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Hyperbolic functions combine the exponential function to describe a hyperbola much as trigonometric functions describe a circle. Their two basic members are the hyperbolic cosine and hyperbolic sine, defined for every real number $t$ byTheir derivatives exchange the two functions:These formulas follow by applying differentiation to the exponential function definitions. ^[ref]

Adding and subtracting gives $\cosh t+\sinh t=e^t$ and $\cosh t-\sinh t=e^{-t}$. Multiplying these equalities proves the hyperbolic Pythagorean identity:Consequently the parametric equation $(x,y)=(\cosh t,\sinh t)$ traverses the entire right branch of the unit hyperbola $x^2-y^2=1$, where $x\ge 1$. For example, $t=0$ gives $(1,0)$, while $t=\pm 1$ gives approximately $(1.543,\pm 1.175)$. Compare $(\cos \theta ,\sin \theta )$ on the unit circle, where the squared coordinates add to one. The minus sign changes the geometry from a closed circle to an unbounded hyperbola.

The other members are ratios and reciprocals: hyperbolic tangent is $\tanh t=\sinh t/\cosh t$, hyperbolic cotangent is $\coth t=\cosh t/\sinh t$, and hyperbolic secant is $\mathrm{sech}t=1/\cosh t$; denominators must be nonzero. The hyperbolic sine and hyperbolic cosine both solve the second-order linear differential equation $f^{\prime \prime }=f$. In fact $f(t)=A\cosh t+B\sinh t$ has $f(0)=A$ and $f^{\prime }(0)=B$, making this basis convenient for an initial value problem. Their geometry also makes hyperbolic substitution useful for integrals containing $\sqrt{1+u^2}$, and the hyperbolic cosine gives the physical catenary of a hanging chain.

An ideal endpoint of a hyperbolic line is one of its two limiting points on the boundary at infinity. In the Poincare disc model these lie on the unit circle; in the upper half-plane model they lie on $\mathbb{R}\cup \{\infty \}$. These are boundary points, not points of the hyperbolic plane. Any two distinct ideal endpoints determine a unique hyperbolic line.

An equation can define a function without giving an explicit formula for it. The implicit function theorem tells us when the constraint $F(x,y)=0$ determines $y$ uniquely from nearby $x$, and makes the resulting function continuously differentiable. It is useful when solving an equation explicitly is cumbersome or impossible: the partial derivatives already reveal whether local recovery is possible.

Let $F$ be a continuously differentiable map from an open subset of ${\mathbb{R}}^m\times {\mathbb{R}}^n$ to ${\mathbb{R}}^n$. Suppose $F(x_0,y_0)=0$ and the partial derivative $D_{y}F(x_0,y_0)$ is an invertible linear map. There are open neighbourhoods $U$ of $x_0$ and $V$ of $y_0$ and a unique continuously differentiable function $g:U\to V$ such that, for $(x,y)\in U\times V$,Thus the nearby level set is a graph of a function. The derivative of this graph isThe matrix inverse acts on the left; the order matters in several dimensions.

For example, take the unit circle, written as the level set $F(x,y)=x^2+y^2-1=0$. At $(x_0,y_0)=(3/5,4/5)$, $F_y=2y_0=8/5\ne 0$, so the local upper branch is $y=g(x)=\sqrt{1-x^2}$. Implicit differentiation givesThis tangent line slope is available directly from the constraint. The negative square root branch is elsewhere on the same circle; it does not destroy uniqueness near the selected upper point.

At $(1,0)$, however, $F_y=0$ and the circle cannot be represented locally as one function $y=g(x)$: just to the left of $x=1$ there are two nearby values of $y$. The circle is still smooth there. Since $F_x=2\ne 0$, use the other coordinate instead: $x=h(y)=\sqrt{1-y^2}$, with $h^{\prime }(0)=0$. A failed coordinate choice is different from a geometric singularity.

The derivative condition is sufficient, not necessary for every possible implicit graph. For $F(x,y)=y^3$, the entire zero set is the smooth graph $y=0$, even though $F_y$ vanishes there. In this case the defining equation itself has a degenerate derivative; the theorem's test gives no conclusion.

The connection to the inverse function theorem is particularly elegant. Augment the constraint to the map $H(x,y)=(x,F(x,y))$, whose Jacobian matrix isIts determinant is $\det D_{y}F$. When this is nonzero, locally invert $H$ and set its second output to zero. The recovered second input is $g(x)$. In this way an implicit constraint becomes an ordinary inverse function problem. ^[ref]

An implicit solution describes a solution of a differential equation through a relation $F(x,y)=C$, instead of isolating $y$ in an explicit formula. The relation describes a level set; an actual scalar solution of a differential equation must be a differentiable function $y(x)$ on an interval whose function graph lies in the equation's domain. A whole level set need not be one such function graph.

For example, considerImplicit differentiation of $x^2+y^2=C$ gives $2x+2y{y}^{\prime }=0$. For $C>0$, the circle contains two solutions of a differential equation,Both satisfy the equation on the stated open interval, and their function graphs stay in $D$. An initial condition $(x_0,y_0)\in D$ fixes $C=x_0^2+y_0^2$ and selects the sign of $y_0$. In particular, $y(0)=2$ gives $C=4$ and the upper branch on $(-2,2)$. The constant alone does not select the branch.^[ref]

The precise local criterion uses the implicit function theorem. Let $F$ be continuously differentiable on an open subset of the plane, with $F(x_0,y_0)=C$. If its partial derivative $F_y(x_0,y_0)\ne 0$, the nearby level set is a unique continuously differentiable function graph $y=g(x)$ through that point. The chain rule givesThis proves that the local function graph solves $y^{\prime }=-F_x/F_y$ there. For a different prescribed equation $y^{\prime }=f(x,y)$, one must also check $F_x+F_{y}f=0$ along that branch and that it stays in the domain of $f$. A level set by itself does not verify an arbitrary ordinary differential equation.^[ref]

For $F=x^2+y^2$, the upper branch through $(1,\sqrt{3})$ has $F_y=2\sqrt{3}$ and tangent line slope $-1/\sqrt{3}$. At $(2,0)$, however, $F_y=0$ and $F_x=4$. The circle remains a smooth regular level set with a vertical tangent line: it can be written locally as $x=\sqrt{4-y^2}$, but not as a single function graph $y=g(x)$ over an open interval containing $2$. Moreover, $-x/y$ is undefined there, so the scalar solution of a differential equation cannot include this point in its domain. Changing the parameter draws the curve through the point but does not extend that scalar initial value problem.

When both partial derivatives vanish, even the geometry needs separate inspection. The level $x^2+y^2=0$ is an isolated point, not a function graph on any nonempty open interval. The zero level set of $G(x,y)=y^2-x^2$ instead consists of the two lines $y=x$ and $y=-x$ crossing at the origin, where $\nabla G=0$. Each line satisfies the implicit ordinary differential equation $y{y}^{\prime }=x$, including at the origin; the explicit form $y^{\prime }=x/y$ excludes the origin. Thus algebraically dividing by a vanishing coefficient changes the domain and can change which solutions of a differential equation are admissible. Vanishing $F_y$ alone is only failure of a sufficient test: $F(x,y)=y^3$ has the smooth zero function graph $y=0$ despite $F_y=0$ everywhere on it.

Implicit solutions are especially useful for exact first-order ordinary differential equations and for finding a first integral of an ordinary differential equation: a conserved relation can reveal allowed states without an explicit formula. For the normalized harmonic oscillator written as $\dot{x}=y$, $\dot{y}=-x$, the energy $H=(x^2+y^2)/2$ has $\dot{H}=x\dot{x}+y\dot{y}=0$. Its positive level sets are the same circles. Where $y\ne 0$, eliminating time gives $dy/dx=-x/y$ and the local implicit solution above. The harmonic oscillator's parametric curve continues through $y=0$ because its vector field governing evolution in time is defined there; treating $x$ as the independent variable fails at that turn. The origin is an equilibrium point of the harmonic oscillator, although it is excluded from the scalar equation $y^{\prime }=-x/y$.

The incircle of a triangle is the circle inside the triangle tangent to all three sides. Its center is the intersection of two internal angle bisectors; its radius is $r=\Delta /s$, where $\Delta$ is the area and $s$ the semiperimeter. To prove it is the largest contained circle, let any candidate have center $P$, radius $\rho$ and distances $d_i\ge \rho$ from the three side lines. Splitting the triangle into three triangles with vertex $P$ gives $\Delta =\frac{1}{2}{\sum }_i{\ell }_i{d}_i\ge s\rho$. The tangent circle attains equality, proving maximality. For a right triangle with legs $a,b$ and hypotenuse $c$, $r=(a+b-c)/2$, since $r=ab/(a+b+c)$ and $a^2+b^2=c^2$.

Let $R>0$ and let $f$ be a holomorphic function on an open neighbourhood of the closed disk $\{z:|z|\le R\}$. Assume $f(0)\ne 0$ and that $f$ has no zeros of a function on the boundary circle $|z|=R$. List its interior zeros of a function as $a_1,\dots ,a_N$, repeating each according to its multiplicity of a zero. With the natural logarithm, Jensen's formula statesThe empty sum is zero. This boundary-zero-free version is the holomorphic special case of the general identity in the source.^[ref]

Every interior zero of a function makes a positive contribution; a zero nearer the centre contributes more. Consequently boundary growth relative to $|f(0)|$ limits how many zeros of a function can lie well inside the disk. The Jensen zero-count bound makes this quantitative, while the integrated zero count retains the individual radial weights. The boundary statistic here is the circle log-mean of a holomorphic function; exponentiating gives a circle geometric mean of a holomorphic function.

For knots, matching bracket smoothing states across a connected sum of knots joins one state circle from each side. Their total circle count is therefore the sum of the two counts minus one. The reduced Kauffman bracket state weights multiply, and writhe of a link diagram adds, proving the formula.

Placing the two link diagrams in disjoint disks keeps all state circles separate. The reduced Kauffman bracket therefore acquires one extra loop factor $-A^2-A^{-2}$. Correcting writhe of a link diagram and setting $t=A^{-4}$ gives the displayed formula. It concerns a split link, not an arbitrary union of linked components.

A state with $k$ B smoothings has at most $s_A+k$ circles, so its largest A exponent is at most $c(D)+2s_A-2$. The dual argument bounds the smallest exponent below by $-c(D)-2s_B+2$. For a connected link diagram, the Turaev surface inequality bounds the resulting breadth by $4c(D)$. For a connected reduced alternating diagram, adequacy prevents cancellation at both extremes and planar Euler characteristic gives equality. A disconnected projection needs the additional term in the Jones polynomial breadth bound for a disconnected diagram.

A knot is the image $K=f(S^1)$ of a smooth embedding $f:S^1\hookrightarrow {\mathbb{R}}^3$, considered up to ambient isotopy. The circle $S^1$ supplies a closed loop with no free ends; the smooth embedding condition rules out self-intersections and requires a nonzero tangent vector in a regular parametrization. The equivalent convention $S^1\hookrightarrow S^3$ uses the three-sphere: removing a point outside the knot identifies the remaining space with three-dimensional Euclidean space by stereographic projection.

More precisely, knots $K_0,K_1$ are equivalent if there is a smooth family of diffeomorphisms $F_s$ of the ambient space, for $0\le s\le 1$, such that $F_0$ is the identity and $F_1(K_0)=K_1$. Every intermediate $F_s(K_0)$ remains a smooth embedding. This ambient isotopy allows bending, stretching and moving a loop, but neither cutting it nor letting strands cross through each other. We compare the embedded images rather than their parametrizations; an oriented knot additionally remembers a direction of travel.

For a worked example, start with $U(t)=(\cos t,\sin t,0)$, where $t$ is taken modulo $2\pi$, and useBoth scale factors stay positive, so each $F_s$ is a diffeomorphism. At $s=1$ the circle becomes the ellipse $(2\cos t,\frac{1}{2}\sin t,0)$, yet its knot type is still the unknot. Exact lengths and angles are therefore irrelevant to knot type; self-intersection is the obstruction that the deformation must avoid.

Knot theory studies closed curves in three-dimensional Euclidean space that cannot intersect themselves, and asks which can be deformed into one another without cutting or passing one strand through another. Its basic object is a knot: a smooth embedding of a circle, considered up to ambient isotopy. Several disjoint knots form a link. More generally, knot theory studies such embeddings in other manifolds and in higher dimensions.

The central surprise is that a flexible circle can carry permanent information about how it sits in space. A trefoil knot and an unknot each consist of just one circle, but no ambient isotopy turns one into the other. A knot diagram translates this spatial question into a finite drawing; Reidemeister moves describe the changes of drawing that preserve the knot. Knot invariants translate it into computations: unequal Alexander polynomials of a knot, Jones polynomials, or knot groups certify different knots. Equal values of a particular knot invariant generally leave the question open; even an isomorphism of knot groups alone is not a complete classification.

An early foundation for working with knot diagrams is Kurt Reidemeister's 1927 paper Elementare Begründung der Knotentheorie, published in volume 5 of Abhandlungen aus dem Mathematischen Seminar der Universität Hamburg, pages 24–32.^[ref] Marc Lackenby's authored notes Elementary knot theory, Section 2.4, explain the Reidemeister move theorem and its role in searching for a certificate that two knot diagrams represent the same knot.^[ref]