Chebyshev polynomials

1 First kind

On the one hand,

${(\cos \theta +i\sin \theta )}^n$ $={\displaystyle \sum _{0\le \nu \le n}}{i}^{\nu }\left({\displaystyle \genfrac{}{}{0pt}{}{n}{\nu }}\right){\cos }^{n-\nu }\theta {\sin }^{\nu }\theta$

$={\displaystyle \sum _{0\le 2k\le n}}{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k}}\right){\cos }^{n-2k}(\theta ){\sin }^{2k}(\theta )$

$+i{\displaystyle \sum _{0\le 2k+1\le n}}{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k+1}}\right){\cos }^{n-2k-1}(\theta ){\sin }^{2k+1}(\theta ).$

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

$\cos n\theta$ $={\displaystyle \sum _{0\le k\le n/2}}{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k}}\right){\cos }^{n-2k}(\theta ){\sin }^{2k}(\theta )$

$={\displaystyle \sum _{0\le k\le n/2}}{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k}}\right){\cos }^{n-2k}(\theta ){(1-{\cos }^2\theta )}^k$

$={\displaystyle \sum _{0\le k\le n/2}}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k}}\right){\cos }^{n-2k}(\theta ){({\cos }^2\theta -1)}^k$

$={\displaystyle \sum _{0\le k\le n/2}}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k}}\right){\cos }^{n-2k}(\theta ){\displaystyle \sum _{0\le j\le k}}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{j}}\right){\cos }^{2k-2j}(\theta ){(-1)}^j$

$={\displaystyle \sum _{0\le j\le n/2}}{(-1)}^j{\cos }^{n-2j}(\theta ){\displaystyle \sum _{j\le k\le n/2}}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{k}{j}}\right).$

Now,

${\displaystyle \sum _{j\le k\le n/2}}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{k}{j}}\right)$ $=2^{n-2j-1}\left({\displaystyle \genfrac{}{}{0pt}{}{n-j}{j}}\right){\displaystyle \frac{n}{n-j}}.$

Hence

$$\cos n\theta =\sum _{0\le j\le n/2}{(-1)}^j{\cos }^{n-2j}(\theta )2^{n-2j-1}\left(\genfrac{}{}{0pt}{}{n-j}{j}\right)\frac{n}{n-j}.$$

For $z\in ℂ$ let

$$T_n(z)=\sum _{0\le j\le n/2}{(-1)}^j{2}^{n-2j-1}\left(\genfrac{}{}{0pt}{}{n-j}{j}\right)\frac{n}{n-j}{z}^{n-2j}.$$ (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 $\theta \in ℝ$,

$$T_n(\cos \theta )=\cos (n\theta ).$$

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.

$$T_n(z)+T_{n-2}(z)=2z{T}_{n-1}(z).$$

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

$T_n(\cos \theta )+T_{n-2}(\cos \theta )$ $=\cos (n\theta )+\cos ((n-2)\theta )$

$=2\cos \theta \cos ((n-1)\theta )$

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

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

$$T_n(z)+T_{n-2}(z)=2z{T}_{n-1}(z),$$

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

2 Second kind

Define

$$n{U}_{n-1}(z)=T_n^{\prime }(z).$$ (2)

Theorem 3.

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

and

$$(1-z^2)T_n^{\prime \prime }(z)=n{U}_n(z)-n(n+1)T_n(z).$$

Proof.

On the one hand,

${(T_n(\cos \theta ))}^{\prime }$ $=-\sin \theta \cdot T_n^{\prime }(\cos \theta ).$

On the other hand,

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

Hence

$$T_n^{\prime }(\cos \theta )=n\frac{\sin (n\theta )}{\sin \theta },U_{n-1}(\cos \theta )=\frac{\sin (n\theta )}{\sin \theta }.$$

Now,

$${(T_n^{\prime }(\cos \theta ))}^{\prime }=-\sin \theta \cdot T_n^{\prime \prime }(\cos \theta )$$

and

${(T_n^{\prime }(\cos \theta ))}^{\prime }$ $=n{\displaystyle \frac{n\cos (n\theta )\sin \theta -\sin (n\theta )\cos \theta }{{\sin }^2\theta }}$

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

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

$=-n{\displaystyle \frac{U_n(\cos \theta )}{\sin \theta }}+n(n+1){\displaystyle \frac{T_n(\cos \theta )}{\sin \theta }}.$

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,

$$(1-z^2)T_n^{\prime \prime }(z)=n{U}_n(z)-n(n+1)T_n(z).$$

∎

Theorem 4.

$$T_{n+1}(z)=z{T}_n(z)-(1-z^2)U_{n-1}(z).$$

Proof.

$T_{n+1}(\cos \theta )$ $=\cos (n\theta +\theta )$

$=\cos (n\theta )\cos \theta -\sin (n\theta )\sin \theta$

$=T_n(\cos \theta )\cos \theta -U_{n-1}(\cos \theta ){\sin }^2\theta$

$=T_n(\cos \theta )\cos \theta -U_{n-1}(\cos \theta )(1-{\cos }^2\theta ).$

Therefore by analytic continuation,

$$T_{n+1}(z)=z{T}_n(z)-(1-z^2)U_{n-1}(z).$$

∎

Theorem 5.

$$U_n(z)=T_n(z)+z{U}_{n-1}(z).$$

Proof.

$U_n(\cos \theta )$ $={\displaystyle \frac{\sin (n\theta +\theta )}{\sin \theta }}$

$={\displaystyle \frac{\cos (n\theta )\sin \theta +\cos \theta \sin (n\theta )}{\sin \theta }}$

$=T_n(\cos \theta )+\cos \theta \cdot U_{n-1}(\cos \theta ).$

Therefore by analytic continuation,

$$U_n(z)=T_n(z)+z{U}_{n-1}(z).$$

∎

Theorem 6.

$$U_n(z)=2z{U}_{n-1}(z)+U_{n-2}(z).$$

Proof.

Using Theorem 4 and Theorem 5,

$U_n(z)$ $=T_n(z)+z{U}_{n-1}(z)$

$=z{T}_{n-1}(z)-(1-z^2)U_{n-2}(z)+z{U}_{n-1}(z)$

$=z\left[U_{n-1}(z)-z{U}_{n-2}(z)\right]-(1-z^2)U_{n-2}(z)+z{U}_{n-1}(z)$

$=2z{U}_{n-1}(z)+U_{n-2}(z).$

∎

Theorem 7.

$$(1-z^2)T_n^{\prime \prime }(z)-z{T}_n^{\prime }(z)+n^2{T}_n(z)=0.$$

Proof.

Using Theorem 3, and Theorem 5,

∎

From Theorem 1

$$T_n(1)=T_n(\cos 0)=\cos (n\cdot 0)=1.$$

From Theorem 3,

$$T_n^{\prime }(1)=n{U}_{n-1}(1)=n^2.$$

Thus, $T_n$ is the unique solution of the initial value problem

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

Theorem 8.

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

Proof.

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

$T_n{(z)}^2-(z^2-1)U_{n-1}{(z)}^2$ $=T_n{(\cos \theta )}^2+({\sin }^2\theta )U_{n-1}{(\cos \theta )}^2$

$={\cos }^2(n\theta )+({\sin }^2\theta ){\displaystyle \frac{{\sin }^2(n\theta )}{{\sin }^2\theta }}$

$={\cos }^2(n\theta )+{\sin }^2(n\theta )$

$=1.$

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

3 Inner products

For $0\le \theta \le \pi$ let $y_n(\theta )=\cos (n\theta )$.

$$y_n^{\prime \prime }+n^2{y}_n=0,y_n^{\prime }(0)=0,y_n^{\prime }(\pi )=0.$$

Theorem 9.

$${\int }_{-1}^1\frac{T_m(x)T_n(x)}{\sqrt{1-x^2}}𝑑x={\int }_0^{\pi }{y}_m{y}_n𝑑\theta =\frac{\pi }{2}\cdot {\delta }_{m,n}.$$

Proof.

Let $W=y_m{y}_n^{\prime }-y_n{y}_m^{\prime }$. We calculate

$W^{\prime }$ $=y_m^{\prime }{y}_n^{\prime }+y_m{y}_n^{\prime \prime }-y_n^{\prime }{y}_m^{\prime }-y_n{y}_m^{\prime \prime }{y}_m{y}_n^{\prime \prime }-y_n{y}_m^{\prime \prime }$

$=y_m{y}_n^{\prime \prime }-y_n{y}_m^{\prime \prime }$

$=y_m(-n^2{y}_n)-y_n(-m^2{y}_m)$

$=(m^2-n^2)y_m{y}_n.$

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

$${\int }_0^{\pi }{W}^{\prime }(\theta )𝑑\theta =W(\pi )-W(0)=0.$$

Then

$${\int }_0^{\pi }(m^2-n^2)y_m{y}_n𝑑\theta =0.$$

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

${\displaystyle {\int }_0^{\pi }}{y}_n^2𝑑\theta$ $={\displaystyle {\int }_0^{\pi }}{\cos }^2(n\theta )𝑑\theta$

$={\displaystyle {\int }_0^{\pi }}{\displaystyle \frac{1+\cos (2n\theta )}{2}}𝑑\theta$

$={\displaystyle \frac{\pi }{2}}.$

Therefore

$${\int }_0^{\pi }{y}_m{y}_n𝑑\theta =\frac{\pi }{2}\cdot {\delta }_{m,n}.$$

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

${\displaystyle {\int }_0^{\pi }}{y}_m{y}_n𝑑\theta$ $={\displaystyle {\int }_0^{\pi }}\cos (m\theta )\cos (n\theta )𝑑\theta$

$={\displaystyle {\int }_0^{\pi }}\cos (m\theta )\cos (n\theta ){\displaystyle \frac{-\sin \theta d\theta }{-\sin \theta }}$

$={\displaystyle {\int }_0^{\pi }}{\displaystyle \frac{\cos (m\theta )\cos (n\theta )}{-\sqrt{1-{\cos }^2\theta }}}(-\sin \theta )𝑑\theta$

$={\displaystyle {\int }_1^{-1}}{\displaystyle \frac{T_m(x)T_n(x)}{-\sqrt{1-x^2}}}𝑑x$

$={\displaystyle {\int }_{-1}^1}{\displaystyle \frac{T_m(x)T_n(x)}{\sqrt{1-x^2}}}𝑑x.$

∎