User:Egm6341.s11.team2.cleveland/HW3

=Problem 3.2 Proof That G(x) Is Zero At All Nodes=

Given
Three functions $$ \displaystyle G(x)$$, $$ \displaystyle e_n^L(f;x)$$ and $$ \displaystyle q_{n+1}(x)$$ that are $$ \displaystyle (n + 1) $$ times continuously differentiable on the the interval $$ \displaystyle I \in [a,b]$$. Then for some $$ \displaystyle t \in I $$ such that $$ \displaystyle t $$ is NOT equal to any of the nodes $$ \displaystyle x_i $$


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$$ \displaystyle i.e \quad t \neq x_i \quad i=0,1,...,n $$
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Then let $$ \displaystyle G(x) $$ be defined as,
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$$ \displaystyle G(x):=e_{n}^{L}(f;x) - \frac{q_{n+1}(x)}{q_{n+1}(t)}e_{n}^{L}(f;t) $$     (2.1)
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Show
That $$ \displaystyle G(x_i)=0 $$ for $$ \displaystyle i=0,1,...,n $$

Solution
Let us rewrite (2.1) using a shorthand notation where the subscripts, superscripts and functional dependence on $$ \displaystyle x $$ and $$ \displaystyle t $$ are implicitly considered yet not explicitly written. Doing so we arrive at an equivalent expression for (2.1) as shown here,


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$$ \displaystyle G(x)=e(x) - \frac{q_{n+1}(x)}{q_{n+1}(t)}e(t) $$     (2.2)
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For conceptual convenience let us write (2.2) as a ratio of the error at t, $$ \displaystyle e(t) $$ and $$ \displaystyle q_{n+1} $$ evaluated at t.
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$$ \displaystyle G(x)=e(x) - q_{n+1}(x) {\color{blue}\left[ \frac{e(t)}{q_{n+1}(t)}\right] } $$     (2.3)
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Let us next attempt to find the relation shown in (2.3) by considering how one may construct it from scratch. Doing so will aid the understanding of why $$ \displaystyle G(x) =0$$, which we will inevitably show. Suppose we wish to scale the error $$ \displaystyle e(\cdot) $$ relative to $$ \displaystyle q_{n+1}(\cdot) $$ for both $$ \displaystyle x $$ and $$ \displaystyle t $$. Mathematically, this is expressed as follows
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$$ \displaystyle {\color{red}\alpha(x)} = \frac{e(x)}{q_{n+1}(x)} $$     (2.4) And
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$$ \displaystyle {\color{red}\beta(t)} =\frac{e(t)}{q_{n+1}(t)} $$     (2.5)
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Next we will define $$ \displaystyle d(x) $$ as the difference between $$ {\color{red}\alpha(x)} $$ and $$\displaystyle {\color{red}\beta(t)} $$ as shown in (2.4) and (2.5).
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$$ \displaystyle \begin{align} & d(x) = {\color{red}\alpha(x)} & - & {\color{red}\beta(t)} \\ \implies \quad & d(x) = \frac{e(x)}{q_{n+1}(x)} & - & \frac{e(t)}{q_{n+1}(t)} \end{align} $$     (2.6)
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Multiplying (2.6) by $$ \displaystyle q_{n+1}(x) $$ results in the following equation for $$ \displaystyle G(x) $$ that was previously introduced in (2.3).
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$$ \displaystyle \begin{align} q_{n+1}(x)\cdot d(x) &= \cancel{q_{n+1}(x)}\cdot \frac{e(x)}{\cancel{q_{n+1}(x)}} -  q_{n+1}(x)\cdot\frac{e(t)}{q_{n+1}(t)} \\ \implies \quad G(x)&= e(x) - q_{n+1}(x)\frac{e(t)}{q_{n+1}(t)} \end{align}$$ (2.7)
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Next we must consider two important questions in order to show that $$ \displaystyle G(x) $$ is zero.

1. What is the value of $$ e(x) $$ at $$ x_i $$?

The value of $$ e(x) $$ at $$ x_i $$ must be zero since there is no error when the point being consider is a node.

2. What is the value of $$ q_{n+1}(x) $$ at $$ x_i $$?

The value of $$ q_{n+1}(x) $$ at $$ x_i $$ is also zero at all node $$ i.e. \ x_i \ \ni \ i=0,1,..,n $$.

Mathematically, one can see that 1 and 2 have the following consequences.
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$$ \displaystyle G(x_i)= {\canceltoe(x_i)} - {\canceltoq_{n+1}(x_i)}\frac{e(t)}{q_{n+1}(t)} $$     (2.8)
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=Problem 3.7: Deriving the Error in Newton-Cotes=

Given: The Convergence of Numerical Integration
The convergence of numerical integration when applied to the error associated with the Newton-Cotes formula as seen in [[media:nm1.s11.mtg17.djvu|Lecture 17]] is given by


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$$ \displaystyle $$    (7.1)
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 * E_n| \leq \frac{M_{n+1}}{(n+1)!} \int_a^b{|q_{n+1}(x)|dx}
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Where $$ \displaystyle M_{n+1}(x):= $$ be defined as,
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$$ \displaystyle M_{n+1}(x):=\underset{a\le x\le b}{\mathop{\max }}\,f^{(n+1)}(x)$$ (7.2)
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Derive: The Error in the Simple Trapezoid Rule
Derive the error $$ \displaystyle |E_1| $$ for the simple trapezoid rule such that $$ \displaystyle n=1 $$ and where
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$$\displaystyle n=1 \quad \implies \quad q_2(x)=(x-x_0)(x-x_1)=(x-a)(x-b) $$     (7.3)
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Then go on to show that
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$$ \displaystyle $$    (7.4)
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 * E_1| \leq \frac{M_{2}}{2!} \int_a^b{|q_{2}(x)|dx} = \frac{(b-a)^3}{12}M_2
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Solution: By Integration
To find the bounded error for $$ \displaystyle |E_1| $$ on can simply expand the quadratic in the integrand of (7.4) and integrate, as shown here.
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$$\displaystyle \begin{align} &= \frac{M_{2}}{2!} \int_a^b{(x-a)(b-x)dx}\\ &= \frac{M_{2}}{2!} \int_a^b{(-x^2+ax+bx-ab)dx}\\ &= \frac{M_{2}}{2!}\left[-\frac{x^3}{3}+\frac{a}{2}x^2+\frac{b}{2}x^2-abx \right]\bigg|_{x = a}^{x=b} \\ &= \frac{M_{2}}{2!}\left[\left(-\frac{b^3}{3}+\frac{ab^2}{2}+\frac{b^3}{2}-ab^2 \right) - \left(-\frac{a^3}{3}+\frac{a^3}{2}+\frac{a^2b}{2}-a^2b \right)\right] \\ &= \frac{M_{2}}{12}\left[{\color{red}-2b^3}+{\color{green}3ab^2}+{\color{red}3b^3}-{\color{green}6ab^2} + 2a^3-3a^3-{\color{cyan}3a^2b}+{\color{cyan}6a^2b}\right] \\ &= \frac{M_{2}}{12}\left[{\color{red}b^3}-{\color{green}3ab^2}+{\color{cyan}3a^2b}-a^3\right] \\ &= \frac{M_{2}}{12}\left(b-a\right)^3 \end{align}$$ (7.5)
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 * E_1| &\leq \frac{M_{2}}{2!} \int_a^b{|q_{2}(x)|dx}\\
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The final solution is

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