# $L^2$ product of Chebyshev polynomials and Legendre polynomials

The following was a problem in a recent numerical analysis exam:

Let $k \in \mathbb{N}\setminus\{0\}$. Prove or disprove: $$\int_{-1}^{1} cos\left(k \operatorname{arccos}(x)\right) \cdot \left(\frac{\partial^{k+1}}{\partial x^{k+1}}(x^2 - 1)^{k+1} \right) dx = 0$$

Our professor said this was obviously true, since the term on the left is a Chebyshev polynomial and the term on the right is a Legendre polynomial. I know that Legendre polynomials are orthogonal with respect to the $L^2$ inner product (when $-1 \leq x \leq 1$), but what is the relation here? Is there any reason for using partials in the Legendre polynomials here (besides trying to confuse your students)?

The Legendre polynomial $P_{k}=C_{k}\frac{d^{k}}{dx^{k}}(x^{2}-1)^{k}$ ($C_k$ is a normalization constant) is orthogonal to all polynomials of lower order with respect to the inner product $(f,g)=\int_{-1}^{1}f(x)g(x)dx$. In fact, the Legendre polynomials can be derived by applying the Graham-Schmidt procedure to $\{ 1,x,x^{2},x^{3},\cdots \}$. So your Professor is correct.
• Thanks. So, since the given Legendre polynomial $P_{k+1}$ is a polynomial of degree $k+1$ and the Chebyshev polynomial $T_k$ is of degree $k$ it can be written as a linear combination of Legendre polynomials (because they span the space of $k+1$-dim. polynomials). However, the $P_i$ are pairwise orthogonal and thus $(T_k, P_{k+1}) = 0$ by bilinearity of the inner product. Am I correct? – jazzinsilhouette Aug 24 '15 at 23:14
• (and especially $P_{k+1}$ is not part of this linear combination...) – jazzinsilhouette Aug 24 '15 at 23:44
• @jazzinsilhouette : Yes, in short, $P_{k+1}$ is orthogonal to all polynomials of order $k$ or less. – Disintegrating By Parts Aug 25 '15 at 0:00