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

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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 ?


1 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 $\varphi_0,\dots,\varphi_{\ell+1}$.

  • $\begingroup$ Okay, glad to hear I'm making it too complicated haha. But okay, this innerproduct is $\int_a^b w(x) [\varphi_{j+1}(x) - C_j x \varphi_j (x)] \varphi_k(x) dx$. But I don't see how this helps me determining $\alpha_{j,k}$ ? $\endgroup$
    – Kasper
    Sep 21, 2013 at 22:08
  • $\begingroup$ What does orthogonal system mean? :) $\endgroup$ Sep 21, 2013 at 22:11
  • $\begingroup$ Aah, I loved that hint :) Haha, so $\langle \varphi_{j+1}(x)-C_jx\varphi_j(x), \varphi_k(x) \rangle = \langle \sum_{i=0}^j a_{ji} \varphi_i(x), \varphi_k(x) \rangle$. Writing this out, gives a bunch of zeros and $\langle a_{jk} \varphi_k(x), \varphi_k(x) \rangle$. So $$a_{jk} = \frac{\langle \varphi_{j+1}(x)-C_jx\varphi_j(x), \varphi_k(x) \rangle}{\langle \varphi_k(x), \varphi_k(x) \rangle}$$ $\endgroup$
    – Kasper
    Sep 21, 2013 at 22:34
  • $\begingroup$ Now $k \leq j$, and so $\langle \varphi_{j+1}, \varphi_k \rangle = 0$. I don't see directly tough why $\langle x \varphi_j(x), \varphi_k (x)\rangle$ would be $0$ if $k < j-1$. $\endgroup$
    – Kasper
    Sep 21, 2013 at 22:40
  • $\begingroup$ @Kasper: I've edited. Note that your instructor asks for $\alpha_{jk}=0$ for $k<j-1$, not $k\le j-1$. :) $\endgroup$ Sep 21, 2013 at 23:11

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