# Chebyshev polynomials

## Definitions

Symbol: ChebyshevT $T_{n}\!\left(x\right)$ Chebyshev polynomial of the first kind
Symbol: ChebyshevU $U_{n}\!\left(x\right)$ Chebyshev polynomial of the second kind

## Tables

Table of $T_{n}\!\left(x\right)$ for $0 \le n \le 15$
Table of $U_{n}\!\left(x\right)$ for $0 \le n \le 15$

## Specific values

$T_{0}\!\left(x\right) = 1$
$T_{1}\!\left(x\right) = x$
$U_{0}\!\left(x\right) = 1$
$U_{1}\!\left(x\right) = 2 x$
$U_{-1}\!\left(x\right) = 0$
$T_{n}\!\left(1\right) = 1$
$T_{n}\!\left(-1\right) = {\left(-1\right)}^{n}$
$T_{2 n}\!\left(0\right) = {\left(-1\right)}^{n}$
$T_{2 n + 1}\!\left(0\right) = 0$
$U_{n}\!\left(1\right) = n + 1$
$U_{n}\!\left(-1\right) = {\left(-1\right)}^{n} \left(n + 1\right)$
$U_{2 n}\!\left(0\right) = {\left(-1\right)}^{n}$
$U_{2 n + 1}\!\left(0\right) = 0$

## Zeros and extrema

$\mathop{\operatorname{zeros}\,}\limits_{x \in \mathbb{C}} T_{n}\!\left(x\right) = \left\{ \cos\!\left(\frac{2 k - 1}{2 n} \pi\right) : k \in \{1, 2, \ldots, n\} \right\}$
$\mathop{\operatorname{zeros}\,}\limits_{x \in \mathbb{C}} U_{n}\!\left(x\right) = \left\{ \cos\!\left(\frac{k}{n + 1} \pi\right) : k \in \{1, 2, \ldots, n\} \right\}$
$\mathop{\operatorname{solutions}\,}\limits_{x \in \mathbb{C}} \left[T_{n}\!\left(x\right) \in \left\{-1, 1\right\}\right] = \left\{ \cos\!\left(\frac{k}{n} \pi\right) : k \in \{0, 1, \ldots, n\} \right\}$
$\mathop{\operatorname{solutions}\,}\limits_{x \in \mathbb{C}} \left[T_{n}\!\left(x\right) = 1\right] = \left\{ \cos\!\left(\frac{2 k}{n} \pi\right) : k \in \{0, 1, \ldots, \left\lfloor \frac{n}{2} \right\rfloor\} \right\}$
$\mathop{\operatorname{solutions}\,}\limits_{x \in \mathbb{C}} \left[T_{n}\!\left(x\right) = -1\right] = \left\{ \cos\!\left(\frac{2 k - 1}{n} \pi\right) : k \in \{1, 2, \ldots, \left\lfloor \frac{n + 1}{2} \right\rfloor\} \right\}$

## Symmetries

$T_{n}\!\left(-x\right) = {\left(-1\right)}^{n} T_{n}\!\left(x\right)$
$U_{n}\!\left(-x\right) = {\left(-1\right)}^{n} U_{n}\!\left(x\right)$
$T_{-n}\!\left(x\right) = T_{n}\!\left(x\right)$
$U_{-n}\!\left(x\right) = -U_{n - 2}\!\left(x\right)$

## Orthogonality

$\int_{-1}^{1} T_{n}\!\left(x\right) T_{m}\!\left(x\right) \frac{1}{\sqrt{1 - {x}^{2}}} \, dx = \begin{cases} 0, & n \ne m\\\pi, & n = m = 0\\\frac{\pi}{2}, & n = m \,\mathbin{\operatorname{and}}\, n \ne 0\\ \end{cases}$
$\int_{-1}^{1} U_{n}\!\left(x\right) U_{m}\!\left(x\right) \sqrt{1 - {x}^{2}} \, dx = \begin{cases} 0, & n \ne m\\\frac{\pi}{2}, & n = m\\ \end{cases}$

## Differential equations

$\left(1 - {x}^{2}\right) y''(x) - x y'(x) + {n}^{2} y(x) = 0\; \text{ where } y(x) = {c}_{1} T_{n}\!\left(x\right) + {c}_{2} U_{n - 1}\!\left(x\right) \sqrt{1 - {x}^{2}}$
$\left(1 - {x}^{2}\right) y''(x) - 3 x y'(x) + n \left(n + 2\right) y(x) = 0\; \text{ where } y(x) = {c}_{1} U_{n}\!\left(x\right) + {c}_{2} \frac{T_{n + 1}\!\left(x\right)}{\sqrt{1 - {x}^{2}}}$

## Recurrence relations

$T_{n}\!\left(x\right) = 2 x T_{n - 1}\!\left(x\right) - T_{n - 2}\!\left(x\right)$
$U_{n}\!\left(x\right) = 2 x U_{n - 1}\!\left(x\right) - U_{n - 2}\!\left(x\right)$
$T_{n}\!\left(x\right) = 2 x T_{n + 1}\!\left(x\right) - T_{n + 2}\!\left(x\right)$
$U_{n}\!\left(x\right) = 2 x U_{n + 1}\!\left(x\right) - U_{n + 2}\!\left(x\right)$
$T_{n}\!\left(x\right) = x T_{n - 1}\!\left(x\right) - \left(1 - {x}^{2}\right) U_{n - 2}\!\left(x\right)$
$U_{n}\!\left(x\right) = x U_{n - 1}\!\left(x\right) + T_{n}\!\left(x\right)$
$T_{n}\!\left(x\right) = \frac{U_{n}\!\left(x\right) - U_{n - 2}\!\left(x\right)}{2}$
$T_{n}\!\left(x\right) = U_{n}\!\left(x\right) - x U_{n - 1}\!\left(x\right)$

## Order transformations

$T_{m}\!\left(T_{n}\!\left(x\right)\right) = T_{m n}\!\left(x\right)$
$T_{m}\!\left(x\right) T_{n}\!\left(x\right) = \frac{T_{m + n}\!\left(x\right) + T_{\left|m - n\right|}\!\left(x\right)}{2}$
$T_{2 n}\!\left(x\right) = 2 T_{n}^{2}\!\left(x\right) - 1$
$T_{2 n + 1}\!\left(x\right) = 2 T_{n + 1}\!\left(x\right) T_{n}\!\left(x\right) - x$
$T_{2 n}\!\left(x\right) = T_{n}\!\left(2 {x}^{2} - 1\right)$
$U_{2 n}\!\left(x\right) = T_{n}\!\left(2 {x}^{2} - 1\right) + U_{n - 1}\!\left(2 {x}^{2} - 1\right)$

## Trigonometric formulas

$T_{n}\!\left(x\right) = \cos\!\left(n \operatorname{acos}(x)\right)$
$T_{n}\!\left(x\right) = \cosh\!\left(n \operatorname{acosh}(x)\right)$
$U_{n - 1}\!\left(x\right) \sqrt{1 - {x}^{2}} = \sin\!\left(n \operatorname{acos}(x)\right)$
$T_{n}\!\left(\cos(x)\right) = \cos\!\left(n x\right)$
$U_{n}\!\left(\cos(x)\right) \sin(x) = \sin\!\left(n x\right)$
$T_{2 n + 1}\!\left(\sin(x)\right) = {\left(-1\right)}^{n} \sin\!\left(\left(2 n + 1\right) x\right)$

## Power formulas

$T_{n}\!\left(x\right) = \frac{1}{2} \left({\left(x + \sqrt{{x}^{2} - 1}\right)}^{n} + {\left(x - \sqrt{{x}^{2} - 1}\right)}^{n}\right)$
$U_{n - 1}\!\left(x\right) \sqrt{{x}^{2} - 1} = \frac{1}{2} \left({\left(x + \sqrt{{x}^{2} - 1}\right)}^{n} - {\left(x - \sqrt{{x}^{2} - 1}\right)}^{n}\right)$
$T_{n}\!\left(x\right) + U_{n - 1}\!\left(x\right) \sqrt{{x}^{2} - 1} = {\left(x + \sqrt{{x}^{2} - 1}\right)}^{n}$
$T_{n}^{2}\!\left(x\right) + \left({x}^{2} - 1\right) U_{n - 1}^{2}\!\left(x\right) = 1$
$T_{n}\!\left(\frac{x + {x}^{-1}}{2}\right) = \frac{{x}^{n} + {x}^{-n}}{2}$

## Product representations

$T_{n}\!\left(x\right) = {2}^{n - 1} \prod_{k=1}^{n} \left(x - \cos\!\left(\frac{2 k - 1}{2 n} \pi\right)\right)$
$U_{n}\!\left(x\right) = {2}^{n} \prod_{k=1}^{n} \left(x - \cos\!\left(\frac{k}{n + 1} \pi\right)\right)$

## Sum representations

$T_{n}\!\left(x\right) = \frac{n}{2} \sum_{k=0}^{\left\lfloor n / 2 \right\rfloor} \frac{{\left(-1\right)}^{k}}{n - k} {n - k \choose k} {\left(2 x\right)}^{n - 2 k}$
$U_{n}\!\left(x\right) = \sum_{k=0}^{\left\lfloor n / 2 \right\rfloor} {\left(-1\right)}^{k} {n - k \choose k} {\left(2 x\right)}^{n - 2 k}$
$T_{n}\!\left(x\right) = \sum_{k=0}^{\left\lfloor n / 2 \right\rfloor} {n \choose 2 k} {\left({x}^{2} - 1\right)}^{k} {x}^{n - 2 k}$
$T_{n}\!\left(x\right) = \frac{n}{2} \sum_{k=0}^{\left\lfloor n / 2 \right\rfloor} \frac{{\left(-1\right)}^{k} \left(n - k - 1\right)!}{k ! \left(n - 2 k\right)!} {\left(2 x\right)}^{n - 2 k}$
$T_{n}\!\left(x\right) = n \sum_{k=0}^{n} \frac{{2}^{k} \left(n + k - 1\right)!}{\left(n - k\right)! \left(2 k\right)!} {\left(x - 1\right)}^{k}$
$U_{n}\!\left(x\right) = \sum_{k=0}^{\left\lfloor n / 2 \right\rfloor} {n + 1 \choose 2 k + 1} {\left({x}^{2} - 1\right)}^{k} {x}^{n - 2 k}$
$U_{n}\!\left(x\right) = \sum_{k=0}^{n} \frac{{2}^{k} \left(n + k + 1\right)!}{\left(n - k\right)! \left(2 k + 1\right)!} {\left(x - 1\right)}^{k}$

## Hypergeometric representations

$T_{n}\!\left(x\right) = \,{}_2F_1\!\left(-n, n, \frac{1}{2}, \frac{1 - x}{2}\right)$
$U_{n}\!\left(x\right) = \left(n + 1\right) \,{}_2F_1\!\left(-n, n + 2, \frac{3}{2}, \frac{1 - x}{2}\right)$

## Generating functions

$\sum_{n=0}^{\infty} T_{n}\!\left(x\right) {z}^{n} = \frac{1 - x z}{1 - 2 x z + {z}^{2}}$
$\sum_{n=0}^{\infty} U_{n}\!\left(x\right) {z}^{n} = \frac{1}{1 - 2 x z + {z}^{2}}$
$\sum_{n=1}^{\infty} T_{n}\!\left(x\right) \frac{{z}^{n}}{n} = -\frac{1}{2} \log\!\left(1 - 2 x z + {z}^{2}\right)$
$\sum_{n=0}^{\infty} T_{n}\!\left(x\right) \frac{{z}^{n}}{n !} = {e}^{z x} \cosh\!\left(z \sqrt{{x}^{2} - 1}\right)$
$\sum_{n=0}^{\infty} U_{n}\!\left(x\right) \frac{{z}^{n}}{n !} = {e}^{z x} \left(\cosh\!\left(z \sqrt{{x}^{2} - 1}\right) + z x \operatorname{sinc}\!\left(i z \sqrt{{x}^{2} - 1}\right)\right)$