Editing Chebyshev polynomials (section)

==Explicit expressions==

Using the complex number exponentiation definition of the Chebyshev polynomial, one can derive the following expression:
<math display="block"> T_n(x) = \dfrac{1}{2} \bigg( \Big(x-\sqrt{x^2-1} \Big)^n + \Big(x+\sqrt{x^2-1} \Big)^n \bigg) \quad \text{ for } x \in \mathbb {R},</math>
<math display="block"> T_n(x) = \dfrac{1}{2} \bigg( \Big(x-\sqrt{x^2-1} \Big)^n + \Big(x-\sqrt{x^2-1} \Big)^{-n} \bigg) \quad \text{ for } x \in \mathbb {R}.</math>
The two are equivalent because <math>(x + \sqrt{x^2 - 1})(x - \sqrt{x^2 - 1}) = 1</math>.

An explicit form of the Chebyshev polynomial in terms of monomials {{math|''x''<sup>''k''</sup>}} follows from [[de Moivre's formula]]:
<math display="block">T_n(\cos(\theta)) = \operatorname{Re}(\cos n \theta + i \sin n \theta) = \operatorname{Re}((\cos \theta + i \sin \theta)^n),</math>
where {{math|Re}} denotes the [[Complex number#Notation|real part]] of a complex number. Expanding the formula, one gets
<math display="block">(\cos \theta + i \sin \theta)^n = \sum\limits_{j=0}^n \binom{n}{j} i^j \sin^j \theta \cos^{n-j} \theta.</math>
The real part of the expression is obtained from summands corresponding to even indices. Noting <math>i^{2j} = (-1)^j</math> and <math>\sin^{2j} \theta = (1-\cos^2 \theta)^j</math>, one gets the explicit formula:
<math display="block">\cos n \theta = \sum\limits_{j=0}^{\lfloor n / 2 \rfloor} \binom{n}{2j} (\cos^2 \theta - 1)^j \cos^{n-2j} \theta,</math>
which in turn means that
<math display="block">T_n(x) = \sum\limits_{j=0}^{\lfloor n / 2 \rfloor} \binom{n}{2j} (x^2-1)^j x^{n-2j}.</math>
This can be written as a {{math|<sub>2</sub>''F''<sub>1</sub>}} [[hypergeometric function]]:
<math display="block">\begin{align}
T_n(x) & = \sum_{k=0}^{\left \lfloor \frac{n}{2} \right \rfloor} \binom{n}{2k} \left (x^2-1 \right )^k x^{n-2k} \\
& = x^n \sum_{k=0}^{\left  \lfloor \frac{n}{2} \right \rfloor} \binom{n}{2k} \left (1 - x^{-2} \right )^k \\
& = \frac{n}{2} \sum_{k=0}^{\left \lfloor \frac{n}{2} \right \rfloor}(-1)^k \frac{(n-k-1)!}{k!(n-2k)!}~(2x)^{n-2k} \quad \text{ for } n > 0 \\
\\
& = n \sum_{k=0}^{n}(-2)^{k} \frac{(n+k-1)!} {(n-k)!(2k)!}(1 - x)^k \quad \text{ for } n > 0 \\
\\
& = {}_2F_1\!\left(-n,n;\tfrac 1 2; \tfrac{1}{2}(1-x)\right) \\
\end{align}</math>
with inverse<ref name=Cody>{{cite journal |first1=W. J. |last1=Cody |title=A survey of practical rational and polynomial approximation of functions |year=1970 |journal=SIAM Review |volume=12 |number=3 |pages=400–423 |doi=10.1137/1012082}}</ref><ref name=Mathar>{{cite journal
 | last=Mathar | first=Richard J. 
 | year=2006 
 | title=Chebyshev series expansion of inverse polynomials
 | journal=Journal of Computational and Applied Mathematics
 | volume=196 | issue=2 | pages=596–607
 | doi=10.1016/j.cam.2005.10.013 | doi-access=free
 | arxiv=math/0403344
 }}</ref>
<math display="block">x^n = 2^{1-n}\mathop{{\sum}'}^n_{j=0\atop j \equiv n \pmod 2} \!\!\binom{n}{\tfrac{n-j}{2}}\!\;T_j(x),</math>
where the prime at the summation symbol indicates that the contribution of {{math|''j'' {{=}} 0}} needs to be halved if it appears.

A related expression for {{math|''T''<sub>''n''</sub>}} as a sum of monomials with binomial coefficients and powers of two is
<math display="block">
 T_n(x) = \sum\limits_{m=0}^{\left\lfloor \frac{n}{2} \right\rfloor} (-1)^m \left(\binom{n - m}{m} + \binom{n - m - 1}{n - 2m}\right) \cdot 2^{n-2m-1} \cdot x^{n-2m}.</math>

Similarly, {{math|''U''<sub>''n''</sub>}} can be expressed in terms of hypergeometric functions:
<math display="block">\begin{align}
U_n(x) &= \frac{\left (x+\sqrt{x^2-1} \right )^{n+1} - \left (x-\sqrt{x^2-1} \right )^{n+1}}{2\sqrt{x^2-1}} \\
 &= \sum_{k=0}^{\left \lfloor {n}/{2} \right \rfloor} \binom{n+1}{2k+1} \left (x^2-1 \right )^k x^{n-2k} \\
 &= x^n \sum_{k=0}^{\left \lfloor {n}/{2} \right \rfloor} \binom{n+1}{2k+1} \left (1 - x^{-2} \right )^k \\
 &= \sum_{k=0}^{\left \lfloor {n}/{2} \right \rfloor} \binom{2k-(n+1)}{k}~(2x)^{n-2k} & \text{ for } n > 0 \\
 &= \sum_{k=0}^{\left \lfloor {n}/{2} \right \rfloor} (-1)^k \binom{n-k}{k}~(2x)^{n-2k} & \text{ for } n > 0 \\
 &= \sum_{k=0}^{n}(-2)^{k} \frac{(n+k+1)!} {(n-k)!(2k+1)!}(1 - x)^k & \text{ for } n > 0 \\
 &= (n + 1)\, {}_2F_1\big(-n, n + 2; \tfrac{3}{2}; \tfrac{1}{2}(1 - x)\big).
\end{align}</math>