Integrating a Tschebyshev polynomial?

**BaconPancakes** · February 18th 2013, 01:17 PM

1. For a whole number n ≥0
define

Integrate

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

**ILikeSerena** · February 18th 2013, 01:56 PM

Originally Posted by BaconPancakes

1. For a whole number n ≥0
define

Integrate

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

Hi BaconPancakes!

How about the substitution $u = \arccos x$ ?

**BaconPancakes** · February 18th 2013, 03:10 PM

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!

**ILikeSerena** · February 18th 2013, 03:36 PM

Originally Posted by BaconPancakes

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?

**BaconPancakes** · February 18th 2013, 04:05 PM

Thanks so much! I found my mistake hahaa. I really appreciate the help!

**BaconPancakes** · February 18th 2013, 04:27 PM

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.

**BaconPancakes** · February 18th 2013, 06:30 PM

Actually I solved it, but thank you again for your help ILikeSerena!

**MarkFL** · February 18th 2013, 08:15 PM

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 $\neq$ 0, and (c) n $\neq$ 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?

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:

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

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

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

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