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Chebyshev Polynomial Problem

  1. May 27, 2015 #1
    This is something Chebyshev polynomial problems. I need to show that:

    ##\sum_{r=0}^{n}T_{2r}(x)=\frac{1}{2}\big ( 1+\frac{U_{2n+1}(x)}{\sqrt{1-x^2}}\big )##

    by using two type of solution :
    ##T_n(x)=\cos(n \cos^{-1}x)## and ##U_n(x)=\sin(n \cos^{-1}x)## with ##x=\cos\theta##,

    I have form the complex superposition:

    ##T_n(x)+iU_n(x)=(x+i\sqrt{1-x^2})^n##

    and expand it by binomial theorem to get :

    ##T_n(x)=x^n-\dbinom{n}{2}x^{n-2}(1-x^2)+\dbinom{n}{4}x^{n-4}(1-x^2)^2-...##

    and

    ##U_n(x)=\sqrt{1-x^2}\big[ \dbinom{n}{1}x^{n-1}-\dbinom{n}{3}(1-x^2)+... \big]##

    I try to change ##T_n(x)## to ##T_{2r}(x)## and ##U_n(x)## to ##U_{2n+1}(x)##, but still stuck and can't solve the problem.

    Any one can help solve this?
     
  2. jcsd
  3. May 28, 2015 #2

    Ray Vickson

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    Note that ##\sum_{r=0}^n T_{2r}(x) = \sum_{r=0}^n \cos(2 r \theta)##, where ##x = \cos(\theta)##. Can you evaluate that last summation?
     
  4. May 28, 2015 #3
    Thanks for your respond. I've made it by choosing an arbitrary ##n## and then evaluate both ##\sum_{r=0}^n T_{2r}(x)## and ##U_{2n+1}(x)##, for example ##n=1##, and, voila... :woot:
     
  5. May 29, 2015 #4

    Ray Vickson

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    How did you evaluate ##\sum_{r=0}^n T_{2r}(x)##? For example, are you able to evaluate this for ##n = 10,000## or ##n = 5,000,000##? The problem requires that you do it for all possible finite values of ##n##.
     
  6. May 29, 2015 #5
    I see mr. that's my problem actually. But for accomplishing an assignment in the short of time, I fall to just using a deductive reasoning.
    Well, let use your advice, so
    ##\sum_{r=0}^n T_{2r}(x)=\sum_{r=0}^n\cos(2r\theta)=1+\cos(2\theta)+\cos(4\theta)+...+\cos(2n\theta)##
    Then... Can't see the pattern.
     
  7. May 29, 2015 #6
    Last edited: May 29, 2015
  8. May 29, 2015 #7

    Ray Vickson

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    See. eg., http://mathworld.wolfram.com/Cosine.html
     
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