User:Egm6341.s11.team4.yang/hw3

=Homework 3.5=

Refer to lecture slide [[media:nm1.s11.mtg16.djvu|mtg-16]] for the problem statement.

Given

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$$ \displaystyle
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{{e}_{n}}(x)=f(x)-{{f}_{n}}(x)

$$     (Eq 1) Where, $$\displaystyle {{f}_{n}}(x)=\sum\limits_{i=0}^{n}{{{l}_{i,n}}(x)f({{x}_{i}})}$$
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Objectives
Prove :

Solutions
Substituting Eq. (2) into Eq. (1) yields to:
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$$ \displaystyle
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{{e}_{n}}(x)=f(x)-{{f}_{n}}(x)=f(x)-\sum\limits_{i=0}^{n}{{{l}_{i,n}}(x)f({{x}_{i}})}

$$     (Eq 3) Where, $$\displaystyle {{l}_{i,n}}(x)=\prod\limits_{j=0,j\ne i}^{n}{\frac{x-{{x}_{i}}}{{{x}_{i}}-{{x}_{j}}}}$$ is a n-order polynomial. Next, take (n+1) –order derivative on both sides of Eq.(4), it becomes:
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$$ \displaystyle
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{{e}^{(n+1)}}(x)={{f}^{(n+1)}}(x)-{{[{{f}_{n}}(x)]}^{(n+1)}}={{f}^{(n+1)}}(x)-{{[\sum\limits_{i=0}^{n}{{{l}_{i,n}}(x)f({{x}_{i}})}]}^{(n+1)}}

$$     (Eq 4) Because $$\displaystyle {{l}_{i,n}}(x)=\prod\limits_{j=0,j\ne i}^{n}{\frac{x-{{x}_{i}}}{{{x}_{i}}-{{x}_{j}}}}$$ is n-order polynomial, $$\displaystyle {{f}_{n}}(x)=\sum\limits_{i=0}^{n}{{{l}_{i,n}}(x)f({{x}_{i}})}$$is also a n-order polynomial, thus taking n+1-order derivative yields:
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$$ \displaystyle
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{{[{{f}_{n}}(x)]}^{(n+1)}}={{[\sum\limits_{i=0}^{n}{{{l}_{i,n}}(x)f({{x}_{i}})}]}^{(n+1)}}=0

$$     (Eq 5) So,
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$$ \displaystyle
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{{e}^{(n+1)}}(x)={{f}^{(n+1)}}(x)-{{[{{f}_{n}}(x)]}^{(n+1)}}={{f}^{(n+1)}}(x)-0

$$     (Eq 6)
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