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The power of a point $P$ relative to a circle with center $O$ and radius $r$ is $O{P}^2-r^2$. If a line through $P$ meets the circle at $A$ and $B$, the product of the directed distances $PA$ and $PB$ equals this quantity.

In a complete decomposing sphere system for a knot, a disk inside a prime-factor ball that meets the knotted arc once cuts its factor into a connected sum of knots. One side must be an unknot. Replacing the ball by the side carrying the prime factor therefore preserves the unordered factor list. A disk missing the arc cuts off an empty ball and gives the same conclusion. After a small transverse perturbation, such a replacement removes an innermost intersection circle with another sphere system. If a chosen innermost disk contains both marked points, another innermost disk, or its complementary disk when it is the only intersection, contains neither. Iterating produces a prime-factor ball disjoint from both systems. The replacement changes only the location of a joining arc in the intervening punctured ball; capping the other boundaries still recovers the same oriented factors.

Every nontrivial oriented knot is a finite connected sum of knots of prime knots, uniquely up to permutation and oriented link isotopy. Existence follows by induction from additivity of Seifert genus. For uniqueness, compare complete decomposing sphere systems for a knot and remove their intersection circles by the prime-ball exchange lemma. Once a prime ball is disjoint from both systems, its factor can be removed from both; induction then matches all factors.

The principal axes of an ellipse are the two perpendicular symmetry axes through its center of an ellipse. For a symmetric positive-definite matrix $M$, its orthonormal eigenvectors give those directions and its positive eigenvalues ${\lambda }_1,{\lambda }_2$ give semiaxis lengths ${\lambda }_1^{-1/2},{\lambda }_2^{-1/2}$. This follows by orthogonal diagonalization of the defining quadratic form. In ${\xi }^2/A^2+{\upsilon }^2/B^2=1$, the axes are already horizontal and vertical, with semiaxes $|A|,|B|$ and full lengths $2|A|,2|B|$. Equal semiaxes give a circle, for which every perpendicular pair through the center can serve.

The zero level set of $F_R$ is the union of the coordinate axes and the circle of radius $R$. The gradient isIts nine stationary points are the origin, the four axis points at distance $R$, and the four points with $x^2=y^2=R^2/4$. The factored gradient proves completeness: a zero coordinate gives the five axis cases; two nonzero coordinates give a linear system in $x^2,y^2$.

The Hessian matrix has diagonal entries $6xy$ and off-diagonal entries $3(x^2+y^2)-R^2$. At the origin its eigenvalues are $\pm R^2$; at the four unit-direction axis points they are $\pm 2R^2$. These are saddle points of a scalar function. At the two interior points with equal coordinate signs, the eigenvalues are $R^2,2R^2$, giving strict local minima of value $-R^4/8$. Opposite coordinate signs give the negative eigenvalues and strict local maxima of value $R^4/8$.

On the zero circle, $\nabla F_R=2xy(x,y{)}^T$. Hence its direction is radial outward for $xy>0$, inward for $xy<0$, and zero at the four axis crossings. This simple geometric sign rule links the zero level set to the gradient sketch without requiring numerical differentiation.

A quotient topological space is a quotient set equipped with the quotient topology. For a concrete gluing, give the closed interval $I=[0,1]$ its usual subspace topology, identify $0\sim 1$, and leave every interior point in its own singleton equivalence class. The resulting quotient space is a circle: the two endpoints become one point, while interior points remain distinct.

The exact continuous map to the unit circle in the plane isHere sine and cosine give $p(0)=p(1)=(1,0)$, and $p(s)=p(t)$ holds exactly when $s=t$ or $\{s,t\}=\{0,1\}$. Thus $p$ induces a bijection $\bar{p}:I/\sim \to S^1$. ^[ref] The universal property of the quotient topology makes $\bar{p}$ continuous. The quotient of a compact space is compact, and the unit circle is Hausdorff, so the compact-to-Hausdorff continuous bijection theorem makes $\bar{p}$ a homeomorphism. This verifies the topology, not merely the point identifications.

To inspect an open neighbourhood of the seam, choose $0<\epsilon <1/2$ and let $U_{\epsilon }=p([0,\epsilon )\cup (1-\epsilon ,1])$. Its full preimage isBoth pieces are open in the subspace topology on $I$, so $U_{\epsilon }$ is open in the circle. In contrast, $A=[0,\epsilon )$ is open in $I$, butis not open in $I$: it contains $1$ without any neighbourhood of $1$. Hence $p(A)$ is not open in the circle. A quotient map can therefore fail to be an open map.

For a continuous function $f:I\to \mathbb{R}$, the only gluing condition is $f(0)=f(1)$. For example, $f(t)=\cos (2\pi t)$ defines a continuous function on the circle, whereas $f(t)=t$ does not. This lets us use an interval as a coordinate for a periodic function while making the seam invisible. More general quotient spaces encode glued geometric pieces in the same way: describe continuous maps before gluing and check that their values agree on every equivalence class.

The Hausdorff property need not survive a quotient topology. In the non-Hausdorff quotient of the real line by rational translation, $x\sim y$ means $x-y\in \mathbb{Q}$. A nonempty open preimage $W$ must contain all translates by rational numbers of any open interval inside it. The density of the rational numbers then forces $W=\mathbb{R}$: for any $y$, some rational number $r$ has $y-r$ in that interval. Thus this quotient space has the indiscrete topology. The irrationality of the square root of two gives distinct equivalence classes for $0$ and $\sqrt{2}$, so this quotient space is not Hausdorff. Unlike the endpoint-gluing circle, this gluing identifies a dense subset with each point.

An angle in radians is the ratio of its subtended arc length to the radius of a circle. A full turn is $2\pi$ radians and a right angle is $\pi /2$ radians.

The radius of a circle or sphere is the Euclidean distance from its centre to its boundary. The diameter is twice the radius.

For a finite connected graph, collapse a spanning tree. Its $v-1$ edges disappear, and each of the remaining $e-v+1$ edges gives a circle in a bouquet of circles. The fundamental group of a bouquet of circles is free group, with one member of a free basis of a group for each circle. This gives the formula in terms of the Euler characteristic $\chi (X)=v-e$.

For $H_R=\lambda (k_x{\sigma }_y-k_y{\sigma }_x)$ with $\lambda >0$, the lower- and upper-band spinors of unit norm can be chosen asTheir spin expectation values are $⟨\mathbf{S}{⟩}_{\pm }=\pm (\hslash /2)(-k_y,k_x,0)/k$. They are tangential to circles of constant momentum magnitude, opposite between branches, and reversed at opposite momenta as required by time-reversal symmetry. When $E_F>0$, the outer contour belongs to the lower branch and the inner one to the upper branch. For $E_F<0$ the two Fermi contours both belong to the lower branch and have the same tangential spin orientation at a given angle in polar coordinates. At $k=0$ the Hamiltonian has quantum degeneracy in spin, and no unique spin texture direction is selected.

A weighted quadratic Bézier curve with positive middle weight $w$ retains endpoint interpolation and endpoint tangent directions while changing the bulge. For a circular arc of angle $\theta$ less than $180^{\circ }$, choose endpoints on the circle, their tangent intersection as the middle control point, and $w=\cos (\theta /2)$. Unlike an ordinary nonconstant polynomial Bézier curve, this rational representation can describe a circular arc exactly.

The Rayleigh distribution is the probability distribution of the Euclidean norm of a two-dimensional random vector with independent zero-mean normal random variables of equal variance as components. If $X,Y\sim N(0,{\sigma }^2)$ are independent, thenHere $\sigma$ is the standard deviation of each component, not of the radius $R$. At $\sigma =1$ this gives the unit-scale probability density function $r{e}^{-r^2/2}$.^[ref]

This geometry turns two signed errors into one nonnegative distance. For example, suppose horizontal and vertical positioning errors are independent $N(0,4{\mathrm{cm}}^2)$ random variables. Their means both vanish, but the distance from the intended point is usually positive. The Rayleigh distribution with $\sigma =2\mathrm{cm}$ gives the probability of landing inside any centered disk. These are chosen model assumptions, rather than measurements of a particular instrument.

An apparently paradoxical feature is that the most likely location per unit area is the origin, while the radial mode is $\sigma$. A narrow annulus farther out has more area. Near zero that growing circumference outweighs the falling planar density; farther out the exponential function wins. Distinguishing densities measured against different elements of area or length resolves the paradox.

For $r\ge 0$, integration gives the survival function and cumulative distribution functionSolving $F_R(r_p)=p$ gives the quantile function^[ref]

For the chosen positioning model $\sigma =2\mathrm{cm}$,The disk of radius one component standard deviation therefore contains about 39.35% of outcomes; the familiar one-dimensional normal interval percentage does not apply to a two-dimensional disk. The 95% and 99% radial thresholds areA circle defined by $R=r_p$ itself still has probability zero; its enclosed disk carries the stated probability.

The closed-form quantile function also gives inverse transform sampling: if $U$ has uniform distribution on $(0,1)$, then $\sigma \sqrt{-2\log U}$ has Rayleigh distribution. Using $1-U$ instead gives the same law. Endpoints are excluded so that the logarithm stays finite.

A real line bundle is a rank-one real vector bundle. Over a paracompact Hausdorff space, or over a smooth manifold in the smooth category, it is orientable exactly when it is a trivial vector bundle.

To prove the nontrivial implication, choose a fiber metric. An orientation of a vector bundle selects one of the two unit vectors in each fiber. These choices agree continuously in oriented local frames, giving a global nowhere-zero section. The map $(x,t)\mapsto ts(x)$ is then a fiber-linear product isomorphism. Conversely, a trivial vector bundle has the positive standard basis as a continuous orientation of a vector bundle.

The Möbius line bundle over a circle is the basic counterexample to triviality: one circuit reverses the fiber orientation, so no global positive unit section of a vector bundle can be chosen.

The real projective line ${\mathbb{RP}}^1$ consists of one-dimensional vector subspaces of ${\mathbb{R}}^2$, written $[x:y]$ with $(x,y)\ne (0,0)$ and $[x:y]=[tx:ty]$ for every nonzero real $t$. The chart $y\ne 0$ gives the affine coordinate $x/y$, while $[1:0]$ is one added point at infinity.

A unit vector $(\cos \theta ,\sin \theta )$ and its negative determine the same line, so the angular coordinate is taken modulo $\pi$. This quotient is itself a circle: the map sending that line to $(\cos 2\theta ,\sin 2\theta )$ is well defined and gives a homeomorphism with $S^1$.

Hence the real projective line is compact and can be viewed as the one-point compactification of the real line. Both $+\infty$ and $-\infty$ approach the same added projective point, unlike the two distinct endpoints of an extended closed real interval.

Write a generalized circle equation as $a|z{|}^2+b\bar{z}+\bar{b}z+c=0$, where $a,c$ are real. Under the reciprocal Möbius transformation $z=1/w$, with $w\ne 0$, multiply by $|w{|}^2$ to obtainThus the coefficient triple transforms as $(a,b,c)\mapsto (c,\bar{b},a)$. A nondegenerate circle through zero has $c=0$ and becomes a straight line; a straight line missing zero becomes a circle through zero on the Riemann sphere. When working only with finite nonzero coordinates, retain the excluded point $w=0$ even if it satisfies the multiplied equation.

This reciprocal transformation is different from inversion in a circle, whose unit-circle formula is $z\mapsto 1/\bar{z}$. The latter reverses orientation; the reciprocal transformation is holomorphic away from its pole.

For real $\alpha \ne 0,\gamma$ and complex $\beta \ne 0$, completing the square in $\alpha |z{|}^2-\beta \bar{z}-\bar{\beta }z+\gamma =0$ gives center $\beta /\alpha$ and radius $\sqrt{|\beta {|}^2-\alpha \gamma }/|\alpha |$, when the radicand is nonnegative. The displayed map is reflection in the line through $0$ and $\beta$: it preserves the complex modulus, fixes the center, and squares to the identity. Hence it maps the circle onto itself, including a radius-zero degeneration.

The residue of a function $f$ at an isolated singularity $a$ is its Laurent coefficient of degree $-1$. More explicitly, if $f$ is holomorphic for $0<|z-a|<\rho$, write its Laurent series asThis one Laurent coefficient determines the contour integral around a sufficiently small counterclockwise circle about $a$. A complicated local Laurent series can therefore leave a remarkably simple answer after one complete trip around the point.

For a simple pole, multiplying by $z-a$ removes the pole, leaving a holomorphic extension whose value at $a$ is the residue:There is a useful shortcut for a quotient $f=g/h$. If $g$ and $h$ are holomorphic near $a$, $h$ has a simple zero there, and $g(a)\ne 0$, then $f$ has a simple pole andThe first-order Taylor series of $h$ explains the denominator: dividing $h(z)$ by $z-a$ leaves its derivative at $a$ in the limit. This computes a residue without finding the full Laurent series, which is especially useful when the numerator or denominator is complicated. ^[ref]

For example, the rational function $f(z)=1/(z^2+1)$ has simple poles at $i$ and $-i$. Factoring the denominator makes both residues immediately visible:The signs matter: these residues cancel when both poles are enclosed, but either one contributes when enclosed alone. The residue theorem below turns this distinction into different contour integrals of the same function.

A residue is not the same thing as a pole, nor does it measure the order of a pole. The function $1/z^2$ has a double pole at zero but residue zero. Conversely,has an essential singularity at zero and residue one. Infinitely many negative powers cause no difficulty in reading off that particular Laurent coefficient. At a removable singularity the residue is zero. ^[ref]

Why does only this Laurent coefficient survive? Parametrize a counterclockwise circle by $z=a+r{e}^{it}$, $0\le t\le 2\pi$, with $0<r<\rho$. For every integer $n$,When $n\ne -1$, the complex exponential completes a whole number of turns and its contributions cancel; when $n=-1$, it is constant. Uniform convergence of the Laurent series on this circle justifies termwise integration, giving $\oint f(z)dz=2\pi i\mathrm{Res}(f,a)$. Thus the residue isolates exactly the local information that a closed contour integral detects. ^[ref]

This selection rule explains why a very large value near an isolated singularity need not produce a large contour integral. The double pole $1/z^2$ grows faster near zero than $1/z$, yet its contributions around a circle cancel completely. What matters is the arrangement of complex values along the path, including the differential $dz$, rather than their moduli alone. A residue records that surviving contribution, not the size of the function near the point.

Let $\Gamma$ be a piecewise smooth contour forming a simple closed curve, positively oriented: travel counterclockwise, with the interior on the left. Suppose $f$ is holomorphic on an open set containing $\Gamma$ and its interior, except at finitely many isolated singularities $a_1,\dots ,a_m$ strictly inside. ThenNo isolated singularity may lie on $\Gamma$: the ordinary contour integral must be defined there. Reversing the direction of the contour reverses the sign of the answer. The theorem applies to essential singularities as well as poles. ^[ref]

Return to $f(z)=1/(z^2+1)$. A counterclockwise circle $|z|=2$ encloses both simple poles, soYet $f$ is nonzero everywhere it is defined. A zero contour integral does not imply a zero function, or even absence of poles: the enclosed residues can cancel. By contrast, the counterclockwise circle $|z-i|=1/2$ encloses only $i$, givingThe picture compares these two selections. Changing the shape of a contour while keeping the same enclosed poles leaves the contour integral unchanged, provided the contour deformation stays where $f$ is holomorphic.

This is the practical appeal of the residue theorem: local Laurent coefficients determine a global contour integral. For a rational function, a few simple pole computations can replace a long parametrized calculation. It also makes real improper integrals accessible by treating the real line as part of a closed contour in the complex plane.

To compute $I={\int }_{-\infty }^{\infty }dx/(1+x^2)$, take $R>1$ and travel along the real line from $-R$ to $R$, then return along the upper semicircle $A_R$, parametrized by $z=R{e}^{it}$ for $0\le t\le \pi$. This contour is positively oriented and encloses the simple pole $i$ only; $-i$ is outside. HenceThe added arc must be controlled before taking a limit. On $A_R$, the reverse triangle inequality gives $|z^2+1|\ge R^2-1$, and the arc length is $\pi R$. The estimation lemma therefore yieldsThe denominator grows quadratically while the arc length grows only linearly. This quantitative comparison, rather than a sketch alone, explains why closing the contour does not alter the limiting real improper integral. It even supplies an error bound: the difference between $\pi$ and the finite real integral is at most $\pi R/(R^2-1)$.

For a numerical check, $R=10$ gives the finite real integral $2\arctan (10)\approx 2.94226$, leaving about $0.19934$ to reach $\pi$. The arc bound is $10\pi /99\approx 0.31733$, comfortably larger than that error. The graph shows both quantities approaching zero. The exact tail here can be checked with the antiderivative $\arctan x$; the residue theorem method remains useful for improper integrals where a convenient real antiderivative is unavailable.

Finally, both one-sided improper integrals converge absolutely: for $|x|\ge 1$, $0<1/(1+x^2)\le 1/x^2$, whose tails are integrable. Thus the symmetric limit computes the ordinary improper integral, not merely a Cauchy principal value, andClosing below instead would enclose the simple pole $-i$. With the same left-to-right traversal of the real line, that contour is clockwise, so its contour integral is $-2\pi i\mathrm{Res}(f,-i)=\pi$ again. This provides a useful sign check: the opposite residue is accompanied by the opposite direction around the enclosed region.

The same strategy extends to many other rational functions: choose a contour, identify its enclosed isolated singularities, compute their residues, and bound the added pieces. The choice of contour and the bound are as much part of the calculation as the residues. ^[ref]

Every supporting centre of an arc between fixed endpoints lies on their perpendicular bisector. Multiplicity at least $k$ therefore makes that line contain at least $k$ of the original points. For each such line containing $s$ points, its $s$ centres each support at most $t$ circles; on one circle at most two consecutive-point arcs have that symmetry axis. The rich line bound summed over dyadic occupancies gives the displayed bound.