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A homogeneous linear differential equation is a linear differential equation with identically zero forcing term, $L[y]=0$. On a common interval, its solutions of a differential equation form the vector space $\ker L$, the kernel of a linear map. They include the zero function and are closed under addition and scalar multiplication. For any two solutions of a differential equation $u,v$ and constant scalars $\alpha ,\beta$, linearity gives $L[\alpha u+\beta v]=\alpha 0+\beta 0=0$: this is the superposition principle. The scalars must be constant; variable multipliers introduce extra derivatives.

For example, $y^{\prime }+2y=0$ has general solution $y=C{e}^{-2t}$, where $C$ is any constant real number. A short verification uses the product rule: $(e^{2t}y{)}^{\prime }=e^{2t}(y^{\prime }+2y)=0$, so $e^{2t}y$ is constant. For $y_1=e^{-2t}$ and $y_2=2e^{-2t}$, the sum is $3e^{-2t}$ and also solves the same homogeneous linear differential equation. At $t=1$, their values are approximately $0.13534$, $0.27067$ and $0.40601$. Every curve is a multiple of one exponential function; the solution space of a homogeneous linear differential equation has dimension one. Both $t$ and $y$ in this example and the following diagram are dimensionless mathematical variables.

Now keep the same differential operator $L[y]=y^{\prime }+2y$ but prescribe the forcing term $1$. The inhomogeneous linear differential equation $y^{\prime }+2y=1$ has particular solution $y_p=1/2$ and general solution $y=1/2+C{e}^{-2t}$. The initial conditions $y(0)=0,1,2$ select, respectively,At $t=1$, their values are approximately $0.43233$, $0.56767$ and $0.70300$. The midpoint $\frac{1}{2}(u_0+u_2)=u_1$ solves the same inhomogeneous linear differential equation, because $L[\frac{1}{2}(u_0+u_2)]=\frac{1}{2}(1+1)=1$. The sum $s=u_0+u_2=1+e^{-2t}$ instead satisfies $s^{\prime }+2s=2$, so it fails the equation with forcing term $1$. More generally, a linear combination of solutions of a differential equation for the same fixed nonzero forcing term retains that forcing term precisely when its scalar weights add to one. This is the affine space structure, rather than unrestricted superposition within that fixed family.

On the left, all three curves solve the same homogeneous linear differential equation. On the right, $u_0,u_1,u_2$ share the particular solution $1/2$, and the shaded difference $u_2-1/2$ is the homogeneous solution $\frac{3}{2}{e}^{-2t}$. The dashed sum has twice the forcing term, even though its starting value coincides with that of $u_2$. Adding the two initial conditions also adds their inputs; taking their midpoint averages both. This distinction makes the superposition principle useful for assembling responses to different inputs without mistaking that assembly for a solution of a differential equation with the original input. See MIT's superposition notes, including the particular solution one half.

For a real-valued function drawn in the plane, the horizontal line test asks whether every horizontal line intersects its function graph at most once. This is exactly injectivity, since two intersections at the same height mean two inputs have the same output. A missing height is compatible with injectivity, so an inverse function is then defined on the image of a function; it is an inverse on the entire specified codomain only when surjectivity also holds.

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.

The identity function on a set $X$ maps each $x\in X$ to itself: ${\mathrm{id}}_X(x)=x$. It changes no information. For every function $f:X\to Y$, function composition satisfies $f\circ {\mathrm{id}}_X=f$ and ${\mathrm{id}}_Y\circ f=f$. The subscripts specify the appropriate source or target set. An inverse function reverses a bijection exactly when the two possible compositions are these identity functions.

The image of a function asks which outputs a chosen set of inputs reaches. A preimage asks which inputs reach a chosen set of outputs. Both constructions apply to every function, including one that loses information by sending several inputs to the same output. Their difference is useful whenever a transformation must be followed forward or a condition on its result must be traced back.

Consider the function $f:A\to B$ withEvery input has exactly one output, but zero has two inputs and three has none. The full image of a function is $f(A)=\{0,1,2\}$, a proper subset of the codomain $B$. Selecting inputs $\{a,c\}$ gives image $\{0,1\}$. Selecting outputs $\{0,2\}$ instead gives preimage $\{a,b,d\}$.

The function is not injective, because $f(a)=f(b)$, and is not surjective onto $B$, because three is absent from its image. Restricting only the codomain to $\{0,1,2\}$ makes the same rule surjective; it remains noninjective. This separates the output set actually reached from the target set specified in the definition. The definitions of images of a function and preimages, including the distinction from an inverse function, are given in Richard Hammack, Book of Proof, §12.6: Image and Preimage.

For a function $f:X\to Y$ and a subset $E\subseteq X$, its image isThe full image of a function is $f(X)$. The set notation ignores repeated outputs: in the finite example above, $f(\{a,b\})=\{0\}$, rather than a list containing zero twice. A function collision can therefore reduce the cardinality of an image.

Images of a function preserve unions:An output reached from the union is reached from at least one of its two input sets. For intersections, however, onlyholds in general. In the finite example, take $E=\{a\}$ and $F=\{b\}$. Their intersection is the empty set, so $f(E\cap F)=\varnothing$. But $f(E)=f(F)=\{0\}$, giving $f(E)\cap f(F)=\{0\}$. Distinct witnesses in the two input sets can produce the same output.

For an injective function, an output shared by $f(E)$ and $f(F)$ must come from the same input. That input lies in $E\cap F$, so equality holds. Conversely, preserving intersections for every pair of input subsets forces injectivity: two distinct inputs with the same output would give the singleton counterexample above. The image and preimage laws are stated in Jiří Lebl, Basic Analysis I, Propositions 0.3.15–0.3.16; the finite example makes the missing reverse inclusion visible.

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]

Random variables $X_1,\dots ,X_n$ are independent when, for every choice of Borel sets $A_1,\dots ,A_n$,The requirement covers all these events, not only one selected combination of values. Equivalently, the generated sigma-algebras are independent sigma-algebras, or the joint probability distribution is the product measure of the marginal distributions. For an infinite family, the same condition must hold for every finite subfamily. For random elements, replace Borel sets by measurable sets in their specified target spaces.

For discrete random variables, checking every entry of the joint probability mass function is enough:Summing this identity over the desired sets proves the event formulation. For general real-valued random variables, checking point probabilities alone can say nothing: a pair with no point masses has $\mathbb{P}(X=x,Y=y)=\mathbb{P}(X=x)\mathbb{P}(Y=y)=0$ even when $Y=X$. The all-set definition avoids that trap. See Gautam Iyer, Conditional Probability and Independence, definitions 12, 16 and 18 and proposition 19, with the continuous case treated in his ^[ref].

For a concrete comparison, let $X,Y$ be independent fair bits: each has a Bernoulli distribution with parameter $1/2$. Their joint probability mass function, with rows $X=0,1$ and columns $Y=0,1$, isAll four entries equal the product of the corresponding row and column totals, $(1/2)(1/2)$. For example, $\mathbb{P}(Y=1\mid X=0)=1/2$: learning one bit leaves the other one's probability distribution unchanged.

Now keep a fair $X$ but set $Y=X$. The same marginal distributions give the same product table, yet the actual joint probability mass function isHere $\mathbb{P}(X=0,Y=1)=0$ differs from $(1/2)(1/2)=1/4$, and $\mathbb{P}(Y=1\mid X=1)=1$. Thus matching marginal distributions does not establish independence. Grinstead and Snell formulate the discrete product criterion in Introduction to Probability, section 4.1, definition 4.4, printed page 143.

Independence is stronger than zero covariance. For square-integrable independent random variables, $\mathbb{E}[XY]=\mathbb{E}[X]\mathbb{E}[Y]$, so they are uncorrelated random variables. The converse fails: take $X$ with a discrete uniform distribution on $\{-1,0,1\}$ and $Y=X^2$. Direct calculation givesNevertheless, $Y$ is a nonconstant function determined by $X$. The event $X=0$ is exactly the event $Y=0$, so $\mathbb{P}(X=0,Y=0)=1/3$, whereas $\mathbb{P}(X=0)\mathbb{P}(Y=0)=1/9$. The positive and negative contributions to the covariance cancel without erasing the statistical dependence.

The usefulness of independence is that a large joint probability distribution can be assembled from small pieces. For $n$ fair independent bits, each of the $2^n$ patterns has probability $2^{-n}$. Products also simplify expected values, and vanishing cross covariances simplify the variance of a sum. Separate measurable functions of independent random variables remain independent: each output event is already an input event, so the defining product rule still applies. These are consequences of a model's independence assumption; whether the assumption describes actual observations is a question for statistical inference.

For a field evolving by a Hamiltonian partial differential equation, infinite-dimensional integrability combines sufficiently rich commuting conservation laws with a mechanism such as the inverse scattering transform or a spectral coordinate system that linearizes evolution. Lax equations supply spectral invariants; Lenard-Magri recursions supply commuting Hamiltonians when their solvability conditions hold. Infinite many conserved quantities alone do not prove completeness or an infinite-dimensional action-angle theorem: the function space, boundary conditions and analytic independence all matter.

An infinite path is a binary function all of whose finite prefixes belong to the tree. Its path space $[T]$ is closed in Cantor space, since failure is witnessed by one finite prefix outside $T$.

An influence-function representer is a centered square-integrable function representing derivatives of a statistical functional along all admissible score functions. Representers may differ by a function orthogonal to the statistical tangent space. This pathwise definition is distinct from defining an influence function solely by point-mass contamination paths.

An initial value problem combines an ordinary differential equation with an initial condition, for example $y^{\prime }=f(t,y)$ and $y(t_0)=y_0$. The task is to find a solution of a differential equation on an interval containing $t_0$ which has that starting value. For a continuous linear differential equation $u^{\prime }=A(t)u$, every prescribed finite vector $u_0$ determines a unique solution on the coefficient interval. A boundary value problem can instead prescribe values at distinct endpoints.

For the cooling initial value problem $y^{\prime }=-k(y-T_a)$, $y(t_0)=y_0$, where $k>0$ and $T_a$ are constant, the unique solution isTake $T_a=20$ degrees Celsius, $y_0=80$ degrees Celsius, $t_0=0$, and $k=0.1$ per minute. Then $y(10)=20+60e^{-1}\approx 42.07$ degrees Celsius. Reaching $30$ degrees Celsius requires $60e^{-0.1t}=10$, so $t=10\log 6\approx 17.92$ minutes. The time constant $1/k=10$ minutes reduces the excess temperature by a factor $e^{-1}$; it does not reduce the absolute temperature by that factor.

Uniqueness needs assumptions. If $f$ is a continuous function on an open neighborhood of $(t_0,y_0)$ and satisfies a local Lipschitz condition in $y$, uniformly in $t$, the Picard-Lindelöf theorem gives a unique local solution. Concretely, a finite $L$ bounds $|f(t,u)-f(t,v)|\le L|u-v|$ in a sufficiently small rectangle. Cooling satisfies this with $L=k$. Local existence does not promise existence for all future time: $y^{\prime }=y^2$, $y(0)=1$, has $y(t)=1/(1-t)$ on $(-\infty ,1)$ and finite-time blowup at $t=1$.

Continuity alone is insufficient for uniqueness. Consider the dimensionless initial value problem involving a square root and an absolute value:Besides $y(t)\equiv 0$, each delay $a\ge 0$ gives a solution for $t\ge 0$:At $t=4$, delays $a=0,1,2$ give $y=4,9/4,1$, despite identical starting data. These are valid solutions of a differential equation: for $a>0$, both one-sided derivatives at the join are zero; for $a=0$, the right derivative at the starting time is zero. After the join, $y_a^{\prime }=(t-a)/2=\sqrt{|y_a|}$ because $t>a$. Squaring alone would also suggest a parabola before the join, but its negative derivative there would violate the equation. Since $y^{\prime }\ge 0$, a forward solution starting from zero stays nonnegative.

The right-hand side is a continuous function but fails the Lipschitz condition at zero: for $h>0$, $|\sqrt{|h|}-0|/|h|=1/\sqrt{h}$ is unbounded. Thus the Peano existence theorem still supplies local existence, while the Picard-Lindelöf theorem does not supply uniqueness. Related waiting-time examples and a sufficient uniqueness condition appear in MIT 18.034, Lecture 2: Fundamental Principles.

A brief proof outline explains the mechanism: rewrite the initial value problem as the integral equation $y(t)=y_0+{\int }_{t_0}^{t}f(s,y(s))ds$. On a short enough interval, the Lipschitz condition makes the right-hand side a contraction mapping on a closed set of continuous functions near $y_0$. The contraction mapping theorem gives a unique fixed point; the fundamental theorem of calculus turns it back into a solution.

An inner product assigns a scalar $⟨u,v⟩$ to two vectors and supplies their lengths and orthogonality. Using the convention linear in the first argument, it satisfiesThe resulting norm is $\Vert v\Vert =\sqrt{⟨v,v⟩}$. Over the real numbers, conjugation has no effect and the form is symmetric and bilinear. Over the complex numbers, it is sesquilinear; the convention linear in the second argument is also common, so formulas for coefficients must match the stated convention.

The Euclidean example is $⟨u,v⟩={\sum }_i{u}_i\overline{v_i}$. For real $u=(2,1)$ and $v=(1,3)$ this gives $⟨u,v⟩=5$, while $\Vert u{\Vert }^2=5$ and $\Vert v{\Vert }^2=10$. If $r=(-1,2)$, then $⟨r,u⟩=0$: perpendicularity becomes an algebraic equation. The Cauchy-Schwarz inequality gives $|⟨u,v⟩|\le \Vert u\Vert \Vert v\Vert$, and in a real inner product space it makes the angle formula $\cos \theta =⟨u,v⟩/(\Vert u\Vert \Vert v\Vert )$ meaningful for nonzero vectors.

Different inner products express different measurements. The weighted example $⟨u,v⟩=4u_1{v}_1+u_2{v}_2$ on ${\mathbb{R}}^2$ gives $\Vert (x,y){\Vert }^2=4x^2+y^2$, so its unit-length vectors lie on an ellipse. For functions, an integral such as $⟨f,g⟩=\int f\overline{g}$ supplies the geometry of the L2 space. This lets orthogonal projection and least-squares solutions apply to curves as well as coordinate lists.