Calculus notations

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Calculus

$\displaystyle \max_{x \in [a,b]} f(x)$ The maximum of the continuous function $f$ in the range $a$ to $b$.
$\displaystyle \min_{x \in [a,b]} f(x)$ The minimum of the continuous function $f$ in the range $a$ to $b$.
$\displaystyle\sup_{x \in [a,b]} f(x)$ The supremum of the function $f$ over the interval $[a,b]$, i.e. the smallest real number that is greater than or equal to every value of $f(x)$ with $x \in [a,b]$.
$\displaystyle \inf_{x \in [a,b]} f(x)$ The infimum of the function $f$ over the interval $[a,b]$, i.e. the biggest real number that is smaller than or equal to every value of $f(x)$ with $x \in [a,b]$.
$ \displaystyle (x_n)_{n \geq 0}$ This denotes an infinite sequence of real numbers $x_0$, $x_1$, $x_2$, ... that are given through some particular formula depending on the index of each term. For example the sequence $(\sin(n \alpha))_{n \geq 0}$ is the sequence of terms \[ \sin(0 \cdot \alpha),\, \sin(1 \cdot \alpha),\, \sin(2 \cdot \alpha),\, \sin(3 \cdot \alpha),\, \ldots \]
$ \displaystyle \lim_{n \rightarrow \infty} x_n$ This denotes the limit of the sequence $(x_n)_{n \geq 0}$, whenever it exists. Intuitively this is the value that the term $x_n$ approaches as the index $n$ gets closer and closer to infinity. It can be shown for example that \[ \lim_{n \rightarrow \infty} \frac{\sin n + \cos n}{n} = 0. \]
$ \displaystyle \sum_{i=0}^{\infty} x_i$ This is called an infinite series of the sequence $(x_n)_{n \geq 0}$ and it is defined through a sequence of partial sums $(S_n)_{n \geq 0}$ whose terms are given by the formula \[ S_k = \sum_{i=0}^k x_i \qquad \forall k \geq 0. \] Whenever it exists, the limit of the sequence $(S_n)_{n \geq 0}$ is called the value of the infinite series and thus \[ \sum_{i=0}^{\infty} x_i = \lim_{n \rightarrow \infty} S_n. \]
$ \displaystyle \prod_{i=0}^{\infty} x_i$ This is called an infinite product of the sequence $(x_n)_{n \geq 0}$ and it is defined through a sequence of partial products $(P_n)_{n \geq 0}$ whose terms are given by the formula \[ P_k = \prod_{i=0}^k x_i \qquad \forall k \geq 0. \] Whenever it exists, the limit of the sequence $(P_n)_{n \geq 0}$ is called the value of the infinite product and thus \[ \prod_{i=0}^{\infty} x_i = \lim_{n \rightarrow \infty} P_n. \]
$ \displaystyle \lim_{x \rightarrow a} f(x)$ This denotes the limit of the function $f(x)$ at point $a$, whenever it exists. Intuitively this is the value that the function $f$ approaches as the argument $x$ gets closer and closer to $a$. It can be shown for example that \[ \lim_{x \rightarrow 0} \frac{\sin x}{x} = 1. \]
$ \displaystyle f`(\alpha), \quad \frac{df}{dx}(\alpha)$ This denotes the first derivative of the function $f:(a,b) \rightarrow \mathbb{R}$ at point $\alpha \in (a,b)$, whenever it exists. It can be defined through the following formula using limits \[ \frac{df}{dx}(\alpha) = \lim_{h \rightarrow 0} \frac{f(\alpha+h) - f(\alpha)}{h}. \]
$ \displaystyle f^{(n)}(\alpha), \quad \frac{d^n f}{dx^n}(\alpha)$ This denotes the $n$-th derivative of the function $f:(a,b) \rightarrow \mathbb{R}$ at point $\alpha \in (a,b)$, whenever it exists. It can be defined recurrently through the following formulae \[ \frac{d^n f}{dx^n}(\alpha) = \lim_{h \rightarrow 0} \frac{1}{h} \left(\frac{d^{n-1}f}{dx^{n-1}}(\alpha+h) - \frac{d^{n-1}f}{dx^{n-1}}(\alpha)\right), \qquad \frac{d^1 f}{dx^1}(\alpha) = \frac{df}{dx}(\alpha). \]
$ \displaystyle \frac{\partial}{\partial x_k} f(a)$ The partial derivative of a function $f:U \rightarrow \mathbb{R}$, $U \subseteq \mathbb{R}^n$ with respect to the $k$-th variable, at a point $a = (a_1, a_2, \ldots, a_n) \in \mathbb{R}^n$. This is defined through the following formula \[ \frac{\partial}{\partial x_k} f(a) = \lim_{h \rightarrow 0} \frac{f(a_1, a_2, \ldots, a_{k-1}, a_k + h, a_{k+1}, \ldots, a_n) - f(a_1, a_2, \ldots, a_n)}{h}. \] Basically this evaluates the first derivative of the function $g_k:I \rightarrow \mathbb{R}$, $I \subseteq \mathbb{R}$ defined by \[ g_k(x) = f(a_1, a_2, \ldots, a_{k-1}, x, a_{k+1}, \ldots, a_n). \] Therefore it will also be valid to write \[ \frac{\partial}{\partial x_k} f(a) = \lim_{h \rightarrow 0} \frac{g_k(a_k + h) - g_k(a_k)}{h} = g_k`(a_k). \] As an example consider the function $f:\mathbb{R}^3 \rightarrow \mathbb{R}$ given by $f(x, y, z) = \sin x + \cos y + \tan z$. Using the basic differentiation rules it is easy to see that \[ \frac{\partial f}{\partial x} = \cos x \qquad \frac{\partial f}{\partial y} = -\sin y \qquad \frac{\partial f}{\partial z} = 1 + (\tan z)^2. \]
$ \displaystyle \frac{\partial^n f}{\partial x_k^n}$ If $f: U \subseteq \mathbb{R}^p \rightarrow \mathbb{R}$ is a function then the $n$-th order partial derivative of $f$ with respect to the $k$-th variable is defined through the following recurrence relation \[ \frac{\partial^n f}{\partial x_k^n} = \frac{\partial}{\partial x_k} \left( \frac{\partial^{n-1}}{\partial x_k^{n-1}} f \right), \qquad \frac{\partial^1}{\partial x_k^1} f = \frac{\partial f}{\partial x_k}. \] For example if $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ is defined by $f(x,y) = x^3 + xy^3$, then \[ \frac{\partial f}{\partial x} = 3x^2 + y^3 \qquad \frac{\partial^2 f}{\partial x^2} = 6x \qquad \frac{\partial^3 f}{\partial x^3} = 6 \qquad \frac{\partial^4 f}{\partial x^4} = 0 \] while \[ \frac{\partial f}{\partial y} = 3xy^2 \qquad \frac{\partial^2 f}{\partial y^2} = 6xy \qquad \frac{\partial^3 f}{\partial y^3} = 6x \qquad \frac{\partial^4 f}{\partial y^4} = 0. \]
$ \displaystyle \int_a^b f(x) \,dx$ The Riemann integral of the non-negative real-valued function $f$ on the interval $[a,b]$. This basically gives the area below the graph of $f$ calculated from point $a$ to point $b$.
$ \displaystyle \int_a^{\infty} f(x) \,dx$ The improper integral of the non-negative real-valued function $f$ defined through \[ \int_a^{\infty} f(x) \,dx = \lim_{b \rightarrow \infty} \int_a^b f(x) \,dx. \] This gives the area below the graph of $f$ calculated from point $a$ to infinity.
$ \displaystyle \int_{-\infty}^b f(x) \,dx$ The improper integral of the non-negative real-valued function $f$ defined through \[ \int_{-\infty}^b f(x) \,dx = \lim_{a \rightarrow -\infty} \int_a^b f(x) \,dx. \] This gives the area below the graph of $f$ calculated from minus infinity to point $b$.