Use radians and take $T(x)=\cos x$ on the closed interval $[0,1]$. This is a complete metric space with distance $|x-y|$, and $T([0,1])=[\cos 1,1]\subset [0,1]$. The mean value theorem and $|T^{\prime }(x)|=\sin x\le \sin 1$ give a contraction mapping with $L=\sin 1\approx 0.841471<1$. Both the interval and its preservation matter; a small derivative by itself would not ensure that all subsequent approximations stay in the chosen space.

Starting at $x_0=0$ gives $x_1=1$, $x_2\approx 0.540302$, $x_3\approx 0.857553$, and $x_4\approx 0.654290$. The cosine is decreasing here, so successive iterates alternate around the unique fixed point $x_{\ast }\approx 0.739085133215161$. The contraction iteration error bound needs no prior knowledge of this numerical value. At step 40, $x_{40}\approx 0.739085078689123$ and the latest-step bound is below $7.20\times 10^{-7}$, certifying error below one millionth for exact iteration. The initial-step bound requires 91 iterations for the same tolerance. The observed step-40 error is only about $5.45\times 10^{-8}$; the guarantees allow less favorable behavior than this example actually exhibits.

This is the beauty of the contraction mapping theorem: a uniform local improvement controls the entire infinite process. Completeness is essential. On the incomplete interval $(0,1)$, the contraction mapping $T(x)=x/2$ keeps every iterate inside but has no fixed point there; its missing limit is zero. Banach's 1922 paper developed abstract methods for different spaces of functions with applications to integral equations, a motivation still visible whenever repeated approximation is used to construct a function.^[ref]

Successive steps satisfy $d(x_{n+1},x_n)\le L^{n}d(x_1,x_0)$. Summing the remaining steps with the triangle inequality gives a bound available before the iteration starts:After computing a new step, the contraction mapping gives the often sharper boundEquivalently, the residual at any $z\in X$ satisfies $d(z,x_{\ast })\le d(z,Tz)/(1-L)$. These estimates convert visible step sizes into guaranteed distances to an unknown fixed point. When $L$ is close to one, the factor $1/(1-L)$ can be large: a small step alone needs interpretation through the contraction constant.

Let $a>0$ and let $g$ be continuous on the upper semicircle ${\gamma }_R:R{e}^{i\theta }$, $0\le \theta \le \pi$. If $M_R={\sup }_{{\gamma }_R}|g|\to 0$, thenIndeed, $|e^{iaz}|=e^{-aR\sin \theta }$. Symmetry and $\sin \theta \ge 2\theta /\pi$ on $[0,\pi /2]$ bound the arc integral by $2R{M}_R{\int }_0^{\pi /2}{e}^{-2aR\theta /\pi }d\theta \le \pi M_R/a$.

A rational factor behaving like $1/z$ therefore suffices, even though the crude length-times-supremum bound would only be of constant order. For $a<0$, use a lower semicircle instead. At $a=0$, the conclusion can fail: the integral of $1/z$ over the upper semicircle is $i\pi$ for every radius. The exponential's sign and the decay assumption are both part of the lemma.

For a simple closed curve in the standard oriented complex plane, positive orientation keeps the bounded interior on the left while traversing the curve. A circle is therefore traversed counterclockwise; reversing direction changes the sign of its contour integral.

A complex integration contour is an oriented piecewise continuously differentiable curve in the complex plane. For a parametrization $\gamma (s)$, its contour integral is ${\int }_{\Gamma }f(z)dz=\int f(\gamma (s)){\gamma }^{\prime }(s)ds$. Reversing orientation changes the sign. Unbounded contours require a convergent improper or explicitly regularized limit.

A commutator contour around two branch points cancels the total monodromy of its integrand. For the contour circling $1,0$ clockwise and then $1,0$ anticlockwise, with initial positive-real branches, the Beta function obeysThe contour integral is entire in $z$, giving a meromorphic continuation of the Beta function.

This contour integral defines an analytic function away from the oriented integration curve, under the relevant integrability assumptions. Its limiting values across the curve are related by the Sokhotski–Plemelj theorem. On an oriented real axis, with $Hf(\rho )={\pi }^{-1}\mathrm{PV}\int f(s)/(\rho -s)ds$, the signed Cauchy boundary operators are $C^{\pm }f=\pm f/2+iHf/2$.

A Bode plot displays the gain and oscillation phase of a frequency response against a logarithmic frequency axis. Gain is often shown as $20{\log }_{10}|G(i\omega )|$ in decibels; phase can use degrees or radians. The first-order low-pass filter loses approximately $20$ decibels per factor of ten in angular frequency well above its corner. ^[ref]

Every function in continuous functions on the p-adic integers has a unique uniformly convergent expansion in the binomial polynomials, with $a_n\to 0$ and $a_n=({\Delta }^{n}f)(0)$. Conversely, any coefficient sequence tending to zero defines such a continuous function. To prove the expansion once $a_n\to 0$ is known, use $|(\genfrac{}{}{0ex}{}{x}{n}){|}_p\le 1$ for uniform convergence. Finite binomial inversion gives agreement with $f$ at each nonnegative integer, and density gives agreement everywhere. The coefficients recover successively from these integer values, proving uniqueness. Moreover $\Vert f{\Vert }_{\infty }={\sup }_n|a_n{|}_p$, because finite differences bound each coefficient by the norm and the expansion gives the reverse inequality.

For a nonzero vector in a real or complex normed vector space, a norming functional is a continuous linear functional satisfying the displayed conditions. The Hahn-Banach theorem extends the functional sending $av$ to $a\Vert v\Vert$ on its one-dimensional span. Consequently $\Vert w\Vert \ge \mathrm{Re}\ell (w)$, with equality at $v$, giving a supporting linear lower bound for the norm. At zero the zero functional can be used as a supporting functional.

The finite negative continued fraction isFor $a_i\ge 2$, it describes a negative-framed chain of unknots: the diagonal framings are $-a_i$ and adjacent linking numbers are one. Repeated slam-dunk moves give rational surgery coefficient $-[a_1,\dots ,a_r{]}^{-}$. The corresponding chain matrix is negative definite. To check this, the successive determinants of its negative have recurrence $d_j=a_j{d}_{j-1}-d_{j-2}$, starting with $d_0=1$, $d_1=a_1$; they are positive and increasing, so Sylvester's criterion applies.

For a quadratic irrational $\sqrt{D}$, write its complete quotient as $(\sqrt{D}+m_j)/d_j$. The recurrence $m_{j+1}=d_j{a}_j-m_j$, $d_{j+1}=(D-m_{j+1}^2)/d_j$, $a_{j+1}=\lfloor (\sqrt{D}+m_{j+1})/d_{j+1}\rfloor$ computes its continued fraction. For $D=19$, the successive pairs after the initial quotient are $(4,3),(2,5),(3,2),(3,5),(2,3),(4,1)$, and then the cycle repeats. They give the displayed period of length six.

For a parametric surface $F(u,v)$ and a scalar implicit equation $f(P)=0$, set $g=f\circ F$. At a regular transversal intersection, $\nabla g\ne 0$, and $(-g_v,g_u)$ is a parameter tangent vector. Normalize its image under $DF$ to choose physical arc length steps. A predictor followed by Newton method correction of $g=0$ and a transverse plane equation gives a local continuation method. Certified patch bounds are useful for finding seeds; corner signs alone do not detect a closed intersection entirely inside a patch.

Let the survival time $T$ have an exponential distribution with rate $\lambda =0.2$ per year. Its hazard function is $h(t)=0.2$ per year, its cumulative hazard is $H(t)=0.2t$ for $t$ in years, and its survival function is $S(t)=e^{-0.2t}$. For a subject still event-free after five years, the conditional probability of an occurrence within the next two years isThe original population's probability of an occurrence between those same endpoints is instead $S(5)-S(7)=e^{-1}-e^{-1.4}\approx 0.121282$. Dividing by $S(5)\approx 0.367879$ recovers the conditional answer. The memoryless property makes the two-year conditional answer independent of the preceding five years.

For just $\Delta =0.01$ year, the exact conditional event probability is $1-e^{-0.002}\approx 0.001998$, close to the instantaneous approximation $h(5)\Delta =0.002$. For two years that approximation gives $0.4$, whereas the exact answer is about $0.329680$. At the single exact time five, the event probability is zero. These are three different quantities: an instantaneous hazard function with inverse-time units, a short-interval approximation, and an exact finite-interval probability.

In a plane set with $x,y>0$ and rational ratio $x/y$, any irrational number $a$ lying between two attained ratios separates the set into the nonempty relatively open sets $x/y<a$ and $x/y>a$. This exhibits a failure of connectedness using the continuous slope map, even though each individual ray is path-connected.

Suppose connected subsets $U_j$ all meet one connected member $U_1$. A separation of their union would place $U_1$ entirely on one side. Each $U_j$ is also entirely on one side and intersects $U_1$, so every $U_j$ lies on the same side, contradicting nonempty separation. Thus the union is connected. If the subsets are path-connected, join two points by a path to $U_1$, a path inside $U_1$, and a path back out. This needs pairwise contact with the hub, not one point shared by every subset.

The identity component is the connected component containing the identity element. It is a closed normal subgroup: products and inverses of connected sets through the identity remain in that component, and conjugation preserves it. It need not be open; in the additive group $\mathbb{Q}$ with its usual topology it is the nonopen singleton $\{0\}$.

For fixed $\ell$ and $g$, the exact conical pendulum oscillation period satisfies $P/P_0=\sqrt{\cos \theta }$. In the small-angle approximation, with $\theta$ in radians,Thus the limiting revolution period equals the small-amplitude oscillation period of a planar simple pendulum. This equality has a geometric explanation: to first order, $x(t)=r\cos (\omega t)$ and $y(t)=r\sin (\omega t)$ are perpendicular simple harmonic motions of equal amplitude, a quarter period apart, with $\omega \to \sqrt{g/\ell }$. At finite angle, however, the conical pendulum's oscillation period decreases with $\theta$, whereas a planar simple pendulum's oscillation period increases with its angular amplitude.

The endpoint $\theta =0$ itself is a stationary bob with $r=0$ and no observable revolution; $P_0$ is a limit of nonzero circular motions, not a measured revolution period at rest. For fixed $\ell$, nonzero conical pendulum motion also requires $\omega >\sqrt{g/\ell }$, because $\cos \theta =g/(\ell {\omega }^2)<1$. At the other endpoint, $\theta \to \pi /2$ gives $h\to 0$, $P\to 0$, and unbounded tension and speed. These mathematical limits explain why the ideal string cannot sustain a horizontal orbit at finite tension; a real experiment ceases to satisfy the ideal assumptions before that limit.

The motivating observation is a bob going around at almost constant height while its string traces a cone. The University of Iowa's demonstration places a camera above a short pendulum to reveal this motion. ^[ref] The following proposed measurement procedure develops that observation into a test of the conical pendulum model; it is not a report of collected data.

Suspend a compact bob from a rigid, fixed support using a light string, and measure $\ell$ from the pivot to the bob's centre. Put a stationary marker beside the orbit as a timing reference. Displace the bob sideways and give it a horizontal push tangent to the intended circle, then let go. For a chosen starting angle $\theta$, the predicted initial speed is $v=\sqrt{g\ell {\sin }^2\theta /\cos \theta }$. Adjust the launch between trials until a short observation interval has nearly constant orbit radius and height. Keep the pivot fixed while recording: circling the support by hand changes the problem, and a maintained drive would introduce another force.

Record from above with the camera looking approximately along the vertical axis. Check that the bob's centre follows a circle and that successive returns to the marker take nearly equal times. A separate side recording checks whether the bob's height stays constant. A markedly elliptical orbit or visibly rising and falling bob is not the uniform circular motion assumed by the conical pendulum equations. Measure the orbit diameter with a scale at the orbit's height; in a level side view, use the two extreme horizontal positions, not one arbitrary frame. Halve the diameter to obtain $r$, or measure the vertical drop $h$ directly in a calibrated side view. An oblique camera view and a scale at a different depth distort apparent lengths, so arrange the geometry before interpreting the images.

Time $N$ complete returns to the same marker in the same direction and estimate $P=\Delta t/N$. Counting ten revolutions, when the orbit stays sufficiently steady, spreads an endpoint timing error over ten periods. Repeat the timing and compare the first and last few revolutions to detect drift; using a longer interval helps only if the motion remains close to uniform. From independently measured geometry, compute $\sin \theta =r/\ell$ or $\cos \theta =h/\ell$, and check $r^2+h^2\simeq {\ell }^2$ if both were measured. Then compare the measured oscillation period with $P^2=4{\pi }^{2}h/g$. MIT's 8.01X experimental assignment explicitly asks students to compare the conical pendulum's period, length, and orbit radius with theory. ^[ref]

Repeat at several launch radii with the same $\ell$. Larger $r$ means smaller $h=\sqrt{{\ell }^2-r^2}$, so the ideal oscillation period decreases. A plot of measured $P^2$ against independently measured $h$ should follow a straight line through the origin with slope $4{\pi }^2/g$; conversely, $g=4{\pi }^{2}h/P^2$ estimates gravitational acceleration. Do not infer $h$ from the timing equation and then use that same equation as a test. Compare discrepancies with the spread of repeated timings and the resolution of the length measurements. String stretch, a moving pivot, and appreciable bob size challenge the stated geometry; drag force and pivot friction challenge the assumption of undriven uniform motion. Observe a short interval and record any shrinking or drifting orbit rather than silently treating it as an exact conical pendulum.

For an illustrative calculation, assume $\ell =1.00\mathrm{m}$, $m=0.050\mathrm{kg}$, $g=9.81m/{\mathrm{s}}^2$, and $\theta =40^{\circ }$. The model gives $r\approx 0.643\mathrm{m}$, $h\approx 0.766\mathrm{m}$, $P\approx 1.756\mathrm{s}$, $v\approx 2.300m/s$, and $T\approx 0.640\mathrm{N}$. Its vertical tension component is $0.4905\mathrm{N}$, equal to the gravitational weight; its inward component is approximately $0.412\mathrm{N}$. These are predictions from chosen parameters, not measured data or measurements of local $g$. The next figure uses the same stipulated $\ell$ and $g$, including illustrative angles $15^{\circ }$, $40^{\circ }$, and $60^{\circ }$.

In the focus-directrix description of a noncircular conic section, the focus is the fixed point $F$ for which every point $X$ of the curve satisfies $|X-F|=e\mathrm{dist}(X,L)$, where $L$ is a directrix of a conic and $e$ is the eccentricity of a conic. For $y^2=4ax$ with $a>0$, the focus of a conic is $(a,0)$, the directrix of a conic is $x=-a$, and $e=1$.