Chebyshev polynomials

Jordan Bell

December 7, 2016

1 First kind

On the one hand,

	$\displaystyle(\cos\theta+i\sin\theta)^{n}$	$\displaystyle=\sum_{0\leq\nu\leq n}i^{\nu}\binom{n}{\nu}\cos^{n-\nu}\theta\sin%^{\nu}\theta$
		$\displaystyle=\sum_{0\leq 2k\leq n}(-1)^{k}\binom{n}{2k}\cos^{n-2k}(\theta)%\sin^{2k}(\theta)$
		$\displaystyle+i\sum_{0\leq 2k+1\leq n}(-1)^{k}\binom{n}{2k+1}\cos^{n-2k-1}(%\theta)\sin^{2k+1}(\theta).$

On the other hand,

(\cos\theta+i\sin\theta)^{n}=(e^{i\theta})^{n}=e^{in\theta}=\cos n\theta+i\sinn\theta.

Therefore

	$\displaystyle\cos n\theta$	$\displaystyle=\sum_{0\leq k\leq n/2}(-1)^{k}\binom{n}{2k}\cos^{n-2k}(\theta)%\sin^{2k}(\theta)$
		$\displaystyle=\sum_{0\leq k\leq n/2}(-1)^{k}\binom{n}{2k}\cos^{n-2k}(\theta)(1%-\cos^{2}\theta)^{k}$
		$\displaystyle=\sum_{0\leq k\leq n/2}\binom{n}{2k}\cos^{n-2k}(\theta)(\cos^{2}%\theta-1)^{k}$
		$\displaystyle=\sum_{0\leq k\leq n/2}\binom{n}{2k}\cos^{n-2k}(\theta)\sum_{0%\leq j\leq k}\binom{k}{j}\cos^{2k-2j}(\theta)(-1)^{j}$
		$\displaystyle=\sum_{0\leq j\leq n/2}(-1)^{j}\cos^{n-2j}(\theta)\sum_{j\leq k%\leq n/2}\binom{n}{2k}\binom{k}{j}.$

Now,

\displaystyle\sum_{j\leq k\leq n/2}\binom{n}{2k}\binom{k}{j}

\displaystyle=2^{n-2j-1}\binom{n-j}{j}\frac{n}{n-j}.

Hence

\cos n\theta=\sum_{0\leq j\leq n/2}(-1)^{j}\cos^{n-2j}(\theta)2^{n-2j-1}\binom%{n-j}{j}\frac{n}{n-j}.

For $z\in\mathbb{C}$ let

T_{n}(z)=\sum_{0\leq j\leq n/2}(-1)^{j}2^{n-2j-1}\binom{n-j}{j}\frac{n}{n-j}z^%{n-2j}.

(1)

Note

T_{n}(z)[z^{n}]=2^{n-1}z^{n}.

Theorem 1.

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

and

T_{m}\circ T_{n}=T_{mn}.

Proof.

For $\theta\in\mathbb{R}$ ,

T_{n}(\cos\theta)=\cos(n\theta).

Then

T_{m}(T_{n}(\cos\theta))=T_{m}(\cos(n\theta))=\cos(mn\theta)=T_{mn}(\theta).

That is, for $z\in[-1,1]$ we have $T_{m}(T_{n}(z))=T_{mn}(z)$ . Then by analytic continuation it follows that this is true for all $z$ . ∎

Theorem 2.

T_{n}(z)+T_{n-2}(z)=2zT_{n-1}(z).

Proof.

Using $\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$ ,

\cos(n\theta)=\cos(\theta+(n-1)\theta)=\cos\theta\cos((n-1)\theta)-\sin\theta%\sin((n-1)\theta)

and

\cos((n-2)\theta)=\cos(-\theta+(n-1)\theta)=\cos\theta\cos((n-1)\theta)+\sin%\theta\sin((n-1)\theta).

Then

\cos(n\theta)+\cos((n-2)\theta)=2\cos\theta\cos((n-1)\theta).

Therefore

	$\displaystyle T_{n}(\cos\theta)+T_{n-2}(\cos\theta)$	$\displaystyle=\cos(n\theta)+\cos((n-2)\theta)$
		$\displaystyle=2\cos\theta\cos((n-1)\theta)$
		$\displaystyle=2\cos\theta\cdot T_{n-1}(\cos\theta).$

That is, for $z\in[-1,1]$ ,

T_{n}(z)+T_{n-2}(z)=2zT_{n-1}(z),

and by analytic continuation this is true for all $z\in\mathbb{C}$ . ∎

2 Second kind

Define

nU_{n-1}(z)=T_{n}^{\prime}(z).

(2)

Theorem 3.

U_{n-1}(\cos\theta)=\frac{\sin(n\theta)}{\sin\theta}

and

(1-z^{2})T_{n}^{\prime\prime}(z)=nU_{n}(z)-n(n+1)T_{n}(z).

Proof.

On the one hand,

\displaystyle(T_{n}(\cos\theta))^{\prime}

\displaystyle=-\sin\theta\cdot T_{n}^{\prime}(\cos\theta).

On the other hand,

(T_{n}(\cos\theta))^{\prime}=(\cos(n\theta))^{\prime}=-n\sin(n\theta).

Hence

T_{n}^{\prime}(\cos\theta)=n\frac{\sin(n\theta)}{\sin\theta},\qquad U_{n-1}(%\cos\theta)=\frac{\sin(n\theta)}{\sin\theta}.

Now,

(T_{n}^{\prime}(\cos\theta))^{\prime}=-\sin\theta\cdot T_{n}^{\prime\prime}(%\cos\theta)

and

	$\displaystyle(T_{n}^{\prime}(\cos\theta))^{\prime}$	$\displaystyle=n\frac{n\cos(n\theta)\sin\theta-\sin(n\theta)\cos\theta}{\sin^{2%}\theta}$
		$\displaystyle=-n\frac{\cos(n\theta)\sin\theta+\sin(n\theta)\cos\theta}{\sin^{2%}\theta}+n(n+1)\frac{\cos(n\theta)\sin\theta}{\sin^{2}\theta}$
		$\displaystyle=-n\frac{\sin((n+1)\theta)}{\sin^{2}\theta}+n(n+1)\frac{\cos(n%\theta)}{\sin\theta}$
		$\displaystyle=-n\frac{U_{n}(\cos\theta)}{\sin\theta}+n(n+1)\frac{T_{n}(\cos%\theta)}{\sin\theta}.$

Hence

T_{n}^{\prime\prime}(\cos\theta)=n\frac{U_{n}(\cos\theta)}{\sin^{2}\theta}-n(n%+1)\frac{T_{n}(\cos\theta)}{\sin^{2}\theta}

and then

T_{n}^{\prime\prime}(\cos\theta)=n\frac{U_{n}(\cos\theta)}{1-\cos^{2}\theta}-n%(n+1)\frac{T_{n}(\cos\theta)}{1-\cos^{2}\theta}.

By analytic continuation,

(1-z^{2})T_{n}^{\prime\prime}(z)=nU_{n}(z)-n(n+1)T_{n}(z).

∎

Theorem 4.

T_{n+1}(z)=zT_{n}(z)-(1-z^{2})U_{n-1}(z).

Proof.

	$\displaystyle T_{n+1}(\cos\theta)$	$\displaystyle=\cos(n\theta+\theta)$
		$\displaystyle=\cos(n\theta)\cos\theta-\sin(n\theta)\sin\theta$
		$\displaystyle=T_{n}(\cos\theta)\cos\theta-U_{n-1}(\cos\theta)\sin^{2}\theta$
		$\displaystyle=T_{n}(\cos\theta)\cos\theta-U_{n-1}(\cos\theta)(1-\cos^{2}\theta).$

Therefore by analytic continuation,

T_{n+1}(z)=zT_{n}(z)-(1-z^{2})U_{n-1}(z).

∎

Theorem 5.

U_{n}(z)=T_{n}(z)+zU_{n-1}(z).

Proof.

	$\displaystyle U_{n}(\cos\theta)$	$\displaystyle=\frac{\sin(n\theta+\theta)}{\sin\theta}$
		$\displaystyle=\frac{\cos(n\theta)\sin\theta+\cos\theta\sin(n\theta)}{\sin\theta}$
		$\displaystyle=T_{n}(\cos\theta)+\cos\theta\cdot U_{n-1}(\cos\theta).$

Therefore by analytic continuation,

U_{n}(z)=T_{n}(z)+zU_{n-1}(z).

∎

Theorem 6.

U_{n}(z)=2zU_{n-1}(z)+U_{n-2}(z).

Proof.

Using Theorem 4 and Theorem 5,

	$\displaystyle U_{n}(z)$	$\displaystyle=T_{n}(z)+zU_{n-1}(z)$
		$\displaystyle=zT_{n-1}(z)-(1-z^{2})U_{n-2}(z)+zU_{n-1}(z)$
		$\displaystyle=z\bigg{[}U_{n-1}(z)-zU_{n-2}(z)\bigg{]}-(1-z^{2})U_{n-2}(z)+zU_{%n-1}(z)$
		$\displaystyle=2zU_{n-1}(z)+U_{n-2}(z).$

∎

Theorem 7.

(1-z^{2})T_{n}^{\prime\prime}(z)-zT_{n}^{\prime}(z)+n^{2}T_{n}(z)=0.

Proof.

Using Theorem 3, and Theorem 5,

\begin{split}&\displaystyle(1-z^{2})T_{n}^{\prime\prime}(z)-zT_{n}^{\prime}(z)%+n^{2}T_{n}(z)\\\displaystyle=&\displaystyle nU_{n}(z)-n(n+1)T_{n}(z)-nzU_{n-1}(z)+n^{2}T_{n}(%z)\\\displaystyle=&\displaystyle n(T_{n}(z)+zU_{n-1}(z))-n(n+1)T_{n}(z)-nzU_{n-1}(%z)+n^{2}T_{n}(z)\\\displaystyle=&\displaystyle 0.\end{split}

∎

From Theorem 1

T_{n}(1)=T_{n}(\cos 0)=\cos(n\cdot 0)=1.

From Theorem 3,

T_{n}^{\prime}(1)=nU_{n-1}(1)=n^{2}.

Thus, $T_{n}$ is the unique solution of the initial value problem

(1-x^{2})y^{\prime\prime}(x)-xy^{\prime}(x)+n^{2}y(x)=0,\qquad y(1)=1,y^{%\prime}(1)=n^{2}.

Theorem 8.

T_{n}(z)^{2}-(z^{2}-1)U_{n-1}(z)^{2}=1.

Proof.

Using Theorem 1 and Theorem 3, for $z=\cos\theta$ ,

	$\displaystyle T_{n}(z)^{2}-(z^{2}-1)U_{n-1}(z)^{2}$	$\displaystyle=T_{n}(\cos\theta)^{2}+(\sin^{2}\theta)U_{n-1}(\cos\theta)^{2}$
		$\displaystyle=\cos^{2}(n\theta)+(\sin^{2}\theta)\frac{\sin^{2}(n\theta)}{\sin^%{2}\theta}$
		$\displaystyle=\cos^{2}(n\theta)+\sin^{2}(n\theta)$
		$\displaystyle=1.$

By analytic continuation, this is true for all $z$ . ∎

3 Inner products

For $0\leq\theta\leq\pi$ let $y_{n}(\theta)=\cos(n\theta)$ .

y_{n}^{\prime\prime}+n^{2}y_{n}=0,\qquad y_{n}^{\prime}(0)=0,y_{n}^{\prime}(%\pi)=0.

Theorem 9.

\int_{-1}^{1}\frac{T_{m}(x)T_{n}(x)}{\sqrt{1-x^{2}}}dx=\int_{0}^{\pi}y_{m}y_{n%}d\theta=\frac{\pi}{2}\cdot\delta_{m,n}.

Proof.

Let $W=y_{m}y_{n}^{\prime}-y_{n}y_{m}^{\prime}$ . We calculate

	$\displaystyle W^{\prime}$	$\displaystyle=y_{m}^{\prime}y_{n}^{\prime}+y_{m}y_{n}^{\prime\prime}-y_{n}^{%\prime}y_{m}^{\prime}-y_{n}y_{m}^{\prime\prime}y_{m}y_{n}^{\prime\prime}-y_{n}%y_{m}^{\prime\prime}$
		$\displaystyle=y_{m}y_{n}^{\prime\prime}-y_{n}y_{m}^{\prime\prime}$
		$\displaystyle=y_{m}(-n^{2}y_{n})-y_{n}(-m^{2}y_{m})$
		$\displaystyle=(m^{2}-n^{2})y_{m}y_{n}.$

Using $W(0)=0$ and $W(\pi)=0$ ,

\int_{0}^{\pi}W^{\prime}(\theta)d\theta=W(\pi)-W(0)=0.

Then

\int_{0}^{\pi}(m^{2}-n^{2})y_{m}y_{n}d\theta=0.

Doing the substitution $\phi=n\theta$ ,

	$\displaystyle\int_{0}^{\pi}y_{n}^{2}d\theta$	$\displaystyle=\int_{0}^{\pi}\cos^{2}(n\theta)d\theta$
		$\displaystyle=\int_{0}^{\pi}\frac{1+\cos(2n\theta)}{2}d\theta$
		$\displaystyle=\frac{\pi}{2}.$

Therefore

\int_{0}^{\pi}y_{m}y_{n}d\theta=\frac{\pi}{2}\cdot\delta_{m,n}.

For $0\leq\theta\leq\pi$ , $\sqrt{1-\cos^{2}\theta}=\sin\theta$ . Then doing the substitution $x=\cos\theta$ , $dx=-\sin\theta d\theta$ ,

	$\displaystyle\int_{0}^{\pi}y_{m}y_{n}d\theta$	$\displaystyle=\int_{0}^{\pi}\cos(m\theta)\cos(n\theta)d\theta$
		$\displaystyle=\int_{0}^{\pi}\cos(m\theta)\cos(n\theta)\frac{-\sin\theta d%\theta}{-\sin\theta}$
		$\displaystyle=\int_{0}^{\pi}\frac{\cos(m\theta)\cos(n\theta)}{-\sqrt{1-\cos^{2%}\theta}}(-\sin\theta)d\theta$
		$\displaystyle=\int_{1}^{-1}\frac{T_{m}(x)T_{n}(x)}{-\sqrt{1-x^{2}}}dx$
		$\displaystyle=\int_{-1}^{1}\frac{T_{m}(x)T_{n}(x)}{\sqrt{1-x^{2}}}dx.$

∎