Chebyshev polynomial of order 11

• Dec 16th 2009, 04:54 PM
spoord
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 $\displaystyle P_{11}(x)$ such that $\displaystyle cos 11x = P_{11}(cosx)$

This is what I have:
$\displaystyle (1-x^{2})y'' - xy' +n^2 y = 0$

$\displaystyle t= \pi/2$

$\displaystyle P_{11}(cost)=cos(11t)$

$\displaystyle P_{11}(0)=0$

$\displaystyle y=P_n(x)$

$\displaystyle cosnx = P_n(cos(x))$

Thanks.
• Dec 16th 2009, 10:14 PM
Drexel28
Quote:

Originally Posted by spoord
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 $\displaystyle P_{11}(x)$ such that $\displaystyle cos 11x = P_{11}(cosx)$

This is what I have:
$\displaystyle (1-x^{2})y'' - xy' +n^2 y = 0$

$\displaystyle t= \pi/2$

$\displaystyle P_{11}(cost)=cos(11t)$

$\displaystyle P_{11}(0)=0$

$\displaystyle y=P_n(x)$

$\displaystyle cosnx = P_n(cos(x))$

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 $\displaystyle P(x)$ such that $\displaystyle P\left(\cos(x)\right)=\cos\left(11x\right)$.

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

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

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

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

3. $\displaystyle \sin\left(3x\right)=\sin(x)\cos\left(2x\right)+\co s(x)\sin\left(2x\right)=$$\displaystyle \sin\left(x\right)\left[2\cos^2(x)-1\right]+2\sin(x)\cos^2(x)=\sin\left(x\right)\left[4\cos^2(x)-1\right] 4. \displaystyle \sin\left(8x\right)=2\sin\left(4x\right)\cos(4x)$$\displaystyle =4\sin(2x)\cos(2x)\cos(4x)=8\sin(x)\cos(x)\cos(2x) \cos(4x)$$\displaystyle =8\sin(x)\cos(x)\left[2\cos^2(x)-1\right]\left[2\left[2\cos^2(x)-1\right]^2-1\right] Now putting it all together (remember to replace \displaystyle \sin^2(x) by \displaystyle 1-\cos^2(x) when multiplying 3. by 4.) and replacing \displaystyle \cos(x) by \displaystyle x we get. \displaystyle P(x)=x\left[4x^2-3\right]\left[2\left[2\left[2x^2-1\right]^2-1\right]^2-1\right]-$$\displaystyle 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]$
• Dec 16th 2009, 10:33 PM
CaptainBlack
Quote:

Originally Posted by spoord
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 $\displaystyle P_{11}(x)$ such that $\displaystyle cos 11x = P_{11}(cosx)$

This is what I have:
$\displaystyle (1-x^{2})y'' - xy' +n^2 y = 0$

$\displaystyle t= \pi/2$

$\displaystyle P_{11}(cost)=cos(11t)$

$\displaystyle P_{11}(0)=0$

$\displaystyle y=P_n(x)$

$\displaystyle cosnx = P_n(cos(x))$

Thanks.

If you know it I would use the recurrence:

$\displaystyle P_0(x)=1$

$\displaystyle P_1(x)=x$

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

CB
• Dec 17th 2009, 04:05 AM
spoord
Thank you so much! That's exactly what I needed to get myself in the right direction.