1. For a whole number n ≥0
define
Integrate
Can someone explain how to integrate Tn(x)? Thanks!
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!
Good!
I'm not sure how you got that... it doesn't seem quite right to me.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?
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.
A similar problem was posted on another forum on which I participate:
This was my response: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:
$\displaystyle I=\int_{0}^{\pi} \cos(n\theta)\cos(m\theta)\,d\theta=$
$\displaystyle \left[\frac{m\sin(m\theta)\cos(n\theta)-n\cos(m\theta)\sin(n\theta)}{m^2-n^2} \right]_{0}^{\pi}=$
$\displaystyle \frac{1}{m^n-n^2}(0-0-0+0)=0$