1. For a whole numbern≥0

define

http://www.webassign.net/latexImages...35acc53c24.gif

Integrate http://www.webassign.net/latexImages...c11b67833f.gif

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

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- Feb 18th 2013, 01:17 PMBaconPancakes[SOLVEDIntegrating a Tschebyshev polynomial?
1. For a whole number

*n*≥0

define

http://www.webassign.net/latexImages...35acc53c24.gif

Integrate http://www.webassign.net/latexImages...c11b67833f.gif

Can someone explain how to integrate Tn(x)? Thanks! - Feb 18th 2013, 01:56 PMILikeSerenaRe: Integrating a Tschebyshev polynomial?
- Feb 18th 2013, 03:10 PMBaconPancakesRe: 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! - Feb 18th 2013, 03:36 PMILikeSerenaRe: Integrating a Tschebyshev polynomial?
Good! ;)

Quote:

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!

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? - Feb 18th 2013, 04:05 PMBaconPancakesRe: Integrating a Tschebyshev polynomial?
Thanks so much! I found my mistake hahaa. I really appreciate the help!

- Feb 18th 2013, 04:27 PMBaconPancakesRe: 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. - Feb 18th 2013, 06:30 PMBaconPancakesRe: Integrating a Tschebyshev polynomial?
Actually I solved it, but thank you again for your help ILikeSerena!

- Feb 18th 2013, 08:15 PMMarkFLRe: Integrating a Tschebyshev polynomial?
A similar problem was posted on another forum on which I participate:

Quote:

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

$\displaystyle T_n(x)=cos(n cos^-^1x)$

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

(b) give equivalent polynomial definitions for $\displaystyle T_n(x)$ when n=0,1,2,3.

That is show that the definition for $\displaystyle T_n$ above really is a polynomial

(c) Compute

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

(d) Compute

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

when (a) n=m=0, (b) n=m$\displaystyle \neq$0, and (c) n$\displaystyle \neq$m.

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?

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

(b)

$\displaystyle T_0(x)=1$

$\displaystyle T_1(x)=x$

$\displaystyle T_2(x)=2x^2-1$

$\displaystyle T_3(x)=4x^3-3x$

Using the Chebyshev method, we find:

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

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

Using the multiple-angle formula for cosine, we have

$\displaystyle 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)$

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

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

From the recurrence relation, we may compute:

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

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

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

For even n:

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

For odd n:

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

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

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

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

For n = m = 0:

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

For n = m ≠ 0:

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

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

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

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

For n ≠ m:

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

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

$\displaystyle \frac{1}{m^n-n^2}(0-0-0+0)=0$