Question on Chebyshev polynomials?

[mathnoob]

Question
A Chebyshev polynomial is Tn(x) = cos(arccos^(-1)(x))

My questions are:

1. what are the domain(s) and range(s) of this function?

2. Give equivalent polynomial definitions for Tn(x) when n = 0; 1; 2; 3. That
is: show that the definition for Tn above really is a polynomial.

3. Compute integral(-1 to 1) Tn(x)dx

4. Compute integral (-1 to 1) [Tn(x)Tm(x) / sqrt(1-x^2)] dx, when
a.) n = m = 0
b.) n = m =/= 0
c.) n =/= m

My attempt at a solution

1. I graphed it on a graphing calculator and I know for sure that both its domain and range = [-1,1]. I'm guessing this is because the domain and range of cos(x) = [-1,1]?

2. I found an equation online that T(n+1)(x) = (2x)Tn(x) - T(n-1)(x) when n >= 1. Hence:
T(o)x = 1
T(1)x = x
T(2)x = 2x^2-1
T(3)x = 4x^3-3x
Would this be enough to prove that Tn(x) is a polynomial?

Lastly, for #3 and #4, How are you supposed to find the integral when there are two variables (n and x)? I got this question for my Calculus II class (mostly integration). Thanks!

 

[mighty2000]

1. You are correct, the domain and range of Tn(x) are both [-1,1] because the inverse cosine function has a domain and range of [0,π] and the cosine function has a domain and range of [-1,1].

2. Your method of finding equivalent polynomial definitions for Tn(x) is correct. Another way to prove that Tn(x) is a polynomial is by using the definition of Chebyshev polynomials, which states that Tn(x) = cos(n arccos(x)). Using this definition, we can see that Tn(x) is a polynomial because it is a composition of polynomials (arccos(x) and cos(x)).

3. The integral of Tn(x) from -1 to 1 can be found by using the substitution u = cos(x). This will give us the integral of Tn(u) from -1 to 1, which is simply the area under the curve of Tn(u) from -1 to 1. This can be found by using the formula for the area under a curve, which is (b-a)f(x)dx, where b and a are the bounds of integration and f(x) is the function. So, substituting back in for u, we get the integral of Tn(u) from -1 to 1 = (1-(-1))Tn(u)du = 2Tn(u)du.

4. To compute the integral of Tn(x)Tm(x)/sqrt(1-x^2), we can use the same substitution as before (u = cos(x)). This will give us the integral of Tn(u)Tm(u)/sqrt(1-u^2) from -1 to 1. For part a, when n = m = 0, the integral becomes 2/(sqrt(1-u^2))du, which can be solved using a trigonometric substitution. For parts b and c, the integral cannot be solved without knowing the specific values of n and m. However, we can use the orthogonality property of Chebyshev polynomials to simplify the integral. This property states that the integral of Tn(x)Tm(x)/sqrt(1-x^2) from -1 to 1 is equal to 0 if n and m are not equal, and is equal to π/2 if n = m. This property