Chebyshev polynomial - induction problem

[Ryuky]

Homework Statement

Let Tn(x)=cos(narccosx) where x is real and belongs to [-1,1] and n E Z+
Find T1(x).
Show that T2(x)=2x^2 - 1.
Show that Tn+1 (x) + Tn-1 (x) = 2xTn(x)
Hence, prove by induction that Tn(x) is a polynomial of degree n.

The Attempt at a Solution

Since cosθ=x and arccosx=θ

we have: T1(x)=cos(arccosx)=cosθ=x.
T2(x)=cos(2arccosx)=cos2θ=cos^2 θ - sin^2 θ = 2cos^2 - 1 = 2x^2 - 1.

Tn+1 (x) + Tn-1(x) = cos((n+1)arccosx) + cos((n-1)arccosx) = 2cos(narccosx)cos(arccosx)=2Tn(x)*x.

As for the induction we have:

For n=1 T1(x)=x a polynomial of degree 1.
Assuming for n to be true, i.e.Tn(x)=cos(narccosx) is a polynomial of degree n, prove
T(n+1)(x)=cos((n+1)arccosx) is of degree n+1.

So, cos((n+1)arccosx)=cos(nθ + θ)=cosnθ*x - sinnθsinx ... Obviously this is not the way to approach the problem.

Anyway, I would prefer to hear tips from you rather than a complete solution.

Thanks in advance.

 

[LCKurtz]

Homework Statement

Let Tn(x)=cos(narccosx) where x is real and belongs to [-1,1] and n E Z+
Find T1(x).
Show that T2(x)=2x^2 - 1.
Show that Tn+1 (x) + Tn-1 (x) = 2xTn(x)
Hence, prove by induction that Tn(x) is a polynomial of degree n.

The Attempt at a Solution

Since cosθ=x and arccosx=θ

we have: T1(x)=cos(arccosx)=cosθ=x.
T2(x)=cos(2arccosx)=cos2θ=cos^2 θ - sin^2 θ = 2cos^2 - 1 = 2x^2 - 1.

Tn+1 (x) + Tn-1(x) = cos((n+1)arccosx) + cos((n-1)arccosx) = 2cos(narccosx)cos(arccosx)=2Tn(x)*x.

As for the induction we have:

For n=1 T1(x)=x a polynomial of degree 1.
Assuming for n to be true, i.e.Tn(x)=cos(narccosx) is a polynomial of degree n, prove
T(n+1)(x)=cos((n+1)arccosx) is of degree n+1.

So, cos((n+1)arccosx)=cos(nθ + θ)=cosnθ*x - sinnθsinx ... Obviously this is not the way to approach the problem.

Anyway, I would prefer to hear tips from you rather than a complete solution.

Thanks in advance.

But you already have Tn+1 (x) + Tn-1 (x) = 2xTn(x) which expresses T_n+1 in terms of lower degree polynomials. Use that.

 

[Ryuky]

Thanks for your response.

So we have Tn+1 (x) = 2xTn(x) + Tn-1(x). So, the only thing to prove is that Tn-1(x) will be of degree k<n+1 (2xTn(x) will be of degree 1 + n by the assumption). This seems obvious, but still, how could I possibly prove it?

 

[LCKurtz]

Thanks for your response.

So we have Tn+1 (x) = 2xTn(x) + Tn-1(x). So, the only thing to prove is that Tn-1(x) will be of degree k<n+1 (2xTn(x) will be of degree 1 + n by the assumption). This seems obvious, but still, how could I possibly prove it?

Your induction hypothesis is T_n is a polynomial of degree n. Demonstrated for n = 1 and 2 already. All you have to do is show for your induction step is that if T_n is a polynomial of degree n, then T_n+1 is a polynomial of degree n+1. It is very easy given what you have to work with.