Chebyshev polynomials

Jordan Bell
December 7, 2016

1 First kind

On the one hand,

(cosθ+isinθ)n =0νniν(nν)cosn-νθsinνθ
=02kn(-1)k(n2k)cosn-2k(θ)sin2k(θ)
+i02k+1n(-1)k(n2k+1)cosn-2k-1(θ)sin2k+1(θ).

On the other hand,

(cosθ+isinθ)n=(eiθ)n=einθ=cosnθ+isinnθ.

Therefore

cosnθ =0kn/2(-1)k(n2k)cosn-2k(θ)sin2k(θ)
=0kn/2(-1)k(n2k)cosn-2k(θ)(1-cos2θ)k
=0kn/2(n2k)cosn-2k(θ)(cos2θ-1)k
=0kn/2(n2k)cosn-2k(θ)0jk(kj)cos2k-2j(θ)(-1)j
=0jn/2(-1)jcosn-2j(θ)jkn/2(n2k)(kj).

Now,

jkn/2(n2k)(kj) =2n-2j-1(n-jj)nn-j.

Hence

cosnθ=0jn/2(-1)jcosn-2j(θ)2n-2j-1(n-jj)nn-j.

For z let

Tn(z)=0jn/2(-1)j2n-2j-1(n-jj)nn-jzn-2j. (1)

Note

Tn(z)[zn]=2n-1zn.
Theorem 1.
Tn(cosθ)=cos(nθ)

and

TmTn=Tmn.
Proof.

For θ,

Tn(cosθ)=cos(nθ).

Then

Tm(Tn(cosθ))=Tm(cos(nθ))=cos(mnθ)=Tmn(θ).

That is, for z[-1,1] we have Tm(Tn(z))=Tmn(z). Then by analytic continuation it follows that this is true for all z. ∎

Theorem 2.
Tn(z)+Tn-2(z)=2zTn-1(z).
Proof.

Using cos(α+β)=cosαcosβ-sinαsinβ,

cos(nθ)=cos(θ+(n-1)θ)=cosθcos((n-1)θ)-sinθsin((n-1)θ)

and

cos((n-2)θ)=cos(-θ+(n-1)θ)=cosθcos((n-1)θ)+sinθsin((n-1)θ).

Then

cos(nθ)+cos((n-2)θ)=2cosθcos((n-1)θ).

Therefore

Tn(cosθ)+Tn-2(cosθ) =cos(nθ)+cos((n-2)θ)
=2cosθcos((n-1)θ)
=2cosθTn-1(cosθ).

That is, for z[-1,1],

Tn(z)+Tn-2(z)=2zTn-1(z),

and by analytic continuation this is true for all z. ∎

2 Second kind

Define

nUn-1(z)=Tn(z). (2)
Theorem 3.
Un-1(cosθ)=sin(nθ)sinθ

and

(1-z2)Tn′′(z)=nUn(z)-n(n+1)Tn(z).
Proof.

On the one hand,

(Tn(cosθ)) =-sinθTn(cosθ).

On the other hand,

(Tn(cosθ))=(cos(nθ))=-nsin(nθ).

Hence

Tn(cosθ)=nsin(nθ)sinθ,Un-1(cosθ)=sin(nθ)sinθ.

Now,

(Tn(cosθ))=-sinθTn′′(cosθ)

and

(Tn(cosθ)) =nncos(nθ)sinθ-sin(nθ)cosθsin2θ
=-ncos(nθ)sinθ+sin(nθ)cosθsin2θ+n(n+1)cos(nθ)sinθsin2θ
=-nsin((n+1)θ)sin2θ+n(n+1)cos(nθ)sinθ
=-nUn(cosθ)sinθ+n(n+1)Tn(cosθ)sinθ.

Hence

Tn′′(cosθ)=nUn(cosθ)sin2θ-n(n+1)Tn(cosθ)sin2θ

and then

Tn′′(cosθ)=nUn(cosθ)1-cos2θ-n(n+1)Tn(cosθ)1-cos2θ.

By analytic continuation,

(1-z2)Tn′′(z)=nUn(z)-n(n+1)Tn(z).

Theorem 4.
Tn+1(z)=zTn(z)-(1-z2)Un-1(z).
Proof.
Tn+1(cosθ) =cos(nθ+θ)
=cos(nθ)cosθ-sin(nθ)sinθ
=Tn(cosθ)cosθ-Un-1(cosθ)sin2θ
=Tn(cosθ)cosθ-Un-1(cosθ)(1-cos2θ).

Therefore by analytic continuation,

Tn+1(z)=zTn(z)-(1-z2)Un-1(z).

Theorem 5.
Un(z)=Tn(z)+zUn-1(z).
Proof.
Un(cosθ) =sin(nθ+θ)sinθ
=cos(nθ)sinθ+cosθsin(nθ)sinθ
=Tn(cosθ)+cosθUn-1(cosθ).

Therefore by analytic continuation,

Un(z)=Tn(z)+zUn-1(z).

Theorem 6.
Un(z)=2zUn-1(z)+Un-2(z).
Proof.

Using Theorem 4 and Theorem 5,

Un(z) =Tn(z)+zUn-1(z)
=zTn-1(z)-(1-z2)Un-2(z)+zUn-1(z)
=z[Un-1(z)-zUn-2(z)]-(1-z2)Un-2(z)+zUn-1(z)
=2zUn-1(z)+Un-2(z).

Theorem 7.
(1-z2)Tn′′(z)-zTn(z)+n2Tn(z)=0.
Proof.

Using Theorem 3, and Theorem 5,

(1-z2)Tn′′(z)-zTn(z)+n2Tn(z)=nUn(z)-n(n+1)Tn(z)-nzUn-1(z)+n2Tn(z)=n(Tn(z)+zUn-1(z))-n(n+1)Tn(z)-nzUn-1(z)+n2Tn(z)=0.

From Theorem 1

Tn(1)=Tn(cos0)=cos(n0)=1.

From Theorem 3,

Tn(1)=nUn-1(1)=n2.

Thus, Tn is the unique solution of the initial value problem

(1-x2)y′′(x)-xy(x)+n2y(x)=0,y(1)=1,y(1)=n2.
Theorem 8.
Tn(z)2-(z2-1)Un-1(z)2=1.
Proof.

Using Theorem 1 and Theorem 3, for z=cosθ,

Tn(z)2-(z2-1)Un-1(z)2 =Tn(cosθ)2+(sin2θ)Un-1(cosθ)2
=cos2(nθ)+(sin2θ)sin2(nθ)sin2θ
=cos2(nθ)+sin2(nθ)
=1.

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

3 Inner products

For 0θπ let yn(θ)=cos(nθ).

yn′′+n2yn=0,yn(0)=0,yn(π)=0.
Theorem 9.
-11Tm(x)Tn(x)1-x2𝑑x=0πymyn𝑑θ=π2δm,n.
Proof.

Let W=ymyn-ynym. We calculate

W =ymyn+ymyn′′-ynym-ynym′′ymyn′′-ynym′′
=ymyn′′-ynym′′
=ym(-n2yn)-yn(-m2ym)
=(m2-n2)ymyn.

Using W(0)=0 and W(π)=0,

0πW(θ)𝑑θ=W(π)-W(0)=0.

Then

0π(m2-n2)ymyn𝑑θ=0.

Doing the substitution ϕ=nθ,

0πyn2𝑑θ =0πcos2(nθ)𝑑θ
=0π1+cos(2nθ)2𝑑θ
=π2.

Therefore

0πymyn𝑑θ=π2δm,n.

For 0θπ, 1-cos2θ=sinθ. Then doing the substitution x=cosθ, dx=-sinθdθ,

0πymyn𝑑θ =0πcos(mθ)cos(nθ)𝑑θ
=0πcos(mθ)cos(nθ)-sinθdθ-sinθ
=0πcos(mθ)cos(nθ)-1-cos2θ(-sinθ)𝑑θ
=1-1Tm(x)Tn(x)-1-x2𝑑x
=-11Tm(x)Tn(x)1-x2𝑑x.