Chebyshev polynomial of order 11

• Dec 16th 2009, 05:54 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 $P_{11}(x)$ such that $cos 11x = P_{11}(cosx)$

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

$t= \pi/2$

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

$P_{11}(0)=0$

$y=P_n(x)$

$cosnx = P_n(cos(x))$

Thanks.
• Dec 16th 2009, 11:14 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 $P_{11}(x)$ such that $cos 11x = P_{11}(cosx)$

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

$t= \pi/2$

$
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.

$P(x)=x\left[4x^2-3\right]\left[2\left[2\left[2x^2-1\right]^2-1\right]^2-1\right]-$ $8x\left(1-x^2\right)\left[4x^2-1\right]\left[2x^2-1\right]\left[2\left[2x^2-1\right]^2-1\right]$
• Dec 16th 2009, 11:33 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 $P_{11}(x)$ such that $cos 11x = P_{11}(cosx)$

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

$t= \pi/2$

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