User:Egm6341.s11.team2.cleveland/HW4

=Problem 4.7: Determining the Upper Bound of the Error in the Composite Simpson's Rule =

Given: The Error by the Difference between Actual and Numeric
The error $$ \displaystyle E_n $$ in the Composite Simpson's rule


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$$ \displaystyle \begin{align} {E_n} &= I-I_n \\ &= \int\limits_{a}^{b} f(x)dx - \dfrac{h}{3}[f(x_0)+4f(x_1)+2f(x_2)+...+2f(x_{n-2})+4f(x_{n-1})+f(x_n)]\\ \end{align}$$ (7.0)
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Show: The Error for the the Composite Simpson's Rule is Bounded Above
That the error for the Composite Simpson's Rule is given by,
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$$ \displaystyle \left| {E}_{n}^{2} \right| \leq \frac{(b-a)^{5}}{2880 n^{4}} M_{4} = \frac{(b-a)h^{4}}{2880} M_{4} \quad where \quad M_{4} := max \left| f^{4}(\xi) \right| \quad s.t. \quad \xi \in [a,b] $$     (7.1)
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Solution: Relating the Error in the Simple Simpson's Rule to the Partitioning of Intervals in the Composite Simpson's Rule
Consider the error $$ \displaystyle E_n $$ which is defined as the difference between the actual value of the integral, $$ \displaystyle I $$ and the value obtained from numerically integrating using the Composite Simpson's Rule, $$ \displaystyle I_n $$ where $$ \displaystyle n \geq 2 $$ is an even integer,


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$$ \displaystyle \begin{align} {E_n} &= I-I_n \\ &= \int\limits_{a}^{b} f(x)dx - \dfrac{h}{3}[f(x_0)+4f(x_1)+2f(x_2)+...+2f(x_{n-2})+4f(x_{n-1})+f(x_n)]\\ \end{align}$$ (7.2) Observe that the following pattern arises from the terms $$ \displaystyle I_n $$ in (7.2)
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$$ \begin{align}\displaystyle & f(x_0)+4f(x_1)+{\color{red}2f(x_2)}+...+{\color{green}2f(x_{n-2})}+4f(x_{n-1})+f(x_n) = \\ & \left[ f(x_0)+4f(x_1) + {\color{red}f(x_2)} \right] + \left[ {\color{red} f(x_2)}+4f(x_3) + f(x_4) \right] + \cdots \\ & \left[ f(x_{n-4})+4f(x_{n-3}) + {\color{green}f(x_{n-2})} \right] + \left[ {\color{green}f(x_{n-2})}+4f(x_{n-1}) + f(x_{n}) \right] \end{align} $$ The above relation allows us to write the terms inside of the summation as what is shown in (7.3). We must also take precautions to ensure that the partitioning of not only the summation, but also the partitioning of the integral contain the same subintervals namely, $$ \displaystyle [x_{2i-2},x_{2i}] $$. Both are accounted for defining $$ \displaystyle h:=\frac{b-a}{n} $$ and letting $$ \displaystyle x_i = a + ih $$ where $$ \displaystyle i = 0,1,2,...,n $$.
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$$ \displaystyle \begin{align} {E_n} &= \sum_{i=1}^{n/2}\left[ \int\limits_{x_{2i-2}}^{x_{2i}} f(x)dx \right] - \dfrac{h}{3}\sum_{i=1}^{n/2} \left[ f(x_{2i-2})+4f(x_{2i-1})+f(x_{2i}) \right]\\ &= \sum_{i=1}^{n/2} \left[ \int\limits_{x_{2i-2}}^{x_{2i}} f(x)dx-\dfrac{h}{3}[f(x_{2i-2})+4f(x_{2i-1})+f(x_{2i})]\right]\\ where \quad h&=\frac{b-a}{n} = \frac{x_i - x_{i-1}}{n} \end{align}$$ (7.3)
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We know that for the Simple Simson's Rule the error associated with the Lagrange Interpolation in, $$ \displaystyle E_2 $$ is,
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$$ \displaystyle {E_2} = \frac{x_{i+1}-x_{i-1}}{2}f^{(4)}(\xi) \quad s.t. \quad \xi \in [x_{i-1},x_{i+1}] $$     (7.4)
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Which implies for the Composite Simpson's Rule
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$$ \displaystyle \begin{align} \left| {E_n} \right| &\leq  \sum_{i=1}^{n/2} max \left| \frac{(x_{i+1}-x_{i-1})^5}{90\cdot 2^5}f^{(4)}(\xi) \right| \quad s.t. \quad \xi \in [x_{i-1},x_{i+1}]\\ &= \frac{(x_{i+1}-x_{i-1})^5}{90} \cdot \sum_{i=1}^{n/2} {\color{red}\underbrace{max \left| f^{(4)}(\xi)\right|}_{M_4}} \quad s.t. \quad \xi \in [x_{i-1},x_{i+1}] \end{align} $$ (7.5)
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Where $$ M_{4} $$ is defined on $$ \displaystyle [a,b] $$ as
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$$ \displaystyle M_{4} := max \left| f^{(4)}(\xi)\right| \quad s.t. \quad \xi \in [a,b] $$
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(7.6) Next we will consider $$ \displaystyle \bar{M}_{4} \leq n \cdot M_{4} $$ allowing us to write $$ \displaystyle E_n $$ as,
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$$ \begin{align} \displaystyle \left| {E_n} \right| & \leq \left| \frac{(b-a)^{5}}{2880n^{5}}nM_{4} \right| \\ &= \left| \left(\frac{(b-a)}{2880}\right) {\color{red} \left( \frac{(b-a)^4} {n^4}\right)} M_{4}\right| \\ &= \left| \frac{(b-a){\color{red}h^{4}}}{2880}M_{4} \right| \end{align} $$ (7.7)
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Therefore showing that the error associated with the Composite Simpson's Rule is

=Problem 4.9: Determining the Conditions For Which Upper Bound of the Error in the Simple Simpson's Rule Exists=

Given: The Error in the Simple Simpson's Rule, the IMVT and Rolle's Theorem
It was previously show in | Problem 3.8 that the error in the Simple Simpson's Rule is given by

Introduced in | Lecture 16-1 was the Derivative Mean Value Theorem(DMVT). It is shown below and is required for this proof.

Derivative Mean Value Theorem:

Given a continuously differentiable function $$\displaystyle f(x) \ s.t. \ f: \mathbb{R} \rightarrow \mathbb{R} $$, then there exists at least one point $$\displaystyle \xi \in [a,b] $$ such that

Another important theorem introduced in | Lecture 16-1 is Rolle's Theorem.

Rolle's Theorem:

If $$\displaystyle f(a)=f(b)=constant$$, then there exists at least one $$\displaystyle \xi [a,b] $$ such that $$\displaystyle f'(\xi)=0 $$.

Part A: For the case with $$ t^4 $$
That the following holds for the case when $$\displaystyle t^4 $$

Part B: For the Case With $$ t^6 $$
That the following holds for the case when $$\displaystyle t^6 $$

Show: What Happens For the Case With $$ t^5 $$
Find $$ \displaystyle G^{(3)}(0) $$ and follow same steps in the proof to see what happens.

Solution: For $$t^4 $$ and $$ t^6 $$
For the purpose of this proof let us define

Next let $$ \displaystyle e(t) $$ shown in (9.6) be defined as

Part A: For the case with $$ t^4 $$
Let us first differentiate $$ \displaystyle G(t) $$ defined in (9.4) three times.

Using (9.7) we know that,

We must next apply Rolle's Theorem three times to see that

Substituting (9.11) and (9.12) into (9.10)

Applying the DMVT shown in (9.2) to the third derivative of F

Substituting the LHS of (9.14) into (9.13) gives

We also know that the error associated with the Simple Simpson's Rule, $$ E_2 $$ is

From (9.15) and (9.16) we get

The relationship between $$ \displaystyle \xi $$ and $$ \displaystyle \zeta_4 $$ was established in Problem 4.5.

Substituting (9.18) into (9.17) gives

This result is interesting because it contains a $$\displaystyle \zeta_{3} $$ term which one would not expect. This is a direct consequence of $$\displaystyle \zeta_{3} $$ having order one and order two in the resulting equation of (9.15). As a result we do not the anticipated cancellation.

Part B: For the Case With $$ t^6 $$
For this part we will consider the case with $$ \displaystyle t^6 $$ shown below,

Let us first differentiate $$ \displaystyle G(t) $$ defined in (9.5) three times.

Applying Rolle's Theorem, the DMVT and making the same substitutions as what was done in (9.11) to (9.14) gives

Using the error associated with the Simple Simpson's Rule given in (9.16) results in

The relationship between $$ \displaystyle \xi $$ and $$ \displaystyle \zeta_4 $$ was established in Problem 4.5.

Substituting (9.18) into (9.24) gives

Solution: For $$ t^5 $$
It follow from (9.11) that the three derivatives of interest are

Now lets focus our attention to what happens when $$\displaystyle G^{(3)}(t = 0)$$ in (9.28)

This means that the only way $$ \displaystyle G^{(3)}(t=0)=0 $$ is if $$ \displaystyle t=\zeta_3 $$ which a consequence of Rolle's Theorem (9.3).

Differentiating (9.28) one more time to obtain $$ \displaystyle G^{(4)} $$ is

Letting $$ \displaystyle t=\zeta_4 $$ results in

Next we will solve for $$\displaystyle e(1)$$ to get

Using the previously established relationship between $$\displaystyle \xi$$ and $$\displaystyle \zeta_4$$ gives

Where

In summary this means that as long as we have $$\displaystyle f(x) \in \mathcal{P}_{3}$$ we get that $$\displaystyle f^{(4)}(x) = 0 $$.

And therefore the Simple Simpson's rule is exact for polynomials of degree less than 3,   $$\displaystyle \quad i.e \quad e(1) = 0 \Rightarrow E_{2} = 0 $$.