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Math Help - Chebyshev polynomial of order 11

  1. #1
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    Chebyshev polynomial of order 11

    I've spent hours working on this 8 problem assignment and have only 1 done. I've searched, read the textbook, and looked for video tutorials, but nothing can help me understand what to do. Here is the question:

    Calculate the Chebyshev Polynomial of order 11. (I.E. find the polynomial P_{11}(x) such that cos 11x = P_{11}(cosx)

    This is what I have:
    (1-x^{2})y'' - xy' +n^2 y = 0<br />

    t= \pi/2

    <br />
P_{11}(cost)=cos(11t)

    P_{11}(0)=0

    y=P_n(x)

    cosnx = P_n(cos(x))

    Thanks.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by spoord View Post
    I've spent hours working on this 8 problem assignment and have only 1 done. I've searched, read the textbook, and looked for video tutorials, but nothing can help me understand what to do. Here is the question:

    Calculate the Chebyshev Polynomial of order 11. (I.E. find the polynomial P_{11}(x) such that cos 11x = P_{11}(cosx)

    This is what I have:
    (1-x^{2})y'' - xy' +n^2 y = 0<br />

    t= \pi/2

    <br />
P_{11}(cost)=cos(11t)

    P_{11}(0)=0

    y=P_n(x)

    cosnx = P_n(cos(x))

    Thanks.
    I am not sure how helpful this will be, considering that this is the DE section but I just feel like doing this.

    Problem: Find P(x) such that P\left(\cos(x)\right)=\cos\left(11x\right).

    Solution: Note that \cos\left(11x\right)=\cos\left(3x+8x\right)=\cos\l  eft(3x\right)\cos\left(8x\right)-\sin\left(3x\right)\sin\left(8x\right).

    Let us take care of each of these four terms individually.

    1. \cos\left(3x\right)=\cos(x)\cos\left(2x\right)-\sin(x)\sin\left(2x\right)= \cos(x)\left[2\cos^2(x)-1\right]-2\sin^2(x)\cos(x)= \cos(x)\left[2\cos^2(x)-1\right]-2\left(1-\cos^2(x)\right)\cos(x)

    2. \cos\left(8x\right)=2\cos^2\left(4x\right)-1=2\left[2\cos^2\left(2x\right)-1\right]^2-1=2\left[2\left[2\cos^2(x)-1\right]^2-1\right]^2-1

    3. \sin\left(3x\right)=\sin(x)\cos\left(2x\right)+\co  s(x)\sin\left(2x\right)= \sin\left(x\right)\left[2\cos^2(x)-1\right]+2\sin(x)\cos^2(x)=\sin\left(x\right)\left[4\cos^2(x)-1\right]

    4. \sin\left(8x\right)=2\sin\left(4x\right)\cos(4x) =4\sin(2x)\cos(2x)\cos(4x)=8\sin(x)\cos(x)\cos(2x)  \cos(4x) =8\sin(x)\cos(x)\left[2\cos^2(x)-1\right]\left[2\left[2\cos^2(x)-1\right]^2-1\right]

    Now putting it all together (remember to replace \sin^2(x) by 1-\cos^2(x) when multiplying 3. by 4.) and replacing \cos(x) by x we get.

    P(x)=x\left[4x^2-3\right]\left[2\left[2\left[2x^2-1\right]^2-1\right]^2-1\right]- 8x\left(1-x^2\right)\left[4x^2-1\right]\left[2x^2-1\right]\left[2\left[2x^2-1\right]^2-1\right]
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by spoord View Post
    I've spent hours working on this 8 problem assignment and have only 1 done. I've searched, read the textbook, and looked for video tutorials, but nothing can help me understand what to do. Here is the question:

    Calculate the Chebyshev Polynomial of order 11. (I.E. find the polynomial P_{11}(x) such that cos 11x = P_{11}(cosx)

    This is what I have:
    (1-x^{2})y'' - xy' +n^2 y = 0<br />

    t= \pi/2

    <br />
P_{11}(cost)=cos(11t)

    P_{11}(0)=0

    y=P_n(x)

    cosnx = P_n(cos(x))

    Thanks.
    If you know it I would use the recurrence:

    P_0(x)=1

    P_1(x)=x

    P_{n+1}(x)=2xP_n(x)-P_{n-1}(x)

    CB
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  4. #4
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    Thank you so much! That's exactly what I needed to get myself in the right direction.
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