User:Egm6322.s12.team1.zheng.zx/R13

Problem 6: Prove that a Polynomial belongs to a Real Set of Numbers
 Solved without assistance 

Find:
Show that:

Solution:
For
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\displaystyle \begin{align} p_n&\in \mathcal P_n\\ p_n(x)&= \sum_{i=0}^n k_i^nx^i \end{align} $$
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Consider the polynomial:
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\displaystyle \begin{align} s(x): &= p_{n+1}(x)-A_nxp_n(x)\in \mathcal P_{n+1}\\ &= \sum_{i=0}^{n+1}k_i^{n+1}x^i-A_n \sum_{i=0}^n k_i^nx^{i+1}\\ &= \sum_{j=-1}^n k_{j+1}^{n+1}x^{j+1}-A_n \sum_{i=0}^n k_i^nx^{i+1}\\ &= \sum_{j=-1}^{n-1}k_{j+1}^{n+1}x^{j+1}+k_{n+1}^{n+1}-A_n \sum_{i=0}^{n-1}k_i^nx^{i+1}-A_nk_n^nx^{n+1}\\ &= \sum_{j=-1}^{n-1}k_{j+1}^{n+1}x^{j+1}-A_n \sum_{i=0}^{n-1}k_i^nx^{i+1}+(k_{n+1}^{n+1}x^{n+1}-A_nk_n^nx^{n+1}) \end{align} $$
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If
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\displaystyle A_n=\frac{k_{n+1}^{n+1}}{k_n^n}=:\frac{k_{n+1}}{k_n} $$
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Then
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\displaystyle k_{n+1}^{n+1}x^{n+1}-A_nk_n^nx^{n+1}=(k_{n+1}^{n+1}-A_nk_n^n)x^{n+1}=0 $$
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\displaystyle \begin{align} s(x) &= \sum_{j=-1}^{n-1}k_{j+1}^{n+1}x^{j+1}-A_n \sum_{i=1}^{n-1}k_i^nx^{i+1}+0\\ &= \sum_{i=0}^n k_i^{n+1}x^i-A_n \sum_{j=2}^nk_{j-1}^nx^j \in \mathcal P_n \end{align} $$
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Thus:
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\displaystyle s(x)=p_{n+1}(x)-A_nxp_n(x)=\sum_{i=0}^n d_ip_i(x) $$
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Problem 7: Show Orthogonality using Scalar Products
 Solved without assistance 

Find
Show that: for $$ j \ne n-1 \ $$ and for $$ j = n-1 \ $$ Also using Eqs 7.1 and 7.3, show that: for $$ j < \ n-1 \ $$

Solution:

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\displaystyle \langle p_j,p_{n+1}-A_nxp_n(x)\rangle= \sum_{i=0}^n d_i \underbrace{\langle p_j,p_i \rangle}_{\displaystyle h_j\delta_{ij} } =h_jd_j $$
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For the equation above to be satisfied
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\displaystyle j\leq n $$
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\displaystyle q_{j+1}:=xp_j \in \mathcal P_{j+1} $$
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When$$j=n$$
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\displaystyle q_{j+1}=q_{n+1}=xp_n=\sum _{i=0}^{n+1}d_ip_i(x) $$
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\displaystyle \langle q_{n+1},p_n \rangle =d_n\langle p_n,p_n \rangle =d_nh_n \neq 0 $$
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When$$j=n-1$$
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\displaystyle q_{j+1}=q_n=xp_{n-1}=\sum_{i=0}^nd_ip_i(x) $$
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\displaystyle \langle q_n,p_n \rangle =d_n \langle p_n,p_n \rangle =d_nh_n \neq 0 $$
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When$$j<n-1$$
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\displaystyle j+1<n $$
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And
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\displaystyle q_{j+1}\in \mathcal P_{j+1} $$
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For $$p_n$$ is perpendicular to $$\mathcal P_{j+1}$$ So
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\displaystyle \langle q_{j+1},p_n\rangle =0 $$
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\displaystyle \begin{align} & \ \langle p_j, p_{n+1}-A_nxp_n(x) \rangle \\ &= \langle p_j, p_{n+1} \rangle - \langle p_j, A_nxp_n \rangle\\ &= 0-\langle A_n x p_j ,p_n \rangle \\ &= \langle q_{j+1}, p_n \rangle\\ &= 0 \end{align} $$
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