User:Egm6341.s11.team2/hwk3

=Problem 3.1 Compare Taylor Series Approximation with Lagrangian Interpolation=

Problem Statement: Taylor Series and Lagrangian Interpolation
Use two methods to approximate the function $$\displaystyle f(x)=\sin x$$,

1. Taylor series expaned with respect to $$\displaystyle \hat{x}=\frac{3\pi }{8}$$ to the order $$\displaystyle n$$,

2. Lagrangian interpolation to the order 4

Compare the error by two methods at $$\displaystyle x=\frac{7\pi }{8}$$, find $$\displaystyle n$$ such that the error of Taylor series expansion is less than 4th order Lagrangian interpolation. And then give the maximum value of Lagrangian interpolation error.

1. Taylor series expansion
The Taylor expansion to $$\displaystyle f(x)=\sin x$$ is

2. Lagrangian interpolation
Use equidistance discretion, the 4th order Lagrangian interpolation requires 5 nodes: While the Lagrangian Interpolation evaluated at $$\displaystyle x=\frac{7\pi }{8}$$ is,

where,

Plug them back into the Eq 1.4 we have,

Comparison of the Taylor and Lagrange Methods
The exact value of $$\displaystyle f(\frac{7\pi }{8})$$ is,

Then we have the error of Lagrangian interpolation,

To find the order n which makes the Taylor series produce small error, we need to try different order.

From the calculation above, the 6th order of Taylor expansion can give better result than Lagrangian interpolation.

The Maximum Value of Lagrangian Interpolation Error
By Lagrangian interpolation error theorem,

Thus the maximum error is contained by $$\displaystyle \left| {{q}_{5}}(t) \right|$$. If $$\displaystyle t$$ is confined within $$\displaystyle (0,\pi )$$, the maximal error is given by

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

Given: The Definition of G(x)
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: G(x)=0 for all n
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 3.3 - Lagrange Interpolation and the Runge Phenomenon=

Given: A Function f(x)
- The function f(x) defined on [-1,1]


 * $$f(x)=\frac{e^{x}-1}{x}$$

- Uniform discretization with n = 4, so the nodes are $$\left \{ x_0,...,x_4 \right \} = \left \{ -1,...,1 \right \}$$

- Fixed points: t = 5, x = 0.75

Find: $$e_{4}^{L}(f,t)$$ and $$e_{4}^{L}(f,x)$$
- Generate plots as in p.14-2

- For middle plot, use i = 2 and generate $$l_{2,4}(.)$$

- Find $$e_{4}^{L}(f,t)$$ and $$e_{4}^{L}(f,x)$$

Solution
Figure 3.3-1 shows the requested plots on the interval [-1,1], and Figure 3.3-2 shows the plots on the interval [-1,5]. The decision was made to generate two separate sets of plots because expanding the domain to [-1,5] to show the evaluation point t (which is desired in the problem statement) greatly distorts the shape of the plots. In order to properly observe and discuss the characteristics of the plotted functions, two sets of graphs were made and will be discussed separately.


 * Egm6341.s11.hw3_3a.jpg]

On the interval [-1,1], the approximation using Lagrange interpolation has a small error as shown in the top graph of Figure 3.3-1. The error at the point x=0.75, defined as

is -1.6274 x 10-4. As can be seen in the top graph of Figure 3.3-2, the error grows if the domain is expanded to [-1,5] to include point t. The error at point t using Equation 3.3.1 was calculated to be e4L(f,t)=11.0239. This is an unacceptably large error, and thus it is recommended that approximations using Lagrange interpolation be used only in the domain in which the nodes are contained. The third plots in FIgures 3.3-1 and 3.3-2 show the values of qn+1(.). According to Equation 3.3.1, the error at a point x is proportional to qn+1(x). The error is zero at the nodes and attains its largest values between the nodes.


 * Egm6341.s11.hw3_3b.jpg]

Given:A Function f(x)
- The function f(x) defined on [-5,5]


 * $$f(x)=\frac{1}{1+4x^2}$$

- Uniform discretization with n = 8, so the nodes are $$\left \{ x_0,...,x_8 \right \} = \left \{ -5,...,5 \right \}$$

- Fixed points: t = 4.5, x = 3

Find: $$e_{8}^{L}(f,t)$$ and $$ e_{8}^{L}(f,x) $$
- Generate plots as in p.14-2

- For middle plot, use i = 3 and generate $$l_{3,8}(.)$$

- Find $$e_{8}^{L}(f,t)$$ and $$e_{8}^{L}(f,x)$$

Solution
In this problem, both of the evaluation points t and x are in the domain of the nodes [-5,5], so there are no large errors caused like in Problem 1. However, this problem is an example of larger errors caused by the Runge phenomenon. In brief, the Runge phenomenon is an error that occurs when approximating some functions using high-order polynomials. The error is worse as higher-order polynomials are used, and increases near the edges of the interval.


 * Egm6341.s11.hw3_3c.jpg]

The red and blue lines in the top plot of Figure 3.3-3 show the differences between the actual function (blue) and the 8th-order polynomial approximation (red). As shown in the third plot of Figure 3.3-3, the qn+1(.) term in the error equation (3.3.1) is on the order of 104 between the first and last pairs of nodes. The error at point x was found to be e8L(f,x)=-0.3786 and the error at point t was found to be e8L(f,x)=1.7840. In order to avoid the Runge phenomenon, non-uniformly distributed nodes must be selected.

MATLAB Code
=Problem 3.4 "Area of Bifolium"=

Given: A Classical Bifolium Curve
The classic curve bifolium is shown as following


 * egm6341.s11.team2.hw3.4.1.png

The exact area of this leaf which can be calculated using WolframAlpha is π/16.

Find: Historical Reference/Application and the Area of One Bifolium Pedal Using 3 Different Methods
1. Do literature search to find history and applications (if any) of this classic curve.

2. Find area in one leaf to $$ {10^{ - 6}} $$ accuracy using following methods.

2.1. Complex Trapezoidal Rule.

2.2. Complex Simpson's Rule.

2.3) Sum the area of triangles which composite the leaf. If we know two edges a and b and the angle t between them, the area of one triangle is

1. Literature Search Containing Historical Importance of the Bifolium
First described by Johannes Kepler in 1609, a bifolium is a curve that has the respective Cartesian and polor form,
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$$ \displaystyle (x^2+y^2)(y^2+x(x+b))=4axy^2 $$    (4.0a) and
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$$ \displaystyle r=-bcos(\theta)+4acos(\theta)sin^2(\theta) $$    (4.0b) The etymology of bifolium stems from the Latin prefix [bi-] meaning "two" and the Latin root word [folium] meaning "leaved", hence the meaning "two-leaved". The of Descartes] was first discovered by Rene Descartes in 1638 and is represented by the equation,
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$$ \displaystyle (x^3+y^3)=3axy $$    (4.0c) Unique to the folium of Descartes is the asymptote located at
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$$ \displaystyle x+y+a=0 $$    (4.0d)
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2.1 & 2.2: Determination of the Area Using The Complex Trapezoid and Complex Simpson's Rule
First we have to divide the curve into two parts like the red and green curves shown in Figure 4.1. Since we can not explicit show the function relationship in terms x, we can write the function relation in terms of y as following:

Because equation (4.4) and (4.5) are describe the green and red curve separately, we can use equation (4.1) and (4.2) to integrate the area between these two curve and y axis and the area difference is just the area of this leaf.

2.3 Determination of the Area By Summing The Triangles
When using summation of triangles that constitute the leaf, we can divide [0,π/2] into several equal angles and the parametric equations are needed to calculate the edge length of each triangles:

Fortran Code for these three methods
In order to simplify the procedure and make it clear, Fortran codes which are written to calculate the leaf area using the above formulation are as following:

Result
Using the above codes, we can have the following output on the screen:

From the above result, the integrated results are calculated exactly. Since we use $$ {10^{ - 6}} $$ accuracy compared to the given exact solution, we also generate the error convergence as figure 4.2 and figure 4.3.


 * egm6341.s11.team2.hw3.4.2.png


 * egm6341.s11.team2.hw3.4.3.png

From figure 4.2 we can find both Complex Trapezoidal and Simpson's method can convergent at the same time, but Complex Trapezoidal integration gives less error for equal block number.

From figure 4.3 we notice that using summation of triangles is very easy to find convergence when just cut a leaf in to more than 1000 triangles which are very easy to calculate its area.

= Problem 3.5: Evaluate the $$\displaystyle (n+1)^{th}$$ Derivative of Lagrange Interpolation Error =

''' This problem was solved without referring to S10 homework. '''

From the lecture slide Mtg 16-3

Given: The Error for the Lagrangian Interpolation
The Lagrange Interpolation error can be expressed as
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E(x)=f(x)-f_n(x) , $$     (5.1)
 * $$\displaystyle
 * $$\displaystyle
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Find: The Error at the (n+1)-th term
Verify the following:


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E^{(n+1)}(x)=f^{(n+1)}(x)-0. $$     (5.2)
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Solution
The Lagrange Interpolation error can be expressed as
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E(x)=f(x)-f_n(x) , $$     (5.3)
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where


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f_n(x) \epsilon \mathcal{P}_n. $$     (5.4)
 * $$\displaystyle
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Because $$\displaystyle f_n(x) \epsilon \mathcal{P}_n $$, it is of the form


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f_n(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0. $$     (5.5)
 * $$\displaystyle
 * $$\displaystyle
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Differentiating $$ \displaystyle f_n $$ n times gives


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f^{(n)}_n(x)=(n!)a_n , $$     (5.6)
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which is a constant.

Differentiating the expression $$\displaystyle (n+1)$$ times gives


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E^{(n+1)}(x)=f^{(n+1)}(x)-f_n^{(n+1)}(x) , $$
 * $$\displaystyle
 * $$\displaystyle
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where


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f_n^{(n+1)}(x) = 0. $$
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Therefore,

= Problem 3.6: Error In a Lagrange Interpolation= From the lecture slide Mtg 16-3

Ref:problem 7, Team 4, S10

Given: A Polynomial q(x)
Defined as
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$$\displaystyle {q_{n + 1}}(x) = \prod\limits_{j = 0}^n {(x - {x_j})} = (x - {x_0})(x - {x_1})...(x - {x_{n - 1}})(x - {x_n}) $$    (6.1)
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Verify: The Error in the Lagrange Term

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q_{n + 1}^{n + 1}(x) = (n + 1)! $$     (6.2)
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Solution
We know that $$ \displaystyle{q_{n + 1}}(x) \in {p_{n + 1}}$$, so (6.1) can be written as:


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{q_{n + 1}}(x) = \underbrace _{f(x)} + \underbrace {{a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} + ... + {a_1}{x^1} + {a_0}{x^0}}_{{p_n}(x)} $$     (6.3)
 * $$\displaystyle
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note that in (6.3), $$ {a_{n + 1}} = 1\displaystyle $$.


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q_{n + 1}^{n + 1}(x) = {f^{n + 1}}(x) + p_n^{n + 1}(x) $$     (6.4)
 * $$\displaystyle
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where $$\displaystyle p_n^{n + 1}(x) = 0 $$ and $$\displaystyle {f^{n + 1}}(x) $$ can be calculated as follows:


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$$\displaystyle\begin{align} f(x)        &= {x^{n + 1}} \\ f^1(x)      &= (n + 1){x^n} \\ {f^2}(x)      &= (n + 1)(n){x^{n - 1}} \\ {f^3}(x)      &= (n + 1)(n)(n - 1){x^{n - 2}} \\ & \vdots \\ {f^{n - 1}}(x) &= (n + 1)(n)(n - 1) \cdots (2){x^{n - (n - 1)}} \\ {f^n}(x)      &= (n + 1)(n)(n - 1)\cdots (2)(1){x^{n - n}} \\ {f^{n + 1}}(x) &= (n + 1)! \end{align} $$
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and consequently

=Problem 3.7: Deriving the Error in Newton-Cotes=

''' This problem was solved without referring to S10 homework. '''

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

=Problem 3.8: Deriving the Error in Newton-Cotes=

''' This problem was solved without referring to S10 homework. '''

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


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$$ \displaystyle $$    (8.1)
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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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$$ \displaystyle M_{n+1}(x):=\underset{a\le x\le b}{\mathop{\max }}\,f^{(n+1)}(x)$$ (8.2)
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Derive: The Error in the Simple Simpson's Rule
Derive the error $$ \displaystyle |E_2| $$ for the simple Simpson's rule such that $$ \displaystyle n=2 $$ and where


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$$\displaystyle n=2 \quad \implies \quad q_3(x)=(x-x_0)(x-x_1)(x-x_2)=(x-a)(x-\frac{a+b}{2})(x-b) $$     (8.3)
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Then go on to show that


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$$ \displaystyle (8.4)
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 * E_2| \leq \frac{M_{3}}{3!} \int_a^b{|q_{3}(x)|dx} = \frac{192}{{M}_{3}}$$
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Solution
To find the bounded error for $$\displaystyle \left| {{E}_{2}} \right|$$ one can simply expand the polynominal in the integrand of (8.4) and integrate, as shown here.

Since function $$\displaystyle g(x)=\left| (x-a)(x-\frac{a+b}{2})(x-b) \right|$$ is symmetric with respect to $$\displaystyle x=\frac{a+b}{2}$$(from the fact $$\displaystyle g(\frac{a+b}{2}+t)=g(\frac{a+b}{2}-t)$$), then we use this property to eliminate the absolute value sign.

Then we have ,

The length of the panel is $$\displaystyle h=\frac{b-a}{2}$$, so we can write the solution like,

=References=

=Contributing Members & Referenced Lecture=