Functions in one real variable

This section is a brief review of function theory in one-variable calculus.

A real-valued function ${\displaystyle f:\mathbb {R} \to \mathbb {R} }$ is continuous at ${\displaystyle a}$ if it is approximately constant near ${\displaystyle a}$; i.e.,

${\displaystyle \lim _{h\to 0}(f(a+h)-f(a))=0.}$

In contrast, the function ${\displaystyle f}$ is differentiable at ${\displaystyle a}$ if it is approximately linear near ${\displaystyle a}$; i.e., there is some real number ${\displaystyle \lambda }$ such that

${\displaystyle \lim _{h\to 0}{\frac {f(a+h)-f(a)-\lambda h}{h}}=0.}$^[1]

(For simplicity, suppose ${\displaystyle f(a)=0}$. Then the above means that ${\displaystyle f(a+h)=\lambda h+g(a,h)}$ where ${\displaystyle g(a,h)}$ goes to 0 faster than h going to 0 and, in that sense, ${\displaystyle f(a+h)}$ behaves like ${\displaystyle \lambda h}$.)

The number ${\displaystyle \lambda }$ depends on ${\displaystyle a}$ and thus is denoted as ${\displaystyle f'(a)}$. If ${\displaystyle f}$ is differentiable on an open interval ${\displaystyle U}$ and if ${\displaystyle f'}$ is a continuous function on ${\displaystyle U}$, then ${\displaystyle f}$ is called a C¹ function. More generally, ${\displaystyle f}$ is called a C^k function if its derivative ${\displaystyle f'}$ is C^k-1 function. Taylor's theorem states that a C^k function is precisely a function that can be approximated by a polynomial of degree k.

If ${\displaystyle f:\mathbb {R} \to \mathbb {R} }$ is a C¹ function and ${\displaystyle f'(a)\neq 0}$ for some ${\displaystyle a}$, then either ${\displaystyle f'(a)>0}$ or ${\displaystyle f'(a)<0}$; i.e., either ${\displaystyle f}$ is strictly increasing or strictly decreasing in some open interval containing a. In particular, ${\displaystyle f:f^{-1}(U)\to U}$ is bijective for some open interval ${\displaystyle U}$ containing ${\displaystyle f(a)}$. The inverse function theorem then says that the inverse function ${\displaystyle f^{-1}}$ is differentiable on U with the derivatives: for ${\displaystyle y\in U}$

${\displaystyle (f^{-1})'(y)={1 \over f'(f^{-1}(y))}.}$

Derivative of a map and chain rule

For functions ${\displaystyle f}$ defined in the plane or more generally on an Euclidean space ${\displaystyle \mathbb {R} ^{n}}$, it is necessary to consider functions that are vector-valued or matrix-valued. It is also conceptually helpful to do this in an invariant manner (i.e., a coordinate-free way). Derivatives of such maps at a point are then vectors or linear maps, not real numbers.

Let ${\displaystyle f:X\to Y}$ be a map from an open subset ${\displaystyle X}$ of ${\displaystyle \mathbb {R} ^{n}}$ to an open subset ${\displaystyle Y}$ of ${\displaystyle \mathbb {R} ^{m}}$. Then the map ${\displaystyle f}$ is said to be differentiable at a point ${\displaystyle x}$ in ${\displaystyle X}$ if there exists a (necessarily unique) linear transformation ${\displaystyle f'(x):\mathbb {R} ^{n}\to \mathbb {R} ^{m}}$, called the derivative of ${\displaystyle f}$ at ${\displaystyle x}$, such that

${\displaystyle \lim _{h\to 0}{\frac {1}{|h|}}|f(x+h)-f(x)-f'(x)h|=0}$

where ${\displaystyle f'(x)h}$ is the application of the linear transformation ${\displaystyle f'(x)}$ to ${\displaystyle h}$.^[2] If ${\displaystyle f}$ is differentiable at ${\displaystyle x}$, then it is continuous at ${\displaystyle x}$ since

${\displaystyle |f(x+h)-f(x)|\leq (|h|^{-1}|f(x+h)-f(x)-f'(x)h|)|h|+|f'(x)h|\to 0}$ as ${\displaystyle h\to 0}$.

As in the one-variable case, there is

This is proved exactly as for functions in one variable. Indeed, with the notation ${\displaystyle {\widetilde {h}}=f(x+h)-f(x)}$, we have:

${\displaystyle {\begin{aligned}&{\frac {1}{|h|}}|g(f(x+h))-g(y)-g'(y)f'(x)h|\\&\leq {\frac {1}{|h|}}|g(y+{\widetilde {h}})-g(y)-g'(y){\widetilde {h}}|+{\frac {1}{|h|}}|g'(y)(f(x+h)-f(x)-f'(x)h)|.\end{aligned}}}$

Here, since ${\displaystyle f}$ is differentiable at ${\displaystyle x}$, the second term on the right goes to zero as ${\displaystyle h\to 0}$. As for the first term, it can be written as:

${\displaystyle {\begin{cases}{\frac {|{\widetilde {h}}|}{|h|}}|g(y+{\widetilde {h}})-g(y)-g'(y){\widetilde {h}}|/|{\widetilde {h}}|,&{\widetilde {h}}\neq 0,\\0,&{\widetilde {h}}=0.\end{cases}}}$

Now, by the argument showing the continuity of ${\displaystyle f}$ at ${\displaystyle x}$, we see ${\displaystyle {\frac {|{\widetilde {h}}|}{|h|}}}$ is bounded. Also, ${\displaystyle {\widetilde {h}}\to 0}$ as ${\displaystyle h\to 0}$ since ${\displaystyle f}$ is continuous at ${\displaystyle x}$. Hence, the first term also goes to zero as ${\displaystyle h\to 0}$ by the differentiability of ${\displaystyle g}$ at ${\displaystyle y}$. ${\displaystyle \square }$

The map ${\displaystyle f}$ as above is called continuously differentiable or ${\displaystyle C^{1}}$ if it is differentiable on the domain and also the derivatives vary continuously; i.e., ${\displaystyle x\mapsto f'(x)}$ is continuous.

As a linear transformation, ${\displaystyle f'(x)}$ is represented by an ${\displaystyle m\times n}$-matrix, called the Jacobian matrix ${\displaystyle Jf(x)}$ of ${\displaystyle f}$ at ${\displaystyle x}$ and we write it as:

${\displaystyle (Jf)(x)={\begin{bmatrix}{\frac {\partial f_{1}}{\partial x_{1}}}(x)&\cdots &{\frac {\partial f_{1}}{\partial x_{n}}}(x)\\\vdots &\ddots &\vdots \\{\frac {\partial f_{m}}{\partial x_{1}}}(x)&\cdots &{\frac {\partial f_{m}}{\partial x_{n}}}(x)\end{bmatrix}}.}$

Taking ${\displaystyle h}$ to be ${\displaystyle he_{j}}$, ${\displaystyle h}$ a real number and ${\displaystyle e_{j}=(0,\cdots ,1,\cdots ,0)}$ the j-th standard basis element, we see that the differentiability of ${\displaystyle f}$ at ${\displaystyle x}$ implies:

${\displaystyle \lim _{h\to 0}{\frac {f_{i}(x+he_{j})-f_{i}(x)}{h}}={\frac {\partial f_{i}}{\partial x_{j}}}(x)}$

where ${\displaystyle f_{i}}$ denotes the i-th component of ${\displaystyle f}$. That is, each component of ${\displaystyle f}$ is differentiable at ${\displaystyle x}$ in each variable with the derivative ${\displaystyle {\frac {\partial f_{i}}{\partial x_{j}}}(x)}$. In terms of Jacobian matrices, the chain rule says ${\displaystyle J(g\circ f)(x)=Jg(y)Jf(x)}$; i.e., as ${\displaystyle (g\circ f)_{i}=g_{i}\circ f}$,

${\displaystyle {\frac {\partial (g_{i}\circ f)}{\partial x_{j}}}(x)={\frac {\partial g_{i}}{\partial y_{1}}}(y){\frac {\partial f_{1}}{\partial x_{j}}}(x)+\cdots +{\frac {\partial g_{i}}{\partial y_{m}}}(y){\frac {\partial f_{m}}{\partial x_{j}}}(x),}$

which is the form of the chain rule that is often stated.

A partial converse to the above holds. Namely, if the partial derivatives ${\displaystyle {\partial f_{i}}/{\partial x_{j}}}$ are all defined and continuous, then ${\displaystyle f}$ is continuously differentiable.^[4] This is a consequence of the mean value inequality:

(This version of mean value inequality follows from mean value inequality in Mean value theorem § Mean value theorem for vector-valued functions applied to the function ${\displaystyle [0,1]\to \mathbb {R} ^{m},\,t\mapsto f(x+ty)-tv}$, where the proof on mean value inequality is given.)

Indeed, let ${\displaystyle g(x)=(Jf)(x)}$. We note that, if ${\displaystyle y=y_{i}e_{i}}$, then

${\displaystyle {\frac {d}{dt}}f(x+ty)={\frac {\partial f}{\partial x_{i}}}(x+ty)y=g(x+ty)(y_{i}e_{i}).}$

For simplicity, assume ${\displaystyle n=2}$ (the argument for the general case is similar). Then, by mean value inequality, with the operator norm ${\displaystyle \|\cdot \|}$,

${\displaystyle {\begin{aligned}&|\Delta _{y}f(x)-g(x)y|\\&\leq |\Delta _{y_{1}e_{1}}f(x_{1},x_{2}+y_{2})-g(x)(y_{1}e_{1})|+|\Delta _{y_{2}e_{2}}f(x_{1},x_{2})-g(x)(y_{2}e_{2})|\\&\leq |y_{1}|\sup _{0<t<1}\|g(x_{1}+ty_{1},x_{2}+y_{2})-g(x)\|+|y_{2}|\sup _{0<t<1}\|g(x_{1},x_{2}+ty_{2})-g(x)\|,\end{aligned}}}$

which implies ${\displaystyle |\Delta _{y}f(x)-g(x)y|/|y|\to 0}$ as required. ${\displaystyle \square }$

Example: Let ${\displaystyle U}$ be the set of all invertible real square matrices of size n. Note ${\displaystyle U}$ can be identified as an open subset of ${\displaystyle \mathbb {R} ^{n^{2}}}$ with coordinates ${\displaystyle x_{ij},0\leq i,j\neq n}$. Consider the function ${\displaystyle f(g)=g^{-1}}$ = the inverse matrix of ${\displaystyle g}$ defined on ${\displaystyle U}$. To guess its derivatives, assume ${\displaystyle f}$ is differentiable and consider the curve ${\displaystyle c(t)=ge^{tg^{-1}h}}$ where ${\displaystyle e^{A}}$ means the matrix exponential of ${\displaystyle A}$. By the chain rule applied to ${\displaystyle f(c(t))=e^{-tg^{-1}h}g^{-1}}$, we have:

${\displaystyle f'(c(t))\circ c'(t)=-g^{-1}he^{-tg^{-1}h}g^{-1}}$.

Taking ${\displaystyle t=0}$, we get:

${\displaystyle f'(g)h=-g^{-1}hg^{-1}}$.

Now, we then have:^[6]

${\displaystyle \|(g+h)^{-1}-g^{-1}+g^{-1}hg^{-1}\|\leq \|(g+h)^{-1}\|\|h\|\|g^{-1}hg^{-1}\|.}$

Since the operator norm is equivalent to the Euclidean norm on ${\displaystyle \mathbb {R} ^{n^{2}}}$ (any norms are equivalent to each other), this implies ${\displaystyle f}$ is differentiable. Finally, from the formula for ${\displaystyle f'}$, we see the partial derivatives of ${\displaystyle f}$ are smooth (infinitely differentiable); whence, ${\displaystyle f}$ is smooth too.

Higher derivatives and Taylor formula

If ${\displaystyle f:X\to \mathbb {R} ^{m}}$ is differentiable where ${\displaystyle X\subset \mathbb {R} ^{n}}$ is an open subset, then the derivatives determine the map ${\displaystyle f':X\to \operatorname {Hom} (\mathbb {R} ^{n},\mathbb {R} ^{m})}$, where ${\displaystyle \operatorname {Hom} }$ stands for homomorphisms between vector spaces; i.e., linear maps. If ${\displaystyle f'}$ is differentiable, then ${\displaystyle f'':X\to \operatorname {Hom} (\mathbb {R} ^{n},\operatorname {Hom} (\mathbb {R} ^{n},\mathbb {R} ^{m}))}$. Here, the codomain of ${\displaystyle f''}$ can be identified with the space of bilinear maps by:

${\displaystyle \operatorname {Hom} (\mathbb {R} ^{n},\operatorname {Hom} (\mathbb {R} ^{n},\mathbb {R} ^{m})){\overset {\varphi }{\underset {\sim }{\to }}}\{(\mathbb {R} ^{n})^{2}\to \mathbb {R} ^{m}{\text{ bilinear}}\}}$

where ${\displaystyle \varphi (g)(x,y)=g(x)y}$ and ${\displaystyle \varphi }$ is bijective with the inverse ${\displaystyle \psi }$ given by ${\displaystyle (\psi (g)x)y=g(x,y)}$.^{[lower-alpha 1]} In general, ${\displaystyle f^{(k)}=(f^{(k-1)})'}$ is a map from ${\displaystyle X}$ to the space of ${\displaystyle k}$-multilinear maps ${\displaystyle (\mathbb {R} ^{n})^{k}\to \mathbb {R} ^{m}}$.

Just as ${\displaystyle f'(x)}$ is represented by a matrix (Jacobian matrix), when ${\displaystyle m=1}$ (a bilinear map is a bilinear form), the bilinear form ${\displaystyle f''(x)}$ is represented by a matrix called the Hessian matrix of ${\displaystyle f}$ at ${\displaystyle x}$; namely, the square matrix ${\displaystyle H}$ of size ${\displaystyle n}$ such that ${\displaystyle f''(x)(y,z)=(Hy,z)}$, where the paring refers to an inner product of ${\displaystyle \mathbb {R} ^{n}}$, and ${\displaystyle H}$ is none other than the Jacobian matrix of ${\displaystyle f':X\to (\mathbb {R} ^{n})^{*}\simeq \mathbb {R} ^{n}}$. The ${\displaystyle (i,j)}$-th entry of ${\displaystyle H}$ is thus given explicitly as ${\displaystyle H_{ij}={\frac {\partial ^{2}f}{\partial x_{i}\partial x_{j}}}(x)}$.

Moreover, if ${\displaystyle f''}$ exists and is continuous, then the matrix ${\displaystyle H}$ is symmetric, the fact known as the symmetry of second derivatives.^[7] This is seen using the mean value inequality. For vectors ${\displaystyle u,v}$ in ${\displaystyle \mathbb {R} ^{n}}$, using mean value inequality twice, we have:

${\displaystyle |\Delta _{v}\Delta _{u}f(x)-f''(x)(u,v)|\leq \sup _{0<t_{1},t_{2}<1}|f''(x+t_{1}u+t_{2}v)(u,v)-f''(x)(u,v)|,}$

which says

${\displaystyle f''(x)(u,v)=\lim _{s,t\to 0}(\Delta _{tv}\Delta _{su}f(x)-f(x))/(st).}$

Since the right-hand side is symmetric in ${\displaystyle u,v}$, so is the left-hand side: ${\displaystyle f''(x)(u,v)=f''(x)(v,u)}$. By induction, if ${\displaystyle f}$ is ${\displaystyle C^{k}}$, then the k-multilinear map ${\displaystyle f^{(k)}(x)}$ is symmetric; i.e., the order of taking partial derivatives does not matter.^[7]

As in the case of one variable, the Taylor series expansion can then be proved by integration by parts:

${\displaystyle f(z+(h,k))=\sum _{a+b<n}\partial _{x}^{a}\partial _{y}^{b}f(z){h^{a}k^{b} \over a!b!}+n\int _{0}^{1}(1-t)^{n-1}\sum _{a+b=n}\partial _{x}^{a}\partial _{y}^{b}f(z+t(h,k)){h^{a}k^{b} \over a!b!}\,dt.}$

Taylor's formula has an effect of dividing a function by variables, which can be illustrated by the next typical theoretical use of the formula.

Example:^[8] Let ${\displaystyle T:{\mathcal {S}}\to {\mathcal {S}}}$ be a linear map between the vector space ${\displaystyle {\mathcal {S}}}$ of smooth functions on ${\displaystyle \mathbb {R} ^{n}}$ with rapidly decreasing derivatives; i.e., ${\displaystyle \sup |x^{\beta }\partial ^{\alpha }\varphi |<\infty }$ for any multi-index ${\displaystyle \alpha ,\beta }$. (The space ${\displaystyle {\mathcal {S}}}$ is called a Schwartz space.) For each ${\displaystyle \varphi }$ in ${\displaystyle {\mathcal {S}}}$, Taylor's formula implies we can write:

${\displaystyle \varphi -\psi \varphi (y)=\sum _{j=1}^{n}(x_{j}-y_{j})\varphi _{j}}$

with ${\displaystyle \varphi _{j}\in {\mathcal {S}}}$, where ${\displaystyle \psi }$ is a smooth function with compact support and ${\displaystyle \psi (y)=1}$. Now, assume ${\displaystyle T}$ commutes with coordinates; i.e., ${\displaystyle T(x_{j}\varphi )=x_{j}T\varphi }$. Then

${\displaystyle T\varphi -\varphi (y)T\psi =\sum _{j=1}^{n}(x_{j}-y_{j})T\varphi _{j}}$.

Evaluating the above at ${\displaystyle y}$, we get ${\displaystyle T\varphi (y)=\varphi (y)T\psi (y).}$ In other words, ${\displaystyle T}$ is a multiplication by some function ${\displaystyle m}$; i.e., ${\displaystyle T\varphi =m\varphi }$. Now, assume further that ${\displaystyle T}$ commutes with partial differentiations. We then easily see that ${\displaystyle m}$ is a constant; ${\displaystyle T}$ is a multiplication by a constant.

(Aside: the above discussion almost proves the Fourier inversion formula. Indeed, let ${\displaystyle F,R:{\mathcal {S}}\to {\mathcal {S}}}$ be the Fourier transform and the reflection; i.e., ${\displaystyle (R\varphi )(x)=\varphi (-x)}$. Then, dealing directly with the integral that is involved, one can see ${\displaystyle T=RF^{2}}$ commutes with coordinates and partial differentiations; hence, ${\displaystyle T}$ is a multiplication by a constant. This is almost a proof since one still has to compute this constant.)

A partial converse to the Taylor formula also holds; see Borel's lemma and Whitney extension theorem.

Inverse function theorem and submersion theorem

A ${\displaystyle C^{k}}$-map with the ${\displaystyle C^{k}}$-inverse is called a ${\displaystyle C^{k}}$-diffeomorphism. Thus, the theorem says that, for a map ${\displaystyle f}$ satisfying the hypothesis at a point ${\displaystyle x}$, ${\displaystyle f}$ is a diffeomorphism near ${\displaystyle x,f(x).}$ For a proof, see Inverse function theorem § A proof using successive approximation.

The implicit function theorem says:^[9] given a map ${\displaystyle f:\mathbb {R} ^{n}\times \mathbb {R} ^{m}\to \mathbb {R} ^{m}}$, if ${\displaystyle f(a,b)=0}$, ${\displaystyle f}$ is ${\displaystyle C^{k}}$ in a neighborhood of ${\displaystyle (a,b)}$ and the derivative of ${\displaystyle y\mapsto f(a,y)}$ at ${\displaystyle b}$ is invertible, then there exists a differentiable map ${\displaystyle g:U\to V}$ for some neighborhoods ${\displaystyle U,V}$ of ${\displaystyle a,b}$ such that ${\displaystyle f(x,g(x))=0}$. The theorem follows from the inverse function theorem; see Inverse function theorem § Implicit function theorem.

Another consequence is the submersion theorem.

Integrable functions on Euclidean spaces

A partition of an interval ${\displaystyle [a,b]}$ is a finite sequence ${\displaystyle a=t_{0}\leq t_{1}\leq \cdots \leq t_{k}=b}$. A partition ${\displaystyle P}$ of a rectangle ${\displaystyle D}$ (product of intervals) in ${\displaystyle \mathbb {R} ^{n}}$ then consists of partitions of the sides of ${\displaystyle D}$; i.e., if ${\displaystyle D=\prod _{1}^{n}[a_{i},b_{i}]}$, then ${\displaystyle P}$ consists of ${\displaystyle P_{1},\dots ,P_{n}}$ such that ${\displaystyle P_{i}}$ is a partition of ${\displaystyle [a_{i},b_{i}]}$.^[10]

Given a function ${\displaystyle f}$ on ${\displaystyle D}$, we then define the upper Riemann sum of it as:

${\displaystyle U(f,P)=\sum _{Q\in P}(\sup _{Q}f)\operatorname {vol} (Q)}$

where

• ${\displaystyle Q}$ is a partition element of ${\displaystyle P}$; i.e., ${\displaystyle Q=\prod _{i=1}^{n}[t_{i,j_{i}},t_{i,j_{i}+1}]}$ when ${\displaystyle P_{i}:a_{i}=t_{i,0}\leq \dots \cdots \leq t_{i,k_{i}}=b_{i}}$ is a partition of ${\displaystyle [a_{i},b_{i}]}$.^[11]
• The volume ${\displaystyle \operatorname {vol} (Q)}$ of ${\displaystyle Q}$ is the usual Euclidean volume; i.e., ${\displaystyle \operatorname {vol} (Q)=\prod _{1}^{n}(t_{i,j_{i}+1}-t_{i,j_{i}})}$.

The lower Riemann sum ${\displaystyle L(f,P)}$ of ${\displaystyle f}$ is then defined by replacing ${\displaystyle \sup }$ by ${\displaystyle \inf }$. Finally, the function ${\displaystyle f}$ is called integrable if it is bounded and ${\displaystyle \sup\{L(f,P)\mid P\}=\inf\{U(f,P)\mid P\}}$. In that case, the common value is denoted as ${\displaystyle \int _{D}f\,dx}$.^[12]

A subset of ${\displaystyle \mathbb {R} ^{n}}$ is said to have measure zero if for each ${\displaystyle \epsilon >0}$, there are some possibly infinitely many rectangles ${\displaystyle D_{1},D_{2},\dots ,}$ whose union contains the set and ${\displaystyle \sum _{i}\operatorname {vol} (D_{i})<\epsilon .}$^[13]

A key theorem is

The next theorem allows us to compute the integral of a function as the iteration of the integrals of the function in one-variables:

In particular, the order of integrations can be changed.

Finally, if ${\displaystyle M\subset \mathbb {R} ^{n}}$ is a bounded open subset and ${\displaystyle f}$ a function on ${\displaystyle M}$, then we define ${\displaystyle \int _{M}f\,dx:=\int _{D}\chi _{M}f\,dx}$ where ${\displaystyle D}$ is a closed rectangle containing ${\displaystyle M}$ and ${\displaystyle \chi _{M}}$ is the characteristic function on ${\displaystyle M}$; i.e., ${\displaystyle \chi _{M}(x)=1}$ if ${\displaystyle x\in M}$ and ${\displaystyle =0}$ if ${\displaystyle x\not \in M,}$ provided ${\displaystyle \chi _{M}f}$ is integrable.^[15]

Surface integral

If a bounded surface ${\displaystyle M}$ in ${\displaystyle \mathbb {R} ^{3}}$ is parametrized by ${\displaystyle {\textbf {r}}={\textbf {r}}(u,v)}$ with domain ${\displaystyle D}$, then the surface integral of a measurable function ${\displaystyle F}$ on ${\displaystyle M}$ is defined and denoted as:

${\displaystyle \int _{M}F\,dS:=\int \int _{D}(F\circ {\textbf {r}})|{\textbf {r}}_{u}\times {\textbf {r}}_{v}|\,dudv}$

If ${\displaystyle F:M\to \mathbb {R} ^{3}}$ is vector-valued, then we define

${\displaystyle \int _{M}F\cdot dS:=\int _{M}(F\cdot {\textbf {n}})\,dS}$

where ${\displaystyle {\textbf {n}}}$ is an outward unit normal vector to ${\displaystyle M}$. Since ${\displaystyle {\textbf {n}}={\frac {{\textbf {r}}_{u}\times {\textbf {r}}_{v}}{|{\textbf {r}}_{u}\times {\textbf {r}}_{v}|}}}$, we have:

${\displaystyle \int _{M}F\cdot dS=\int \int _{D}(F\circ {\textbf {r}})\cdot ({\textbf {r}}_{u}\times {\textbf {r}}_{v})\,dudv=\int \int _{D}\det(F\circ {\textbf {r}},{\textbf {r}}_{u},{\textbf {r}}_{v})\,dudv.}$

Tangent vectors and vector fields

Let ${\displaystyle c:[0,1]\to \mathbb {R} ^{n}}$ be a differentiable curve. Then the tangent vector to the curve ${\displaystyle c}$ at ${\displaystyle t}$ is a vector ${\displaystyle v}$ at the point ${\displaystyle c(t)}$ whose components are given as:

${\displaystyle v=(c_{1}'(t),\dots ,c_{n}'(t))}$.^[16]

For example, if ${\displaystyle c(t)=(a\cos(t),a\sin(t),bt),a>0,b>0}$ is a helix, then the tangent vector at t is:

${\displaystyle c'(t)=(-a\sin(t),a\cos(t),b).}$

It corresponds to the intuition that the a point on the helix moves up in a constant speed.

If ${\displaystyle M\subset \mathbb {R} ^{n}}$ is a differentiable curve or surface, then the tangent space to ${\displaystyle M}$ at a point p is the set of all tangent vectors to the differentiable curves ${\displaystyle c:[0,1]\to M}$ with ${\displaystyle c(0)=p}$.

A vector field X is an assignment to each point p in M a tangent vector ${\displaystyle X_{p}}$ to M at p such that the assignment varies smoothly.

Differential forms

The dual notion of a vector field is a differential form. Given an open subset ${\displaystyle M}$ in ${\displaystyle \mathbb {R} ^{n}}$, by definition, a differential 1-form (often just 1-form) ${\displaystyle \omega }$ is an assignment to a point ${\displaystyle p}$ in ${\displaystyle M}$ a linear functional ${\displaystyle \omega _{p}}$ on the tangent space ${\displaystyle T_{p}M}$ to ${\displaystyle M}$ at ${\displaystyle p}$ such that the assignment varies smoothly. For a (real or complex-valued) smooth function ${\displaystyle f}$, define the 1-form ${\displaystyle df}$ by: for a tangent vector ${\displaystyle v}$ at ${\displaystyle p}$,

${\displaystyle df_{p}(v)=v(f)}$

where ${\displaystyle v(f)}$ denotes the directional derivative of ${\displaystyle f}$ in the direction ${\displaystyle v}$ at ${\displaystyle p}$.^[17] For example, if ${\displaystyle x_{i}}$ is the ${\displaystyle i}$-th coordinate function, then ${\displaystyle dx_{i,p}(v)=v_{i}}$; i.e., ${\displaystyle dx_{i,p}}$ are the dual basis to the standard basis on ${\displaystyle T_{p}M}$. Then every differential 1-form ${\displaystyle \omega }$ can be written uniquely as

${\displaystyle \omega =f_{1}\,dx_{1}+\cdots +f_{n}\,dx_{n}}$

for some smooth functions ${\displaystyle f_{1},\dots ,f_{n}}$ on ${\displaystyle M}$ (since, for every point ${\displaystyle p}$, the linear functional ${\displaystyle \omega _{p}}$ is a unique linear combination of ${\displaystyle dx_{i}}$ over real numbers). More generally, a differential k-form is an assignment to a point ${\displaystyle p}$ in ${\displaystyle M}$ a vector ${\displaystyle \omega _{p}}$ in the ${\displaystyle k}$-th exterior power ${\displaystyle \bigwedge ^{k}T_{p}^{*}M}$ of the dual space ${\displaystyle T_{p}^{*}M}$ of ${\displaystyle T_{p}M}$ such that the assignment varies smoothly.^[17] In particular, a 0-form is the same as a smooth function. Also, any ${\displaystyle k}$-form ${\displaystyle \omega }$ can be written uniquely as:

${\displaystyle \omega =\sum _{i_{1}<\cdots <i_{k}}f_{i_{1}\dots i_{k}}\,dx_{i_{1}}\wedge \cdots \wedge dx_{i_{k}}}$

for some smooth functions ${\displaystyle f_{i_{1}\dots i_{k}}}$.^[17]

Like a smooth function, we can differentiate and integrate differential forms. If ${\displaystyle f}$ is a smooth function, then ${\displaystyle df}$ can be written as:^[18]

${\displaystyle df=\sum _{i=1}^{n}{\frac {\partial f}{\partial x_{i}}}\,dx_{i}}$

since, for ${\displaystyle v=\partial /\partial x_{j}|_{p}}$, we have: ${\displaystyle df_{p}(v)={\frac {\partial f}{\partial x_{j}}}(p)=\sum _{i=1}^{n}{\frac {\partial f}{\partial x_{i}}}(p)\,dx_{i}(v)}$. Note that, in the above expression, the left-hand side (whence the right-hand side) is independent of coordinates ${\displaystyle x_{1},\dots ,x_{n}}$; this property is called the invariance of differential.

The operation ${\displaystyle d}$ is called the exterior derivative and it extends to any differential forms inductively by the requirement (Leibniz rule)

${\displaystyle d(\alpha \wedge \beta )=d\alpha \wedge \beta +(-1)^{p}\alpha \wedge d\beta .}$

where ${\displaystyle \alpha ,\beta }$ are a p-form and a q-form.

The exterior derivative has the important property that ${\displaystyle d\circ d=0}$; that is, the exterior derivative ${\displaystyle d}$ of a differential form ${\displaystyle d\omega }$ is zero. This property is a consequence of the symmetry of second derivatives (mixed partials are equal).

Boundary and orientation

A circle can be oriented clockwise or counterclockwise. Mathematically, we say that a subset ${\displaystyle M}$ of ${\displaystyle \mathbb {R} ^{n}}$ is oriented if there is a consistent choice of normal vectors to ${\displaystyle M}$ that varies continuously. For example, a circle or, more generally, an n-sphere can be oriented; i.e., orientable. On the other hand, a Möbius strip (a surface obtained by identified by two opposite sides of the rectangle in a twisted way) cannot oriented: if we start with a normal vector and travel around the strip, the normal vector at end will point to the opposite direction.

The proposition is useful because it allows us to give an orientation by giving a volume form.

Integration of differential forms

If ${\displaystyle \omega =f\,dx_{1}\wedge \cdots \wedge dx_{n}}$ is a differential n-form on an open subset M in ${\displaystyle \mathbb {R} ^{n}}$ (any n-form is that form), then the integration of it over ${\displaystyle M}$ with the standard orientation is defined as:

${\displaystyle \int _{M}\omega =\int _{M}f\,dx_{1}\cdots dx_{n}.}$

If M is given the orientation opposite to the standard one, then ${\displaystyle \int _{M}\omega }$ is defined as the negative of the right-hand side.

Then we have the fundamental formula relating exterior derivative and integration:

Here is a sketch of proof of the formula.^[19] If ${\displaystyle f}$ is a smooth function on ${\displaystyle \mathbb {R} ^{n}}$ with compact support, then we have:

${\displaystyle \int d(f\omega )=0}$

(since, by the fundamental theorem of calculus, the above can be evaluated on boundaries of the set containing the support.) On the other hand,

${\displaystyle \int d(f\omega )=\int df\wedge \omega +\int f\,d\omega .}$

Let ${\displaystyle f}$ approach the characteristic function on ${\displaystyle M}$. Then the second term on the right goes to ${\displaystyle \int _{M}d\omega }$ while the first goes to ${\displaystyle -\int _{\partial M}\omega }$, by the argument similar to proving the fundamental theorem of calculus. ${\displaystyle \square }$

The formula generalizes the fundamental theorem of calculus as well as Stokes' theorem in multivariable calculus. Indeed, if ${\displaystyle M=[a,b]}$ is an interval and ${\displaystyle \omega =f}$, then ${\displaystyle d\omega =f'\,dx}$ and the formula says:

${\displaystyle \int _{M}f'\,dx=f(b)-f(a)}$.

Similarly, if ${\displaystyle M}$ is an oriented bounded surface in ${\displaystyle \mathbb {R} ^{3}}$ and ${\displaystyle \omega =f\,dx+g\,dy+h\,dz}$, then ${\displaystyle d(f\,dx)=df\wedge dx={\frac {\partial f}{\partial y}}\,dy\wedge dx+{\frac {\partial f}{\partial z}}\,dz\wedge dx}$ and similarly for ${\displaystyle d(g\,dy)}$ and ${\displaystyle d(g\,dy)}$. Collecting the terms, we thus get:

${\displaystyle d\omega =\left({\frac {\partial h}{\partial y}}-{\frac {\partial g}{\partial z}}\right)dy\wedge dz+\left({\frac {\partial f}{\partial z}}-{\frac {\partial h}{\partial x}}\right)dz\wedge dx+\left({\frac {\partial g}{\partial x}}-{\frac {\partial f}{\partial y}}\right)dx\wedge dy.}$