Show that $\varphi_{j+1}(x)-C_j x \varphi_j (x) = \sum_{k=0}^j \alpha_{jk} \varphi_k (x)$ where $\{\varphi_j \}$ is a syst. of orth. polynom.

Question

This is a homework exercise. I'm only asking for hints, please don't give a full solution. This is the exercise:

This is my attempt to solve this problem:

If $C_j$ is chosen to be equal to the ratio of the leading coefficients of $\varphi_{j+1}(x)$ and $\varphi_j(x)$ then the coefficient of the leading term is zero and the result is a polynomial of degree $j$ only. Hence it can be expressed as a linear combination of the polynomials $\varphi_k(x)$ in the form

$$\varphi_{j+1}(x)-C_j x \varphi_j (x) = \sum_{k=0}^j \alpha_{jk} \varphi_k (x)$$

Now I need to show (using the orthogonality properties), that $\alpha_{jk} = 0$. But I don't see how I can determine $\alpha_{jk}$. I know that $\varphi_{j+1}(x)$ is a nonzero-constant multiple of $q$ where $q(x)=x^{j+1} - a_0 \varphi_0(x) - ... - a_j \varphi_j(x)$ where $$a_k = \frac{\int_a ^b w(x) x^{j+1} \varphi_k (x) dx}{\int_a^b w(x) [\varphi_k (x)]^2 dx}$$

Similarly I could write $\varphi_j(x)$ this way. $\varphi_{j}(x)$ is a nonzero-constant multiple of $r$ where $r(x)=x^{j} - b_0 \varphi_0(x) - ... - b_{j-1} \varphi_{j-1}(x)$ where $$b_k = \frac{\int_a ^b w(x) x^{j} \varphi_k (x) dx}{\int_a^b w(x) [\varphi_k (x)]^2 dx}$$

Now, $C_j x \varphi_{j}(x)$ is a nonzero-constant multiple of $C_jxr(x)=C_jx^{j+1} - c_0 \varphi_0(x) - ... - c_{j-1} \varphi_{j-1}(x)$

where $$c_k = \frac{C_j x\int_a ^b w(x) x^{j} \varphi_k (x) dx}{\int_a^b w(x) [\varphi_k (x)]^2 dx}$$

Now, $\alpha_{jk} = a_k - c_k$. But I think this should be wrong, as this is now a function of $x$ and not a real number. Am I doing something wrong ? Any help to find the right direction to solve this problem ?

Answer 1

It seems you're making this way too complicated. Just take the inner product of $\varphi_{j+1}(x)-C_jx\varphi_j(x)$ with $\varphi_k(x)$ for $k=0,\dots,j-2$ without micromanaging so much (i.e., without writing out explicit formulas).

Edit: You will need to make the observation that $\langle x\varphi_k,\varphi_\ell\rangle = \langle\varphi_k,x\varphi_\ell\rangle$ and use the fact that $x\varphi_\ell$ is a linear combination of $\varphi_0,\dots,\varphi_{\ell+1}$.