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MathsNotations

basic

Basic notations
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Basic

\displaystyle f:A \rightarrow B This defines a function f whose arguments take values from the set A and which returns a unique value from the set B for each corresponding argument. For example consider the function f:\mathbb{R} \rightarrow \mathbb{R}_+, given by f(x) = 10x. By evaluating this function at various values of the argument the following results are obtained, f(10.2) = 102, f(-5.8) = -58, f(0) = 0.
\displaystyle f \circ g This defines the composite function f \circ g defined by the relation f \circ g (x) = f(g(x)) for all values of x, where f and g are apropriately chosen functions.
\displaystyle |x| The absolute value function defined as:

|x| = \left\{ \begin{array}{rl} x, & \text{for } x \geq 0 \\ -x, & \text{for } x < 0 \\ \end{array} \right.
For example |1.23| = 1.23, |0| = 0, |-3.14| = 3.14.
\displaystyle x \approx \alpha This tells that x has an approximate value of \alpha, in other words |x - \alpha| < \varepsilon where \varepsilon is a very small positive value. For example \pi \approx 3.1415926535898.
\displaystyle x \gg \alpha This tells that x has a much greater value than \alpha, for example 10000 \gg 5.34.
\displaystyle x \ll \alpha This tells that x has a much smaller value than \alpha, for example -31000 \ll 0.
\displaystyle {\rm sign}(x) The signum function defined as:
\rm sign(x) = \left\{ \begin{array}{rl} 1, & \text{for } x > 0 \\ 0, & \text{for } x = 0 \\ -1, & \text{for } x < 0 \end{array} \right .
For example {\rm sign}(-1034.5) = -1, {\rm sign}(0) = 0, {\rm sign}(11.3) = 1.
\displaystyle \lfloor x \rfloor The floor function which gives the largest integer less than or equal to x. For example \lfloor 2.8 \rfloor = 2, \lfloor -3.3 \rfloor = -4.
\displaystyle \lceil x \rceil The ceiling function which gives the smallest integer not less than x. For example \lceil 5.4 \rceil = 6, \lceil -2.8 \rceil = -2.
\displaystyle \sqrt{x} The square root of x is a non-negative real number y such that y^2 = x. Obviously x always has to be non-negative. For example \sqrt{144} = 12, \sqrt{256} = 16 while \sqrt{-16} is not a proper expression since -16 is negative.
\displaystyle \sqrt[n]{x} The n-th root of x is a real number y such that y^n = x. If n is even x needs to be non-negative, as is the case for the square root with n = 2. For example \sqrt[3]{-8} = -2, \sqrt[5]{1024} = 4, while \sqrt[4]{-256} is not a proper expression since the order of the root n is even but -256 is negative.
\displaystyle \sum_{i=1}^n E(i) If E(i) is an expression that depends on the value of i, this evaluates the following sum
\sum_{i=1}^n E(i) = E(1) + E(2) + E(3) + \ldots + E(n).
For example if E(i) = \sin(i\alpha) then the above sum becomes
\sum_{i=1}^n \sin(i\alpha) = \sin(\alpha) + \sin(2\alpha) + \sin(3\alpha) + \ldots + \sin(n\alpha).
\displaystyle \sum_{P(i)} E(i) If E(i) is an expression that depends on the value of i, this evaluates the sum over those indices i for which the predicate P(i) becomes true. For example if E(i) = \sin(i) and the predicate is P(i): "i is a prime and is less than 12" then the sum becomes
\sum_{\displaystyle i \mbox{ is a prime} \atop{\displaystyle i < 12}} \sin(i) =   \sin(2) + \sin(3) + \sin(5) + \sin(7) + \sin(11).
Notice that if the predicate consists of several conditions, these are written one below the other as above. By letting P(i): "1 \leq i \leq n" this generalised sum becomes the previous easier sum
\sum_{1 \leq i \leq n} E(i) = \sum_{i=1}^n E(i).
\displaystyle \prod_{i=1}^n E(i) If E(i) is an expression that depends on the value of i, this evaluates the following product
\prod_{i=1}^n E(i) = E(1) \cdot E(2) \cdot E(3) \cdot \ldots \cdot E(n).
For example if E(i) = i + \sqrt{2 \cdot i} then the above product becomes
\prod_{i=1}^n i + \sqrt{2 \cdot i} = (1 + \sqrt{2 \cdot 1}) \cdot (2 + \sqrt{2 \cdot 2}) \cdot (3 + \sqrt{2 \cdot 3}) \cdot \ldots \cdot (n + \sqrt{2 \cdot n}).
\displaystyle \prod_{P(i)} E(i) If E(i) is an expression that depends on the value of i, this evaluates the product over those indices i for which the predicate P(i) becomes true. For example if E(i) = i and the predicate is P(i): "i is a prime and is less than 16" then the product becomes Notice that if the predicate consists of several conditions, these are written one below the other as above. By letting P(i): "1 \leq i \leq n" this generalised product becomes the previous easier product
\displaystyle n! The factorial function of argument n \in \mathbb{N} defined through
n! = \prod_{i=1}^n i \qquad 0! = 1.
\displaystyle \binom{n}{k} The binomial coefficient of the natural numbers n and k defined through
\binom{n}{k} = \frac{n!}{k!(n-k)!} \qquad n \geq k \geq 0.