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Chebyshev polynomial - induction problem

  1. Dec 13, 2011 #1
    1. The problem statement, all variables and given/known data

    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.

    3. 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.
     
  2. jcsd
  3. Dec 13, 2011 #2

    LCKurtz

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    But you already have Tn+1 (x) + Tn-1 (x) = 2xTn(x) which expresses Tn+1 in terms of lower degree polynomials. Use that.
     
  4. Dec 13, 2011 #3
    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?
     
  5. Dec 13, 2011 #4

    LCKurtz

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    Your induction hypothesis is Tn 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 Tn is a polynomial of degree n, then Tn+1 is a polynomial of degree n+1. It is very easy given what you have to work with.
     
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