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The $k$ order-$k$ B-splines active on a simple-knot cell form a basis of its degree-$k-1$ polynomials. To see this from Marsden's identity, use the dual polynomials ${\psi }_i(y)={\prod }_{\ell =1}^{k-1}(y-t_{i+\ell })$. On $t_j<x<t_{j+1}$ the identity is $(y-x{)}^{k-1}={\sum }_{i=j-k+1}^j{N}_i(x){\psi }_i(y)$. Evaluation of the dual polynomials at $t_j,t_{j-1},\dots ,t_{j-k+1}$ gives a triangular coefficient matrix with nonzero diagonal. They are independent, and comparison of coefficients of $y$ in the identity gives independence of the active splines. Consequently a spline vanishes on a whole open cell exactly when all its active coefficients vanish.

Matrix similarity invariants are quantities unchanged by replacing $A$ with $S^{-1}AS$. Important examples are the characteristic polynomial, trace, determinant, matrix rank, and eigenvalues with their algebraic multiplicities. The characteristic polynomial convention here is ${\chi }_A(t)=\det (tI-A)$, where $I$ is the identity matrix.^[ref]

The short determinant proof works over any field: because $tI-B=S^{-1}(tI-A)S$, multiplicativity of the determinant givesThus its roots, the eigenvalues, retain their algebraic multiplicities in any field extension where the polynomial splits. Also $\det B=\det A$. The cyclic property of the trace gives $\mathrm{tr}(S^{-1}AS)=\mathrm{tr}(AS{S}^{-1})=\mathrm{tr}A$; equivalently, the trace is determined by the coefficient of $t^{n-1}$ in the characteristic polynomial for an $n$-by-$n$ matrix.^[ref]

For the two matrices in the opening example,Their entries differ, yet these quantities agree. Agreement is a useful necessary test for matrix similarity; it is not sufficient in general, as same eigenvalues do not imply similarity illustrates.

The relation also preserves repeated application. Adjacent factors $S{S}^{-1}$ cancel to giveFor the opening example, matrix powers therefore have the explicit formAt $k=2$, this is $\left(\begin{array}{cc}4& -21\\ 0& 25\end{array}\right)$. Acting on $y=(1,1{)}^T$ gives $(-17,25{)}^T$, which reconstructs as $S(-17,25{)}^T=(8,25{)}^T=A^{2}x$. More generally $q(B)=S^{-1}q(A)S$ for every polynomial $q$, so the minimal polynomial is also unchanged. Multiplication by invertible matrices preserves matrix rank, giving another invariant without finding eigenvalues.

Modified Gram-Schmidt starts with $w=v_j$ and, for each retained orthonormal vector $q_i$ in order, replaces $w$ by $w-⟨w,q_i⟩q_i$. Each orthogonal projection coefficient uses the current remainder. In exact arithmetic, subtracting earlier orthogonal directions does not change the inner product with $q_i$, so the final $w$ equals the Classical Gram-Schmidt remainder $u_j$; then divide by $\Vert w\Vert$ if it is nonzero.

The Newton polynomial expresses the interpolation polynomial in products that successively vanish at all previously used interpolation nodes. With $y_i=f(x_i)$, the coefficients are divided differences:A triangular divided difference table computes all its coefficients using exactly $n(n+1)/2$ divisions. The Newton polynomial and the Lagrange polynomial have the same polynomial degree bound and the same nodal values, hence are equal by uniqueness of polynomial interpolation.

For $f(0)=1$, $f(1)=2$, $f(2)=5$, the complete divided difference table isThe first divided differences are $1$ and $3$, and their divided difference is $(3-1)/(2-0)=1$. Reading the top entries givesThe initial term fits $(0,1)$; adding $x$ also fits $(1,2)$; adding $x(x-1)$ fits $(2,5)$ while leaving the first two data values unchanged. This is the practical beauty of the Newton polynomial: each new correction respects all previous interpolation constraints.

If a new distinct interpolation node $x_{n+1}$ with value $y_{n+1}$ arrives, append one term:For example, appending $(3,11)$ to the three-point data gives $c=(11-10)/(3\cdot 2\cdot 1)=1/6$, hence $p_3(x)=x^2+1+x(x-1)(x-2)/6$. Changing the order of the interpolation nodes changes the Newton polynomial coefficients and products, but not the final interpolation polynomial in exact arithmetic.

The products also permit nested evaluation analogous to Horner's method. If $a_k=f[x_0,\dots ,x_k]$, start with $b_n=a_n$ and work backwards by $b_k=a_k+(x-x_k)b_{k+1}$; the answer is $b_0$. The three-point example becomes $1+x[1+(x-1)]$, needing two multiplications for each evaluation. This organization is useful when adding observations incrementally or evaluating a stored Newton polynomial many times; the Runge phenomenon still depends on the interpolation nodes rather than on which representation is used.

To justify the update formula, its added product vanishes at every previous interpolation node. Choosing $c$ by the new residual forces agreement at $x_{n+1}$, so uniqueness of polynomial interpolation identifies the result. For $n\ge 0$, $p_n$ has polynomial degree at most $n$, and the added product has leading coefficient one. Thus the new coefficient $c$ is exactly $f[x_0,\dots ,x_{n+1}]$. The initial state is $p_0(x)=f(x_0)$, with the empty product interpreted as one.^[ref]

Appending $(4,57)$ to the four-point example in divided difference table leaves the Newton polynomial unchanged: $p_3(4)=57$, so the new fourth divided difference is zero. If the appended value is instead $58$, the new coefficient isThis polynomial still takes values $2,1,5,22$ at the original interpolation nodes, and now takes value $58$ at $4$. A zero update is a valid completed step: adding another exact sample of the original polynomial need not increase the polynomial degree.

Newton's identities let two different summaries of the same polynomial roots talk to each other. Elementary symmetric polynomials record products of distinct root positions; power sums record sums of individual powers, $p_r={\sum }_{i=1}^n{x}_i^r$. Either can be easier to obtain in a particular problem. The identities turn one into the other, often avoiding any need to find the individual polynomial roots.

Write $e_0=1$ and $e_m=0$ for $m>n$. For every positive integer $m$, the identity isIt is a polynomial identity over the integers, so it remains true on substitution in any commutative ring. To solve it for $e_m$ by division, however, $m$ must have a multiplicative inverse. Over a field of characteristic zero, the first $n$ power sums recover every elementary symmetric polynomial. The same recovery works when the positive characteristic of a field is greater than $n$; small characteristic needs care.^[ref]

The first three relations are $e_1=p_1$, $2e_2=e_1{p}_1-p_2$, and $3e_3=e_2{p}_1-e_1{p}_2+p_3$. When division by two and three is permitted, this becomesFor the three roots $1,2,3$, their power sums are $p_1=6$, $p_2=14$, and $p_3=36$. The formulas give $e_2=(36-14)/2=11$ and $e_3=(216-252+72)/6=6$, recovering the polynomial $z^3-6z^2+11z-6$ from aggregate data.

The beauty is that the summaries do not depend on the order of the polynomial roots. In symmetric polynomials, one can work with quantities that survive every relabeling; in linear algebra, traces of matrix powers give the same kind of aggregate information about eigenvalues. The identities convert those summaries into coefficients of a characteristic polynomial.

A short derivation packages the identities into a formal power series. Let $E(t)={\prod }_i(1+x_{i}t)={\sum }_{m\ge 0}{e}_m{t}^m$. Since its constant coefficient is one, it has an inverse as a formal power series, and its formal derivative satisfiesMultiply by $E(t)$ and compare the coefficient of $t^{m-1}$. This gives Newton's identities without a convergence assumption or division by $m$. The division issue arises only when using them to recover $e_m$. The logarithmic-derivative approach is also developed in Newton's Identities (January 2019 notes) by Milan Mossé (§4, pp. 6–8).

The alternate name Newton-Girard identities recalls Albert Girard's earlier work. A modern historical account credits formulas through sixth powers to Albert Girard in 1629 and discusses their later appearance in Isaac Newton's Arithmetica universalis.^[ref] This citation refers to that account; the original historical editions have not been inspected here.

A nonlinear ordinary differential equation depends nonlinearly on the unknown function or its derivatives: it cannot be written in the form defining a linear differential equation. For example, $y^{\prime }=y^2$ contains a square of the unknown function, and $y^{\prime }=y(1-y)$ is a logistic differential equation. A coefficient such as $t^2$ depending only on the independent variable does not make an ordinary differential equation nonlinear. This distinction concerns dependence on the unknown, rather than the shape of its function graph.

For a concrete initial value problem, consider $y^{\prime }=y^2$, $y(0)=y_0>0$. Its solution of a differential equation isTaking its derivative gives $y^{\prime }=y_0^2/(1-y_{0}t{)}^2=y^2$, and evaluating at zero checks the initial condition. The stated interval contains zero and ends at finite-time blowup: $y$ tends to positive infinity as $t$ approaches $1/y_0$ from below. Increasing the positive initial condition brings finite-time blowup closer. This is a simple mathematical example with dimensionless $t$ and $y$, rather than a calibrated physical model.

Choose $y_0=1$ and $y_0=2$, giving $y_1=1/(1-t)$ and $y_2=2/(1-2t)$. At $t=1/4$ they have values $4/3$ and $4$, so their sum $s=y_1+y_2$ has value $16/3$. On the common interval $0\le t\le 0.4$, both are valid solutions of a differential equation, but their sum fails the same ordinary differential equation:Already at $t=0$, $s=3$ and its derivative is $1+4=5$, whereas $y^{\prime }=y^2$ would require $9$. At $t=1/4$, the defect $s^{\prime }-s^2$ is $-32/3$. The cross term explains the failure of the superposition principle, even though each individual solution of a differential equation is a smooth function.

The first panel shows $y_1$, $y_2$ and their invalid sum only up to $t=0.4$, before either finite-time blowup. The second measures the defect in the ordinary differential equation, including the computed value at $t=1/4$. For a homogeneous linear differential equation, the corresponding defect of a sum is zero. Nonlinear ordinary differential equations can describe state-dependent growth such as the logistic differential equation, whose negative quadratic term limits growth instead of accelerating it without bound. Recognizing the distinction tells us when we may assemble solutions of a differential equation from simpler pieces and when we must account for their interaction. See MIT's worked counterexample to nonlinear superposition.

The equality holds when an alternating unit-norm spline $s_{\ast }$ has as many ordered extremal sites as the dimension, and those sites satisfy the Schoenberg–Whitney theorem. For every admissible set, sampling has operator norm at most one, giving $\kappa (\mathcal{S})\le \Vert A_{\mathbf{x}}^{-1}{\Vert }_{\infty }$. At the alternating sites the checkerboard inverse of a totally nonnegative matrix identifies each inverse absolute row sum with the magnitude of the corresponding coefficient of $s_{\ast }$. The Chebyshev spline coefficient and dual norm equality therefore gives equality. This minimizes amplification into B-spline coefficients; it does not by itself prove optimality for the distinct Lebesgue constant of interpolation.

Optimal truncation chooses how many terms to retain for a useful finite approximation. In a divergent asymptotic expansion, increasing the truncation order eventually makes the terms larger rather than smaller. Stopping near a smallest term often gives the best direct approximation, but the actual remainder must be checked. A small last term alone is not a universal error bound.^[ref]

For a divergent factorial series can approximate a convergent integral, $T_k=k!/z^k$ decreases while $k+1<z$ and increases while $k+1>z$. Thus the useful cutoff is near $k=z$, rather than at arbitrarily large order. At $z=10$, $T_9=T_{10}=0.00036288$ are the two smallest terms. Retaining $N=10$ terms givesThe actual error is about $0.0001770$. Retaining twenty terms instead gives $S_{20}(10)\approx 0.9076137084$, with error about $0.008020$: more work produces a worse answer.

For this example the bound $N!/z^N$ is explicit, so a requested tolerance can be tested before using a truncation. For a different integral, a nearby saddle point or simple pole can change the relevant remainder scale. This is why the smallest-term rule is a useful guide, rather than a replacement for remainder analysis.

The cutoff for factorial terms $n!z^{-n}$ has $n=|z|+O(1)$. Better analysis can expose contributions smaller than every retained inverse power, linking optimal truncation to the Stokes phenomenon. Such contributions can be important even when many asymptotic coefficients agree. Directly summing ever more terms does not automatically recover them.

For scalar coefficients, leading symbols commute. If $P=a{\partial }^p+\cdots$ and $Q=b{\partial }^q+\cdots$, their product has leading coefficient $ab$, and their commutator has coefficient $pa{b}^{\prime }-qb{a}^{\prime }$ at order $p+q-1$. Thus product order is at most $p+q$, with equality when $ab$ is not identically zero; commutator order is at most $p+q-1$, with equality when its displayed coefficient is not identically zero. Constant-coefficient operators commute. Nonzero smooth functions with disjoint supports show why product orders need not add over $C^{\infty }(\mathbb{R})$.

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.

The Young diagrams of partitions in a rectangle with at most $k$ rows and at most $n-k$ cells in each row correspond to the lattice paths in the lattice-path interpretation of Gaussian binomial coefficients. Read the shaded row lengths from top to bottom; the result is a nonincreasing list of nonnegative integers. This bijection preserves the number $|\lambda |$ of cells.

In a two-by-two rectangle, the possibilities are the empty integer partition, $(1)$, $(2)$, $(1,1)$, $(2,1)$, and $(2,2)$. Their sizes are $0,1,2,2,3,4$, giving $G_{4,2}(t)$. In particular, there are two allowed diagrams of size two, explaining the middle coefficient.^[ref]

Row $n$ of Pascal's triangle consists of the binomial coefficients $\left(\genfrac{}{}{0ex}{}{n}{0}\right),\dots ,\left(\genfrac{}{}{0ex}{}{n}{n}\right)$, beginning with row $0$. Boundary entries are one, and Pascal's identity says each interior entry is the sum of the two entries above it. The hockey-stick identity evaluates sums along its diagonals.

The highlighted entry in the binomial coefficient illustration is $\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$. Split two-person teams according to whether they contain the fifth person: six choose both members from the other four, and four contain the fifth person together with one of those four. Thus $10=6+4$. All entries can be generated by the same local rule, so a large table of counts needs no repeated enumeration of its subsets.

Reading row seven. Label the entries in each row from $k=0$ on the left through $k=n$ on the right. Rows zero through seven are shown below. The highlighted entry $T_{7,3}=35$ comes from $T_{6,2}=15$ and $T_{6,3}=20$. Starting with row zero $(1)$, the edges remain one and Pascal's rule determines every interior entry, so mathematical induction uniquely determines the whole array.

The triangular and rectangular layouts describe the same entries. For a north-east lattice path ending at $(a,b)$, set $n=a+b$ and $k=b$, so $P(a,b)=T_{n,k}$. The predecessor $(a-1,b)$ has indices $(n-1,k)$, while $(a,b-1)$ has indices $(n-1,k-1)$. Thus the grid's west and south arrivals are exactly the triangle's two parents. This change of indices explains why the north-east lattice path recurrence and Pascal's rule agree.

There is also a global check on the local construction: row $n$ sums to $2^n$. Every subset of an $n$-element set appears exactly once among the classes of sizes $0,\dots ,n$, and each element has two inclusion choices by the rule of product. In row seven, $1+7+21+35+35+21+7+1=128=2^7$. Replacing every subset by its complement reverses the row, explaining $T_{n,k}=T_{n,n-k}$. The binomial theorem packages the whole row as the coefficients of $(1+x{)}^n$.

Blaise Pascal's Traité du triangle arithmétique, published in 1665, presents an arithmetic array and connects it to combinations, games of chance, and powers of binomial expressions. This makes the connection between a counting table and several applications visible in the original work. ^[ref] The University of Cambridge's account of the treatise explains its 1654 composition and the organization of these applications. ^[ref]

On an interval with $x>0$, use the power-law ansatz $y=x^r$. Substitution gives $r(r-1)+2r+1=r^2+r+1=0$, whose roots are $r=-1/2\pm i\sqrt{3}/2$. Taking real and imaginary parts of $x^r$ givesThese two real solutions are linearly independent: their Wronskian is $(\sqrt{3}/2)x^{-2}\ne 0$. They therefore span the solution space of the second-order linear ordinary differential equation on that interval.

For $x<0$, replace $x^{-1/2}$ by $|x{|}^{-1/2}$ and $\log x$ by $\log |x|$, with independent real constants on the negative interval. This follows equally by the change of variable $t=\log |x|$, which turns the Cauchy-Euler equation into $d^{2}y/d{t}^2+dy/dt+y=0$. The coefficient of $y^{\prime \prime }$ vanishes at zero, so the general formula is stated separately on either side. Every nonzero mode has an unbounded oscillatory envelope as $x\to 0$; a continuous extension through zero requires the solution to vanish identically on both sides.

Divide by $2x$ away from zero to obtain $y^{\prime \prime }+P(x)y^{\prime }+Q(x)y=0$, with $P(x)=x/2-1/(2x)$ and $Q(x)=\lambda /2$. At zero, the quantities $xP(x)=(x^2-1)/2$ and $x^{2}Q(x)=\lambda x^2/2$ are analytic. The pole in $P$ makes zero singular, so $x=0$ is a regular singular point. Its indicial equation isAt $x=1$, both normalized coefficients are analytic functions, so $x=1$ is an ordinary point. These are the standard normalized-coefficient and indicial tests.^[ref]

For infinity, set $t=1/x$ and $Y(t)=y(1/t)$. The chain rule gives $y^{\prime }=-t^2{Y}^{\prime }$ and $y^{\prime \prime }=t^4{Y}^{\prime \prime }+2t^3{Y}^{\prime }$, so the transformed normalized equation isThe product of the first-derivative coefficient with $t$ still has a second-order pole at zero. Hence the criterion for a regular singular point fails and $x=\infty$ is an irregular singular point, including when $\lambda =0$. The primes on $Y$ here refer to differentiation with respect to $t$.

Seek a Frobenius series $y=x^r{\sum }_{n\ge 0}{a}_n{x}^n$, with $a_0\ne 0$, on the positive side of zero. The lowest coefficient gives the indicial equation from part (a); the next gives $(r+1)(2r-1)a_1=0$. For either permitted indicial exponent, $a_1=0$. Matching the subsequent powers givesThe two indicial exponents differ by a noninteger, so there are two independent Frobenius solutions without a logarithmic companion. Normalizing each leading coefficient to one, their forms areThus the general local solution is $C_0{f}_0+C_1{f}_{3/2}$. The endpoint condition at zero forces $C_0=0$. For a nontrivial solution, $C_1\ne 0$, so $y(x)\sim C_1{x}^{3/2}$ and $\alpha =3/2$, with the arbitrary nonzero amplitude understood.

The condition at one selects values of $\lambda$ for which this branch also vanishes there; it does not change the selected local exponent. The zero solution, available for every $\lambda$, has no nonzero leading coefficient and is excluded when asking for a leading exponent. This is a direct application of the Frobenius method and the noninteger-separated-root case.^[ref]

First classify nonzero polynomial solutions of the differential equation itself. If $y$ has degree $d$ and leading coefficient $a_d\ne 0$, the coefficient of $x^{d+1}$ is $(d+\lambda )a_d$. Hence $\lambda =-d$ is necessary.

An analytic solution at zero belongs to the exponent-zero Frobenius branch. Its odd-power coefficients vanish, because $a_1=0$ and the recurrence relation preserves parity. For the even-power coefficients,All denominators are nonzero for $j\ge 0$. With $\lambda =-2m$, the coefficients through $a_{2m}$ are nonzero when $a_0\ne 0$, the next coefficient vanishes, and every subsequent coefficient vanishes. This proves sufficiency as well as the restriction to even polynomial degree:For example, setting $a_0=1$ gives $1$ when $\lambda =0$, $1+x^2$ when $\lambda =-2$, and $1+2x^2+x^4/5$ when $\lambda =-4$. A polynomial cannot instead begin with a positive integer power: its lowest power must have an indicial exponent, and the only integer indicial exponent is zero. This also proves that no odd-degree family was missed. The finite-series mechanism is a terminating Frobenius series.

If the additional endpoint conditions from part (b) are imposed on this part too, there is no nonzero polynomial solution: every nonzero polynomial just classified has $y(0)=a_0\ne 0$. The displayed family and degree answer therefore concern the differential equation without those extra endpoint conditions.

Take $ϵ\to 0^{+}$. A root on the smaller scale has $x=ϵu$, giving $u+2+ϵu^3=0$. Put $u=u_0+ϵu_1+O({ϵ}^2)$. Successive coefficients give $u_0=-2$ and $u_1+u_0^3=0$, hence $u_1=8$. The real root is thereforeThe other dominant balance is between the cubic and linear terms, requiring $x={ϵ}^{1/2}z$. Set $t={ϵ}^{1/2}$; the equation becomes $z^3+z+2t=0$. At $t=0$ the two nonzero roots are $z_0=\pm i$. With $z=z_0+t{z}_1+O(t^2)$, the first correction satisfies $(3z_0^2+1)z_1+2=0$, so $-2z_1+2=0$ and $z_1=1$. ThusThe limiting roots in both rescalings are simple, so the implicit function theorem justifies these regular expansions in their respective small parameters. The fractional powers in the original variable are a fractional-power root expansion. The real derivative $3x^2+ϵ$ is strictly positive; the cubic therefore has exactly one real root. Its other roots are the displayed complex conjugate pair, accounting for all three roots. Their leading real corrections sum to $-2ϵ+ϵ+ϵ=0$, as required by the absence of a quadratic coefficient.

Label the polygon vertices of the nondegenerate triangle $K$ as $a_1,a_2,a_3$, and put $m_{ij}=(a_i+a_j)/2$ for its three edge midpoints. The six monomials $1,x_1,x_2,x_1^2,x_1{x}_2,x_2^2$ form a basis of ${\mathcal{P}}_2(K)$, so its dimension is six. Nondegeneracy matters: $K$ has nonempty interior, and a polynomial vanishing on that interior is the zero polynomial.

Let ${\lambda }_1,{\lambda }_2,{\lambda }_3$ be the affine functions of the barycentric coordinate system, characterized by ${\lambda }_i(a_j)={\delta }_{ij}$ and ${\sum }_i{\lambda }_i=1$. Define six quadratic polynomialsAt $a_k$, $N_i(a_k)={\delta }_{ik}$ and all $N_{ij}(a_k)$ vanish. At $m_{kl}$, exactly ${\lambda }_k$ and ${\lambda }_l$ equal $1/2$, while the third is zero. Thus every $N_i$ vanishes at all three edge midpoints, and $N_{ij}(m_{kl})$ is one for the matching unordered pair and zero otherwise.

These six polynomials have the cardinal property for the six point evaluation functionals. Evaluating any vanishing linear combination at the nodes forces each of its coefficients to vanish. They are therefore six linearly independent members of a six-dimensional vector space, hence a basis.

In particular, every $p\in {\mathcal{P}}_2(K)$ has the unique expansionThis proves both existence and uniqueness for arbitrary six prescribed nodal values, which is precisely an unisolvent finite element. It also identifies the finite element degrees of freedom and shape functions of the quadratic Lagrange triangle explicitly.

Alexandre-Théophile Vandermonde presents a programme in elimination theory: remove unknowns from simultaneous polynomial equations in a way that works uniformly as the number of equations and their polynomial degrees vary. The immediate achievement is narrower than that programme: a general construction for linear equations. The academy year 1772 used in the exam is not the printing date of the memoir, which appeared in 1776. Its compact indices distinguished the equation containing a coefficient from the unknown it multiplied; they were not powers. That notational separation is crucial to interpreting the work.^[ref][ref]

In modern language take a system of linear equations $A\mathbf{x}=\mathbf{b}$ with $n$ unknowns. Its determinant is an alternating expression in the coefficients:Each product selects one coefficient from each equation and each unknown's column. The sign of a permutation determines whether it is added or subtracted. This makes the number $n$ a genuine parameter of a general rule, rather than merely supplying separately memorised formulas for two or three unknowns. The notation here is modern, not a transcription of Alexandre-Théophile Vandermonde's symbols.

For two unknowns, multiply $ax+by=u$ by $d$ and $cx+dy=v$ by $b$, then subtract. Similarly eliminate $x$. This givesThe same principle in arbitrary size is Cramer's rule. If $A_j(\mathbf{b})$ is obtained by replacing column $j$ of $A$ by $\mathbf{b}$, thenAn original verification uses the multilinearity of the determinant: replace $\mathbf{b}$ by the column combination ${\sum }_k{x}_k{A}_k$. Every resulting determinant vanishes through repeated columns except the $k=j$ term, which is $x_j\det A$. This explains how an elimination expression isolates one unknown.

The condition $D\ne 0$ is substantive. If $D=0$, multiplying by an expression which vanishes cannot justify division. For instance, $x+y=1$ and $2x+2y=3$ are inconsistent, whereas replacing $3$ by $2$ gives infinitely many solutions. Thus an elimination formula must distinguish exceptional coefficients and preserve the connection to the original system of linear equations. Nor does the displayed rule solve the much broader problem of arbitrary nonlinear equations.

The passage's importance is the move from particular elimination calculations towards a theory of alternating expressions. It would be misleading to turn this into the claim that the familiar Vandermonde determinant, formed from powers of variables, is what this extract displays. A positional upper index and an exponent have different meanings; the history of that eponym requires separate evidence. Likewise, modern matrix algebra is a useful explanatory language here, not the institutional or conceptual setting of an eighteenth-century memoir.^[ref]

Joseph Fourier's Théorie analytique de la chaleur by Joseph Fourier (1822) makes trigonometric series instruments for solving problems of heat conduction. A prescribed temperature distribution need not be given by one power series: it may describe different heating on different parts of a body. The extract occurs in Chapter III, §233, where the earlier assumption of an odd-power expansion is relaxed. The innovation concerns both representation and what can count as a function in an analytical calculation.^[ref]

Use a separate integration variable $t$ to unpack the printed notation. The proposed Fourier sine series has coefficientsThe integral signs in the extract denote definite integrals over $[0,\pi ]$, not antiderivatives; the author explicitly supplies these bounds in the surrounding sentence. Their absent printed bounds are a notational convention, not permission to choose arbitrary constants of integration.

The key calculation is the orthogonality relationThe product-to-sum formulas prove this by integrating cosines of integer multiples of $t$. If an expansion can be integrated term by term, multiplication by $\sin (mt)$ followed by integration selects exactly one coefficient. This derives the only possible coefficients, but does not by itself prove the expansion converges to the proposed function. That distinction is central to assessing the extract's claim.

For a revealing example take $f(x)=1$ on $0<x<\pi$. Direct integration gives $b_n=2(1-(-1{)}^n)/(\pi n)$, and thereforeEvery term is zero at $0$ and $\pi$, so the series cannot take the value $1$ at those endpoints. Its natural interpretation uses the odd extension followed by periodic extension; the endpoints become jump discontinuities. More generally, for a piecewise continuously differentiable function, the Fourier series converges at an interior jump to the average of the two one-sided limits, irrespective of an individually assigned value there. Changing one value also leaves the integrals unchanged. Consequently the unqualified assertion of representation for every arbitrary function cannot be a modern pointwise theorem. Dirichlet's theorem on Fourier series supplies sufficient conditions and the correct distinction between continuity points and jumps; its 1829 source explicitly addresses the proof problem.^[ref]

The physical payoff is explicit. For a rod scaled to $0<x<\pi$ with zero end temperature, separation of variables in the heat equation producesEach mode solves $u_{\tau }=\kappa u_{xx}$ and the end boundary conditions; the Fourier sine coefficients encode the initial temperature. This rod is an explanatory example, distinct from the stationary rectangular-solid problem opening Joseph Fourier's Chapter III. The factor $e^{-\kappa n^2\tau }$ damps high modes rapidly for positive time. The representation of arbitrary data makes the physical solution possible; the proof of that representation remains a separate task.