Problem II.3.a Prove: $P_ng$ can be represented as $(P_ng)(x) = \sum_{i=1}^n \alpha_i T_{i-1}(y(x)),$ with $$\alpha_{i+1} := \sum_{j=1}^n \frac{g(\tau_j^C) T_i(\tau_j^*)}{\left< T_i,T_i \right>}.$$ Proof. There's exactly one polynomial $P_ng$ that matches $g$ at $\tau = [\tau_1, ..., \tau_n]$. Hence, $$(P_ng)(x) = \sum_{i=1}^n \alpha_i T_{i-1}(y(x)) \qquad \longleftrightarrow \qquad (\forall \tau_k^C) \: (P_ng)(\tau_k^C) = \sum_{i=1}^n T_{i-1}(\tau_k^*) \sum_{j=1}^n \frac{g(\tau_j^C) T_{i-1}(\tau_j^*)}{\left< T_{i-1},T_{i-1} \right>} = \sum_{j=1}^n g(\tau_j^C) \sum_{i=1}^n \frac{T_{i-1}(\tau_j^*)T_{i-1}(\tau_k^*)}{\left< T_{i-1},T_{i-1} \right>}.$$ Since $g(\tau_j^C)$ is arbitrary, the above identity holds iff $$\gamma_{jk} := \sum_{i=1}^n \frac{T_{i-1}(\tau_j^*)T_{i-1}(\tau_k^*)}{\left< T_{i-1},T_{i-1} \right>} = \begin{cases} 0 & \text{if } j \neq k \\ 1 & \text{if } j = k. \end{cases}$$ From the definition $\tau_j^* := \cos\left( \frac{(2j-1)\pi}{2n} \right)$, the fact that $T_n(\cos(x)) = \cos(nx)$, $\left< T_{i},T_{i} \right> = n/2$ if $i=1...n-1$ and $\left< T_0,T_0 \right> = n$, we have that $$\gamma_{jk} = \frac{1}{n} \left( 1 + 2 \sum_{i=1}^{n-1} \cos\left(\frac{i\pi}{2n}(2j-1)\right)\cos\left(\frac{i\pi}{2n}(2k-1)\right) \right).$$ From $\cos(x)\cos(y) = \frac{1}{2}(\cos(x-y)+\cos(x+y))$ it follows that $$\gamma_{jk} = \frac{1}{n} \left( 1 + \sum_{i=1}^{n-1} \cos\left(\frac{i\pi}{n}(j-k)\right) + \cos\left(\frac{i\pi}{n}(j+k-1)\right) \right).$$ Note that if $j-k$ is odd, $j+k-1$ is even and vice versa.
The sum in the definition of $\gamma_{jk}$ is evaluated as follows: $$\sum_{i=1}^{n-1} \cos\left(\frac{i\pi}{n}\iota\right) =: s \qquad \longleftrightarrow \qquad \sum_{i=1}^{n-1} 2\cos\left(\frac{i\pi}{n}\iota\right)\sin\left(\frac{\pi}{2n}\iota\right) = 2s\sin\left(\frac{\pi}{2n}\iota\right).$$ The identity $2\cos(\theta)\sin(\phi) = \sin(\theta + \phi) - \sin(\theta - \phi)$ yields $$\sum_{i=1}^{n-1} \sin\left(\frac{\iota\pi}{n}\left(i+\frac{1}{2}\right)\right) - \sin\left(\frac{\iota\pi}{n}\left(i-\frac{1}{2}\right)\right) = 2s\sin\left(\frac{i\pi}{2n}\iota\right),$$ such that only two terms of the sum remain: $$- \sin\left(\frac{\iota\pi}{2n}\right) + \sin\left(\frac{\iota\pi}{n}\left(n - 1 + \frac{1}{2}\right)\right) = 2s\sin\left(\frac{i\pi}{2n}\iota\right).$$ Now observe that $$\sin\left(\frac{\iota\pi}{n}\left(n - 1 + \frac{1}{2}\right)\right) = \sin\left(\frac{\iota\pi}{2n}\left(2n - 2 + 1\right)\right) = \sin\left( \iota\pi - \frac{\iota\pi}{2n} \right) = \begin{cases} - \sin\left(\frac{\iota\pi}{2n}\right) & \text{if } \iota \text{ odd} \\ \sin\left(\frac{\iota\pi}{2n}\right) & \text{if } \iota \text{ even}, \\ \end{cases}$$ which implies $$\sum_{i=1}^{n-1} \cos\left(\frac{i\pi}{n}\iota\right) = \begin{cases} -1 & \text{if } \iota \text{ odd} \\ 0 & \text{if } \iota \text{ even} \\ n-1 & \text{if } \iota = 0. \end{cases}$$ This, eventually, leads to $$\gamma_{jk} = \begin{cases} 0 & \text{if } j \neq k \\ 1 & \text{if } j = k. \\ \end{cases}$$ $\Box$
2015 APR 19 (v1.0)
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