dx$ are finite for every integer $k\ge 0$. These conditions make the integrals finite and give every nonzero polynomial positive squared norm. A sequence $p_0,p_1,\dots$ is an orthogonal polynomial family when $p_n$ has real coefficients and polynomial degree exactly $n$, and $⟨p_n,p_k{⟩}_w=0$ for $n\ne k$.^[ref]

Here $h_n=\Vert p_n{\Vert }_w^2>0$ and ${\delta }_{nk}$ is the Kronecker delta. Orthogonality does not require $h_n=1$: an orthonormal set is obtained only after replacing $p_n$ by $p_n/\sqrt{h_n}$. Multiplying each member by any nonzero real constant preserves orthogonality, but changes its norm and the corresponding expansion coefficient. The monic orthogonal polynomial convention instead fixes each leading coefficient to one.

The first four Legendre polynomials, named for Adrien-Marie Legendre, use $w=1$ on $[-1,1]$ and the normalization $P_n(1)=1$:Their orthogonality and squared norms areThus their Gram matrix has a positive diagonal and zero entries elsewhere, although their function graphs cross and overlap on the page. Orthogonality measures an integral of products, rather than separation of function graphs.^[ref]

For these four polynomials, a short verification expands each product and uses ${\int }_{-1}^1{x}^{2j+1}dx=0$ and ${\int }_{-1}^1{x}^{2j}dx=2/(2j+1)$. For example, $⟨P_0,P_2⟩=(3(2/3)-2)/2=0$ and $⟨P_1,P_3⟩=(5(2/5)-3(2/3))/2=0$. The same calculation gives the diagonal entries $2,2/3,2/5,2/7$. More generally, apply the Gram-Schmidt process to the monomial basis $1,x,x^2,\dots$: at step $n$, subtract projections onto the earlier polynomials. The result retains its leading term, has degree $n$, and has positive norm, proving existence without a differential equation. Different degrees also give linear independence, so $p_0,\dots ,p_m$ form a basis of the polynomials of degree at most $m$.

This basis turns least-squares solutions for functions into independent scalar calculations, as the decoupled coefficients of orthogonal polynomial projections show. In a spectral method, increasing the degree enlarges the approximation space while keeping earlier projection coefficients fixed. Weighted integrals also connect the same orthogonal polynomials to Gaussian quadrature: its nodes are polynomial roots of a member of the family. The projection problem minimizes an integrated squared error; Gaussian quadrature uses the family to choose evaluation points for an integral.

Use the quantum plane $A=k⟨x,y⟩/(yx-qxy)$ and the quantum enveloping algebra of sl2 comultiplication from Question 2. The generator actions are $Kx=qx$, $Ky=q^{-1}y$, $Ex=0$, $Ey=x$, $Fx=y$, $Fy=0$. Extend them by the module algebra rulesThese descend to $A$: $E(yx)=q{x}^2=E(qxy)$ and $F(yx)=q{y}^2=F(qxy)$, while $K$ preserves $yx-qxy$. For the ordered monomial basis, induction with $[s+1]=q^{-1}[s]+q^s$ gives the quantum enveloping algebra action on the quantum planeTerms with a negative exponent have zero coefficients and are interpreted as zero.

We now verify the defining relations explicitly on every basis vector, rather than infer them merely from the generator actions. The $K$ weight changes by $q^2$ under $E$ and $q^{-2}$ under $F$, so $KE=q^{2}EK$ and $KF=q^{-2}FK$. AlsoThe middle identity follows by inserting $[j]=(q^j-q^{-j})/(q-q^{-1})$ and cancelling the terms $q^{r+s+1}$ and $q^{-r-s-1}$; the remaining numerator is $(q-q^{-1})(q^{r-s}-q^{s-r})$. Finally $K{K}^{-1}=1$. Thus all the rank-one defining relations hold. The phrase “Serre relations” here refers to these rank-one presentation relations; there are no additional relations involving distinct simple roots for ${\mathfrak{sl}}_2$.

The general modular form, the nonuniform Frankl-Wilson theorem, can be stated as follows. Let $p$ be a prime number, and let $L\subseteq {\mathbb{F}}_p$ have $s$ elements. Suppose $\mathcal{F}\subseteq \mathcal{P}([n])$ has $|A|\mathrm{mod}p\notin L$ for every member, while $|A\cap B|\mathrm{mod}p\in L$ for any distinct members. ThenThe uniform Frankl-Wilson theorem sharpens this to $|\mathcal{F}|\le \left(\genfrac{}{}{0ex}{}{n}{s}\right)$ when all members have size $k$ and $0\le s\le \min \{k,n-k\}$. Both forms require a prime modulus and exclusion of the self-intersection residue.

For the general proof, work over the finite field ${\mathbb{F}}_p$. Associate to each member $A$ the intersection polynomialAt the characteristic vectors of sets $1_B$, this is zero if $B\ne A$, and nonzero if $B=A$. Thus these polynomials, regarded as functions on the Boolean hypercube, are linearly independent: evaluating any relation at $1_A$ isolates its coefficient. Apply multilinear reduction on the Boolean cube, replacing every positive power of a variable by that variable. Values on the Boolean hypercube remain unchanged, and the resulting multilinear polynomials have degree at most $s$. Their ambient space has a monomial basis consisting of $x_S={\prod }_{i\in S}{x}_i$ for $|S|\le s$, with dimension ${\sum }_{j=0}^{\min (s,n)}\left(\genfrac{}{}{0ex}{}{n}{j}\right)$. This proves the general bound, including $s=0$.

For completeness, obtain the uniform sharpening without dividing by factorials in a finite field. For $s\ge 1$ put $a=k\mathrm{mod}p$ and $g(x)={\sum }_i{x}_i-a$. The functions $x_{S}g$ with $|S|<s$ are linearly independent. Indeed, a relation gives $gh=0$ with $\mathrm{deg}h<s$, so $h$ is supported only on weights in $H=\{j\in [0,n]:j\equiv a(\mathrm{mod}p)\}$. With boundary weights $-1,n+1$, this set has a gap of at least $s+1$: either consecutive allowed weights differ by $p\ge s+1$, or its terminal gap does, since $n\ge k+s\ge a+s$. The alternating sum of $h$ over any interval of $s$ free coordinates vanishes by its degree. Across that gap only one endpoint level can contribute, forcing $h$ to vanish there. Delete that level and repeat across the enlarged gap until $H$ is empty. Hence $h=0$, and independence of the square-free monomials gives the claim. Adjoin these functions to the $f_A$. Evaluation at each family vector eliminates the coefficients of $f_A$, since $g$ vanishes there; the claim eliminates all remaining coefficients. Counting dimensions yieldsThe case $s=0$ permits at most one member directly. The gap argument is an instance of the modular layer vanishing lemma.

For the final application, enumerate the distinct sets as $A_1,\dots ,A_m$ and let $v_i=1_{A_i}\in {\mathbb{R}}^n$ be their characteristic vectors of sets. If $m\le 1$, the desired bound is immediate for $n\ge 1$. Otherwise $|A_i|\ge k$, because $A_i$ intersects another member in $k$ points. At most one member can have size $k$: two distinct $k$-sets cannot have intersection size $k$. The Gram matrix of these vectors isFor real coefficients $c_i$,All terms are nonnegative, with $k>0$. If the expression vanishes, every coefficient corresponding to a set of size greater than $k$ is zero. At most one coefficient remains, and the first term forces it to be zero as well. Thus the characteristic vectors of sets are linearly independent in ${\mathbb{R}}^n$, proving the constant-intersection family boundThis proof explicitly covers the possible member of size exactly $k$; assuming all diagonal corrections were strictly positive would miss that case. Positivity of $k$ is essential: when $k=0$, the empty set and all $n$ singleton sets give $n+1$ members.

Give $V_n$ the action $(g\cdot P)(x,y)=P((x,y)g)$, where $(x,y)$ is a row vector. Substitution preserves homogeneous degree, and composition agrees with matrix multiplication, so this is a complex representation of SU(2) group. The diagonal torus $\mathrm{diag}(z,z^{-1})$ acts on $x^{n-j}{y}^j$ with weight $z^{n-2j}$; these weights are distinct.

A nonzero invariant subspace is invariant under the torus, so finite-dimensional Fourier projection onto its weight spaces supplies a nonzero monomial. Differentiating the group action and complexifying supplies the operators $E=x{\partial }_y$ and $F=y{\partial }_x$. Their repeated application moves between every successive monomial, with nonzero coefficients until an endpoint is reached:The invariant subspace therefore contains the whole monomial basis. ThusFor $n=0$ this is the one-dimensional trivial representation. Torus weight projections are legitimate because an invariant complex subspace is closed in finite dimension.

Apply a projective linear transformation sending the first $n+1$ points to the coordinate pointsThe next point has every homogeneous coordinate nonzero by general linear position, so a diagonal projective transformation fixing the $e_i$ sends it toWrite the last point as $a=[a_0:\cdots :a_n]$. Again general linear position implies that every $a_i$ is nonzero. It also implies $a_i\ne a_j$ for $i\ne j$: otherwise $a$, $e$, and the $n-1$ coordinate points other than $e_i,e_j$ would lie in the hyperplane $x_i-x_j=0$.

Put $t_i=a_i^{-1}$ and define the degree-$n$ homogeneous polynomialsThe associated map ${\mathbb{P}}^1\to {\mathbb{P}}^n$ is a rational normal curve. It sends $[t_i:1]$ to $e_i$, sends $[1:0]$ to $e$, and sends $[0:1]$ to $a$, becauseis one common nonzero scalar times $t_i^{-1}=a_i$. This proves existence.

For uniqueness, parametrize any rational normal curve through these points and choose a coordinate $[S:T]$ on its source ${\mathbb{P}}^1$ so that the preimages of $e$ and $a$ are respectively $[1:0]$ and $[0:1]$. If the preimage of $e_i$ is $[s_i:1]$, then its $i$th coordinate polynomial must have all the other $s_j$ as roots. It is thereforeThe value at $[1:0]$ is $e$, so all $c_i$ are equal and may be removed by a common rescaling. The value at $[0:1]$ is $a$, so $s_i^{-1}$ is proportional to $a_i$. Thus all $s_i$ are obtained from the $t_i$ by one common rescaling of the coordinate on ${\mathbb{P}}^1$. This merely reparametrizes the curve, proving uniqueness.

For $n=3$, choose the monomial basis $S^3,S^{2}T,S{T}^2,T^3$. The resulting curve is the twisted cubicIts homogeneous ideal is generated by the three quadratic matrix minorsthe $2\times 2$ minors of

A vector is an element of a vector space. Its defining operations are vector addition and scalar multiplication. An arrow in a Euclidean space is one example; a polynomial or a function can also be a vector when the chosen vector space uses the corresponding addition and scalar multiplication.

For instance, the polynomials $p(x)=1+x$ and $q(x)=x-x^2$ belong to the real vector space of polynomials of degree at most two. Their vector addition gives $(p+q)(x)=1+2x-x^2$, and scalar multiplication gives $(2p)(x)=2+2x$. In the monomial basis $(1,x,x^2)$ these are the coordinate vectors $(1,1,0)$, $(0,1,-1)$, $(1,2,-1)$ and $(2,2,0)$ respectively. The vectors are the polynomials; these tuples are their representations in a chosen basis.

A general vector space does not itself prescribe lengths or angles. A norm supplies a size, while an inner product also supplies angle information and orthogonality. The familiar arrow geometry below belongs to Euclidean vectors, where this extra structure is available.