# 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,

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

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 $x=\cos\theta$, $dx=-\sin\theta d\theta$,

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