Results 1 to 8 of 8

Math Help - Integrating a Tschebyshev polynomial?

  1. #1
    Newbie
    Joined
    Feb 2013
    From
    NYC
    Posts
    5

    [SOLVEDIntegrating a Tschebyshev polynomial?

    1. For a whole number n ≥0
    define

    Integrate


    Can someone explain how to integrate Tn(x)? Thanks!
    Last edited by BaconPancakes; February 18th 2013 at 07:31 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member ILikeSerena's Avatar
    Joined
    Dec 2011
    Posts
    733
    Thanks
    121

    Re: Integrating a Tschebyshev polynomial?

    Quote Originally Posted by BaconPancakes View Post
    1. For a whole number n ≥0
    define

    Integrate


    Can someone explain how to integrate Tn(x)? Thanks!
    Hi BaconPancakes!

    How about the substitution u = \arccos x?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2013
    From
    NYC
    Posts
    5

    Re: Integrating a Tschebyshev polynomial?

    Thanks ILikeSerena! Do you think you could help explain the other parts? It says integrate Tn(x) from -1,1 if n is even, and if n is odd.
    After integrating the function I got:
    [cos(arccos(x)(1-n)]/2(1-n) + [cos(arccos(x)(1+n)]/2(1+n) and if I integrate from -1 to 1,

    1/(2(1-n)) - cos( pi(1+n))/2(1+n) but I'm not sure where to go from there in terms of the even and odd...

    Thanks so much!
    Last edited by BaconPancakes; February 18th 2013 at 04:13 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member ILikeSerena's Avatar
    Joined
    Dec 2011
    Posts
    733
    Thanks
    121

    Re: Integrating a Tschebyshev polynomial?

    Quote Originally Posted by BaconPancakes View Post
    Thanks ILikeSerena! Do you think you could help explain the other parts? It says integrate Tn(x) from -1,1 if n is even, and if n is odd.
    After integrating the function I got:
    [cos(arccos(x)(1-n)]/2(1-n) + [cos(arccos(x)(1+n)]/2(1+n) and if I integrate from -1 to 1,
    Good!


    1/(2(1-n)) - cos( pi(1+n))/2(1+n) but I'm not sure where to go from there in terms of the even and odd...

    Thanks so much!
    I'm not sure how you got that... it doesn't seem quite right to me.

    It should be:

    {[cos(arccos(1)(1-n)]/2(1-n) + [cos(arccos(1)(1+n))]/2(1+n)} - {[cos(arccos(-1)(1-n))]/2(1-n) + [cos(arccos(-1)(1+n))]/2(1+n)}

    {1/2(1-n) + 1/2(1+n)} - {[cos(pi(1-n)]/2(1-n) + [cos(pi(1+n)]/2(1+n)}

    Now suppose n is odd, what would you get?
    And what would you get if n is even?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Feb 2013
    From
    NYC
    Posts
    5

    Re: Integrating a Tschebyshev polynomial?

    Thanks so much! I found my mistake hahaa. I really appreciate the help!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Feb 2013
    From
    NYC
    Posts
    5

    Re: Integrating a Tschebyshev polynomial?

    This is a continuation of the problem:

    http://www.webassign.net/latexImages...c14433b1dc.gif

    compute for the following cases:

    a) n=m=0
    b) n=m>0
    c) n=/=m


    I figured out a), it's just pi. But for part b i'm kind of stuck, using u = arccosx and du = -1/sqrt(1-x^2)dx so it looks like integral of cos(nu)*cos(mu) du from -1 to 1. Since n=m, I replaced the m with n, and rewrote it as integral of cos^2(nu) du and integrated using the trig identity cos^2 (x) = .5(1+cos2x). I ended up with -pi/2 -sin(2npi)/2 but it appears to be incorrect and i'm not really sure where to go from here.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Feb 2013
    From
    NYC
    Posts
    5

    Re: Integrating a Tschebyshev polynomial?

    Actually I solved it, but thank you again for your help ILikeSerena!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    1,988
    Thanks
    734

    Re: Integrating a Tschebyshev polynomial?

    A similar problem was posted on another forum on which I participate:

    These problems are about a useful class of polynomial called Chebyshev polynomials which are defined as:

    T_n(x)=cos(n cos^-^1x)

    (a) what are the domain(s) and range(s) of the functions?

    (b) give equivalent polynomial definitions for T_n(x) when n=0,1,2,3.
    That is show that the definition for T_n above really is a polynomial

    (c) Compute

    \int^1_{-1}T_n(x)\,dx

    (d) Compute

    \int^1_{-1}T_n(x)T_m(x)\dfrac{dx}{\sqrt{1-x^2} }

    when (a) n=m=0, (b) n=m \neq0, and (c) n \neqm.

    some comments: The summation and double angle formulas for cosine may be useful for these exercises. Also, the integrals can be solved using a good substitution and then looking around in integration tables. There is one important detail in (d) that is being ignored. Can you find it?
    This was my response:

    (a) The domains are [-1,1] and the ranges are [-1,1]

    (b)

    T_0(x)=1

    T_1(x)=x

    T_2(x)=2x^2-1

    T_3(x)=4x^3-3x

    Using the Chebyshev method, we find:

    T_n(x)=2x\cos\left((n-1)cos^{\small{-1}}(x) \right)-\cos\left((n-2)cos^{\small{-1}}(x) \right)

    T_n(x)=2xT_{n-1}(x)-T_{n-2}(x)

    Using the multiple-angle formula for cosine, we have

    T_n(x)=\sum_{k=0}^n{n \choose k}\cos^k\left(\cos^{\small{-1}}\(x\) \right)\left(\sqrt{1-x^2} \right)^{n-k}\cos\left(\frac{\pi}{2}(n-k) \right)

    T_n(x)=\sum_{k=0}^n{n \choose k}x^k\left(\sqrt{1-x^2} \right)^{n-k}\cos\left(\frac{\pi}{2}(n-k) \right)

    Observe that when n - k is odd, the term is zero, from the last factor in the summation. When n - k is a multiple of 4, the last factor is 1 and when n - k is a multiple of 2 but not 4, the last factor is -1. With the terms having a square root eliminated, it is easy to see we have a polynomial.

    When n is odd, all even values for k give a term equal to zero.

    (c) \int_{-1}^1 T_n(x)\,dx

    From the recurrence relation, we may compute:

    T_n(x)=\frac{1}{2}\left(\frac{1}{n+1}\cdot\frac{d}  {dx}T_{n+1}(x)-\frac{1}{n-1}\cdot\frac{d}{dx}T_{n-1}(x) \right)

    \int_{-1}^1 T_n(x)\,dx=\frac{1}{2}\left[\frac{T_{n+1}(x)}{n+1}-\frac{T_{n-1}(x)}{n-1} \right]_{-1}^1

    From the definition, we see that T_n(1)=1 and T_n(-1)=1\text{ n even or }-1\text{ n odd.}, thus:

    For even n:

    \int_{-1}^1 T_n(x)\,dx=\frac{1}{n+1}-\frac{1}{n-1}=\frac{2}{1-n^2}

    For odd n:

    \int_{-1}^1 T_n(x)\,dx=0

    (d) \int_{-1}^1 T_n(x)T_m(x)\frac{dx}{\sqrt{1-x^2}}

    Let x=\cos(\theta)\:\therefore\:dx=-\sin(\theta)\,d\theta

    I=\int_{0}^{\pi} \cos(n\theta)\cos(m\theta)\,d\theta

    For n = m = 0:

    I=\int_{0}^{\pi}\,d\theta=\pi

    For n = m ≠ 0:

    I=\int_{0}^{\pi} \cos^2(n\theta)\,d\theta

    Let u=n\theta\:\therefore\:du=n\,d\theta

    I=\frac{1}{n}\int_{0}^{n\pi} \cos^2(u)\,du=\frac{1}{2n}\int_{0}^{n\pi}\cos(2u)+  1\,du=

    \frac{1}{2n}\left[\frac{1}{2}\sin(2u)+u \right]_0^{n\pi}=\frac{1}{2n}(0+n\pi-0-0)=\frac{\pi}{2}

    For n ≠ m:

    I=\int_{0}^{\pi} \cos(n\theta)\cos(m\theta)\,d\theta=

    \left[\frac{m\sin(m\theta)\cos(n\theta)-n\cos(m\theta)\sin(n\theta)}{m^2-n^2} \right]_{0}^{\pi}=

    \frac{1}{m^n-n^2}(0-0-0+0)=0
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: November 25th 2012, 11:11 AM
  2. Replies: 0
    Last Post: March 15th 2012, 05:31 AM
  3. Integrating a linear polynomial approx
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 23rd 2010, 10:02 AM
  4. Replies: 1
    Last Post: December 15th 2009, 08:26 AM
  5. Integrating the square root of a polynomial
    Posted in the Calculus Forum
    Replies: 17
    Last Post: October 7th 2009, 08:53 PM

Search Tags


/mathhelpforum @mathhelpforum