A teacher has defined a family of polynomials which they denoted T0, T1, T2...Answer questions below?

Question

These polys satisfy these conditions:
1) polynomial Tk is of degree k
2)T0(x)= 1, for k=1, 2, ..., coefficient of x^k in Tk equals 2^(k-1)
3) polynomials T0, T1, T2..., Tn form orthogonal basis in space of polynomials Vn(-1, 1; w(x)= 1/(sqrt(1-x^2))
* Vn(a, b; w(x)) stands for the linear space of polynomials degree (< or = to) n endowed with inner product
(P, Q)= integral from a to b of P(x)Q(x)w(x) dx.
In the solution of this problem use the following identities:
1. integral from -1 to 1 of 1/(sqrt(1-x^2)) dx = pi
2. integral from -1 to 1 of (x^(2m))/(sqrt(1-x^2)) dx= ((1x3x5...(2m-1))/(2^m x m!)) x pi for m= 1, 2, 3, ...
3. integral from -1 to 1 of (x^(2m-1)) / (sqrt(1-x^2)) dx = 0, for m=1, 2, ...

a) find the polynomial T1 of degree 1 of the form T1(x) = x + ...that is orthogonal to T0
b) Find the only quad. polynomial T2 of the form T2(x) = 2x^2..that is orthogonal to both T0 & T1
c) show P(x)= 6x^2-5x + 4 can be a linear combo of T0, T1, T2 as follows P= 3T2- 5T1 +7T0

Answer 1

I was just reviewing theory of inner product spaces today, so I should be able to figure this one out...

I abbreviate sqrt(1-x^2) as U.

T1(x) = x + a, we must find a.

= INT (x + a)/U dx
= INT x/U dx + a INT 1/U dx
= 0 + a pi = a pi = 0
So we conclude a = 0, therefore T1(x) = x.

T2(x) = 2x^2 + ax + b, w must find a,b.

= INT (2x^2 + ax + b)/U dx
= 2 INT x^2/U dx + a INT x/U dx + b INT 1/U dx
= 2 * 1/2 pi + a * 0 + b * pi
= (b + 1) pi = 0
So b = -1; and

= INT (2x^3 + ax^2 + bx)/U dx
= 2 INT x^3/U dx + a INT x^2/U dx + b INT x/U dx
= 2 * 0 + a * 1/2 pi + b * 0 = 1/2 a pi = 0
So a = 0.

Conclusion: T2(x) = 2x^2 - 1.

P(x) = 3*(2x^2 - 1) - 5x + 7 is easily checked.

(Mmmm these polynomials look awfully familiar. Ah, they are Tschebysheff polynomials of the first kind; solutions of (1-x^2)y'' - xy' + n^2y = 0; and can be calculated as Tn(x) = cos (n * inv cos x).)