Chebyshev polynomials

The Chebyshev polynomials of the first kind can be defined by the recurrence relation

$${\displaystyle {\begin{aligned}T_{0}(x)&=1,\\T_{1}(x)&=x,\\T_{n+1}(x)&=2x\,T_{n}(x)-T_{n-1}(x).\end{aligned}}}$$

The Chebyshev polynomials of the second kind can be defined by the recurrence relation

$${\displaystyle {\begin{aligned}U_{0}(x)&=1,\\U_{1}(x)&=2x,\\U_{n+1}(x)&=2x\,U_{n}(x)-U_{n-1}(x),\end{aligned}}}$$ which differs from the above only by the rule for n=1.

The Chebyshev polynomials of the first and second kind can be defined as the unique polynomials satisfying

$${\displaystyle T_{n}(\cos \theta )=\cos(n\theta )}$$

and

$${\displaystyle U_{n}(\cos \theta )={\frac {\sin {\big (}(n+1)\theta {\big )}}{\sin \theta }},}$$

for n = 0, 1, 2, 3, ….

An equivalent way to state this is via exponentiation of a complex number: given a complex number z = a + bi with absolute value of one,

$${\displaystyle z^{n}=T_{n}(a)+ibU_{n-1}(a).}$$

Chebyshev polynomials can also be defined in this form when studying trigonometric polynomials.^[4]

That ${\displaystyle \cos(nx)}$ is an ${\displaystyle n}$th-degree polynomial in ${\displaystyle \cos(x)}$ can be seen by observing that ${\displaystyle \cos(nx)}$ is the real part of one side of de Moivre's formula:

$${\displaystyle \cos n\theta +i\sin n\theta =(\cos \theta +i\sin \theta )^{n}.}$$

The real part of the other side is a polynomial in ${\displaystyle \cos(x)}$ and ${\displaystyle \sin(x)}$, in which all powers of ${\displaystyle \sin(x)}$ are even and thus replaceable through the identity ${\displaystyle \cos ^{2}(x)+\sin ^{2}(x)=1}$. By the same reasoning, ${\displaystyle \sin(nx)}$ is the imaginary part of the polynomial, in which all powers of ${\displaystyle \sin(x)}$ are odd and thus, if one factor of ${\displaystyle \sin(x)}$ is factored out, the remaining factors can be replaced to create a ${\displaystyle n-1}$st-degree polynomial in ${\displaystyle \cos(x)}$.

For ${\displaystyle x}$ outside the interval [-1,1], the above definition implies

$${\displaystyle T_{n}(x)={\begin{cases}\cos(n\arccos x)&{\text{ if }}~|x|\leq 1,\\\cosh(n\operatorname {arcosh} x)&{\text{ if }}~x\geq 1,\\(-1)^{n}\cosh(n\operatorname {arcosh} (-x))&{\text{ if }}~x\leq -1.\end{cases}}}$$

Chebyshev polynomials can also be characterized by the following theorem:^[5]

If ${\displaystyle F_{n}(x)}$ is a family of monic polynomials with coefficients in a field of characteristic ${\displaystyle 0}$ such that ${\displaystyle \deg F_{n}(x)=n}$ and ${\displaystyle F_{m}(F_{n}(x))=F_{n}(F_{m}(x))}$ for all ${\displaystyle m}$ and ${\displaystyle n}$, then, up to a simple change of variables, either ${\displaystyle F_{n}(x)=x^{n}}$ for all ${\displaystyle n}$ or ${\displaystyle F_{n}(x)=2\cdot T_{n}(x/2)}$ for all ${\displaystyle n}$.

The Chebyshev polynomials can also be defined as the solutions to the Pell equation:

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

in a ring ${\displaystyle R[x]}$.^[6] Thus, they can be generated by the standard technique for Pell equations of taking powers of a fundamental solution:

$${\displaystyle T_{n}(x)+U_{n-1}(x)\,{\sqrt {x^{2}-1}}=\left(x+{\sqrt {x^{2}-1}}\right)^{n}~.}$$

The ordinary generating function for ${\displaystyle T_{n}}$ is

$${\displaystyle \sum _{n=0}^{\infty }T_{n}(x)\,t^{n}={\frac {1-tx}{1-2tx+t^{2}}}.}$$

There are several other generating functions for the Chebyshev polynomials; the exponential generating function is

$${\displaystyle {\begin{aligned}\sum _{n=0}^{\infty }T_{n}(x){\frac {t^{n}}{n!}}&={\tfrac {1}{2}}{\Bigl (}{\exp }{\Bigl (}{\textstyle t{\bigl (}x-{\sqrt {x^{2}-1}}~\!{\bigr )}}{\Bigr )}+{\exp }{\Bigl (}{\textstyle t{\bigl (}x+{\sqrt {x^{2}-1}}~\!{\bigr )}}{\Bigr )}{\Bigr )}\\&=e^{tx}\cosh \left({\textstyle t{\sqrt {x^{2}-1}}}~\!\right).\end{aligned}}}$$

The generating function relevant for 2-dimensional potential theory and multipole expansion is

$${\displaystyle \sum \limits _{n=1}^{\infty }T_{n}(x)\,{\frac {t^{n}}{n}}=\ln \left({\frac {1}{\sqrt {1-2tx+t^{2}}}}\right).}$$

The ordinary generating function for U_n is

$${\displaystyle \sum _{n=0}^{\infty }U_{n}(x)\,t^{n}={\frac {1}{1-2tx+t^{2}}},}$$

and the exponential generating function is

$${\displaystyle \sum _{n=0}^{\infty }U_{n}(x){\frac {t^{n}}{n!}}=e^{tx}{\biggl (}\!\cosh \left(t{\sqrt {x^{2}-1}}\right)+{\frac {x}{\sqrt {x^{2}-1}}}\sinh \left(t{\sqrt {x^{2}-1}}\right){\biggr )}.}$$

The Chebyshev polynomials of the first and second kinds correspond to a complementary pair of Lucas sequences ${\displaystyle {\tilde {V}}_{n}(P,Q)}$ and ${\displaystyle {\tilde {U}}_{n}(P,Q)}$ with parameters ${\displaystyle P=2x}$ and ${\displaystyle Q=1}$:

$${\displaystyle {\begin{aligned}{\tilde {U}}_{n}(2x,1)&=U_{n-1}(x),\\{\tilde {V}}_{n}(2x,1)&=2\,T_{n}(x).\end{aligned}}}$$

It follows that they also satisfy a pair of mutual recurrence equations:^[7]

$${\displaystyle {\begin{aligned}T_{n+1}(x)&=x\,T_{n}(x)-(1-x^{2})\,U_{n-1}(x),\\U_{n+1}(x)&=x\,U_{n}(x)+T_{n+1}(x).\end{aligned}}}$$

The second of these may be rearranged using the recurrence definition for the Chebyshev polynomials of the second kind to give:

$${\displaystyle T_{n}(x)={\frac {1}{2}}{\big (}U_{n}(x)-U_{n-2}(x){\big )}.}$$

Using this formula iteratively gives the sum formula:

$${\displaystyle U_{n}(x)={\begin{cases}2\sum _{{\text{ odd }}j>0}^{n}T_{j}(x)&{\text{ for odd }}n.\\2\sum _{{\text{ even }}j\geq 0}^{n}T_{j}(x)-1&{\text{ for even }}n,\end{cases}}}$$

while replacing ${\displaystyle U_{n}(x)}$ and ${\displaystyle U_{n-2}(x)}$ using the derivative formula for ${\displaystyle T_{n}(x)}$ gives the recurrence relationship for the derivative of ${\displaystyle T_{n}}$:

$${\displaystyle 2\,T_{n}(x)={\frac {1}{n+1}}\,{\frac {\mathrm {d} }{\mathrm {d} x}}\,T_{n+1}(x)-{\frac {1}{n-1}}\,{\frac {\mathrm {d} }{\mathrm {d} x}}\,T_{n-1}(x),\qquad n=2,3,\ldots }$$

This relationship is used in the Chebyshev spectral method of solving differential equations.

Turán's inequalities for the Chebyshev polynomials are:^[8]

$${\displaystyle {\begin{aligned}T_{n}(x)^{2}-T_{n-1}(x)\,T_{n+1}(x)&=1-x^{2}>0&&{\text{ for }}-1<x<1&&{\text{ and }}\\U_{n}(x)^{2}-U_{n-1}(x)\,U_{n+1}(x)&=1>0~.\end{aligned}}}$$

The integral relations are^[9][10]

$${\displaystyle {\begin{aligned}\int _{-1}^{1}{\frac {T_{n}(y)}{y-x}}\,{\frac {\mathrm {d} y}{\sqrt {1-y^{2}}}}&=\pi \,U_{n-1}(x)~,\\[1.5ex]\int _{-1}^{1}{\frac {U_{n-1}(y)}{y-x}}\,{\sqrt {1-y^{2}}}\mathrm {d} y&=-\pi \,T_{n}(x)\end{aligned}}}$$

where integrals are considered as principal value.

Using the complex number exponentiation definition of the Chebyshev polynomial, one can derive the following expressions, valid for any real ⁠${\displaystyle x}$⁠:^{[citation needed]}

$${\displaystyle {\begin{aligned}T_{n}(x)&={\tfrac {1}{2}}{\Big (}{\bigl (}{\textstyle x-{\sqrt {x^{2}-1}}\!~}{\bigr )}^{n}+{\bigl (}{\textstyle x+{\sqrt {x^{2}-1}}\!~}{\bigr )}^{n}{\Big )}\\[5mu]&={\tfrac {1}{2}}{\Big (}{\bigl (}{\textstyle x-{\sqrt {x^{2}-1}}\!~}{\bigr )}^{n}+{\bigl (}{\textstyle x-{\sqrt {x^{2}-1}}\!~}{\bigr )}^{-n}{\Big )}.\end{aligned}}}$$

The two are equivalent because ${\displaystyle \textstyle {\bigl (}x+{\sqrt {x^{2}-1}}\!~{\bigr )}{\bigl (}x-{\sqrt {x^{2}-1}}\!~{\bigr )}=1}$.

An explicit form of the Chebyshev polynomial in terms of monomials ${\displaystyle x^{k}}$ follows from de Moivre's formula:

$${\displaystyle T_{n}(\cos(\theta ))=\operatorname {Re} (\cos n\theta +i\sin n\theta )=\operatorname {Re} ((\cos \theta +i\sin \theta )^{n}),}$$

where ${\displaystyle \mathrm {Re} }$ denotes the real part of a complex number. Expanding the formula, one gets

$${\displaystyle (\cos \theta +i\sin \theta )^{n}=\sum \limits _{j=0}^{n}{\binom {n}{j}}i^{j}\sin ^{j}\theta \cos ^{n-j}\theta .}$$

The real part of the expression is obtained from summands corresponding to even indices. Noting ${\displaystyle i^{2j}=(-1)^{j}}$ and ${\displaystyle \sin ^{2j}\theta =(1-\cos ^{2}\theta )^{j}}$, one gets the explicit formula:

$${\displaystyle \cos n\theta =\sum \limits _{j=0}^{\lfloor n/2\rfloor }{\binom {n}{2j}}(\cos ^{2}\theta -1)^{j}\cos ^{n-2j}\theta ,}$$

which in turn means that

$${\displaystyle T_{n}(x)=\sum \limits _{j=0}^{\lfloor n/2\rfloor }{\binom {n}{2j}}(x^{2}-1)^{j}x^{n-2j}.}$$

This can be written as a ₂F₁ hypergeometric function:

$${\displaystyle {\begin{aligned}T_{n}(x)&=\sum _{k=0}^{\left\lfloor {\frac {n}{2}}\right\rfloor }{\binom {n}{2k}}\left(x^{2}-1\right)^{k}x^{n-2k}\\&=x^{n}\sum _{k=0}^{\left\lfloor {\frac {n}{2}}\right\rfloor }{\binom {n}{2k}}\left(1-x^{-2}\right)^{k}\\&={\frac {n}{2}}\sum _{k=0}^{\left\lfloor {\frac {n}{2}}\right\rfloor }(-1)^{k}{\frac {(n-k-1)!}{k!(n-2k)!}}~(2x)^{n-2k}\quad {\text{ for }}n>0\\\\&=n\sum _{k=0}^{n}(-2)^{k}{\frac {(n+k-1)!}{(n-k)!(2k)!}}(1-x)^{k}\quad {\text{ for }}n>0\\\\&={}_{2}F_{1}\!\left(-n,n;{\tfrac {1}{2}};{\tfrac {1}{2}}(1-x)\right)\\\end{aligned}}}$$

with inverse^[11][12]

$${\displaystyle x^{n}=2^{1-n}\mathop {{\sum }'} _{j=0 \atop j\equiv n{\pmod {2}}}^{n}\!\!{\binom {n}{\tfrac {n-j}{2}}}\!\;T_{j}(x),}$$

where the prime at the summation symbol indicates that the contribution of ${\displaystyle j=0}$ needs to be halved if it appears.

A related expression for ${\displaystyle T_{n}}$ as a sum of monomials with binomial coefficients and powers of two is

$${\displaystyle T_{n}(x)=\sum \limits _{m=0}^{\left\lfloor {\frac {n}{2}}\right\rfloor }(-1)^{m}\left({\binom {n-m}{m}}+{\binom {n-m-1}{n-2m}}\right)\cdot 2^{n-2m-1}\cdot x^{n-2m}.}$$

Similarly, ${\displaystyle U_{n}}$ can be expressed in terms of hypergeometric functions:

$${\displaystyle {\begin{aligned}U_{n}(x)&={\frac {\left(x+{\sqrt {x^{2}-1}}\right)^{n+1}-\left(x-{\sqrt {x^{2}-1}}\right)^{n+1}}{2{\sqrt {x^{2}-1}}}}\\&=\sum _{k=0}^{\left\lfloor {n}/{2}\right\rfloor }{\binom {n+1}{2k+1}}\left(x^{2}-1\right)^{k}x^{n-2k}\\&=x^{n}\sum _{k=0}^{\left\lfloor {n}/{2}\right\rfloor }{\binom {n+1}{2k+1}}\left(1-x^{-2}\right)^{k}\\&=\sum _{k=0}^{\left\lfloor {n}/{2}\right\rfloor }{\binom {2k-(n+1)}{k}}~(2x)^{n-2k}&{\text{ for }}n>0\\&=\sum _{k=0}^{\left\lfloor {n}/{2}\right\rfloor }(-1)^{k}{\binom {n-k}{k}}~(2x)^{n-2k}&{\text{ for }}n>0\\&=\sum _{k=0}^{n}(-2)^{k}{\frac {(n+k+1)!}{(n-k)!(2k+1)!}}(1-x)^{k}&{\text{ for }}n>0\\&=(n+1)\,{}_{2}F_{1}{\big (}-n,n+2;{\tfrac {3}{2}};{\tfrac {1}{2}}(1-x){\big )}.\end{aligned}}}$$

$${\displaystyle {\begin{aligned}T_{n}(-x)&=(-1)^{n}\,T_{n}(x),\\[1ex]U_{n}(-x)&=(-1)^{n}\,U_{n}(x).\end{aligned}}}$$

That is, Chebyshev polynomials of even order have even symmetry and therefore contain only even powers of ${\displaystyle x}$. Chebyshev polynomials of odd order have odd symmetry and therefore contain only odd powers of ${\displaystyle x}$.

A Chebyshev polynomial of either kind with degree n has n different simple roots, called Chebyshev roots, in the interval [−1, 1]. The roots of the Chebyshev polynomial of the first kind are sometimes called Chebyshev nodes because they are used as nodes in polynomial interpolation. Using the trigonometric definition and the fact that:

$${\displaystyle \cos \left((2k+1){\frac {\pi }{2}}\right)=0}$$

one can show that the roots of ${\displaystyle T_{n}}$ are:

$${\displaystyle x_{k}=\cos \left({\frac {2k+1}{2n}}\pi \right),\quad k=0,\ldots ,n-1.}$$

Similarly, the roots of ${\displaystyle U_{n}}$ are:

$${\displaystyle x_{k}=\cos \left({\frac {k}{n+1}}\pi \right),\quad k=1,\ldots ,n.}$$

The extrema of ${\displaystyle T_{n}}$ on the interval ${\displaystyle -1\leq x\leq 1}$ are located at:

$${\displaystyle x_{k}=\cos \left({\frac {k}{n}}\pi \right),\quad k=0,\ldots ,n.}$$

One unique property of the Chebyshev polynomials of the first kind is that on the interval ${\displaystyle -1\leq x\leq 1}$ all of the extrema have values that are either −1 or 1. Thus these polynomials have only two finite critical values, the defining property of Shabat polynomials. Both the first and second kinds of Chebyshev polynomial have extrema at the endpoints, given by:

$${\displaystyle {\begin{aligned}T_{n}(1)&=1\\T_{n}(-1)&=(-1)^{n}\\U_{n}(1)&=n+1\\U_{n}(-1)&=(-1)^{n}(n+1).\end{aligned}}}$$

The extrema of ${\displaystyle T_{n}(x)}$ on the interval ${\displaystyle -1\leq x\leq 1}$ where ${\displaystyle n>0}$ are located at ${\displaystyle n+1}$ values of ${\displaystyle x}$. They are ${\displaystyle \pm 1}$, or ${\displaystyle \cos \left({\frac {2\pi k}{d}}\right)}$ where ${\displaystyle d>2}$, ${\displaystyle d\;|\;2n}$, ${\displaystyle 0<k<d/2}$ and ${\displaystyle (k,d)=1}$, i.e., ${\displaystyle k}$ and ${\displaystyle d}$ are relatively prime numbers.

Specifically (Minimal polynomial of 2cos(2pi/n)^[13][14]) when ${\displaystyle n}$ is even:

• ${\displaystyle T_{n}(x)=1}$ if ${\displaystyle x=\pm 1}$, or ${\displaystyle d>2}$ and ${\displaystyle 2n/d}$ is even. There are ${\displaystyle n/2+1}$ such values of ${\displaystyle x}$.
• ${\displaystyle T_{n}(x)=-1}$ if ${\displaystyle d>2}$ and ${\displaystyle 2n/d}$ is odd. There are ${\displaystyle n/2}$ such values of ${\displaystyle x}$.

When ${\displaystyle n}$ is odd:

• ${\displaystyle T_{n}(x)=1}$ if ${\displaystyle x=1}$, or ${\displaystyle d>2}$ and ${\displaystyle 2n/d}$ is even. There are ${\displaystyle (n+1)/2}$ such values of ${\displaystyle x}$.
• ${\displaystyle T_{n}(x)=-1}$ if ${\displaystyle x=-1}$, or ${\displaystyle d>2}$ and ${\displaystyle 2n/d}$ is odd. There are ${\displaystyle (n+1)/2}$ such values of ${\displaystyle x}$.

The derivatives of the polynomials can be less than straightforward. By differentiating the polynomials in their trigonometric forms, it can be shown that:

$${\displaystyle {\begin{aligned}{\frac {\mathrm {d} T_{n}}{\mathrm {d} x}}&=nU_{n-1}\\{\frac {\mathrm {d} U_{n}}{\mathrm {d} x}}&={\frac {(n+1)T_{n+1}-xU_{n}}{x^{2}-1}}\\{\frac {\mathrm {d} ^{2}T_{n}}{\mathrm {d} x^{2}}}&=n\,{\frac {nT_{n}-xU_{n-1}}{x^{2}-1}}=n\,{\frac {(n+1)T_{n}-U_{n}}{x^{2}-1}}.\end{aligned}}}$$

The last two formulas can be numerically troublesome due to the division by zero (⁠0/0⁠ indeterminate form, specifically) at ${\displaystyle x=1}$ and ${\displaystyle x=-1}$. By L'Hôpital's rule:

$${\displaystyle {\begin{aligned}\left.{\frac {\mathrm {d} ^{2}T_{n}}{\mathrm {d} x^{2}}}\right|_{x=1}\!\!&={\frac {n^{4}-n^{2}}{3}},\\\left.{\frac {\mathrm {d} ^{2}T_{n}}{\mathrm {d} x^{2}}}\right|_{x=-1}\!\!&=(-1)^{n}{\frac {n^{4}-n^{2}}{3}}.\end{aligned}}}$$

More generally,

$${\displaystyle \left.{\frac {\mathrm {d} ^{p}T_{n}}{\mathrm {d} x^{p}}}\right|_{x=\pm 1}\!\!=(\pm 1)^{n+p}\prod _{k=0}^{p-1}{\frac {n^{2}-k^{2}}{2k+1}}~,}$$

which is of great use in the numerical solution of eigenvalue problems.

Also, we have:

$${\displaystyle {\frac {\mathrm {d} ^{p}}{\mathrm {d} x^{p}}}\,T_{n}(x)=2^{p}\,n\mathop {{\sum }'} _{0\leq k\leq n-p \atop k\,\equiv \,n-p{\pmod {2}}}{\binom {{\frac {n+p-k}{2}}-1}{\frac {n-p-k}{2}}}{\frac {\left({\frac {n+p+k}{2}}-1\right)!}{\left({\frac {n-p+k}{2}}\right)!}}\,T_{k}(x),~\qquad p\geq 1,}$$

where the prime at the summation symbols means that the term contributed by k = 0 is to be halved, if it appears.

Concerning integration, the first derivative of the T_n implies that:

$${\displaystyle \int U_{n}\,\mathrm {d} x={\frac {T_{n+1}}{n+1}}}$$

and the recurrence relation for the first kind polynomials involving derivatives establishes that for ${\displaystyle n\geq 2}$:

$${\displaystyle \int T_{n}\,\mathrm {d} x={\frac {1}{2}}\,\left({\frac {T_{n+1}}{n+1}}-{\frac {T_{n-1}}{n-1}}\right)={\frac {n\,T_{n+1}}{n^{2}-1}}-{\frac {x\,T_{n}}{n-1}}.}$$

The last formula can be further manipulated to express the integral of ${\displaystyle T_{n}}$ as a function of Chebyshev polynomials of the first kind only:

$${\displaystyle {\begin{aligned}\int T_{n}\,\mathrm {d} x&={\frac {n}{n^{2}-1}}T_{n+1}-{\frac {1}{n-1}}T_{1}T_{n}\\&={\frac {n}{n^{2}-1}}\,T_{n+1}-{\frac {1}{2(n-1)}}\,(T_{n+1}+T_{n-1})\\&={\frac {1}{2(n+1)}}\,T_{n+1}-{\frac {1}{2(n-1)}}\,T_{n-1}.\end{aligned}}}$$

Furthermore, we have:

$${\displaystyle \int _{-1}^{1}T_{n}(x)\,\mathrm {d} x={\begin{cases}{\frac {(-1)^{n}+1}{1-n^{2}}}&{\text{ if }}~n\neq 1\\0&{\text{ if }}~n=1.\end{cases}}}$$

The Chebyshev polynomials of the first kind satisfy the relation:

$${\displaystyle T_{m}(x)\,T_{n}(x)={\tfrac {1}{2}}\!\left(T_{m+n}(x)+T_{|m-n|}(x)\right)\!,\qquad \forall m,n\geq 0,}$$

which is easily proved from the product-to-sum formula for the cosine:

$${\displaystyle 2\cos \alpha \,\cos \beta =\cos(\alpha +\beta )+\cos(\alpha -\beta ).}$$

For ${\displaystyle n=1}$ this results in the already known recurrence formula, just arranged differently, and with ${\displaystyle n=2}$ it forms the recurrence relation for all even or all odd indexed Chebyshev polynomials (depending on the parity of the lowest m) which implies the evenness or oddness of these polynomials. Three more useful formulas for evaluating Chebyshev polynomials can be concluded from this product expansion:

$${\displaystyle {\begin{aligned}T_{2n}(x)&=2\,T_{n}^{2}(x)-T_{0}(x)&&=2T_{n}^{2}(x)-1,\\T_{2n+1}(x)&=2\,T_{n+1}(x)\,T_{n}(x)-T_{1}(x)&&=2\,T_{n+1}(x)\,T_{n}(x)-x,\\T_{2n-1}(x)&=2\,T_{n-1}(x)\,T_{n}(x)-T_{1}(x)&&=2\,T_{n-1}(x)\,T_{n}(x)-x.\end{aligned}}}$$

The polynomials of the second kind satisfy the similar relation:

$${\displaystyle T_{m}(x)\,U_{n}(x)={\begin{cases}{\frac {1}{2}}\left(U_{m+n}(x)+U_{n-m}(x)\right),&~{\text{ if }}~n\geq m-1,\\\\{\frac {1}{2}}\left(U_{m+n}(x)-U_{m-n-2}(x)\right),&~{\text{ if }}~n\leq m-2.\end{cases}}}$$

(with the definition ${\displaystyle U_{-1}\equiv 0}$ by convention ). They also satisfy:

$${\displaystyle U_{m}(x)\,U_{n}(x)=\sum _{k=0}^{n}\,U_{m-n+2k}(x)=\sum _{\underset {\text{ step 2 }}{p=m-n}}^{m+n}U_{p}(x)~.}$$

for ${\displaystyle m\geq n}$. For ${\displaystyle n=2}$ this recurrence reduces to:

$${\displaystyle U_{m+2}(x)=U_{2}(x)\,U_{m}(x)-U_{m}(x)-U_{m-2}(x)=U_{m}(x)\,{\big (}U_{2}(x)-1{\big )}-U_{m-2}(x)~,}$$

which establishes the evenness or oddness of the even or odd indexed Chebyshev polynomials of the second kind depending on whether ${\displaystyle m}$ starts with 2 or 3.

The trigonometric definitions of ${\displaystyle T_{n}}$ and ${\displaystyle U_{n}}$ imply the composition or nesting properties:^[15]

$${\displaystyle {\begin{aligned}T_{mn}(x)&=T_{m}(T_{n}(x)),\\U_{mn-1}(x)&=U_{m-1}(T_{n}(x))U_{n-1}(x).\end{aligned}}}$$

For ${\displaystyle T_{mn}}$ the order of composition may be reversed, making the family of polynomial functions ${\displaystyle T_{n}}$ a commutative semigroup under composition.

Since ${\displaystyle T_{m}(x)}$ is divisible by ${\displaystyle x}$ if ${\displaystyle m}$ is odd, it follows that ${\displaystyle T_{mn}(x)}$ is divisible by ${\displaystyle T_{n}(x)}$ if ${\displaystyle m}$ is odd. Furthermore, ${\displaystyle U_{mn-1}(x)}$ is divisible by ${\displaystyle U_{n-1}(x)}$, and in the case that ${\displaystyle m}$ is even, divisible by ${\displaystyle T_{n}(x)U_{n-1}(x)}$.

Both ${\displaystyle T_{n}}$ and ${\displaystyle U_{n}}$ form a sequence of orthogonal polynomials. The polynomials of the first kind ${\displaystyle T_{n}}$ are orthogonal with respect to the weight:

$${\displaystyle {\frac {1}{\sqrt {1-x^{2}}}},}$$

on the interval [−1, 1], i.e. we have:

$${\displaystyle \int _{-1}^{1}T_{n}(x)\,T_{m}(x)\,{\frac {\mathrm {d} x}{\sqrt {1-x^{2}}}}={\begin{cases}0&~{\text{ if }}~n\neq m,\\[5mu]\pi &~{\text{ if }}~n=m=0,\\[5mu]{\frac {\pi }{2}}&~{\text{ if }}~n=m\neq 0.\end{cases}}}$$

This can be proven by letting ${\displaystyle x=\cos(\theta )}$ and using the defining identity ${\displaystyle T_{n}(\cos(\theta ))=\cos(n\theta )}$.

Similarly, the polynomials of the second kind U_n are orthogonal with respect to the weight:

$${\displaystyle {\sqrt {1-x^{2}}}}$$ on the interval [−1, 1], i.e. we have:

$${\displaystyle \int _{-1}^{1}U_{n}(x)\,U_{m}(x)\,{\sqrt {1-x^{2}}}\,\mathrm {d} x={\begin{cases}0&~{\text{ if }}~n\neq m,\\[5mu]{\frac {\pi }{2}}&~{\text{ if }}~n=m.\end{cases}}}$$

(The measure ${\displaystyle {\sqrt {1-x^{2}}}\,dx}$ is, to within a normalizing constant, the Wigner semicircle distribution.)

These orthogonality properties follow from the fact that the Chebyshev polynomials solve the Chebyshev differential equations:

$${\displaystyle {\begin{aligned}(1-x^{2})T_{n}''-xT_{n}'+n^{2}T_{n}&=0,\\[1ex](1-x^{2})U_{n}''-3xU_{n}'+n(n+2)U_{n}&=0,\end{aligned}}}$$ which are Sturm–Liouville differential equations. It is a general feature of such differential equations that there is a distinguished orthonormal set of solutions. (Another way to define the Chebyshev polynomials is as the solutions to those equations.)

The ${\displaystyle T_{n}}$ also satisfy a discrete orthogonality condition:

$${\displaystyle \sum _{k=0}^{N-1}{T_{i}(x_{k})\,T_{j}(x_{k})}={\begin{cases}0&~{\text{ if }}~i\neq j,\\[5mu]N&~{\text{ if }}~i=j=0,\\[5mu]{\frac {N}{2}}&~{\text{ if }}~i=j\neq 0,\end{cases}}}$$

where ${\displaystyle N}$ is any integer greater than ${\displaystyle \max(i,j)}$,^[10] and the ${\displaystyle x_{k}}$ are the ${\displaystyle N}$ Chebyshev nodes (see above) of ${\displaystyle T_{N}(x)}$:

$${\displaystyle x_{k}=\cos \left(\pi \,{\frac {2k+1}{2N}}\right)\quad ~{\text{ for }}~k=0,1,\dots ,N-1.}$$

For the polynomials of the second kind and any integer ${\displaystyle N>i+j}$ with the same Chebyshev nodes ${\displaystyle x_{k}}$, there are similar sums:

$${\displaystyle \sum _{k=0}^{N-1}{U_{i}(x_{k})\,U_{j}(x_{k})\left(1-x_{k}^{2}\right)}={\begin{cases}0&{\text{ if }}~i\neq j,\\[5mu]{\frac {N}{2}}&{\text{ if }}~i=j,\end{cases}}}$$

and without the weight function:

$${\displaystyle \sum _{k=0}^{N-1}{U_{i}(x_{k})\,U_{j}(x_{k})}={\begin{cases}0&~{\text{ if }}~i\not \equiv j{\pmod {2}},\\[5mu]N\cdot (1+\min\{i,j\})&~{\text{ if }}~i\equiv j{\pmod {2}}.\end{cases}}}$$

For any integer ${\displaystyle N>i+j}$, based on the ${\displaystyle N}$} zeros of ${\displaystyle U_{N}(x)}$:

$${\displaystyle y_{k}=\cos \left(\pi \,{\frac {k+1}{N+1}}\right)\quad ~{\text{ for }}~k=0,1,\dots ,N-1,}$$

one can get the sum:

$${\displaystyle \sum _{k=0}^{N-1}{U_{i}(y_{k})\,U_{j}(y_{k})(1-y_{k}^{2})}={\begin{cases}0&~{\text{ if }}i\neq j,\\[5mu]{\frac {N+1}{2}}&~{\text{ if }}i=j,\end{cases}}}$$