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Complex Variables: Need help with Chebyshek poly and de Moivre Theorem

  1. Jun 3, 2009 #1
    1. The problem statement, all variables and given/known data
    The nth order Chebyshev polynomial is defined by
    Tn(x)= cos( n arccos(x) ) , n is a positive integer; -1<= x <= 1.
    Using the de Moivre theorem, show that Tn(x) has the polynomial representation
    Tn(x)= 1/2 [(x+sqrt(x2-1))n+(x-sqrt(x2-1))n]




    3. The attempt at a solution
    I really have no idea where to begin. Only thing i can come up is to try to simplify cos (n arccos(x)) , but i get stuck.
     
  2. jcsd
  3. Jun 3, 2009 #2

    Dick

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    Apply deMoivre to z=exp(i*n*arccos(x))=exp(i*arccos(x))^n. You want the real part. How do get the real part of z using complex conjugate?
     
  4. Jun 3, 2009 #3
    Re(z) = (z+conjugate z)/2
    z = [cos(arccos(x)) + isin(arccos(x))]^n
    conjugate z = [cos(arccos(x)) - isin(arccos(x))]^n

    now, do i simplify z in to
    z= [x+ isqrt(1-x^2)]^n
    and
    conjugate z = [x- isqrt(1-x^2)]^n
    ?
    if i keep them in polar form, i get
    (z+conjugate z )/2 = 1/2 [2 cos(n arccos(x))]

    That's as far as i can go. I just can't see what to do.
     
  5. Jun 3, 2009 #4

    Dick

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    You are basically done, aren't you? The question is writing sqrt(x^2-1) instead of i*sqrt(1-x^2). But x^2-1 is negative.
     
  6. Jun 3, 2009 #5
    Thank you!!
     
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