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For real polynomial $f(x)=c{x}^n+\cdots$, $c\ne 0$, the zero mode of $H=(p+if)(p-if)/(2m)$ is normalizable on $\mathbb{R}$ exactly when $n$ is odd and $c>0$. Its squared modulus has leading exponential $\exp [-2c{x}^{n+1}/((n+1)\hslash )]$. Odd $n$ gives the same tail sign at both ends; positivity of $c$ makes both tails decay. This kernel is one-dimensional because $(p-if)\psi =0$ is a first-order differential equation. On the operator domain where integration by parts is valid, the energy form is $\Vert (p-if)\psi {\Vert }^2/(2m)\ge 0$.

Finite local subdivision matrices for the scaled difference scheme track derivative variations in nested parameter intervals. A product-norm bound below one proves uniform contraction for every interval and therefore gives a lower bound on derivative continuity. A product eigenvalue of modulus at least one is an obstruction when its mode is reachable from the data and visible in the limit: generic data excite it, so the corresponding derivative differences fail to vanish. Harmless polynomial or unobservable modes must be removed before applying such an upper bound. Borderline smoothness needs direct analysis, not rounding a numerical exponent. Stability and convergence of the underlying scheme must be checked rather than inferred from mask width.

Numerical integration approximates a definite integral using finitely many sampled function values. For an integrand $f$ on a finite interval $[a,b]$, $a<b$, a quadrature rule has the formThe quadrature nodes $x_j$ tell us where to evaluate $f$; the quadrature weights $w_j$ turn those values into an approximation to the accumulated signed area. The exact definite integral depends on the integrand throughout the interval, whereas the quadrature rule sees only these samples. Integrating a chosen interpolant exactly still generally approximates the original integrand's integral.^[ref]

The fundamental theorem of calculus evaluates $I(f)$ as $F(b)-F(a)$ when an antiderivative $F$ is available. Numerical integration is useful when no convenient antiderivative is known, or each evaluation of the integrand requires an expensive computation. The task then becomes choosing informative quadrature nodes and controlling the quadrature error with as few evaluations as possible. Fixed data also permit a quadrature rule, but refinement cannot supply new information between observations.^[ref]

For an exact comparison, take the polynomial $f(x)=x^2$ on $[0,1]$. Its antiderivative $x^3/3$ gives $I(f)=1/3$. On a partition of an interval into $n\ge 1$ equal panels, the composite midpoint quadrature rule uses $n$ midpoint samples, each with quadrature weight $1/n$; the composite trapezoidal rule uses $n+1$ endpoint samples, with outer quadrature weights $1/(2n)$ and interior quadrature weights $1/n$. The existing derivations under these two quadrature rules giveIn particular, $M_1=1/4$ and $T_1=1/2$ already bracket $1/3$; two panels give $M_2=5/16$ and $T_2=3/8$. The convex function $x^2$ explains the opposite signs. A midpoint rectangle can cross the curve within its panel, so its total signed area, rather than pointwise ordering, determines the underestimate.

Define the signed approximation error as $E_n=Q_n(f)-I(f)$. In this example the absolute errors are $1/(12n^2)$ and $1/(6n^2)$: doubling $n$ reduces each by exactly four. A plot with both axes on a logarithmic scale therefore has slope $-2$, expressing order two with respect to panel refinement. This is an exact calculation for this polynomial, including $n=1$, rather than an observed rate extrapolated from a few computed answers.

For a general integrand, the composite midpoint quadrature rule and composite trapezoidal rule give second-order quadrature error bounds when $f$ has two continuous derivatives on $[a,b]$ and a known bound on $|f^{\prime \prime }|$. These hypotheses describe behavior between samples. A formal error formula containing an unknown derivative is not a numerical accuracy certificate or a justified sample count. Finite samples do not certify quadrature accuracy makes the missing information explicit. Errors in the evaluations and roundoff error must also be considered; noise sensitivity of a quadrature rule treats the former.

More elaborate choices exploit other information: Gaussian quadrature obtains high polynomial exactness from carefully chosen quadrature nodes, and Simpson's rule uses a quadratic interpolant. Adaptive numerical integration compares approximations and concentrates samples where they disagree, often saving evaluations. Such a difference is an error estimate, not a certified bound without further assumptions: both approximations can miss the same feature.^[ref]

For $-W^{\prime \prime }+2x{W}^{\prime }=EW$, the power series coefficients obey $(n+2)(n+1)a_{n+2}=(2n-E)a_n$. The condition $W(0)=0$ makes every even coefficient zero. A nonzero odd series terminates precisely when $E=2d$ with $d$ an odd positive integer. A polynomial of degree $d$ must have this value of $E$ by comparison of leading coefficients, so these are all possibilities. With $W^{\prime }(0)=1$, the first three polynomial solutions are $x$, $x-2x^3/3$, and $x-4x^3/3+4x^5/15$, with $E=2,6,10$. They are scalar multiples of the odd Hermite polynomials.

A finite group permutes the associate classes of irreducible polynomial factors of an invariant. The product of one representative from each orbit transforms by a scalar character of the group. If that character is trivial, the product is invariant. Its transitive factor orbit makes it prime in the invariant ring: an invariant polynomial divisible by one orbit factor is divisible by all of them.

If a nonzero continuous real function on an interval is orthogonal to every polynomial of degree below $n$, it has at least $n$ distinct interior zeros. If it had only finitely many zeros and fewer than $n$ sign changes, multiply it by the product of linear factors at its sign-changing zeros. The product has constant sign and is nonzero on an open interval, so its integral cannot vanish. Orthogonality to this degree-below-$n$ polynomial contradicts that conclusion. Infinitely many interior zeros already satisfy the assertion. Orthogonal polynomials of respective degrees $0,\dots ,n-1$ form a basis of the needed polynomial space.

An $(n+1)$-node quadrature exact on degree $n$ is exact through degree $2n+1$ precisely when its node polynomial $p={\prod }_i(x-x_i)$ is orthogonal to every degree-$n$ polynomial for the integration weight. Necessity applies the rule to $pr$. Sufficiency divides an arbitrary degree-$2n+1$ polynomial as $pq+r$, with both $q,r$ of degree at most $n$; the integral and quadrature of $pq$ vanish while both agree on $r$.

The orthogonal projection $P_F$ onto a closed subspace of a Hilbert space $F$ sends $x$ to its component in the orthogonal decompositionThe residual $r$ is orthogonal to every permitted direction. For any $z\in F$, the Pythagorean theorem givesThus $P_{F}x$ uniquely minimizes the distance to $F$. It is a linear map, is idempotent, and is nonexpansive: $\Vert P_{F}x-P_{F}y\Vert \le \Vert x-y\Vert$; taking $y=0$ gives $\Vert P_{F}x\Vert \le \Vert x\Vert$.

In ${\mathbb{R}}^2$, project $v=(1,3)$ onto the line spanned by $u=(2,1)$. The inner product givesThe check $⟨r,u⟩=-2+2=0$ confirms orthogonality, and $\Vert v{\Vert }^2=10=5+5=\Vert P_{F}v{\Vert }^2+\Vert r{\Vert }^2$. Projecting again changes nothing, since the result already lies on the line.

For a finite orthonormal basis $e_1,\dots ,e_m$ of $F$, the same operation is $P_{F}x={\sum }_j⟨x,e_j⟩e_j$, using the convention linear in the first argument. Replacing coordinate vectors by functions gives exactly the same approximation principle. For $f(x)=x^2$ in the real L2 space on $[-1,1]$ and $F=\mathrm{span}\{1,x\}$, the directions are orthogonal, with squared norms $2$ and $2/3$. Their inner products with $f$ are $2/3$ and zero, soDirect integration gives the squared error $\Vert x^2-1/3{\Vert }^2=8/45$. For any competing affine polynomial, the orthogonal residual yieldsEvery deviation from $a=1/3$, $b=0$ adds a nonnegative error. The best function need not pass through any selected data point: it minimizes an integrated error across the whole interval.

Near a non-characteristic plane $x=0$, use the small cap ${\omega }_{\epsilon }=\{x>0,x+y^2<\epsilon \}$. Its curved face is non-characteristic by continuity of the principal symbol. A polynomial coordinate change flattens that face. Uniform real analytic coefficient bounds and the uniform Cauchy radius for polynomial forcing solve $P^{\ast }v=g$ throughout a common cap, with zero boundary jet on the curved face. The complete boundary-jet condition for formal adjoints then gives ${\int }_{{\omega }_{\epsilon }}ug=0$ for every polynomial $g$. Uniform approximation of $\bar{u}$ gives $\int |u{|}^2=0$. Repeating on the other side of the plane proves two-sided local uniqueness.

Parameter continuation changes $\lambda$ by a small amount and solves again for $u$, using the previous solution as the initial guess for Newton method. If $F_u$ is an invertible matrix, the implicit function theorem gives a local solution branch $u(\lambda )$, withLocal convergence of Newton method also needs regularity and a sufficiently small step: the cited result assumes a Lipschitz continuous derivative and an invertible $F_u$ at the seed.^[ref]

For the cubic polynomial,The derivative becomes unbounded at either continuation fold, and the scalar Newton method divides by $F_x=0$ there. Just beyond the upper continuation fold, no nearby real solution exists at the increased $\lambda$, although a distant negative-$x$ solution still exists. Converging to that solution would jump away from the local solution branch. Decreasing the parameter step cannot make $x(\lambda )$ a valid smooth local description at the bend.

Write $J_k(\lambda )$ for the Jordan block with diagonal $\lambda$ and ones on the superdiagonal. Its characteristic polynomial and minimal polynomial are both $(x-\lambda {)}^k$. Therefore one answer to the first request isFor the second request, takeThe algebraic multiplicities are five and two, and the largest block at each eigenvalue has size two, so both prescribed polynomials hold. They are not similar matrices: the eigenspace at three has dimensions three and four respectively, and similarity preserves eigenspace dimension.

There is no third similarity class. At eigenvalue one, the total size two and maximal block size two force the single block $J_2(1)$. At eigenvalue three, the blocks partition five, each has size at most two, and at least one has size two. The only partitions are $2+2+1$ and $2+1+1+1$. The uniqueness of Jordan normal form up to block order therefore leaves exactly the two classes already exhibited.

Let $I$ denote the identity endomorphism. The two-root minimal polynomial gives $(\alpha -{\lambda }_{1}I)(\alpha -{\lambda }_{2}I)=0$. Every vector has the decompositionThe first summand lies in the eigenspace $E_1=\ker (\alpha -{\lambda }_{1}I)$, and the second in $E_2$. If a vector belongs to both, subtracting its two eigenvalue equations gives $({\lambda }_1-{\lambda }_2)v=0$, hence $v=0$. Thus $V=E_1\oplus E_2$.

For distinct real roots ${\lambda }_1,\dots ,{\lambda }_m$, define polynomial projectorsThe Lagrange interpolation polynomials in this expression sum to one: their sum has degree at most $m-1$ and equals one at the $m$ distinct roots. Hence ${\sum }_i{P}_i=I$. Also $(\alpha -{\lambda }_{i}I)P_i=0$, because its numerator is the minimal polynomial evaluated at $\alpha$. Thus every vector is a sum of eigenvectors, $v={\sum }_i{P}_{i}v$. On $E_j$, $P_i$ acts as ${\delta }_{ij}I$, so applying $P_i$ to any zero sum of vectors from the eigenspaces shows that every summand is zero. ThereforeChoosing a basis in each eigenspace proves diagonalizability over the reals. Conversely, if a real matrix is similar over the reals to a diagonal matrix, a polynomial annihilates it exactly when it vanishes at each distinct real diagonal entry. Its minimal polynomial is therefore the product of those distinct linear factors. We have proved real diagonalizability is equivalent to a minimal polynomial with distinct real roots.