An inclined plane tilted slowly from horizontal turns a coefficient of static friction into a critical angle. Assume a nonrotating block on a stationary, rigid, dry surface, with gravitational weight and the contact forces as its only forces. Take $0\le \theta <\pi /2$ to be the angle between the uphill surface direction and horizontal. Let ${\mathbf{e}}_t$ be a unit vector pointing uphill and ${\mathbf{e}}_n$ a perpendicular unit vector pointing outward from the surface. Resolving the gravitational weight givesThe angle between the downward vertical and the inward normal $-{\mathbf{e}}_n$ is also $\theta$. Thus the inward component uses cosine and the downhill component uses sine. These are components of one force, not two additional forces.

Before downhill contact slip, zero acceleration and Newton's second law requireThe required static friction points uphill. Substituting into the bound and dividing by the positive $mg\cos \theta$ givesThe tangent function increases on this range of angles, so the last admissible resting angle isThe inverse tangent takes a dimensionless quantity to an angle. At equality the ideal model still permits a limiting resting state; increasing $\theta$ further excludes it. At a smaller angle, $f_s/N=\tan \theta$ is the demand ratio, not the coefficient of static friction.^[ref]

For the stipulated $m=2\mathrm{kg}$, $g=9.8m/{\mathrm{s}}^2$ and ${\mu }_s=0.40$, the critical angle is approximately $21.8014^{\circ }$. At $\theta =15^{\circ }$, $N\simeq 18.932\mathrm{N}$ and $f_s\simeq 5.073\mathrm{N}$, comfortably below the bound ${\mu }_{s}N\simeq 7.573\mathrm{N}$. At $\theta =25^{\circ }$, maintaining rest would require $f_s\simeq 8.283\mathrm{N}$, but the bound is only $7.105\mathrm{N}$: rest is impossible. At ${\theta }_c$, $N\simeq 18.198\mathrm{N}$ and $f_s\simeq 7.279\mathrm{N}$. Notice that the limiting tangential force here is smaller than the horizontal $7.84\mathrm{N}$ limit because tilting reduces the normal force.

Cancellation of mass makes this relation useful: the ideal critical angle tests a contact coefficient without needing a calibrated force gauge. Doubling the block's mass doubles both demanded static friction and its bound, leaving ${\theta }_c$ unchanged if ${\mu }_s$ is unchanged. This is a testable model prediction. It need not survive a change in surface condition, a direction-dependent coefficient, appreciable deformation, or vibration.

The derivation also assumes continued contact and that sliding occurs before tipping or rolling without slipping. The resultant contact force must act within the supporting footprint so that the block can satisfy torque balance; a tall block may tip first. No extra tension, imposed tangential pull, or lift is included. In a slowly raised apparatus, quasistatic process means that motion of the board contributes negligible inertial forces while locating the threshold. Once downhill contact slip begins, a constant-velocity sliding angle would concern kinetic friction, not the onset of static friction.

A clique records pairwise relationships. If vertices represent sets and an edge means a nonempty intersection, a clique does not by itself guarantee an element common to every set. TakeThen $A\cap B=\{2\}$, $B\cap C=\{3\}$ and $C\cap A=\{1\}$, so their intersection graph is $K_3$. But $A\cap B\cap C=\varnothing$. The graph faithfully records all three pairwise intersections; the stronger claim of a single shared element requires additional hypotheses.

A maximum clique is a clique with the greatest possible number of vertices in the graph. Its size is the clique number. Every maximum clique is a maximal clique, since adding a vertex would produce a larger clique. A graph can have several maximum cliques; the word maximum specifies their common size, rather than uniqueness.

Leonhard Euler's 1751 correspondence already studied triangulations of a convex polygon and gave a closed formula for their counts. Johann Andreas von Segner supplied a combinatorial recurrence relation in a paper written in 1758. Eugène Charles Catalan's nineteenth-century work connected the formula to bracketed products. This history, documented in Igor Pak's account, shows how the same counting structure reappeared through different objects; the modern name alone does not establish discovery priority.^[ref]

A binary parenthesization specifies the order of binary operations on a fixed ordered list of operands, combining exactly two existing expressions at each step. It has a full binary tree as its expression structure: operands are leaves and operations are internal vertices. For four operands $a,b,c,d$, the five possibilities are $((ab)c)d$, $(a(bc))d$, $(ab)(cd)$, $a((bc)d)$, and $a(b(cd))$.

A binary parenthesization of $n+1$ operands contains $n$ binary operations. At the last operation, its left expression can contain $k$ operations and its right expression $n-1-k$. The Catalan recurrence therefore gives $C_n$ possible evaluation structures. The order of operands stays fixed; this is not a count of their permutations. Even for an associative operation whose result is unchanged, different structures can have different evaluation costs, as in the matrix-chain multiplication problem.^[ref]

The first-return decomposition of a Dyck path and the rule of product giveFor a given $k$, choose the inside path in $C_k$ ways and its suffix in $C_{n-1-k}$ ways. Different first-return locations give disjoint groups, so the rule of sum adds their counts. For $n=4$, the terms are $1\cdot 5+1\cdot 2+2\cdot 1+5\cdot 1=14$. Iterating the recurrence relation computes the counts without enumerating every full path.

The carpel region enclosing one or more ovules.

The connection between the stigma and ovary through which a pollen tube can grow.

In the l2 sequence space, let $e_n$ have value one in coordinate $n$ and zero elsewhere. This orthonormal sequence is bounded: $\Vert e_n{\Vert }_2=1$. For distinct indices $n,m$, its terms obey $\Vert e_n-e_m{\Vert }_2=\sqrt{2}$. No subsequence can therefore be a Cauchy sequence, so none converges in the norm. Thus even a complete metric space can have a bounded sequence with no convergent subsequence. Finite dimension is essential to the usual Euclidean space extension, and in a general metric space the relevant replacement is compactness or total boundedness together with completeness.

The rational numbers $q_n=\lfloor 10^n\sqrt{2}\rfloor /10^n$ lie in $[1,2]$ and obey $0\le \sqrt{2}-q_n<10^{-n}$. They form a convergent sequence in the real numbers, with limit of a sequence $\sqrt{2}$. Every subsequence has that same real limit. Because $\sqrt{2}$ is an irrational number, no subsequence converges to a point in the rational numbers with their usual metric. Boundedness alone therefore does not give the corresponding theorem in every metric space; completeness of the real numbers is doing real work here. The floor function simply truncates the decimal expansion in this example.

Begin with a closed interval $I_0=[a,b]$ containing the entire bounded sequence. Split the interval at its midpoint. At least one of the two closed halves contains infinitely many indexed terms; choose such a half and repeat. This gives nested intervals $I_j$ with lengths $(b-a)/2^j$. The nested interval theorem, using completeness of the real numbers, supplies their unique common point $\ell$.

Choose $n_j>n_{j-1}$ with $x_{n_j}\in I_j$. Infinitely many indexed terms in $I_j$ guarantee that the next index can be chosen. Both $x_{n_j}$ and $\ell$ belong to $I_j$, soThis is a short existence proof, not necessarily a finite algorithm for an arbitrary input sequence: deciding which half contains infinitely many future terms may require mathematical knowledge about that sequence.

An invertible matrix need not have an invertible chosen diagonal block. For example,has singular matrices as both one-by-one diagonal blocks. Neither diagonal-block formula applies in that partition. Interchanging the two rows produces the identity matrix and a usable pivot element. This is why a failed block pivot is not proof that the full matrix is a singular matrix; Pivoted LU decomposition changes the order before eliminating.

Conversely, invertible diagonal blocks do not suffice: $M=\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)$ has $A=D=1$ but Schur complement $S=0$, so it is a singular matrix. Both hypotheses in the block matrix inverse formula have work to do.

Write a linear system as $Ax+By=f$ and $Cx+Dy=g$, with $A$ an invertible matrix. Solving the first equation for $x$ and substituting in the second givesThus the Schur complement is the coefficient matrix remaining after the $x$ variables have been eliminated. When it is an invertible matrix, solve the reduced linear system for $y$ and recover $x$. In computation, expressions such as $A^{-1}B$ mean solving with each column of $B$, using the same LU decomposition or Cholesky decomposition when applicable.

For the example under block matrix inverse and right-hand side $(3,1,4{)}^T$, take $f=(3,1{)}^T$ and $g=4$. Since $C{A}^{-1}f=13/6$, the reduced equation is $(19/6)y=11/6$. Back substitution givesThe original equations return $3$, $1$ and $4$, respectively. The reduction retained the influence of $x_1,x_2$ on $y$ through the correction $C{A}^{-1}B$.

A short proof outline also explains whether $M$ is an invertible matrix: the exact matrix factorizationhas invertible matrices on the outside. Therefore $M$ is an invertible matrix precisely when the central block diagonal matrix is one. Reversing the factors gives the block matrix inverse, while taking determinants gives the Schur complement determinant identity.

For a finite simple graph $G$ and a matching in a graph $M\subseteq E(G)$, $M$ is a maximum cardinality matching if and only if $G$ has no augmenting path in a matching relative to $M$. The graph need not be a bipartite graph, and the assertion concerns cardinality, not edge benefits.^[ref]

The criterion turns a global comparison with every competing matching in a graph into a search for one repair. Starting from any matching in a graph, repeatedly find and toggle an augmenting path in a matching. Each step adds one selected edge, so finiteness gives termination after at most $\lfloor |V(G)|/2\rfloor -|M|$ steps; absence of any further augmenting path in a matching certifies maximum cardinality. This is an optimality criterion, not by itself a bound on the work needed to find each graph path.

For a short proof outline, an augmenting path in a matching immediately contradicts maximum cardinality. Conversely, if another matching in a graph $N$ has $|N|>|M|$, consider the subgraph whose set of edges is $M△N$. Its vertices have degree of a vertex at most two; its nontrivial connected components are graph paths and even cycles in a graph, alternating between the two matchings in a graph. A cycle in a graph contributes equal numbers from each. Hence some graph path contributes one more edge from $N$ than from $M$ and starts and ends with edges from $N$. Its endpoints are unmatched vertices in a matching relative to $M$, yielding the required augmenting path in a matching.

An actual teaching setup is described in Laboratory Exercise No. 4: Diffraction, MIT 6.237, Fall 2024, for MIT's 6.237 Modern Optics Project Laboratory, Fall 2024. Preparation includes predicting the aperture-size dependence and the distinction between Fraunhofer diffraction and Fresnel diffraction. A helium–neon laser supplies $632.8\mathrm{nm}$ light; the far-field setup uses a planar incident wavefront, a diffracting object and a distant screen, with beam expansion when needed. Figure 1 also marks an optional CCD camera. Exercise 4.7 asks students to sketch the circular-aperture pattern, measure the Airy disk diameter between correctly identified boundaries, and infer the aperture size. No measured values are supplied. As a regime comparison, exercise 4.9 varies the circular opening's position near a microscope objective optical lens, observes near-field patterns and compares them with the far-field case. This tests whether a change in propagation geometry changes the pattern. The coherent illumination measures one aperture's response, rather than the discrimination of two mutually incoherent optical point sources.^[ref]

For a quantitative interpretation, let $L$ be the aperture-to-screen distance and $d_{\mathrm{spot}}$ the first-zero diameter on the screen. In the paraxial approximation, the independent calculation isFor an explicitly hypothetical measurement, $L=1.80\mathrm{m}$ and $d_{\mathrm{spot}}=5.60\mathrm{mm}$ at $632.8\mathrm{nm}$ would imply $D\simeq 0.496\mathrm{mm}$. The scale $D^2/\lambda \simeq 0.389\mathrm{m}$ is smaller than $L$, but only by a factor of about $4.6$; this alone gives limited confidence in a high-precision Fraunhofer diffraction approximation. It is not a reported MIT measurement.

Additional recommended controls are to keep wavelength fixed, record $L$, centre and uniformly illuminate the aperture, and repeat with another $D$. Compare $d_{\mathrm{spot}}/L$, which should vary inversely with $D$. A camera readout should include blocked-beam background and avoid saturation. A scaled repetition at larger $L$ tests residual Fresnel diffraction; nonuniform illumination, optical aberrations, detector sampling and uncertainty in locating a dark ring limit the inferred $D$. The screen pattern's apparent bright edge depends on display brightness and is not a substitute for the first-zero boundary.

Peak normalization $P(0)=1$ does not make the integral of a point spread function equal to one. In the paraxial approximation to the focal plane, define $b=\pi D/(\lambda f)$ and $u=br$. The unit-integral point spread function normalization isHere $p$ has units of inverse area; multiplying by the detected radiant flux gives image-plane irradiance. Its encircled energy, meaning the fraction inside radius $R=U/b$, isThis is an integral over area, so its radial weight is $u$. An unweighted integral of a line through the centre is a different quantity.

A short proof uses the Bessel function of the first kind identities $J_0^{\prime }=-J_1$ and $J_1^{\prime }=J_0-J_1/u$. They giveBoth expressions for $E$ vanish at zero. Since these Bessel functions of the first kind tend to zero as $U\to \infty$, $E(\infty )=1$, establishing the normalization. At the first dark ring,A circular detector region ending at the first zero therefore misses about $16.22\%$ of the ideal radiant flux, even though the surrounding rings have low peak light intensity. This matters when converting measured counts into total source brightness or choosing an image aperture for a photodetector.

Write $a=D/2$ and wavenumber $k=2\pi /\lambda$. The pupil Fourier transform, normalized by pupil area, gives the Fraunhofer diffraction wave amplitude. In pupil polar coordinates,The angular average is $J_0$ because $J_0(z)=(2\pi {)}^{-1}{\int }_0^{2\pi }{e}^{-iz\cos \varphi }d\varphi$. The Bessel function of the first kind identity $d\{z{J}_1(z)\}/dz=z{J}_0(z)$ then reduces the radial integral to $A(u)=2J_1(u)/u$. Taking the squared magnitude yields the Airy disk light intensity profile.^[ref]

The apparent division by zero is removable. The power seriesgives $A(u)=1-u^2/8+u^4/192+O(u^6)$ and $P(u)=1-u^2/4+5u^4/192+O(u^6)$. At the centre, every pupil contribution has the same wave phase, so their normalized sum is one. The first dark ring instead corresponds to $J_1(u)=0$ with $u>0$, rather than to the removable zero of $J_1$ at the origin.

An affine map need not be injective. For $G(x,y)=(x+y,0)$, the direction $(1,-1)$ belongs to the kernel of its linear part of an affine map. Consequently the whole line $y=-x$ maps to $(0,0)$, and every parallel line $y=-x+c$ maps to $(c,0)$. These are collapsed lines, not images with a surviving direction. Assertions that an affine map preserves distinct parallel lines require the appropriate noncollapse or injectivity hypothesis.

For coefficients with sum one, the defining identity follows from linearity:The translation appears once on both sides precisely because ${\sum }_i{\lambda }_i=1$. In particular, midpoints map to midpoints, and a parameterized line segment $(1-t)P+tQ$, $0\le t\le 1$, maps to $(1-t)T(P)+tT(Q)$. A whole line has the same formula with $t\in \mathbb{R}$; its image is a line if $M(Q-P)\ne 0$, and a single point otherwise.