• \({e^x} = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + ... = \sum\limits_{n = 0}^\infty {\frac{x^n}{n!}}, x \in \mathbb {R}\)

• \(\sin x = x - \frac{{{x^3}}}{{3!}} + \frac{{{x^5}}}{{5!}} - ... = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{\left( {2n + 1} \right)!}}} {x^{2n + 1}}, x \in \mathbb {R}\)

• \(\cos x = 1 - \frac{{{x^2}}}{{2!}} + \frac{{{x^4}}}{{4!}} - ... = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{\left( {2n} \right)!}}} {x^{2n}}, x \in \mathbb {R}\)

• \(\frac{1}{{1 - x}} = 1 + x + {x^2} + {x^3} + ... = \sum\limits_{n = 0}^\infty {{x^n}} ,{\rm{ }}\left| x \right| < 1\)

• \(\frac{1}{{1 + x}} = 1 - x + {x^2} - {x^3} + ... = \sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)}^n}{x^n}} ,{\rm{ }}\left| x \right| < 1\)

• \({\mathop{\rm tg}\nolimits} x = x + \frac{{{x^3}}}{3} + \frac{{2{x^5}}}{{15}} + ...\)

• \({\mathop{\rm ch}\nolimits} x = 1 + \frac{{{x^2}}}{{2!}} + \frac{{{x^4}}}{{4!}} + ... = \sum\limits_{n = 0}^\infty {\frac{{{x^{2n}}}}{{\left( {2n} \right)!}}} , x \in \)

• \({\mathop{\rm sh}\nolimits} x = x + \frac{{{x^3}}}{{3!}} + \frac{{{x^5}}}{{5!}} + \cdots = \sum\limits_{n = 0}^\infty {\frac{1}{{(2n + 1)!}}} {x^{2n + 1}},x \in \)

• \({\left( {1 + x} \right)^\alpha } = 1 + \sum\limits_{n = 1}^\infty {\left( \begin{array}{c} \alpha \\ n \end{array} \right)} {x^n},{\rm{ }}{\rm{ }}\left( \begin{array}{c} \alpha \\ n \end{array} \right) = \prod\limits_{k = 1}^n {\frac{{\alpha - k + 1}}{k}} = \frac{{\alpha \left( {\alpha - 1} \right) \cdots \left( {\alpha - n + 1} \right)}}{{n!}},{\rm{ }}\left| x \right| < 1\)

• \({\mathop{\rm arctg}\nolimits} x = x - \frac{{{x^3}}}{3} + \frac{{{x^5}}}{5} - \cdots = \sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^{n - 1}}}}{{2n - 1}}} {x^{2n - 1}} \)

• \(\ln \left( {1 + x} \right) = x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3} - \cdots = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}{x^{n + 1}}}}{{n + 1}}} = \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}{x^n}}}{n}} ,{\rm{ }} - 1 < x \le 1 \)

• \(\sqrt {1 + x} = 1 + \frac{x}{2} - \frac{{{x^2}}}{8} + \frac{{{x^3}}}{{16}} - \cdots = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}\left( {2n} \right)!}}{{\left( {1 - 2n} \right)n{!^2}{4^n}}}} {x^n}\)