text:
User:Egm6341.s11.team2/hwk2

=Problem 2.1 Taylor's Series Proof (Solution 1)=

Problem Statement
2.1.1) Using integration by parts show that the first five terms of the Taylor's Series expansion for $$ f(x) $$ is

2.1.2) Using the IMVT shown below in (2.2) express the remainder in terms of $$ f^5(\xi) $$ such that $$ \xi\in [x_0,x] $$. Integral Mean Value Theorem: Let $$\displaystyle w(x)$$ be a nonnegative integrable function defined on $$\displaystyle [x_0,x] $$

If $$\displaystyle f(x) $$ is a continuous function defined on $$\displaystyle [x_0,x] $$, then

for some $$\displaystyle \xi $$ such that

2.1.3) Assume that (1.5) and (1.6) defining $$ p_{n}(x) $$ and $$ R_{n+1}(x) $$ hold. Perform integration by parts again to obtain the $$ (n+1)th $$ expansion containing $$ R_{n+2}(x) $$. where $$ R_{n+1}(x) $$ will be

2.1.4) Use the IMVT shown in (2.1.2) to show

For n=1
Since $$ f(x) $$ can be written as follows, where we will define $$ {\color{red} I_1(x)} $$ as

Recall the general form for Integration By Parts

Then using integration by parts as shown in (2.9) we can let

Which provides us with the following result

Plugging (2.11) into (2.8) reveals the first term of the Taylor's Series Expansion

For n=2,3,4
Similarly we can carry out the same procedure to elucidate the $$ 2^{nd},3^{rd} and ,4^{th} $$ terms.

Putting all of these terms together results in

Solution to 2.1.2
From the previous part we know the remainder for the Taylor's Series Expansion

Using the IMVT shown in (1.2), (1.3) and (1.4) let us define $$ w(x=t) $$ and $$ f(x=t) $$ as follows

Since we are integrating from our constants of integration x and $$ x_{0} $$ we can use the IMVT, substituting our equations, to obtain the following.

Solution to 2.1.3
Assume that (1.5) and (1.6) defining $$ p_{n}(x) $$ and $$ R_{n+1}(x) $$ hold. Perform integration by parts again to obtain the $$ (n+1)th $$ expansion containing $$ R_{n+2}(x) $$.

where $$ R_{n+1}(x) $$ will be

Performing the integration results in

Since we are integrating from our constants of integration x and $$ x_{0} $$ we can use the IMVT, substituting our equations, to obtain the following.

Solution to 2.1.4
Therefore by applying the IMVT for $$ R_{n+2}(x) $$ will be

where $$ R_{n+2}(x) $$ will be

= 2.1 Generate Higher Order Terms of Taylor Series (Solution 2) =

''' This problem was solved referring to Team 3, Hwk 1, S10 homework. '''

From the lecture slide Mtg 6-2

Given
Taylor Series equation below

 * {| style="width:100%" border="0" align="left"

$$ $$
 * $$\displaystyle f(x) = f(x_0) + \frac{(x-x_0)}{1!}\ f^{(1)}(x_0) + \int_{x_0}^{x} (x-t) f^{(2)}(t)\ dt.
 * $$\displaystyle f(x) = f(x_0) + \frac{(x-x_0)}{1!}\ f^{(1)}(x_0) + \int_{x_0}^{x} (x-t) f^{(2)}(t)\ dt.
 * $$\displaystyle (1.1)
 * }
 * }

Find
(1) Repeat integration by parts on the Taylor Series to reveal the following items with remainder.

 * {| style="width:100%" border="0" align="left"

$$ $$
 * $$\displaystyle \frac{(x-x_0)^2}{2!}\ f^{(2)}(x_0) + \frac{(x-x_0)^3}{3!}\ f^{(3)}(x_0)
 * $$\displaystyle \frac{(x-x_0)^2}{2!}\ f^{(2)}(x_0) + \frac{(x-x_0)^3}{3!}\ f^{(3)}(x_0)
 * $$\displaystyle
 * }
 * }

(2) Assume Eqn 3 and Eqn 4 in [[media:nm1.s11.mtg3.djvu|Mtg3-3]] are true, do integration by parts one more time.

1 Integration by Parts
Integrate $$\int_{x_0}^{x} (x-t){f^{(2)}(t)} dt $$ by parts in Eq 1.

 * {| style="width:100%" border="0" align="left"

\int_{x_0}^{x} \underbrace{(x-t)}_{U^\prime} \underbrace{f^{(2)}(t)}_{V}\, dt & = \left [ UV \right ]_{x_0}^x -  \int_{t=x} UV^\prime\, dt \\ & = \left [ -\frac{(x-t)^2}{2}\ f^{(2)}(t) \right ]_{t=x_0}^{t=x} + \underbrace{\int_{x_0}^{x} \frac{(x-t)^2}{2}f^{(3)}(t)\, dt}_{W1} \\ & = \cancelto{o}{-\frac{(x-x)^2}{2}\ f^{(2)}(x)} + \frac{(x-x_0)^2}{2}\ f^{(2)}(x_0) + W1\\ & = \frac{(x-x_0)^2}{2!}\ f^{(2)}(x_0) + W1\\ \end{align} $$ $$
 * $$ \displaystyle \begin{align}
 * $$ \displaystyle \begin{align}
 * $$\displaystyle (1.2)
 * }
 * }

Use Eq 1.1, Eq 1.2, we have
 * {| style="width:100%" border="0" align="left"

$$ $$
 * $$\Rightarrow \ f(x) = f(x_0) + \frac{(x-x_0)}{1!}\ f^{(1)}(x_0) + \frac{(x-x_0)^2}{2!}\ f^{(2)}(x_0) + \frac{1}{2!}\int_{x_0}^{x} {(x-t)^2}f^{(3)}(t)\ dt.
 * $$\Rightarrow \ f(x) = f(x_0) + \frac{(x-x_0)}{1!}\ f^{(1)}(x_0) + \frac{(x-x_0)^2}{2!}\ f^{(2)}(x_0) + \frac{1}{2!}\int_{x_0}^{x} {(x-t)^2}f^{(3)}(t)\ dt.
 * $$\displaystyle (1.3)
 * }
 * }

Integrate $$\int_{x_0}^{x} {(x-t)^2}f^{(3)}(t)\,dt $$ by part one more time in Eq 1.3.
 * {| style="width:100%" border="0" align="left"

\int_{x_0}^{x} \underbrace{(x-t)^2}_{U^\prime} \underbrace{f^{(3)}(t)}_{V}\, dt & = \left [ UV \right ]_{x_0}^x -  \int_{t=x} UV^\prime\, dt \\ & = \left [ -\frac{(x-t)^3}{3}\ f^{(3)}(t) \right ]_{t=x_0}^{t=x} + \underbrace{\int_{x_0}^{x} \frac{(x-t)^3}{3}f^{(4)}(t)\, dt}_{W2} \\ & = \cancelto{o}{-\frac{(x-x)^3}{3}\ f^{(3)}(x)} + \frac{(x-x_0)^3}{3}\ f^{(3)}(x_0) + W2\\ & = \frac{(x-x_0)^3}{3}\ f^{(3)}(x_0) + W2\\ \end{align} $$ $$
 * $$ \displaystyle \begin{align}
 * $$ \displaystyle \begin{align}
 * $$\displaystyle (1.4)
 * }
 * }

Use Eq 1.3 and Eq 1.4, we have

2 Integration by Parts, Again
Int. $$\int_{x_0}^{x} {(x-t)^n}f^{(n+1)}(t)\,dt $$ by part one more time:

 * {| style="width:100%" border="0" align="left"

\int_{x_0}^{x} \underbrace{(x-t)^n}_{U^\prime} \underbrace{f^{(n+1)}(t)}_{V}\, dt & = \left [ UV \right ]_{x_0}^x -  \int_{t=x} UV^\prime\, dt \\ & = \left [ -\frac{(x-t)^{n+1}}{n+1}\ f^{(n+1)}(t) \right ]_{t=x_0}^{t=x} + \underbrace{\int_{x_0}^{x} \frac{(x-t)^{n+1}}{n+1}f^{(n+2)}(t)\, dt}_{W} \\ & = \cancelto{o}{-\frac{(x-x)^{n+1}}{n+1}\ f^{(n+1)}(x)} + \frac{(x-x_0)^{n+1}}{n+1}\ f^{(n+1)}(x_0) + W\\ & = \frac{(x-x_0)^{n+1}}{n+1}\ f^{(n+1)}(x_0) + W\\ \end{align} $$ $$
 * $$ \displaystyle \begin{align}
 * $$ \displaystyle \begin{align}
 * $$\displaystyle (1.6)
 * }
 * }

Use Eqn 1.6, Eqn 3 and Eqn 4 in [[media:nm1.s11.mtg3.djvu|Mtg3-3]], we have

= Problem 2.2 Taylor series expansion of sin(x) =

From the lecture slide Mtg 6 [[media:nm1.s11.mtg6.djvu|Mtg6]]

Ref: Egm6341.s10.team3.sa/HW1

Given
$$ f(x)=sin x, x \in [0,\pi] $$

Find
A) Construct a Taylor series of $$f(x)$$ around $$x_0 = \frac{\pi}{4}$$ for $$\displaystyle n=0,1,2......,10.$$

B) Construct a Taylor series of $$f(x)$$ around $$x_0 = \frac{3\pi}{8}$$ for $$\displaystyle n=0,1,2......,10.$$

C) Estimate the maximum $$\displaystyle R(x)$$ at $$ x=\frac{3\pi}{4}$$

D) Plot the series for each $$\displaystyle n $$

Solution
A)Considering

$$ {p_n}(x) = f({x_0}) + \frac{f^{(1)}}({x_0}) + \frac{f^{(2)}}({x_0}) + ... + \frac{f^{(n)}}({x_0}) $$

for $$ \displaystyle f({x}) = \sin ({x}) $$ around the point $${x_0} = \frac{\pi }{4} $$ the follwoing equations can be calculated for $$ \displaystyle n= 0,1,2,...,10.$$ Note that $$f({x_0}) = \sin (\frac{\pi }{4}) = \frac{2} $$

$${p_0}(x) = \frac{2}$$

$${p_1}(x) = \frac{2} + \frac\cos (\frac{\pi }{4}) $$

$${p_2}(x) = \frac{2} + \frac\cos (\frac{\pi }{4}) - \frac\sin (\frac{\pi }{4}) $$

$${p_3}(x) = \frac{2} + \frac\cos (\frac{\pi }{4}) - \frac\sin (\frac{\pi }{4}) - \frac\cos (\frac{\pi }{4}) $$

$${p_4}(x) = \frac{2} + \frac\cos (\frac{\pi }{4}) - \frac\sin (\frac{\pi }{4}) - \frac\cos (\frac{\pi }{4}) + \frac\sin (\frac{\pi }{4}) $$

$${p_5}(x) = \frac{2} + \frac\cos (\frac{\pi }{4}) - \frac\sin (\frac{\pi }{4}) - \frac\cos (\frac{\pi }{4}) + \frac\sin (\frac{\pi }{4}) + \frac\cos (\frac{\pi }{4}) $$

$$ {p_6}(x) = \frac{2} + \frac\cos (\frac{\pi }{4}) - \frac\sin (\frac{\pi }{4}) - \frac\cos (\frac{\pi }{4}) + \frac\sin (\frac{\pi }{4}) + \frac\cos (\frac{\pi }{4}) - \frac\sin (\frac{\pi }{4}) $$

$${p_7}(x) = \frac{2} + \frac\cos (\frac{\pi }{4}) - \frac\sin (\frac{\pi }{4}) - \frac\cos (\frac{\pi }{4}) + \frac\sin (\frac{\pi }{4}) + \frac\cos (\frac{\pi }{4}) - \frac\sin (\frac{\pi }{4}) - \frac\cos (\frac{\pi }{4})$$

$$ {p_8}(x) = \frac{2} + \frac\cos (\frac{\pi }{4}) - \frac\sin (\frac{\pi }{4}) - \frac\cos (\frac{\pi }{4}) + \frac\sin (\frac{\pi }{4}) + \frac\cos (\frac{\pi }{4}) - \frac\sin (\frac{\pi }{4}) - \frac\cos (\frac{\pi }{4}) + \frac\sin (\frac{\pi }{4}) $$

$${p_9}(x) = \frac{2} + \frac\cos (\frac{\pi }{4}) - \frac\sin (\frac{\pi }{4}) - \frac\cos (\frac{\pi }{4}) + \frac\sin (\frac{\pi }{4}) + \frac\cos (\frac{\pi }{4}) - \frac\sin (\frac{\pi }{4}) - \frac\cos (\frac{\pi }{4}) + \frac\sin (\frac{\pi }{4}) + \frac\cos (\frac{\pi }{4}) $$

$${p_{10}}(x) = \frac{2} + \frac\cos (\frac{\pi }{4}) - \frac\sin (\frac{\pi }{4}) - \frac\cos (\frac{\pi }{4}) + \frac\sin (\frac{\pi }{4}) + \frac\cos (\frac{\pi }{4}) - \frac\sin (\frac{\pi }{4}) - \frac\cos (\frac{\pi }{4}) + \frac\sin (\frac{\pi }{4}) + \frac\cos (\frac{\pi }{4}) - \frac\sin (\frac{\pi }{4}) $$

which can be simplified to the following equations:

B)Considering

$$ {p_n}(x) = f({x_0}) + \frac{f^{(1)}}({x_0}) + \frac{f^{(2)}}({x_0}) + ... + \frac{f^{(n)}}({x_0}) $$

for $$ \displaystyle f({x}) = \sin ({x}) $$ around the point $${x_0} = \frac{3\pi }{8} $$ the follwoing equations can be calculated for $$ \displaystyle n= 0,1,2,...,10.$$

$${p_0}(x) = \sin (\frac{8}) $$

$${p_1}(x) = \sin (\frac{8}) + \frac\cos (\frac{8})$$

$${p_2}(x) = \sin (\frac{8}) + \frac\cos (\frac{8}) - \frac\sin (\frac{8}) $$

$${p_3}(x) = \sin (\frac{8}) + \frac\cos (\frac{8}) - \frac\sin (\frac{8}) - \frac\cos (\frac{8}) $$

$${p_4}x) = \sin (\frac{8}) + \frac\cos (\frac{8}) - \frac\sin (\frac{8}) - \frac\cos (\frac{8}) + \frac\sin (\frac{8}) $$

$${p_5}(x) = \sin (\frac{8}) + \frac\cos (\frac{8}) - \frac\sin (\frac{8}) - \frac\cos (\frac{8}) + \frac\sin (\frac{8}) + \frac\cos (\frac{8}) $$

$${p_6}(x) = \sin (\frac{8}) + \frac\cos (\frac{8}) - \frac\sin (\frac{8}) - \frac\cos (\frac{8}) + \frac\sin (\frac{8}) + \frac\cos (\frac{8}) - \frac\sin (\frac{8})$$

$${p_7}(x) = \sin (\frac{8}) + \frac\cos (\frac{8}) - \frac\sin (\frac{8}) - \frac\cos (\frac{8}) + \frac\sin (\frac{8}) + \frac\cos (\frac{8}) - \frac\sin (\frac{8}) - \frac\cos (\frac{8}) $$

$${p_8}(x) = \sin (\frac{8}) + \frac\cos (\frac{8}) - \frac\sin (\frac{8}) - \frac\cos (\frac{8}) + \frac\sin (\frac{8}) + \frac\cos (\frac{8}) - \frac\sin (\frac{8}) - \frac\cos (\frac{8}) + \frac\sin (\frac{8}) $$

$${p_9}(x) = \sin (\frac{8}) + \frac\cos (\frac{8}) - \frac\sin (\frac{8}) - \frac\cos (\frac{8}) + \frac\sin (\frac{8}) + \frac\cos (\frac{8}) - \frac\sin (\frac{8}) - \frac\cos (\frac{8}) + \frac\sin (\frac{8}) + \frac\cos (\frac{8}) $$

$${p_{10}}(x) = \sin (\frac{8}) + \frac\cos (\frac{8}) - \frac\sin (\frac{8}) - \frac\cos (\frac{8}) + \frac\sin (\frac{8}) + \frac\cos (\frac{8}) - \frac\sin (\frac{8}) - \frac\cos (\frac{8}) + \frac\sin (\frac{8}) + \frac\cos (\frac{8}) - \frac\sin (\frac{8})$$

which can be simplified to the following equations:

D)Matlab Code and Graphs
And here, the graphs show the results. Note that the Matlab code is the same for part A and B and in order to run the Matlab code for part B, the value for $$ x_0 $$ needs to be changed to $$ \frac{8} $$

 * HW2 P2 A.png
 * HW2 P2 B.png

=Problem 2.3: Finding Taylor series=

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

From the lecture slide Mtg 7

Given
Function $$\displaystyle f(x)$$ is,

Find
A)Expand $$\displaystyle {{e}^{x}}$$ in Taylor series up to order $$\displaystyle n$$ with remainder $$\displaystyle {{R}_{n+1}}[{{e}^{x}};x]$$.

B)Find Taylor series of $$\displaystyle f(x)$$ up to order $$\displaystyle n$$ with remainder $$\displaystyle {{R}_{n+1}}[f;x]$$.

Solution
A)Expand $$\displaystyle {{e}^{x}}$$ in Taylor series up to order $$\displaystyle n$$ with remainder $$\displaystyle {{R}_{n+1}}[{{e}^{x}};x]$$.

According the Taylor series theorem ,

Function $$\displaystyle g(x)={{e}^{x}}$$ can be expanded at $$\displaystyle {{x}_{0}}=0$$ as,

where the remaider $$\displaystyle {{R}_{n+1}}[{{e}^{x}};x]$$ is

where $$\displaystyle \xi \in [0,x]$$.

B)Find Taylor series of $$\displaystyle f(x)$$ up to order $$\displaystyle n$$ with remainder $$\displaystyle {{R}_{n+1}}[f;x]$$.

From Eq(3.2) we have,

Substituting Eq(3.4) into Eq(3.1) yield,

where the the Taylor series expansion is,

and the remainder is

from which we can conclude,

which satisfy Eq(1) in Mtg7.

=Problem 2.4 - Comparision of Numerical Integration Techniques= Note: This problem was studied for $$ [a,b] = [0,1]$$ in s10.

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

Given

 * $$ f(x) = \frac{x} $$ for $$ x \in [a,b] $$

where $$ [a,b] = [-1,1]$$

Find
Plot f(x), then approximate the integral of f(x) as
 * $$ I_{n}\approx I := \int_{a}^{b}f(x)dx$$

using the following techniques: 1) Taylor Series Expansion 2) Composite Trapezoidal Rule 3) Composite Simpson Rule 4) Gauss-Legendre Quadrature

The value of n should be selected as n = 2,4,8,... until $$I_{n}$$ has a maximum error on the order of $$10^{-6}$$.

Plot of f(x)

 * Egm6341.s11.hw2_4a.jpg]

2.4.1 Taylor Series
As shown in Problem 1.2, the Taylor series expansion of f(x) is Because this is a polynomial, each term in the summation can be integrated individually. Thus, the integral approximation can be found as The maximum error of the Tayor series integration is shown in (Atkinson, 250) as The following table shows the results for integration by Taylor series expansion.

2.4.2 Composite Trapezoidal Rule
The formula for integration by composite trapezoidal rule is given by (Atkinson, 253)

The error in the composite trapezoidal integration is given by (Atkinson, 253)

Since we know that the error will achieve its maximum when Thus, the maximum error for integration by the composite trapezoidal rule is

The following table shows the results for integration by composite trapezoidal rule.

2.4.3 Composite Simpson Rule
The formula for integration by composite Simpson rule is given by (Atkinson, 257)

The error in the composite Simpson integration is given by (Atkinson, 257)

Like with the composite trapezoidal rule, since we know that the error will achieve its maximum when

Thus, the maximum error for integration by the composite Simpson rule is

The following table shows the results for integration by composite Simpson rule.

2.4.4 Gauss-Legendre Quadrature
The formula for Gaussian quadrature on the interval [-1,1] is given by (Atkinson, 276) The quadrature weights wj are given by (Atkinson, 276) and the nodes xj are the roots of the nth degree Legendre polynomial Pn(x) on the interval [-1,1]. The numerical values of xj and wj are listed in and verified in Problem 2.5.

The error using Gauss-Legendre quadrature is found as As in the composite trapezoidal and Simpson rules, the maximum value of f(2n)(η) is e. Thus, the maximum error using Gauss-Legendre quadrature is

The following table lists the results for integration by Gauss-Legendre quadrature.

For n = 1
The root of the Legendre Polynomial $$ P_1(x)=0 $$ is found by solving

$$ \displaystyle P_1(x)=x=0 $$

And the weight, $$ w_{1} $$ is

$$ \displaystyle \begin{align} w_{1} & = \frac{-2}{(1+1)[P_{1}^{'}({{x}_{1}}){{P}_{2}}({{x}_{1}})]}  \\ & = \frac{-2}{(1+1)\left[ 1\cdot \frac{3{x_1}^{2}-1}{2}\right] }  \\ & = \frac{-2}{(1+1)\left[ 1\cdot \frac{3\cdot{0}^{2}-1}{2}\right] } \\ & = 2 \end{align} $$

For n = 2
The root of the Legendre Polynomial $$ P_2(x)=0 $$ are found by solving

$$ \displaystyle P_2(x)=(3x^2-1)=0 $$

Where the weights, $$ w_{1} $$ and $$ w_{2} $$ are

$$ \displaystyle \begin{align} w_{1} & = \frac{-2}{(2+1)[P_2^{'}(x_1)P_3(x_1)]}  \\ & = \frac{-2}{(2+1)\left[ 3x_1 \cdot \frac{5{x_1}^{3}-3x_1}{2}\right] }  \\ & = \frac{-2}{(2+1)\left[ 3\left( {\frac{1}{\sqrt 3 }}\right)^3 \cdot \frac{5\left( {\frac{1}{\sqrt 3 }}\right)^3 -3\left(\frac{1}{\sqrt 3 }\right) }{2}\right] } \\ & = 1 \end{align} $$

$$ \displaystyle \begin{align} w_{2} & = \frac{-2}{(2+1)[P_2^{'}(x_2)P_3(x_2)]}  \\ & = \frac{-2}{(2+1)\left[ 3x_2 \cdot \frac{5{x_2}^{3}-3x_2}{2}\right] }  \\ & = \frac{-2}{(2+1)\left[ 3\left( {-\frac{1}{\sqrt 3 }}\right)^3 \cdot \frac{5\left( {-\frac{1}{\sqrt 3 }}\right)^3 -3\left( -\frac{1}{\sqrt 3 }\right) }{2}\right] } \\ & = 1 \end{align} $$

For n = 3
The root of the Legendre Polynomial $$ P_3(x)=0 $$ are found by solving

$$ \displaystyle P_3(x)=(5x^3-3x)=0 $$

Where the weights, $$ w_{1}, w_{2}, $$ and $$ w_{3} $$ are

$${{W}_{1}}=\frac{-2}{(3+1)[P_{3}^{'}({{x}_{1}}){{P}_{4}}({{x}_{1}})]}=\frac{8}{9} $$

$${{W}_{2}}=\frac{-2}{(3+1)[P_{3}^{'}({{x}_{2}}){{P}_{4}}({{x}_{2}})]}=\frac{5}{9} $$

$${{W}_{3}}=\frac{-2}{(3+1)[P_{3}^{'}({{x}_{3}}){{P}_{4}}({{x}_{3}})]}=\frac{5}{9}$$

For n = 4
The root of the Legendre Polynomial $$ P_4(x)=0 $$ are found by solving

$$ \displaystyle P_4(x)=(35x^4-30x^2+3)=0 $$

Where the weights, $$ w_{1}, w_{2}, w_{3}$$ and $$ w_{4} $$ are

$${{W}_{1}}=\frac{-2}{(4+1)[P_{4}^{'}({{x}_{1}}){{P}_{5}}({{x}_{1}})]}=(18-\sqrt{30})/36 $$

$${{W}_{2}}=\frac{-2}{(4+1)[P_{4}^{'}({{x}_{2}}){{P}_{5}}({{x}_{2}})]}=(18-\sqrt{30})/36 $$

$${{W}_{3}}=\frac{-2}{(4+1)[P_{4}^{'}({{x}_{3}}){{P}_{5}}({{x}_{3}})]}=(18+\sqrt{30})/36 $$

$${{W}_{4}}=\frac{-2}{(4+1)[P_{4}^{'}({{x}_{4}}){{P}_{5}}({{x}_{4}})]}=(18+\sqrt{30})/36 $$

For n = 5
The root of the Legendre Polynomial $$ P_5(x)=0 $$ are found by solving

$$ \displaystyle P_5(x)=(63x^5-70x^3+15x)=0 $$

Where the weights, $$ w_{1}, w_{2}, w_{3}, w_{4} $$ and $$ w_{5} $$ are

$${{W}_{1}}=\frac{-2}{(5+1)[P_{5}^{'}({{x}_{1}}){{P}_{6}}({{x}_{1}})]}=\frac{128}{225}$$

$${{W}_{2}}=\frac{-2}{(5+1)[P_{5}^{'}({{x}_{2}}){{P}_{6}}({{x}_{2}})]}=\frac{322-13\sqrt{70}}{900}$$

$${{W}_{3}}=\frac{-2}{(5+1)[P_{5}^{'}({{x}_{3}}){{P}_{6}}({{x}_{3}})]}=\frac{322-13\sqrt{70}}{900}$$

$${{W}_{4}}=\frac{-2}{(5+1)[P_{5}^{'}({{x}_{4}}){{P}_{6}}({{x}_{4}})]}=\frac{322+13\sqrt{70}}{900}$$

$${{W}_{5}}=\frac{-2}{(5+1)[P_{5}^{'}({{x}_{5}}){{P}_{6}}({{x}_{5}})]}=\frac{322+13\sqrt{70}}{900}$$

As can be seen from the values computed here, the table present in lecture has been verified by myself and is in agreement with the NIST publication.

=Problem 2.6: Verifying Lagrange Polynomials=

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

From the lecture slide Mtg 8-1

Given
The equation for the Lagrange Polynomial is,

 * {| style="width:100%" border="0"

{P}_{n}(x) = \sum_{i=0}^{\lfloor n/2 \rfloor} (-1)^{i} \cdot \frac {(2n-2i)! \cdot x^{n-2i}} {2^{n}\cdot i! \cdot (n-i)! \cdot (n-2i)!} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

and the equations for $$\displaystyle P_0\left(x\right) $$ through $$\displaystyle P_4\left(x\right). $$

 * {| style="width:100%" border="0"

\begin{align} &{P}_{0}(x) = 1\\ &{P}_{1}(x) = x\\ &{P}_{2}(x) = \frac {1}{2}(3x^{2}-1)\\ &{P}_{3}(x) = \frac {1}{2}(5x^{3}-3x)\\ &{P}_{4}(x) = \frac {35}{8}x^{4} - \frac {15}{4}x^{2} + \frac {3}{8}\\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Find
Show that the equations for $$\displaystyle P_0\left(x\right) $$ through $$\displaystyle P_4\left(x\right) $$ can be written as $$\displaystyle P_n\left(x\right). $$

Solution
i) n=0
 * {| style="width:100%" border="0"

\begin{align} {P}_{0}(x) &= \sum_{i=0}^{\lfloor 0/2 \rfloor} (-1)^{i} \cdot \frac {(2n-2i)! \cdot x^{n-2i}} {2^{n}\cdot i! \cdot (n-i)! \cdot (n-2i)!}\\ {P}_{0}(x) &= 1 \cdot \frac {1 \cdot 1} {1\cdot 1 \cdot 1 \cdot 1}\\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

ii) n=1
 * {| style="width:100%" border="0"

\begin{align} {P}_{1}(x) &= \sum_{i=0}^{\lfloor 1/2 \rfloor} (-1)^{i} \cdot \frac {(2n-2i)! \cdot x^{n-2i}} {2^{n}\cdot i! \cdot (n-i)! \cdot (n-2i)!}\\ &=(-1)^{0} \cdot \frac {2! \cdot x^{1}} {2^{1}\cdot 0! \cdot 1! \cdot 1!} \\ {P}_{1}(x) &= 1 \cdot \frac {\cancel {2} \cdot x} {\cancel {2} \cdot 1 \cdot 1 \cdot 1} \\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

iii) n=2
 * {| style="width:100%" border="0"

\begin{align} {P}_{2}(x) &= \sum_{i=0}^{\lfloor 2/2 \rfloor} (-1)^{i} \cdot \frac {(2n-2i)! \cdot x^{n-2i}} {2^{n}\cdot i! \cdot (n-i)! \cdot (n-2i)!}\\ &=(-1)^{0} \cdot \frac {(4)! \cdot x^{2}} {2^{2}\cdot 0! \cdot (2)! \cdot (2)!} + (-1)^{1} \cdot \frac {(4-2)! \cdot x^{2-2}} {2^{2}\cdot 1! \cdot (2-1)! \cdot (2-2)!}\\ &= 1 \cdot \frac {4 \cdot 3 \cdot 3 \cdot x^{2}}{4 \cdot 1 \cdot 2 \cdot 2} + (-1) \cdot \frac {2 \cdot 1}{4 \cdot 1 \cdot 1 \cdot 1} \\ {P}_{2}(x) &= \frac {3}{2}x^{2}-\frac {1}{2}\\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

iv) n=3
 * {| style="width:100%" border="0"

\begin{align} {P}_{3}(x) &= \sum_{[i=0]}^{\lfloor 3/2 \rfloor} (-1)^{i} \cdot \frac {(2n-2i)! \cdot x^{n-2i}} {2^{n}\cdot i! \cdot (n-i)! \cdot (n-2i)!} \\ &=(-1)^{0} \cdot \frac {6! \cdot x^{3}} {2^{3}\cdot 0! \cdot 3! \cdot 3!} + (-1)^{1} \cdot \frac {4! \cdot x^{1}} {2^{3}\cdot 1! \cdot 2! \cdot 1!} \\ &= 1 \cdot \frac {6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot x^{3}} {8 \cdot 1 \cdot 3 \cdot 2 \cdot 3 \cdot 2} + (-1) \cdot \frac {4 \cdot 3 \cdot 2 \cdot x^{1}} {8\cdot 1 \cdot 2 \cdot 1} \\ {P}_{3}(x) &= \frac {5}{2}x^{3} - \frac {3}{2} x \\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

v) n=4
 * {| style="width:100%" border="0"

\begin{align} {P}_{4}(x) &= \sum_{[i]=0}^{\lfloor 4/2 \rfloor} (-1)^{i} \cdot \frac {(2n-2i)! \cdot x^{n-2i}} {2^{n}\cdot i! \cdot (n-i)! \cdot (n-2i)!}\\ &= (-1)^{0} \cdot \frac {8! \cdot x^4} {2^{4}\cdot 0! \cdot 4! \cdot 4!} + (-1)^{1} \cdot \frac {6! \cdot x^2} {2^{4}\cdot 1! \cdot 3! \cdot 2!} + (-1)^{2} \cdot \frac {4! \cdot x^0} {2^{4}\cdot 2! \cdot 2! \cdot 0!} \\ {P}_{4}(x) &= 1 \cdot \frac {8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot x^4} {16 \cdot 1 \cdot 4 \cdot 3 \cdot 2 \cdot 4 \cdot 3 \cdot 2} + (-1) \cdot \frac {6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot x^2} {16 \cdot 1 \cdot 3 \cdot 2 \cdot 2} + 1 \cdot \frac {4 \cdot 3 \cdot 2 \cdot 1} {16 \cdot 2 \cdot 2 \cdot 1} \\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

=Problem 2.7: Verification of Orthogonality of Legendre polynominals=

Refer to lecture slide from [[media:nm1.s11.mtg9.djvu|mtg-9]] for the problem statement and table shown below.

Background
Legendre polynominals are given by :

Note that they are all defined on the domain $$\displaystyle (-1,+1)$$.

The first 6 Legendre polynominals are :

While the Gramian matrix is defined as:

Find
A) Construct Gramian Matrix $$\displaystyle \Gamma $$ with respect to Legendre polynominals from order $$\displaystyle 0$$ to $$\displaystyle 5$$ and show that it‘s diagonal.

B) Compute the determinant of $$\displaystyle \Gamma$$.

C)Discuss the orthogonality of Legendre polynominals.

Solution
A) Construct Gramian Matrix $$\displaystyle \Gamma $$ with respect to Legendre polynominals from order $$\displaystyle 0$$ to $$\displaystyle 5$$.

When the order of Legendere polynominals is less than or equal 2, we have,

As the order goes higher we may use WolframAlpha to help us construct the Gramian matrix(click on the link to check the results):

which means the Gramian matrix $$\displaystyle \Gamma $$ is,

which is apparently a diagonal matrix

B) Compute the determinant of $$\displaystyle \Gamma $$.

Since $$\displaystyle \Gamma $$ is a diagonal matrix, the determinant of it is

C)Discuss the orthogonality of Legendre polynominals.

From Eq(7.2) we can conclude that within order of 5 the Legendre polynominals are orthogonal to each other.

In fact Legendre polynominals have such orthogonality property :

where $$\displaystyle {{\delta }_{mn}}$$ is the Kronecker delta.

The proof of the above statement is given by A.K.Lal et al.

=Problem 2.8 - Newton-Cotes Simplification to Trapezoidal Rule=

Find
Show that the Newton-Cotes method for numerical integration with n=1 simplifies to the simple trapezoidal rule.

Given
The Newton-Cotes method using Lagrange interpolating polynomials is given as The simple trapezoidal rule is expressed as

Solution
For n=1, the Newton-Cotes method becomes Let a = x0 and b = x1, and substitute in the Lagrange interpolating polynomials as

The Newton-Cotes equation becomes Manipulation proceeds as follows to yield the simple trapezoidal rule: The last line of 2.8.6 is identical to the simple trapezoidal rule, thus the proof is complete.

=Problem 2.9: "Gauss-Legendre Quadrature and Runge's phenomenon"=

Given
Legendre polynomial

Find
A) Use (9.1) to generate $$ {P_5}(x) $$ and Matlab command "roots" to compute the roots of $$ {P_5}(x) $$ to check values in table on p7-5. Plot the roots on $$ [-1,+1] $$ using Matlab "plot" command.

B) Repeat the above for $$ {P_{10}}(x) $$ and observe the location of roots near end points -1 and +1.

Solution
A) Substitute n=5 in (9.1) and then can get $$ {P_5}(x) $$:

Use the following codes in Matlab :

p=[63 0 -70 0 15 0]/8 roots(p)

we can get the roots for $$ {P_5}(x) $$: ans = 0  -0.9062   -0.5385    0.9062    0.5385

The above answers are exactly the same as the data in table on p7-5.

Use the "plot" command we can get the distribution of roots on $$ [-1,+1] $$ as following:

B) Substitute n=10 in (9.1) and then can get $$ {P_{10}}(x) $$:

Use the following codes in Matlab :

p=[46189 0 -109395 0 90090 0 -30030 0 3465 0 -63]/256 roots(p)

we can get the roots for $$ {P_{10}}(x) $$: ans = -0.9739  -0.8651   -0.6794    0.9739    0.8651    0.6794   -0.4334    0.4334   -0.1489    0.1489

The above answers are exactly the same as the data in table on p7-5.

Use the "plot" command we can get the distribution of roots on $$ [-1,+1] $$ as following:

From the above image, we can observe that quadrature nodes become more dense near the end points. That means the interpolation with high order polynomial may has more oscillation near the edge of interval region because the interpolation only exactly accurate at the quadrature nodes and this phenomenon is so called Runge's phenomenon.

=Problem 2.10: "Proof of Simpson's Rule(simple)"=

Given
In the interval $$ \left[ {a,b} \right] $$ there are three nodes $$ {x_0} = a, {x_1} = \frac{2},  x{_2} = b $$ satisfy that

Find
Use (10.1) to find expression for $$ \left\{ \right\} $$ in terms $$ \left( {{x_i},f({x_i})} \right) $$.

Proof
From (10.1) we can have three equations for 3 unknown $$ \left\{ \right\} $$

$$ {c_2}{x_0}^2 + {c_1}{x_0} + {c_0} = f\left( \right) $$

$$ {c_2}{x_1}^2 + {c_1}{x_1} + {c_0} = f\left( \right) $$

$$ {c_2}{x_2}^2 + {c_1}{x_2} + {c_0} = f\left( \right) $$

And then we write them in matrix form:

where

$$ \left[ K \right] = \left[ {\begin{array}{ccccccccccccccc} 1&&{x_0^2} \\ 1&&{x_1^2} \\ 1&&{x_2^2} \end{array}} \right] $$

Because the determine of the matrix K is

$$ \det \left[ K \right] = - \left( {{x_0} - {x_1}} \right)\left( {{x_0} - {x_2}} \right)\left( {{x_1} - {x_2}} \right) $$

we then compute the inverse of matrix K is

$${\left[ K \right]^{ - 1}} = \left[ {\begin{array}{ccccccccccccccc} {\frac}&{ - \frac}&{\frac} \\ { - \frac}&{\frac}&{ - \frac} \\ {\frac{1}}&{ - \frac{1}}&{\frac{1}} \end{array}} \right]\; $$

Then we can get $$ \left\{ \right\} $$ by multiple $$  {\left[ K \right]^{ - 1}} $$ on both sides of (10.2):

Consider that $$ {x_0} = a, {x_1} = \frac{2},  x{_2} = b $$ we can rewrite $$ \left\{ \right\} $$:

If integrate (10,1) on $$ \left[ {a,b} \right] $$, then

(10.9) is exactly Simpson's Rule for numerical integration.

=Problem 2.11: Proof of the Simple Simpson's Rule =

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

From the lecture slide Mtg 10-4

Given
The Lagrange Interpretation Equation,

 * {| style="width:100%" border="0"

P_2(x_j)=\sum_{k=0}^2 l_{i,2}(x_j)f(x_i). $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Find
Derive the Simple Simpson's rule from the above Lagrange Interpolation Equation of degree 2. The Simple Simpson's rule is as follows,

 * {| style="width:100%" border="0"

I_2=\dfrac{h}{3}[f(x_0)+4f(x_1)+f(x_2)] , $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

where

 * {| style="width:100%" border="0"

h=\dfrac{b-a}{2}. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Solution
By definition,
 * {| style="width:100%" border="0"

I_2 = \int_{a}^{b} P_2(x)\ dx $$
 * $$ \displaystyle
 * $$ \displaystyle
 * <p style="text-align:right;">
 * }
 * }

Evaluating the above integral gives

 * {| style="width:100%" border="0"

I_2 & = \int_{a}^{b} P_2(x)\ dx \\ & = \int_{a}^{b} \sum_{i=0}^{2}l_i(x)f(x_i)\ dx \\ & = \int_{a}^{b} l_0(x)f(x_0) + l_1(x)f(x_1) + l_2(x)f(x_2)\ dx \\ & = f(x_0)\int_{a}^{b} l_0(x)\ dx + f(x_1)\int_{a}^{b} l_1(x)\ dx + f(x_2)\int_{a}^{b} l_2(x)\ dx , \end{align} $$
 * $$ \displaystyle \begin{align}
 * $$ \displaystyle \begin{align}
 * <p style="text-align:right;">(11.1-4)
 * }
 * }

where $$ \displaystyle x_0 = a, x_1 = \frac{a+b}{2}, $$ and $$\displaystyle x_2 = b .$$

Next, integrate $$\displaystyle l_0(x), l_1(x),$$ and $$\displaystyle l_2(x). $$

1)$$\displaystyle I_0 $$

 * {| style="width:100%" border="0"

\int_{a}^{b} l_0(x)\ dx & = \int_{a}^{b}\frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}\ dx \\ & = \int_{a}^{b}\frac{(x-\frac{a+b}{2})(x-b)}{(a-\frac{a+b}{2})(a-b)}\ dx \\ & = \frac{2}{(b-a)^2}\left [ \frac{x^3}{3} - \frac{a+3b}{4}x^2 + \frac{b(a+b)}{2}x \right ]_{a}^b \\ & = \frac{2}{(b-a)^2} \left [ \frac{(b-a)(b^2 + ab + a^2)}{3} - \frac{a+3b}{4}(b-a)(b+a) + \frac{b(a+b)}{2}(b-a) \right ] \\ & = \frac{2}{(b-a)} \left [ \frac{(b^2 + ab + a^2)}{3} - \frac{a+3b}{4}(b+a) + \frac{b(a+b)}{2} \right ] \\ & = \frac{(b-a)^2}{6(b-a)} \\ & = \frac{b-a}{6}\\ \end{align} $$
 * $$ \displaystyle \begin{align}
 * $$ \displaystyle \begin{align}
 * <p style="text-align:right;">(11.4-11)
 * }
 * }

2)$$\displaystyle I_1 $$

 * {| style="width:100%" border="0"

\int_{a}^{b} l_1(x)\ dx & = \int_{a}^{b}\frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)}\ dx \\ & = \int_{a}^{b}\frac{(x-a)(x-b)}{(\frac{a+b}{2}-a)(\frac{a+b}{2}-b)}\ dx \\ & = \frac{4}{(b-a)(a-b)}\left [ \frac{x^3}{3} - \frac{a+b}{2}x^2 + abx \right ]_{a}^b \\ & = \frac{4}{(b-a)(a-b)}\left [ \frac{(b-a)(b^2+ab+a^2)}{3} - \frac{(a+b)(a+b)(a-b)}{2} + ab(b-a) \right ]_{a}^b \\ & = \frac{4}{(a-b)}\left [ \frac{(b^2+ab+a^2)}{3} - \frac{(a+b)(a+b)}{2} + ab \right ]_{a}^b \\ & = \frac{2(b-a)^2}{3(b-a)} \\ & = \frac{2(b-a)}{3}\\ \end{align} $$
 * $$ \displaystyle \begin{align}
 * $$ \displaystyle \begin{align}
 * <p style="text-align:right;">(11.11-18)
 * }
 * }

3)$$\displaystyle I_2 $$

 * {| style="width:100%" border="0"

\int_{a}^{b} l_2(x) \ dx & = \int_{a}^{b}\frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}\ dx \\ & = \int_{a}^{b}\frac{(x-a)(x-\frac{a+b}{2})}{(b-a)(b-\frac{a+b}{2})}\ dx \\ & = \frac{2}{(b-a)^2}\left [ \frac{x^3}{3} - \frac{3a+b}{4}x^2 + \frac{a(a+b)}{2}x \right ]_{a}^b \\ & = \frac{2}{(b-a)^2} \left [ \frac{(b-a)(b^2 + ab + a^2)}{3} - \frac{3a+b}{4}(b-a)(b+a) + \frac{b(a+b)}{2}(b-a) \right ] \\ & = \frac{2}{(b-a)} \left [ \frac{(b^2 + ab + a^2)}{3} - \frac{3a+b}{4}(b+a) + \frac{b(a+b)}{2} \right ] \\ & = \frac{(b-a)^2}{6(b-a)} \\ & = \frac{b-a}{6}\\ \end{align} $$
 * $$ \displaystyle \begin{align}
 * $$ \displaystyle \begin{align}
 * <p style="text-align:right;">(11.18-26)
 * }
 * }

Substituting the values for the integrals of $$\displaystyle l_0(x), l_1(x),$$ and $$\displaystyle l_2(x) $$ into equation 11.4 gives

 * {| style="width:100%" border="0"

\begin{align} I_2 & = f(x_0)\frac{b-a}{6}+f(x_1)\frac{2(b-a)}{3}+f(x_2)\frac{b-a}{6} \\ I_2 & = \frac{b-a}{6}[f(x_0)+4f(x_1)+f(x_2)]\\ \end{align} $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }

where

 * {| style="width:100%" border="0"

h = \frac{b-a}{2}. $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">(11.28)
 * }
 * }

= Problem 2.12 Numerically calculating integral of a function using Lagrange method =

From the lecture slide Mtg 10 [[media:nm1.s11.mtg10.djvu|Mtg10]]

Ref: Egm6341.s10.team3.sa/HW2

Given
$$ f(x)=\frac{e^{x}-1}{x} $$ on $$ x\in[0,1] (a) $$ and $$ x\in[0,1] (b) $$

Find
A) Construct $$ f_{n}(x)= \sum_{i=0}^{n}l_{i,n}(x)f(x_{i}) $$ for $$\displaystyle n = 1, 2, 4, 8, 16 $$ for both intervals.

B) Plot $$\displaystyle f(x) $$ and $$\displaystyle f_{n}(x) $$ for n = 1, 2, 4, 8, 16.

C) Compute $$ I_{n}=\int_{b}^{b}f_{n}(x)dx $$ for n = 1, 2, 4, 8 and compare to $$\displaystyle I $$.

D) For n = 5, plot $$\displaystyle l_{0}, l_{1}, l_{2} $$.

Solution
A) Considering the following equations:

$$  \displaystyle l_{i,n}(x) = \prod_{j=0,i\neq j}^{n}\frac{x-x_{j}}{x_{i}-x_{j}} $$ and $$  \displaystyle

f_{n}=\sum_{i=0}^{n}l_{i,n}(x)f(x_{i}) $$

for different n values and $$ x\in[0,1] $$, the followings can be calculated:

When n = 1

$$ \displaystyle x_{0} = 0, x_{1} = 1 $$.

$$\displaystyle l_{0,1}=\prod_{j=0,j\neq 0}^{1}\frac{x-x_{j}}{x_{0}-x_{j}}=\frac{x-1}{0-1}, \quad f(x_{0})=f(0)=1, $$

$$\displaystyle l_{1,1}=\prod_{j=0,j\neq 1}^{1}\frac{x-x_{j}}{x_{1}-x_{j}}=\frac{x-0}{1-0}, \quad f(x_{1})=f(1)=e-1. $$

When n = 2

$$ x_{0} = 0, x_{1} = \frac{1}{2}, x_{2} = 1 $$.

$$\displaystyle l_{0,2}=\prod_{j=0,j\neq 0}^{2}\frac{x-x_{j}}{x_{0}-x_{j}}=\frac{(x-\frac{1}{2})(x-1)}{(0-\frac{1}{2}(0-1)}, \quad f(x_{0})=f(0)=1, $$

$$\displaystyle l_{1,2}=\prod_{j=0,j\neq 1}^{2}\frac{x-x_{j}}{x_{1}-x_{j}}=\frac{(x-0)(x-1)}{(\frac{1}{2}-0)(\frac{1}{2}-1)}, \quad f(x_{1})=f(\frac{1}{2})=2\left(e^{\frac{1}{2}}-1\right), $$

$$\displaystyle l_{2,2}=\prod_{j=0,j\neq 2}^{2}\frac{x-x_{j}}{x_{2}-x_{j}}=\frac{(x-0)(x-\frac{1}{2})}{(1-0)(1-\frac{1}{2})}, \quad f(x_{2})=f(1)=e-1. $$

When n = 4

$$ x_{0} = 0, x_{1} = \frac{1}{4}, x_{2} = \frac{2}{4}, x_{3} = \frac{3}{4}, x_{4} = 1 $$.

$$\displaystyle l_{0,4}=\prod_{j=0,j\neq 0}^{4}\frac{x-x_{j}}{x_{0}-x_{j}}=\frac{(x-\frac{1}{4})(x-\frac{2}{4})(x-\frac{3}{4})(x-1)}{(0-\frac{1}{4})(0-\frac{2}{4})(0-\frac{3}{4})(0-1)}, \quad f(x_{0})=f(0)=1, $$

$$\displaystyle l_{1,4}=\prod_{j=0,j\neq 1}^{4}\frac{x-x_{j}}{x_{1}-x_{j}}=\frac{(x-0)(x-\frac{2}{4})(x-\frac{3}{4})(x-1)}{(\frac{1}{4}-0)(\frac{1}{4}-\frac{2}{4})(\frac{1}{4}-\frac{3}{4})(\frac{1}{4}-1)}, \quad f(x_{1})=f(\frac{1}{4})=4\left(e^{\frac{1}{4}}-1\right), $$

$$\displaystyle l_{2,4}=\prod_{j=0,j\neq 2}^{4}\frac{x-x_{j}}{x_{2}-x_{j}}=\frac{(x-0)(x-\frac{1}{4})(x-\frac{3}{4})(x-1)}{(\frac{2}{4}-0)(\frac{2}{4}-\frac{1}{4})(\frac{2}{4}-\frac{3}{4})(\frac{2}{4}-1)}, \quad f(x_{2})=f(\frac{2}{4})=2\left(e^{\frac{2}{4}}-1\right), $$

$$\displaystyle l_{3,4}=\prod_{j=0,j\neq 3}^{4}\frac{x-x_{j}}{x_{3}-x_{j}}=\frac{(x-0)(x-\frac{1}{4})(x-\frac{2}{4})(x-1)}{(\frac{3}{4}-0)(\frac{3}{4}-\frac{1}{4})(\frac{3}{4}-\frac{2}{4})(\frac{3}{4}-1)}, \quad f(x_{3})=f(\frac{3}{4})=\frac{4}{3}\left(e^{\frac{3}{4}}-1\right), $$

$$\displaystyle l_{4,4}=\prod_{j=0,j\neq 4}^{4}\frac{x-x_{j}}{x_{4}-x_{j}}=\frac{(x-0)(x-\frac{1}{4})(x-\frac{2}{4})(x-\frac{3}{4})}{(1-0)(1-\frac{1}{4})(1-\frac{2}{4})(1-\frac{3}{4})}, \quad f(x_{4})=f(1)=e-1, $$

When n = 8

$$ x_{0} = 0, x_{1} = \frac{1}{8}, x_{2} = \frac{2}{8}, x_{3} = \frac{3}{8}, x_{4} = \frac{4}{8}, x_{5} = \frac{5}{8}, x_{6} = \frac{6}{8}, x_{7} = \frac{7}{8}, x_{8} = 1 $$.

$$\displaystyle l_{0,8}=\prod_{j=0,j\neq 0}^{8}\frac{x-x_{j}}{x_{0}-x_{j}}=\frac{(x-\frac{1}{8})(x-\frac{2}{8})(x-\frac{3}{8})(x-\frac{4}{8})(x-\frac{5}{8})(x-\frac{6}{8})(x-\frac{7}{8})(x-1)}{(0-\frac{1}{8})(0-\frac{2}{8})(0-\frac{3}{8})(0-\frac{4}{8})(0-\frac{5}{8})(0-\frac{6}{8})(0-\frac{7}{8})(0-1)}, \quad f(x_{0})=f(0)=1, $$

$$\displaystyle l_{1,8}=\prod_{j=0,j\neq 1}^{8}\frac{x-x_{j}}{x_{1}-x_{j}}=\frac{(x-0)(x-\frac{2}{8})(x-\frac{3}{8})(x-\frac{4}{8})(x-\frac{5}{8})(x-\frac{6}{8})(x-\frac{7}{8})(x-1)}{(\frac{1}{8}-0)(\frac{1}{8}-\frac{2}{8})(\frac{1}{8}-\frac{3}{8})(\frac{1}{8}-\frac{4}{8})(\frac{1}{8}-\frac{5}{8})(\frac{1}{8}-\frac{6}{8})(\frac{1}{8}-\frac{7}{8})(\frac{1}{8}-1)}, \quad f(x_{1})=f(\frac{1}{8})=8\left(e^{\frac{1}{8}}-1\right), $$

$$\displaystyle l_{2,8}=\prod_{j=0,j\neq 2}^{8}\frac{x-x_{j}}{x_{2}-x_{j}}=\frac{(x-0)(x-\frac{1}{8})(x-\frac{3}{8})(x-\frac{4}{8})(x-\frac{5}{8})(x-\frac{6}{8})(x-\frac{7}{8})(x-1)}{(\frac{2}{8}-0)(\frac{2}{8}-\frac{1}{8})(\frac{2}{8}-\frac{3}{8})(\frac{2}{8}-\frac{4}{8})(\frac{2}{8}-\frac{5}{8})(\frac{2}{8}-\frac{6}{8})(\frac{2}{8}-\frac{7}{8})(\frac{2}{8}-1)}, \quad f(x_{2})=f(\frac{2}{8})=\frac{8}{2}\left(e^{\frac{2}{8}}-1\right), $$

$$\displaystyle l_{3,8}=\prod_{j=0,j\neq 3}^{8}\frac{x-x_{j}}{x_{3}-x_{j}}=\frac{(x-0)(x-\frac{1}{8})(x-\frac{2}{8})(x-\frac{4}{8})(x-\frac{5}{8})(x-\frac{6}{8})(x-\frac{7}{8})(x-1)}{(\frac{3}{8}-0)(\frac{3}{8}-\frac{1}{8})(\frac{3}{8}-\frac{2}{8})(\frac{3}{8}-\frac{4}{8})(\frac{3}{8}-\frac{5}{8})(\frac{3}{8}-\frac{6}{8})(\frac{3}{8}-\frac{7}{8})(\frac{3}{8}-1)}, \quad f(x_{3})=f(\frac{3}{8})=\frac{8}{3}\left(e^{\frac{3}{8}}-1\right), $$

$$\displaystyle l_{4,8}=\prod_{j=0,j\neq 4}^{8}\frac{x-x_{j}}{x_{4}-x_{j}}=\frac{(x-0)(x-\frac{1}{8})(x-\frac{2}{8})(x-\frac{3}{8})(x-\frac{5}{8})(x-\frac{6}{8})(x-\frac{7}{8})(x-1)}{(\frac{4}{8}-0)(\frac{4}{8}-\frac{1}{8})(\frac{4}{8}-\frac{2}{8})(\frac{4}{8}-\frac{3}{8})(\frac{4}{8}-\frac{5}{8})(\frac{4}{8}-\frac{6}{8})(\frac{4}{8}-\frac{7}{8})(\frac{4}{8}-1)}, \quad f(x_{4})=f(\frac{4}{8})=\frac{8}{4}\left(e^{\frac{4}{8}}-1\right), $$

$$\displaystyle l_{5,8}=\prod_{j=0,j\neq 5}^{8}\frac{x-x_{j}}{x_{5}-x_{j}}=\frac{(x-0)(x-\frac{1}{8})(x-\frac{2}{8})(x-\frac{3}{8})(x-\frac{4}{8})(x-\frac{6}{8})(x-\frac{7}{8})(x-1)}{(\frac{5}{8}-0)(\frac{5}{8}-\frac{1}{8})(\frac{5}{8}-\frac{2}{8})(\frac{5}{8}-\frac{3}{8})(\frac{5}{8}-\frac{4}{8})(\frac{5}{8}-\frac{6}{8})(\frac{5}{8}-\frac{7}{8})(\frac{5}{8}-1)}, \quad f(x_{5})=f(\frac{5}{8})=\frac{8}{5}\left(e^{\frac{5}{8}}-1\right), $$

$$\displaystyle l_{6,8}=\prod_{j=0,j\neq 6}^{8}\frac{x-x_{j}}{x_{6}-x_{j}}=\frac{(x-0)(x-\frac{1}{8})(x-\frac{2}{8})(x-\frac{3}{8})(x-\frac{4}{8})(x-\frac{5}{8})(x-\frac{7}{8})(x-1)}{(\frac{6}{8}-0)(\frac{6}{8}-\frac{1}{8})(\frac{6}{8}-\frac{2}{8})(\frac{6}{8}-\frac{3}{8})(\frac{6}{8}-\frac{4}{8})(\frac{6}{8}-\frac{5}{8})(\frac{6}{8}-\frac{7}{8})(\frac{6}{8}-1)}, \quad f(x_{6})=f(\frac{6}{8})=\frac{8}{6}\left(e^{\frac{6}{8}}-1\right), $$

$$\displaystyle l_{7,8}=\prod_{j=0,j\neq 7}^{8}\frac{x-x_{j}}{x_{7}-x_{j}}=\frac{(x-0)(x-\frac{1}{8})(x-\frac{2}{8})(x-\frac{3}{8})(x-\frac{4}{8})(x-\frac{5}{8})(x-\frac{6}{8})(x-1)}{(\frac{7}{8}-0)(\frac{7}{8}-\frac{1}{8})(\frac{7}{8}-\frac{2}{8})(\frac{7}{8}-\frac{3}{8})(\frac{7}{8}-\frac{4}{8})(\frac{7}{8}-\frac{5}{8})(\frac{7}{8}-\frac{6}{8})(\frac{7}{8}-1)}, \quad f(x_{7})=f(\frac{7}{8})=\frac{8}{7}\left(e^{\frac{7}{8}}-1\right), $$

$$\displaystyle l_{8,8}=\prod_{j=0,j\neq 8}^{8}\frac{x-x_{j}}{x_{8}-x_{j}}=\frac{(x-0)(x-\frac{1}{8})(x-\frac{2}{8})(x-\frac{3}{8})(x-\frac{4}{8})(x-\frac{5}{8})(x-\frac{6}{8})(x-\frac{7}{8})}{(1-0)(1-\frac{1}{8})(1-\frac{2}{8})(1-\frac{3}{8})(1-\frac{4}{8})(1-\frac{5}{8})(1-\frac{6}{8})(1-\frac{7}{8})}, \quad f(x_{8})=f(1)=e-1. $$

When n = 16 $$ x_{0} = 0, x_{1} = \frac{1}{16}, x_{2} = \frac{2}{16}, x_{3} = \frac{3}{16}, x_{4} = \frac{4}{16}, x_{5} = \frac{5}{16}, x_{6} = \frac{6}{16}, x_{7} = \frac{7}{16}, x_{8} = \frac{8}{16}, x_{9} = \frac{9}{16}, x_{10} = \frac{10}{16}, x_{11} = \frac{11}{16}, x_{12} = \frac{12}{16}, x_{13} = \frac{13}{16}, x_{14} = \frac{14}{16}, x_{15} = \frac{15}{16}, x_{16} = 1 $$.

$$\displaystyle l_{0,16}=\prod_{j=0,j\neq 0}^{16}\frac{x-x_{j}}{x_{0}-x_{j}}=\frac{(x-\frac{1}{16})(x-\frac{2}{16})(x-\frac{3}{16})(x-\frac{4}{16})(x-\frac{5}{16})(x-\frac{6}{16})(x-\frac{7}{16})(x-\frac{8}{16})(x-\frac{9}{16})(x-\frac{10}{16})(x-\frac{11}{16})(x-\frac{12}{16})(x-\frac{13}{16})(x-\frac{14}{16})(x-\frac{15}{16})(x-1)}{(0-\frac{1}{16})(0-\frac{2}{16})(0-\frac{3}{16})(0-\frac{4}{16})(0-\frac{5}{16})(0-\frac{6}{16})(0-\frac{7}{16})(0-\frac{8}{16})(0-\frac{9}{16})(0-\frac{10}{16})(0-\frac{11}{16})(0-\frac{12}{16})(0-\frac{13}{16})(0-\frac{14}{16})(0-\frac{15}{16})(0-1)}, \quad f(x_{0})=f(0)=1, $$

$$\displaystyle l_{1,16}=\prod_{j=0,j\neq 1}^{16}\frac{x-x_{j}}{x_{1}-x_{j}}=\frac{(x-0)(x-\frac{2}{16})(x-\frac{3}{16})(x-\frac{4}{16})(x-\frac{5}{16})(x-\frac{6}{16})(x-\frac{7}{16})(x-\frac{8}{16})(x-\frac{9}{16})(x-\frac{10}{16})(x-\frac{11}{16})(x-\frac{12}{16})(x-\frac{13}{16})(x-\frac{14}{16})(x-\frac{15}{16})(x-1)}{(\frac{1}{16}-0)(\frac{1}{16}-\frac{2}{16})(\frac{1}{16}-\frac{3}{16})(\frac{1}{16}-\frac{4}{16})(\frac{1}{16}-\frac{5}{16})(\frac{1}{16}-\frac{6}{16})(\frac{1}{16}-\frac{7}{16})(\frac{1}{16}-\frac{8}{16})(\frac{1}{16}-\frac{9}{16})(\frac{1}{16}-\frac{10}{16})(\frac{1}{16}-\frac{11}{16})(\frac{1}{16}-\frac{12}{16})(\frac{1}{16}-\frac{13}{16})(\frac{1}{16}-\frac{14}{16})(\frac{1}{16}-\frac{15}{16})(\frac{1}{16}-1)}, \quad f(x_{1})=f(\frac{1}{16})=16\left(e^{\frac{1}{16}}-1\right), $$

and it can be calculate up to $$ l_{16,16} $$ in the same manner.

$$\displaystyle l_{16,16}=\prod_{j=0,j\neq 16}^{16}\frac{x-x_{j}}{x_{16}-x_{j}}=\frac{(x-0)(x-\frac{1}{16})(x-\frac{2}{16})(x-\frac{3}{16})(x-\frac{4}{16})(x-\frac{5}{16})(x-\frac{6}{16})(x-\frac{7}{16})(x-\frac{8}{16})(x-\frac{9}{16})(x-\frac{10}{16})(x-\frac{11}{16})(x-\frac{12}{16})(x-\frac{13}{16})(x-\frac{14}{16})(x-\frac{15}{16})}{(1-0)(1-\frac{1}{16})(1-\frac{2}{16})(1-\frac{3}{16})(1-\frac{4}{16})(1-\frac{5}{16})(1-\frac{6}{16})(1-\frac{7}{16})(1-\frac{8}{16})(1-\frac{9}{16})(1-\frac{10}{16})(1-\frac{11}{16})(1-\frac{12}{16})(1-\frac{13}{16})(1-\frac{14}{16})(1-\frac{15}{16})}, \quad f(x_{16})=f(1)=e-1, $$

And for different n values and $$ x\in[-1,1] $$, the followings can be calculated:

When n = 1

$$ \displaystyle x_{0} = -1, x_{1} = 1 $$.

$$\displaystyle l_{0,1} = \prod\limits_{j = 0,j \ne 0}^1 {\frac} = \frac,\quad f({x_0}) = f( - 1) =  - ({e^{ - 1}} - 1) $$

$$ {l_{1,1}} = \prod\limits_{j = 0,j \ne 1}^1 {\frac} = \frac,\quad f({x_1}) = f(1) = e - 1. $$

When n = 2

$$ \displaystyle x_{0} = -1, x_{1} = 0, x_{2} = 1 $$.

$${l_{0,2}} = \prod\limits_{j = 0,j \ne 0}^2 {\frac} = \frac,\quad f({x_0}) = f( - 1) =  - ({e^{ - 1}} - 1), $$

$${l_{1,2}} = \prod\limits_{j = 0,j \ne 1}^2 {\frac} = \frac,\quad f({x_1}) = f(0) = 1 $$

$${l_{2,2}} = \prod\limits_{j = 0,j \ne 2}^2 {\frac} = \frac,\quad f({x_2}) = f(1) = e - 1. $$

When n = 4

$$ \displaystyle x_{0} = -1, x_{1} = -\frac{1}{2}, x_{2} = 0, x_{3} = \frac{1}{2}, x_{4} = 1 $$.

$$ {l_{0,4}} = \prod\limits_{j = 0,j \ne 0}^4 {\frac} = \frac,\quad f({x_0}) = f( - 1) =  - ({e^{ - 1}} - 1)$$

$${l_{1,4}} = \prod\limits_{j = 0,j \ne 1}^4 {\frac} = \frac,\quad f({x_1}) = f( - \frac{1}{2}) =  - 2\left( {{e^{ - \frac{1}{2}}} - 1} \right), $$

$${l_{2,4}} = \prod\limits_{j = 0,j \ne 2}^4 {\frac} = \frac,\quad f({x_2}) = f(0) = 1, $$

$${l_{3,4}} = \prod\limits_{j = 0,j \ne 3}^4 {\frac} = \frac,\quad f({x_3}) = f(\frac{1}{2}) = 2\left( {{e^{\frac{1}{2}}} - 1} \right), $$

$$ {l_{4,4}} = \prod\limits_{j = 0,j \ne 4}^4 {\frac} = \frac,\quad f({x_4}) = f(1) = e - 1, $$

When n = 8

$$ \displaystyle x_{0} = -1, x_{1} = -\frac{3}{4}, x_{2} =- \frac{2}{4}, x_{3} = -\frac{1}{4}, x_{4} = 0, x_{5} = \frac{1}{4}, x_{6} = \frac{2}{4}, x_{7} = \frac{3}{4}, x_{8} = 1 $$.

$$ l_{0,8} = \prod\limits_{j = 0,j \ne 0}^8 {\frac} = \frac,\quad f({x_0}) = f( - 1) =  - ({e^{ - 1}} - 1), $$

$${l_{1,8}} = \prod\limits_{j = 0,j \ne 1}^8 {\frac} = \frac,\quad f({x_1}) = f( - \frac{3}{4}) =  - \frac{4}{3}\left( {{e^{ - \frac{3}{4}}} - 1} \right), $$

$$ {l_{2,8}} = \prod\limits_{j = 0,j \ne 2}^8 {\frac}  = \frac,\quad f({x_2}) = f( - \frac{2}{4}) =  - 2\left( {{e^{ - \frac{2}{4}}} - 1} \right), $$

$$ {l_{3,8}} = \prod\limits_{j = 0,j \ne 3}^8 {\frac}  = \frac,\quad f({x_3}) = f( - \frac{1}{4}) =  - 4\left( {{e^{ - \frac{1}{4}}} - 1} \right)$$

$$ {l_{4,8}} = \prod\limits_{j = 0,j \ne 4}^8 {\frac}  = \frac,\quad f({x_4}) = f(0) = 1, $$

$$ {l_{5,8}} = \prod\limits_{j = 0,j \ne 5}^8 {\frac} = \frac,\quad f({x_5}) = f(\frac{1}{4}) = 4\left( {{e^{\frac{1}{4}}} - 1} \right),$$

$$ {l_{6,8}} = \prod\limits_{j = 0,j \ne 6}^8 {\frac}  = \frac,\quad f({x_6}) = f(\frac{2}{4}) = 2\left( {{e^{\frac{2}{4}}} - 1} \right), $$

$$ {l_{7,8}} = \prod\limits_{j = 0,j \ne 7}^8 {\frac} = \frac,\quad f({x_7}) = f(\frac{3}{4}) = \frac{4}{3}\left( {{e^{\frac{3}{4}}} - 1} \right),$$

$${l_{8,8}} = \prod\limits_{j = 0,j \ne 8}^8 {\frac} = \frac,\quad f({x_8}) = f(1) = e - 1. $$

When n = 16 $$ \displaystyle x_{0} = -1, x_{1} = -\frac{7}{8}, x_{2} = -\frac{6}{8}, x_{3} = -\frac{5}{8}, x_{4} = -\frac{4}{8}, x_{5} = -\frac{3}{8}, x_{6} = -\frac{2}{8}, x_{7} = \frac{1}{8}, x_{8} = 0, x_{9} = \frac{1}{8}, x_{10} = \frac{2}{8}, x_{11} = \frac{3}{8}, x_{12} = \frac{4}{8}, x_{13} = \frac{5}{8}, x_{14} = \frac{6}{8}, x_{15} = \frac{7}{8}, x_{16} = 1 $$.

$$ l_{0,16} = \prod\limits_{j = 0,j \ne 0}^{16} {\frac} = \frac,\quad f({x_0}) = f( - 1) =  - ({e^{ - 1}} - 1), $$

$${l_{1,16}} = \prod\limits_{j = 0,j \ne 1}^{16} {\frac} = \frac,\quad f({x_1}) = f( - \frac{7}{8}) =  - \frac{8}{7}\left( {{e^{ - \frac{7}{8}}} - 1} \right),$$

and it can be calculate up to $$ l_{16,16} $$ in the same manner.

$$ {l_{16,16}} = \prod\limits_{j = 0,j \ne 16}^{16} {\frac}  = \frac,\quad f({x_{16}}) = f(1) = e - 1, $$

B) Considering the following equation, the approximated integral can be calculated:

$$ {I_n} = \int\limits_a^b {\sum\limits_{i = 0}^n {{l_{i,n}}({x_i}).f({x_i})dx} } $$

note that $$ {E_n} = \left| {I - {I_n}} \right| $$ is the difference between the exact integral $$\displaystyle I $$ and approximated integral $$\displaystyle  I_n $$ The exact integral was calculated using wolframalpha

C) Figure 12.1 shows the $$\displaystyle f(x), f_1(x), f_2(x), f_4(x), f_8(x) for x\in[0,1] $$. As it can be seen, other than $$\displaystyle f_1(x)$$ the rest of the higher order approximations lie on the exact function.

 * [[File:HW2 P12 A.png|1000px|center|thumb|Figure 12.1: $$\displaystyle f(x), f_1(x), f_2(x), f_4(x), f_8(x) for x\in[0,1] $$]]

Figure 12.2 shows the $$\displaystyle f(x), f_1(x), f_2(x), f_4(x), f_8(x) for x\in[-1,1] $$. As it can be seen, other than $$\displaystyle f_1(x)$$ the rest of the higher order approximations lie on the exact function.

 * [[File:HW2 P12 B.png|1000px|center|thumb|Figure 12.2: $$\displaystyle f(x), f_1(x), f_2(x), f_4(x), f_8(x) for x\in[-1,1] $$]]

D) Figure 12.3 shows $$\displaystyle l_{0,5},l_{1,5},l_{2,5} for x\in[0,1] $$]].

 * [[File:HW2 P12 C.png|1000px|center|thumb|Figure 12.3: $$\displaystyle f(x), l_{0,5},l_{1,5},l_{2,5} for x\in[0,1] $$]]

= Problem 2.13: Simple to Composite Trapezoidal and Simpson's Rules =

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

From the lecture slide Mtg 10-5

Given
The simple Trapezoidal rule is

 * {| style="width:100%" border="0"

I_1=\dfrac{b-a}{2}[f(a)+f(b)]. $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">(13.1)
 * }
 * }

The simple Simpson's rule is

 * {| style="width:100%" border="0"

I_2=\dfrac{h}{3}[f(x_0)+4f(x_1)+f(x_2)]. $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">(13.2)
 * }
 * }

Find
Show that the simple Trapezoidal and Simpson's rules can be transformed into their respective composite forms.

The composite Trapezoidal rule is as follows

 * {| style="width:100%" border="0"

I_n=h[\dfrac{1}{2}f_0+f_1+...+\dfrac{1}{2}f_{n-1}]. $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">(13.3)
 * }
 * }

The composite Simpson's rule is as follows

 * {| style="width:100%" border="0"

I_n=\dfrac{h}{3}[f_0+4f_1+2f_2+4f_3+2f_4+...+2f_{n-2}+4f_{n-1}+f_n]. $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">(13.4)
 * }
 * }

Trapezoidal Rule
The Trapezoidal rule, $$\displaystyle n=1 $$, $$\displaystyle h $$ is equal to width of the interval $$\displaystyle [a,b] $$ is

 * {| style="width:100%" border="0"

I_1=\dfrac{x_1-x_0}{2}[f(x_0)+f(x_1)]. $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">(13.5)
 * }
 * }

Next, the Trapezoidal rule, $$\displaystyle n=2 $$ is

 * {| style="width:100%" border="0"

I_2=\dfrac{x_1-x_0}{2}(f(x_0)+f(x_1))+\dfrac{x_2-x_1}{2}(f(x_1)+f(x_2)). $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">(13.6)
 * }
 * }

Combine terms.

 * {| style="width:100%" border="0"

I_2=\dfrac{h}{2}[f(x_0)+2f(x_1)+f(x_2)] $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">(13.7)
 * }
 * }

Next, the Trapezoidal rule, $$\displaystyle n=4 $$ is

 * {| style="width:100%" border="0"

I_4=\dfrac{x_1-x_0}{2}[f(x_0)+f(x_1)]+\dfrac{x_2-x_1}{2}[f(x_1)+f(x_2)]+\dfrac{x_3-x_2}{2}[f(x_2)+f(x_3)]+\dfrac{x_4-x_3}{2}[f(x_3)+f(x_4)]. $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">(13.8)
 * }
 * }

Combine terms.

 * {| style="width:100%" border="0"

I_4=\dfrac{h}{2}[f(x_0)+2f(x_1)+2f(x_2)+2f(x_3)+f(x_4)] $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">(13.9)
 * }
 * }

Finally, the composite Trapezoidal rule is

Simpson's Rule
Similarly, Simpson's rule, $$\displaystyle n=2 $$, $$\displaystyle h $$ is equal to width of the interval $$\displaystyle [a,b] $$ is

 * {| style="width:100%" border="0"

I_2=\dfrac{h}{3}[f(x_0)+4f(x_1)+f(x_2)] $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">(13.11)
 * }
 * }

Next, Simpson's rule, $$\displaystyle n=4 $$ is

 * {| style="width:100%" border="0"

I_4=\dfrac{h}{3}[f(x_0)+4f(x_1)+f(x_2)]+\dfrac{h}{3}[f(x_2)+4f(x_3)+f(x_4)] $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">(13.12)
 * }
 * }

Combine terms.

 * {| style="width:100%" border="0"

I_4=\dfrac{h}{3}[f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+f(x_4)] $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">(13.13)
 * }
 * }

Next, Simpson's rule, $$\displaystyle n=6 $$ is

 * {| style="width:100%" border="0"

I_6=\dfrac{h}{3}[f(x_0)+4f(x_1)+f(x_2)]+\dfrac{h}{3}[f(x_2)+4f(x_3)+f(x_4)]+\dfrac{h}{3}[f(x_4)+4f(x_5)+f(x_6)] $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">(13.14)
 * }
 * }

Combine terms.

 * {| style="width:100%" border="0"

I_6=\dfrac{h}{3}[f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+2f(x_4)+4f(x_5)+f(x_6)] $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">(13.15)
 * }
 * }

Finally, the composite Simpson's rule is

=Problem 2.14: Expand the specific interpolation interval to arbitrary domains=

Refer to lecture slide from [[media:nm1.s11.mtg11.djvu|mtg-11]] for the problem statement and table shown below.

Given
Tables have been established for Lagrangian Interpolation method to acquire weights on different points for integrations like:

Find
Expand the applicable domain from $$\displaystyle [-1,+1]$$ to an arbitrary one, i.e. make the tables established applicable to integrations like:

Solution
Integration such as Eq $$\displaystyle \int_{a}^{b}{f(x)dx}$$ can be transformed to $$\displaystyle \int_{-1}^{1}{f(x)dx}$$ by transformation of variables $$\displaystyle x=x(t)$$such that:

where $$\displaystyle \bar{f}(t)=f\left( x(t) \right){x}'(t)$$.

Note that this transformation requires that $$\displaystyle x=x(t)$$ is a continuously differentiable function and that:

For simplicity, let‘s assume $$\displaystyle x(t)$$ to be a linear function such that:

Combine Eqs(14.1) and Eq(14.2) we have,

And,

Then we can transform the integration $$\displaystyle I(f)=\int_{a}^{b}{f(x)dx}$$ into,

which meets the prerequisite of tables for Lagrangian Interpolation in Mtg11.

=References=

=Contributing Members=

title: User:Egm6341.s11.team2/hwk2


text:
User:Egm6341.s11.team2.franklin/hwk1

=Problem 1.1=

Solution
=Problem 1.2=

Solution
=Problem 1.3 Proving the Integral Mean Value Theorom=

''' I solved this problem without referring to S10 homework. '''

Find
Prove the Integral Mean Value Theorem (IMVT) for

a) $$\displaystyle w\left(x\right) \geq 1, \forall x \in \left[a,b\right]$$ and

b) $$\displaystyle w\left(x\right) < 1, \forall x \in \left[a,b\right]$$

using same technique used in lecture 4 to prove IVMT for $$\displaystyle w\left(x\right) = 1 $$.

Given
IVMT states
 * {| style="width:100%" border="0"

\int_a^b{w\left(x\right)f\left(x\right)}dx = f\left(\xi\right)\int_a^b{w\left(x\right)}dx, \forall x \in \left[a,b\right]. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1.3.1)
 * style= |
 * }

a
Let $$\displaystyle w\left(x\right) \geq 0, \forall x \in \left[a,b\right]. $$

Consider $$\displaystyle w\left(x\right) f\left(x\right) $$ continuos on $$\displaystyle\left[a,b\right] .$$

Let

 * {| style="width:100%" border="0"

m:=min f\left(x\right), \forall x \in \left[a,b\right] $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1.3.2)
 * style= |
 * }

and

 * {| style="width:100%" border="0"

M:=max f\left(x\right), \forall x \in \left[a,b\right] , $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1.3.3)
 * style= |
 * }

such that

 * {| style="width:100%" border="0"

m \leq f\left(x\right) \leq M, \forall x \in \left[a,b\right]. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1.3.4)
 * style= |
 * }

Multiplying Eq. 1.3.3 by $$\displaystyle w\left(x\right) $$

 * {| style="width:100%" border="0"

w\left(x\right) m \leq w\left(x\right) f\left(x\right) \leq  w\left(x\right) M, \forall x \in \left[a,b\right]. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1.3.4)
 * style= |
 * }

Integrating Eq. 1.3.4 over $$\displaystyle \left[a,b\right] $$

 * {| style="width:100%" border="0"

\int_a^b{w\left(x\right) m}dx \leq \int_a^b{w\left(x\right) f\left(x\right)}dx \leq \int_a^b{ w\left(x\right) M}dx. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1.3.5)
 * style= |
 * }

$$\displaystyle m $$ and $$\displaystyle M $$ are both constants and can be pulled out of the integrals.

 * {| style="width:100%" border="0"

m \int_a^b{w\left(x\right)}dx \leq \int_a^b{w\left(x\right) f\left(x\right)}dx \leq M \int_a^b{ w\left(x\right) }dx. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1.3.6)
 * style= |
 * }

Let

 * {| style="width:100%" border="0"

\alpha := \int_a^b{ w\left(x\right) }dx. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1.3.7)
 * style= |
 * }

Divide Eq 1.3.6 by $$\displaystyle \alpha $$

 * {| style="width:100%" border="0"

m \leq \underbrace{\frac{1}{\alpha} \int_a^b{w\left(x\right) f\left(x\right)}dx}_{Z:=} \leq M. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1.3.8)
 * style= |
 * }

By the Intermediate Value Theorem (IVT), $$\displaystyle \exists \xi  \in \left[a,b\right] $$ such that $$\displaystyle f\left(\xi\right)=Z. $$

 * {| style="width:100%" border="0"

f\left(\xi\right)=\frac{1}{\alpha} \int_a^b{w\left(x\right) f\left(x\right)}dx $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1.3.9)
 * style= |
 * }

Therefore,

 * {| style="width:100%" border="0"

$$ \int_a^b{w\left(x\right)f\left(x\right)}dx = f\left(\xi\right)\int_a^b{w\left(x\right)}dx, w\left(x\right) \geq 0. $$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:100%" border="0"|
 * $$\,(Eq. 1..3.10)
 * style= |
 * }

b
Let $$\displaystyle w\left(x\right) < 0, \forall x \in \left[a,b\right]. $$

Consider $$\displaystyle w\left(x\right) f\left(x\right) $$ continuos on $$\displaystyle\left[a,b\right] .$$

Let

 * {| style="width:100%" border="0"

m:=min f\left(x\right), \forall x \in \left[a,b\right] $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1.3.11)
 * style= |
 * }

and

 * {| style="width:100%" border="0"

M:=max f\left(x\right), \forall x \in \left[a,b\right] , $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1.3.12)
 * style= |
 * }

such that

 * {| style="width:100%" border="0"

m \leq f\left(x\right) \leq M, \forall x \in \left[a,b\right]. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1.3.13)
 * style= |
 * }

Multiplying Eq. 1.3.3 by $$\displaystyle w\left(x\right) $$

 * {| style="width:100%" border="0"

$$ $$
 * $$\displaystyle
 * w\left(x\right)| m \leq |w\left(x\right))| f\left(x\right) \leq  |w\left(x\right)| M, \forall x \in \left[a,b\right].
 * w\left(x\right)| m \leq |w\left(x\right))| f\left(x\right) \leq  |w\left(x\right)| M, \forall x \in \left[a,b\right].
 * $$\displaystyle (Eq. 1.3.14)
 * style= |
 * }

Integrating Eq. 1.3.4 over $$\displaystyle \left[a,b\right] $$

 * {| style="width:100%" border="0"

\int_a^b{|w\left(x\right)| m}dx \leq \int_a^b{|w\left(x\right)| f\left(x\right)}dx \leq \int_a^b{ |w\left(x\right)| M}dx. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1.3.15)
 * style= |
 * }

$$\displaystyle m $$ and $$\displaystyle M $$ are both constants and can be pulled out of the integrals.

 * {| style="width:100%" border="0"

m \int_a^b{|w\left(x\right)|}dx \leq \int_a^b{|w\left(x\right)| f\left(x\right)}dx \leq M \int_a^b{ |w\left(x\right)| }dx. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 1.3.16)
 * style= |
 * }

Let

 * {| style="width:100%" border="0"

\alpha := \int_a^b{ |w\left(x\right)| }dx. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 1.3.17)
 * style= |
 * }

Divide Eq 1.3.6 by $$\displaystyle \alpha $$

 * {| style="width:100%" border="0"

m \leq \underbrace{\frac{1}{\alpha} \int_a^b{|w\left(x\right)| f\left(x\right)}dx}_{Z:=} \leq M. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 1.3.18)
 * style= |
 * }

By the Intermediate Value Theorem (IVT), $$\displaystyle \exists \xi  \in \left[a,b\right] $$ such that $$\displaystyle f\left(\xi\right)=Z. $$

 * {| style="width:100%" border="0"

f\left(\xi\right)=\frac{1}{\alpha} \int_a^b{w\left(x\right) f\left(x\right)}dx. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 1.3.19)
 * style= |
 * }

Since $$ \displaystyle w\left(x\right) < 0 ,$$

 * {| style="width:100%" border="0"

f\left(\xi\right)=\frac{1}{ \cancel{\left(-1\right)} \int_a^b{ w\left(x\right) }dx} \cancel{\left(-1\right)} \int_a^b{w\left(x\right) f\left(x\right)}dx. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 1.3.20)
 * style= |
 * }

Therefore,

 * {| style="width:100%" border="0"

$$ \int_a^b{w\left(x\right)f\left(x\right)}dx = f\left(\xi\right)\int_a^b{w\left(x\right)}dx, w\left(x\right) < 0. $$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:100%" border="0"|
 *
$$\,(Eq. 1..3.21)
 * style= |
 * }

Contributing Members

 * Solved and posted by Egm6341.s11.team2.franklin 22:22, 23 January 2011 (UTC)

=Problem 1.4=

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: User:Egm6341.s11.team2.franklin/hwk1


text:
User:Egm6341.s11.team2.franklin/hwk2

= 2.1 Generate Higher Order Terms of Taylor Series =

''' This problem was solved referring to Team 3, S10 homework. '''

From the lecture slide Mtg 6-2

Given
Taylor Series equation below

 * {| style="width:100%" border="0" align="left"

$$ $$
 * $$\displaystyle f(x) = f(x_0) + \frac{(x-x_0)}{1!}\ f^{(1)}(x_0) + \int_{x_0}^{x} (x-t) f^{(2)}(t)\ dt.
 * $$\displaystyle f(x) = f(x_0) + \frac{(x-x_0)}{1!}\ f^{(1)}(x_0) + \int_{x_0}^{x} (x-t) f^{(2)}(t)\ dt.
 * $$\displaystyle (1.1)
 * }
 * }

Find
(1) Repeat integration by parts on the Taylor Series to reveal the following items with remainder.

 * {| style="width:100%" border="0" align="left"

$$ $$
 * $$\displaystyle \frac{(x-x_0)^2}{2!}\ f^{(2)}(x_0) + \frac{(x-x_0)^3}{3!}\ f^{(3)}(x_0)
 * $$\displaystyle \frac{(x-x_0)^2}{2!}\ f^{(2)}(x_0) + \frac{(x-x_0)^3}{3!}\ f^{(3)}(x_0)
 * $$\displaystyle
 * }
 * }

(2) Assume Eqn 3 and Eqn 4 in [[media:nm1.s11.mtg3.djvu|Mtg3-3]] are true, do integration by parts one more time.

1 Integration by Parts
Integrate $$\int_{x_0}^{x} (x-t){f^{(2)}(t)} dt $$ by parts in Eq 1.

 * {| style="width:100%" border="0" align="left"

\int_{x_0}^{x} \underbrace{(x-t)}_{U^\prime} \underbrace{f^{(2)}(t)}_{V}\, dt & = \left [ UV \right ]_{x_0}^x -  \int_{t=x} UV^\prime\, dt \\ & = \left [ -\frac{(x-t)^2}{2}\ f^{(2)}(t) \right ]_{t=x_0}^{t=x} + \underbrace{\int_{x_0}^{x} \frac{(x-t)^2}{2}f^{(3)}(t)\, dt}_{W1} \\ & = \cancelto{o}{-\frac{(x-x)^2}{2}\ f^{(2)}(x)} + \frac{(x-x_0)^2}{2}\ f^{(2)}(x_0) + W1\\ & = \frac{(x-x_0)^2}{2!}\ f^{(2)}(x_0) + W1\\ \end{align} $$ $$
 * $$ \displaystyle \begin{align}
 * $$ \displaystyle \begin{align}
 * $$\displaystyle (1.2)
 * }
 * }

Use Eq 1.1, Eq 1.2, we have
 * {| style="width:100%" border="0" align="left"

$$ $$
 * $$\Rightarrow \ f(x) = f(x_0) + \frac{(x-x_0)}{1!}\ f^{(1)}(x_0) + \frac{(x-x_0)^2}{2!}\ f^{(2)}(x_0) + \frac{1}{2!}\int_{x_0}^{x} {(x-t)^2}f^{(3)}(t)\ dt.
 * $$\Rightarrow \ f(x) = f(x_0) + \frac{(x-x_0)}{1!}\ f^{(1)}(x_0) + \frac{(x-x_0)^2}{2!}\ f^{(2)}(x_0) + \frac{1}{2!}\int_{x_0}^{x} {(x-t)^2}f^{(3)}(t)\ dt.
 * $$\displaystyle (1.3)
 * }
 * }

Integrate $$\int_{x_0}^{x} {(x-t)^2}f^{(3)}(t)\,dt $$ by part one more time in Eq 1.3.
 * {| style="width:100%" border="0" align="left"

\int_{x_0}^{x} \underbrace{(x-t)^2}_{U^\prime} \underbrace{f^{(3)}(t)}_{V}\, dt & = \left [ UV \right ]_{x_0}^x -  \int_{t=x} UV^\prime\, dt \\ & = \left [ -\frac{(x-t)^3}{3}\ f^{(3)}(t) \right ]_{t=x_0}^{t=x} + \underbrace{\int_{x_0}^{x} \frac{(x-t)^3}{3}f^{(4)}(t)\, dt}_{W2} \\ & = \cancelto{o}{-\frac{(x-x)^3}{3}\ f^{(3)}(x)} + \frac{(x-x_0)^3}{3}\ f^{(3)}(x_0) + W2\\ & = \frac{(x-x_0)^3}{3}\ f^{(3)}(x_0) + W2\\ \end{align} $$ $$
 * $$ \displaystyle \begin{align}
 * $$ \displaystyle \begin{align}
 * $$\displaystyle (1.4)
 * }
 * }

Use Eq 1.3 and Eq 1.4, we have

2 Integration by Parts, Again
Int. $$\int_{x_0}^{x} {(x-t)^n}f^{(n+1)}(t)\,dt $$ by part one more time:

 * {| style="width:100%" border="0" align="left"

\int_{x_0}^{x} \underbrace{(x-t)^n}_{U^\prime} \underbrace{f^{(n+1)}(t)}_{V}\, dt & = \left [ UV \right ]_{x_0}^x -  \int_{t=x} UV^\prime\, dt \\ & = \left [ -\frac{(x-t)^{n+1}}{n+1}\ f^{(n+1)}(t) \right ]_{t=x_0}^{t=x} + \underbrace{\int_{x_0}^{x} \frac{(x-t)^{n+1}}{n+1}f^{(n+2)}(t)\, dt}_{W} \\ & = \cancelto{o}{-\frac{(x-x)^{n+1}}{n+1}\ f^{(n+1)}(x)} + \frac{(x-x_0)^{n+1}}{n+1}\ f^{(n+1)}(x_0) + W\\ & = \frac{(x-x_0)^{n+1}}{n+1}\ f^{(n+1)}(x_0) + W\\ \end{align} $$ $$
 * $$ \displaystyle \begin{align}
 * $$ \displaystyle \begin{align}
 * $$\displaystyle (1.6)
 * }
 * }

Use Eqn 1.6, Eqn 3 and Eqn 4 in [[media:nm1.s11.mtg3.djvu|Mtg3-3]], we have

=Problem 2.6=

Given
The equation for the Lagrange Polynomial is,

 * {| style="width:100%" border="0"

{P}_{n}(x) = \sum_{i=0}^{[n/2]} (-1)^{i} \cdot \frac {(2n-2i)! \cdot x^{n-2i}} {2^{n}\cdot i! \cdot (n-i)! \cdot (n-2i)!} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

and the equations for $$\displaystyle P_0\left(x\right) $$ through $$\displaystyle P_4\left(x\right). $$

 * {| style="width:100%" border="0"

\begin{align} &{P}_{0}(x) = 1\\ &{P}_{1}(x) = x\\ &{P}_{2}(x) = \frac {1}{2}(3x^{2}-1)\\ &{P}_{3}(x) = \frac {1}{2}(5x^{3}-3x)\\ &{P}_{4}(x) = \frac {35}{8}x^{4} - \frac {15}{4}x^{2} + \frac {3}{8}\\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Find
Show that the equations for $$\displaystyle P_0\left(x\right) $$ through $$\displaystyle P_4\left(x\right) $$ can be written as $$\displaystyle P_n\left(x\right). $$

Solution
i) n=0
 * {| style="width:100%" border="0"

\begin{align} {P}_{0}(x) &= \sum_{i=0}^{[0/2]} (-1)^{i} \cdot \frac {(2n-2i)! \cdot x^{n-2i}} {2^{n}\cdot i! \cdot (n-i)! \cdot (n-2i)!}\\ &= 1 \cdot \frac {1 \cdot 1} {1\cdot 1 \cdot 1 \cdot 1}\\ {P}_{0}(x)&= 1\\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

ii) n=1
 * {| style="width:100%" border="0"

\begin{align} {P}_{1}(x) &= \sum_{i=0}^{[1/2]} (-1)^{i} \cdot \frac {(2n-2i)! \cdot x^{n-2i}} {2^{n}\cdot i! \cdot (n-i)! \cdot (n-2i)!}\\ &=(-1)^{0} \cdot \frac {2! \cdot x^{1}} {2^{1}\cdot 0! \cdot 1! \cdot 1!} = 1 \cdot \frac {\cancel {2} \cdot x} {\cancel {2} \cdot 1 \cdot 1 \cdot 1} \\ {P}_{1}(x)&= x\\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

iii) n=2
 * {| style="width:100%" border="0"

\begin{align} {P}_{2}(x) &= \sum_{i=0}^{[2/2]} (-1)^{i} \cdot \frac {(2n-2i)! \cdot x^{n-2i}} {2^{n}\cdot i! \cdot (n-i)! \cdot (n-2i)!}\\ &=(-1)^{0} \cdot \frac {(4)! \cdot x^{2}} {2^{2}\cdot 0! \cdot (2)! \cdot (2)!} + (-1)^{1} \cdot \frac {(4-2)! \cdot x^{2-2}} {2^{2}\cdot 1! \cdot (2-1)! \cdot (2-2)!}\\ &= 1 \cdot \frac {4 \cdot 3 \cdot 3 \cdot x^{2}}{4 \cdot 1 \cdot 2 \cdot 2} + (-1) \cdot \frac {2 \cdot 1}{4 \cdot 1 \cdot 1 \cdot 1} = \frac {3}{2}x^{2}-\frac {1}{2})\\ {P}_{2}(x)&= \frac {1}{2}(3x^{2}-1)\\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

iv) n=3
 * {| style="width:100%" border="0"

\begin{align} {P}_{3}(x) &= \sum_{[i=0]}^{[3/2]} (-1)^{i} \cdot \frac {(2n-2i)! \cdot x^{n-2i}} {2^{n}\cdot i! \cdot (n-i)! \cdot (n-2i)!} \\ &=(-1)^{0} \cdot \frac {6! \cdot x^{3}} {2^{3}\cdot 0! \cdot 3! \cdot 3!} + (-1)^{1} \cdot \frac {4! \cdot x^{1}} {2^{3}\cdot 1! \cdot 2! \cdot 1!} \\ &= 1 \cdot \frac {6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot x^{3}} {8 \cdot 1 \cdot 3 \cdot 2 \cdot 3 \cdot 2} + (-1) \cdot \frac {4 \cdot 3 \cdot 2 \cdot x^{1}} {8\cdot 1 \cdot 2 \cdot 1} = \frac {5}{2}x^{3} - \frac {3}{2} x \\ {P}_{3}(x)&= \frac {1}{2}(5x^{3}-3x)\\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

v) n=5
 * {| style="width:100%" border="0"

\begin{align} {P}_{4}(x) &= \sum_{[i]=0}^{[4/2]} (-1)^{i} \cdot \frac {(2n-2i)! \cdot x^{n-2i}} {2^{n}\cdot i! \cdot (n-i)! \cdot (n-2i)!}\\ &= (-1)^{0} \cdot \frac {8! \cdot x^4} {2^{4}\cdot 0! \cdot 4! \cdot 4!} + (-1)^{1} \cdot \frac {6! \cdot x^2} {2^{4}\cdot 1! \cdot 3! \cdot 2!} + (-1)^{2} \cdot \frac {4! \cdot x^0} {2^{4}\cdot 2! \cdot 2! \cdot 0!} \\ &= 1 \cdot \frac {8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot x^4} {16 \cdot 1 \cdot 4 \cdot 3 \cdot 2 \cdot 4 \cdot 3 \cdot 2} + (-1) \cdot \frac {6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot x^2} {16 \cdot 1 \cdot 3 \cdot 2 \cdot 2} + 1 \cdot \frac {4 \cdot 3 \cdot 2 \cdot 1} {16 \cdot 2 \cdot 2 \cdot 1} \\ {P}_{4}(x)&= \frac {35}{8}x^{4} - \frac {15}{4}x^{2} + \frac {3}{8}\\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

=Problem 2.11: Proof of the Simple Simpson's Rule =

Given
The Lagrange Interpretation Equation,

 * {| style="width:100%" border="0"

P_2(x_j)=\sum_{k=0}^2 l_{i,2}(x_j)f(x_i). $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Find
Derive the Simple Simpson's rule from the above Lagrange Interpolation Equation of degree 2. The Simple Simpson's rule is as follows,

 * {| style="width:100%" border="0"

I_2=\dfrac{h}{3}[f(x_0)+4f(x_1)+f(x_2)] , $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

where

 * {| style="width:100%" border="0"

h=\dfrac{b-a}{2}. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Solution
By definition,
 * {| style="width:100%" border="0"

I_2 = \int_{a}^{b} P_2(x)\ dx $$
 * $$ \displaystyle
 * $$ \displaystyle
 *

 * }
 * }

Evaluating the above integral gives

 * {| style="width:100%" border="0"

I_2 & = \int_{a}^{b} P_2(x)\ dx \\ & = \int_{a}^{b} \sum_{i=0}^{2}l_i(x)f(x_i)\ dx \\ & = \int_{a}^{b} l_0(x)f(x_0) + l_1(x)f(x_1) + l_2(x)f(x_2)\ dx \\ & = f(x_0)\int_{a}^{b} l_0(x)\ dx + f(x_1)\int_{a}^{b} l_1(x)\ dx + f(x_2)\int_{a}^{b} l_2(x)\ dx , \end{align} $$
 * $$ \displaystyle \begin{align}
 * $$ \displaystyle \begin{align}
 *
(Eqs11.1-4)
 * }
 * }

where $$ \displaystyle x_0 = a, x_1 = \frac{a+b}{2}, $$ and $$\displaystyle x_2 = b .$$

Next, integrate $$\displaystyle l_0(x), l_1(x),$$ and $$\displaystyle l_2(x). $$

1)$$\displaystyle I_0 $$

 * {| style="width:100%" border="0"

\int_{a}^{b} l_0(x)\ dx & = \int_{a}^{b}\frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}\ dx \\ & = \int_{a}^{b}\frac{(x-\frac{a+b}{2})(x-b)}{(a-\frac{a+b}{2})(a-b)}\ dx \\ & = \frac{2}{(b-a)^2}\left [ \frac{x^3}{3} - \frac{a+3b}{4}x^2 + \frac{b(a+b)}{2}x \right ]_{a}^b \\ & = \frac{2}{(b-a)^2} \left [ \frac{(b-a)(b^2 + ab + a^2)}{3} - \frac{a+3b}{4}(b-a)(b+a) + \frac{b(a+b)}{2}(b-a) \right ] \\ & = \frac{2}{(b-a)} \left [ \frac{(b^2 + ab + a^2)}{3} - \frac{a+3b}{4}(b+a) + \frac{b(a+b)}{2} \right ] \\ & = \frac{(b-a)^2}{6(b-a)} \\ & = \frac{b-a}{6}\\ \end{align} $$
 * $$ \displaystyle \begin{align}
 * $$ \displaystyle \begin{align}
 *
(Eqs11.4-11)
 * }
 * }

2)$$\displaystyle I_1 $$

 * {| style="width:100%" border="0"

\int_{a}^{b} l_1(x)\ dx & = \int_{a}^{b}\frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)}\ dx \\ & = \int_{a}^{b}\frac{(x-a)(x-b)}{(\frac{a+b}{2}-a)(\frac{a+b}{2}-b)}\ dx \\ & = \frac{4}{(b-a)(a-b)}\left [ \frac{x^3}{3} - \frac{a+b}{2}x^2 + abx \right ]_{a}^b \\ & = \frac{4}{(b-a)(a-b)}\left [ \frac{(b-a)(b^2+ab+a^2)}{3} - \frac{(a+b)(a+b)(a-b)}{2} + ab(b-a) \right ]_{a}^b \\ & = \frac{4}{(a-b)}\left [ \frac{(b^2+ab+a^2)}{3} - \frac{(a+b)(a+b)}{2} + ab \right ]_{a}^b \\ & = \frac{2(b-a)^2}{3(b-a)} \\ & = \frac{2(b-a)}{3}\\ \end{align} $$
 * $$ \displaystyle \begin{align}
 * $$ \displaystyle \begin{align}
 *
(Eqs11.11-18)
 * }
 * }

3)$$\displaystyle I_2 $$

 * {| style="width:100%" border="0"

\int_{a}^{b} l_2(x) \ dx & = \int_{a}^{b}\frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}\ dx \\ & = \int_{a}^{b}\frac{(x-a)(x-\frac{a+b}{2})}{(b-a)(b-\frac{a+b}{2})}\ dx \\ & = \frac{2}{(b-a)^2}\left [ \frac{x^3}{3} - \frac{3a+b}{4}x^2 + \frac{a(a+b)}{2}x \right ]_{a}^b \\ & = \frac{2}{(b-a)^2} \left [ \frac{(b-a)(b^2 + ab + a^2)}{3} - \frac{3a+b}{4}(b-a)(b+a) + \frac{b(a+b)}{2}(b-a) \right ] \\ & = \frac{2}{(b-a)} \left [ \frac{(b^2 + ab + a^2)}{3} - \frac{3a+b}{4}(b+a) + \frac{b(a+b)}{2} \right ] \\ & = \frac{(b-a)^2}{6(b-a)} \\ & = \frac{b-a}{6}\\ \end{align} $$
 * $$ \displaystyle \begin{align}
 * $$ \displaystyle \begin{align}
 *
(Eqs11.18-26)
 * }
 * }

Substituting the values for the integrals of $$\displaystyle l_0(x), l_1(x),$$ and $$\displaystyle l_2(x). $$ into Eqn 11.4 gives

 * {| style="width:100%" border="0"

\begin{align} I_2 & = f(x_0)\frac{b-a}{6}+f(x_1)\frac{2(b-a)}{3}+f(x_2)\frac{b-a}{6} \\ & = \frac{b-a}{6}[f(x_0)+4f(x_1)+f(x_2)]\\ \Rightarrow I_2 & = \frac{h}{3}[f(x_0)+4f(x_1)+f(x_2)], \\ \end{align} $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }

where

 * {| style="width:100%" border="0"

h = \frac{b-a}{2}. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle
 * }
 * }

= Problem 2.13: Simple to Composite Trapezoidal and Simpson's Rules =

Given
The simple Trapezoidal rule is

 * {| style="width:100%" border="0"

I_1=\dfrac{b-a}{2}[f(a)+f(b)]. $$
 * $$\displaystyle
 * $$\displaystyle
 *
(Eq 13.1)
 * }
 * }

The simple Simpson's rule is

 * {| style="width:100%" border="0"

I_2=\dfrac{h}{3}[f(x_0)+4f(x_1)+f(x_2)]. $$
 * $$\displaystyle
 * $$\displaystyle
 *
(Eq 13.2)
 * }
 * }

Find
Show that the simple Trapezoidal and Simpson's rules can be transformed into their respective composite forms.

The composite Trapezoidal rule is as follows

 * {| style="width:100%" border="0"

I_n=h[\dfrac{1}{2}f_0+f_1+...+\dfrac{1}{2}f_{n-1}]. $$
 * $$\displaystyle
 * $$\displaystyle
 *
(Eq 13.3)
 * }
 * }

The composite Simpson's rule is as follows

 * {| style="width:100%" border="0"

I_n=\dfrac{h}{3}[f_0+4f_1+2f_2+4f_3+2f_4+...+2f_{n-2}+4f_{n-1}+f_n]. $$
 * $$\displaystyle
 * $$\displaystyle
 *
(Eq 13.4)
 * }
 * }

Trapezoidal Rule
The Trapezoidal rule, $$\displaystyle n=1 $$, $$\displaystyle h $$ is equal to width of the interval $$\displaystyle [a,b] $$ is

 * {| style="width:100%" border="0"

I_1=\dfrac{x_1-x_0}{2}[f(x_0)+f(x_1)]. $$
 * $$\displaystyle
 * $$\displaystyle
 *
(Eq 13.5)
 * }
 * }

Next, the Trapezoidal rule, $$\displaystyle n=2 $$ is

 * {| style="width:100%" border="0"

I_2=\dfrac{x_1-x_0}{2}[f(x_0)+f(x_1))+\dfrac{x_2-x_1}{2}(f(x_1)+f(x_2)]. $$
 * $$\displaystyle
 * $$\displaystyle
 *
(Eq 13.6)
 * }
 * }

Combine terms.

 * {| style="width:100%" border="0"

I_2=\dfrac{h}{2}[f(x_0)+2f(x_1)+f(x_2)] $$
 * $$\displaystyle
 * $$\displaystyle
 *
(Eq 13.7)
 * }
 * }

Next, the Trapezoidal rule, $$\displaystyle n=4 $$ is

 * {| style="width:100%" border="0"

I_4=\dfrac{x_1-x_0}{2}[f(x_0)+f(x_1))+\dfrac{x_2-x_1}{2}(f(x_1)+f(x_2))+\dfrac{x_3-x_2}{2}(f(x_2)+f(x_3))+\dfrac{x_4-x_3}{2}(f(x_3)+f(x_4)]. $$
 * $$\displaystyle
 * $$\displaystyle
 *
(Eq 13.8)
 * }
 * }

Combine terms.

 * {| style="width:100%" border="0"

I_4=\dfrac{h}{2}[f(x_0)+2f(x_1)+2f(x_2)+2f(x_3)+f(x_4)] $$
 * $$\displaystyle
 * $$\displaystyle
 *
(Eq 13.9)
 * }
 * }

Finally, the composite Trapezoidal rule is

Simpson's Rule
Similarly, Simpson's rule, $$\displaystyle n=2 $$, $$\displaystyle h $$ is equal to width of the interval $$\displaystyle [a,b] $$ is

 * {| style="width:100%" border="0"

I_2=\dfrac{h}{3}[f(x_0)+4f(x_1)+f(x_2)] $$
 * $$\displaystyle
 * $$\displaystyle
 *
(Eq 13.11)
 * }
 * }

Next, Simpson's rule, $$\displaystyle n=4 $$ is

 * {| style="width:100%" border="0"

I_4=\dfrac{h}{3}[f(x_0)+4f(x_1)+f(x_2))+\dfrac{h}{3}(f(x_2)+4f(x_3)+f(x_4)] $$
 * $$\displaystyle
 * $$\displaystyle
 *
(Eq 13.12)
 * }
 * }

Combine terms.

 * {| style="width:100%" border="0"

I_4=\dfrac{h}{3}[f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+f(x_4)] $$
 * $$\displaystyle
 * $$\displaystyle
 *
(Eq 13.13)
 * }
 * }

Next, Simpson's rule, $$\displaystyle n=6 $$ is

 * {| style="width:100%" border="0"

I_6=\dfrac{h}{3}[f(x_0)+4f(x_1)+f(x_2))+\dfrac{h}{3}(f(x_2)+4f(x_3)+f(x_4))+\dfrac{h}{3}(f(x_4)+4f(x_5)+f(x_6)] $$
 * $$\displaystyle
 * $$\displaystyle
 *
(Eq 13.14)
 * }
 * }

Combine terms.

 * {| style="width:100%" border="0"

I_6=\dfrac{h}{3}[f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+2f(x_4)+4f(x_5)+f(x_6)] $$
 * $$\displaystyle
 * $$\displaystyle
 *
(Eq 13.15)
 * }
 * }

Finally, the composite Simpson's rule is

title: User:Egm6341.s11.team2.franklin/hwk2


text:
User:Egm6341.s11.team3.russo

This is my Numerical Methods wikiversity page. Below are my contributions to team assignments.

=Homework 1 due Wed Jan 26=

Given:

 * {| style="width:100%" border="0"

$$f(x)=3x-2x^3\ $$ (4.1)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$g(x)=sinh(x)=\frac{1}{2}\left(e^{x}-e^{-x}\right)$$ (4.2)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Find:
A) Plot f(x) and g(x)

B) Find the following norms:

 * {| style="width:100%" border="0"

$$||f(x)||_{\infty}, ||g(x)||_{\infty}, ||f(x) - g(x)||_{\infty}$$
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Solution
A) Equations (4.1) and (4.2) are plotted over the domain of x=[0,1] as shown below:

The difference between these two functions will be called Y(x). It is defined below as:

 * {| style="width:100%" border="0"

$$Y(x) = f(x)-g(x)=[3x-2x^3]-[sinh(x)]=3x-2x^3-\frac{1}{2}\left(e^{x}-e^{-x}\right)$$ (4.3)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

The plot of equation (4.3) over the domain of x=[0,1] is shown below:

B) The infinity norms of these three equations using the restricted domain of x=[0,1] are found below:

The infinity norm of a function is simply the maximum value the function achieves over a specified domain. All three of the functions plotted above will approach infinity over the unrestricted domain of $$x=(-\infty,+\infty)$$.

However, since we are restricting the domain to x=[0,1], these norms must be calculated using techniques discussed below:

B.1) Upon observing the plot of equation (4.1), it is obvious that the maximum value of f(x) occurs at the vertex.  So, the infinity norm will be found by solving for the vertex.  This is accomplished by finding the point of the function where the derivative is zero, since f(x) is not changing at the vertex:

 * {| style="width:100%" border="0"

$$f(x)=3x-2x^3\ $$ (4.1)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$ \frac{df(x)}{dx} = (1)3x^{(1-1)} - (3)2x^{(3-1)} = 3-6x^2$$ (4.4)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Solving equation (4.4) for x when the derivative of f(x) is zero yields the location of the vertex:
 * {| style="width:100%" border="0"

$$ \frac{df(x_{vertex})}{dx} = 3-6x_{vertex}^2=0$$ (4.5)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$ x_{vertex} = \sqrt{\frac{(0 - 3)}{-6}} = \frac{\sqrt{2}}{2}$$ (4.6)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Substituting this value of x back into equation (4.1) will yield the value of f(x) at the vertex, which is also the infinity norm on x=[0,1]:

 * {| style="width:100%" border="0"

$$ ||f(x)||_{\infty} = f(x_{vertex}) = 3\left(\frac{\sqrt{2}}{2}\right)-2\left(\frac{\sqrt{2}}{2}\right)^3 = \sqrt{2}$$ (4.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

B.2) Upon observing the plot of equation (4.2), it is obvious that the maximum value over the prescribed domain occurs at the right boundary of the domain, or when x=1.  So, the infinity norm may be found simply by substituting this value of x directly into the function g(x):

 * {| style="width:100%" border="0"

$$g(x)=sinh(x)=\frac{1}{2}\left(e^{x}-e^{-x}\right)$$ (4.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$ ||g(x)||_{\infty} = g(1) = sinh(1) = \frac{1}{2}\left(e^{(1)}-e^{-(1)}\right) = 1.1752$$ (4.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

B.3) Upon observing the plot of Y(x), which is defined above by equation (4.3), it is once again obvious that the maximum value of Y(x) occurs at the vertex.  So, the infinity norm will again be found by solving for the vertex:

 * {| style="width:100%" border="0"

$$Y(x)=3x-2x^3-\frac{1}{2}\left(e^{x}-e^{-x}\right)$$ (4.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$ \frac{dY(x)}{dx} = (1)3x^{(1-1)} - (3)2x^{(3-1)} - \frac{1}{2}\left((1)e^{x}-(-1)e^{-x}\right) = 3-6x^2 - \frac{1}{2}\left(e^{x}+e^{-x}\right)$$ (4.9)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Solving equation (4.9) for x when the derivative of Y(x) equals zero yields the location of the vertex:
 * {| style="width:100%" border="0"

$$ \frac{dY(x_{vertex})}{dx} = 3-6x_{vertex}^2 -\frac{1}{2}\left(e^{x_{vertex}}+e^{-x_{vertex}}\right)=0$$ (4.10)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

This equation cannot be solved analytically. An iterative approach was used, and the solution verified by observation of the plot above:

 * {| style="width:100%" border="0"

$$ x_{vertex} = 0.5541\ $$ (4.11)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Substituting this value of x back into equation (4.3) will yield the value of Y(x) at the vertex, which is also the infinity norm on x=[0,1]:

 * {| style="width:100%" border="0"

$$ ||f(x)-g(x)||_{\infty}=||Y(x)||_{\infty} = Y(x_{vertex}) = 3(0.5541)-2(0.5541)^3-\frac{1}{2}\left(e^{(0.5541)}-e^{-(0.5541)}\right) = 0.7392$$ (4.12)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Thus, each norm has been found and the problem is complete:

When the domain is unrestricted, $$x=(-\infty,+\infty)$$ and the norms are:
 * {| style="width:100%" border="0"

$$||f(x)||_{\infty} = ||g(x)||_{\infty} = ||f(x) - g(x)||_{\infty} = \infty $$ (4.13)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

When the domain is restricted to $$x=[0,1]\ $$, the norms become:

Solution Part 2:
All of the necessary information is given in the problem statement. Therefore, all that is left to do is perform the calculations using the proposed methods. The same MatLAB code from Solution Part 1 may be used, with the exception of the file main.m. This file will be replaced by mainB.m as shown below, where the only difference between these files is that mainB.m uses the angle of attack governed by Equation (7.2.2.B), and mainB does not calculate J or ode45.

Below are the new resulting state variable plots, followed by the new M-File that produced them:

title: User:Egm6341.s11.team3.russo


text:
User:Egm6341.s11.team2.zou

= Self Introduction =

Hi, I’m Xiangyu Zou, an international student from China, from Mechanical and Aerospace Department, pursuing Master’s degree. My study interests are broad, not particular in a chosen field since I’m just a would-be Master – I have no idea whether to write a thesis because I just don’t have an idea. But this semester I want to get myself some knowledge about calculation and mechanics. So I have chosen both NM1 and FE1 and Continuum Mechanics. I’ve taken prof’s course “Principle of Engineering Analysis 1” last semester so I know how to type Mediawiki and Latex. I’m inclined to use MS Word2007 + MathType + some Macros. I think it’s efficient and easy to learn to use. And I have fundamentals in solving ODEs. However my knowledge in FE and NM is very limited, and my coding skills are modest, my English is very poor too. So, I hope we can help each other and have a nice collaboration. By the way, my given name “Xiangyu” is very difficult to pronounce, so if you’re my teammates please just call me my last name “Zou” or anything except “Xiangyu”. Thank you!

= Useful Links =

Lecture Plan

Course Wiki

Team2 info

Math Templates

Team Format

= Homework =

Homewoek#1

Homewoek#2

Homewoek#3

Homewoek#4

Homewoek#5

Homewoek#6

Homewoek#7

title: User:Egm6341.s11.team2.zou


text:
Conservation Entrepreneurship/Acre case study

this is a test


text:
User:Eml5526.sp11.team1.adam

Hi all, my name is Cedric Adam from team 1

Useful Links - Course Wiki
Eml5526 Course Wiki

Lecture Plan
Lecture Plan

Homework - Report Table
Report table

Group Page
Team 1 Page

Homework Figures
[| Homework Figures]

Homeworks
[Homework 2]

[Homework 3]

[Homework 4]

[Homework 5]

title: User:Eml5526.sp11.team1.adam


text:
User talk:MikeLynch

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title: User talk:MikeLynch


text:
User:Sujin

=Spring 2011 Team 1 Homework Archived= Homework 1 Homework 2 Homework 3 Homework 4 Homework 5 Homework 6 Temp

Wiki Editing Help Link: Help:Editing Wiki Formula Editing Help Link: Help:Formula How to link the archived version of your HW report to the Report table. https://docs.google.com/Doc?docid=0AY5m1lfNbLroZGM4Mm5mZ2JfMTQzZDVkNnFkZGc&hl=en link to "ideal" reports http://en.wikiversity.org/wiki/User:Egm6341.s10/Report_table/Good_model_reports Link to class HW Report Table for Fall 2011 http://en.wikiversity.org/wiki/User:Egm6341.s11/Report_table

Homework 1.4

title: User:Sujin


title: File:DRB1-0101.gif


title: File:DRB1-0102.gif


title: File:DRB1-0103.gif


title: File:DRB1-0301.gif


text:
User:Eml5526.s11.team2.reiss

eml5526.s11.team2.reiss.HW

eml5526.s11.team2.reiss.HW3 HW3

title: User:Eml5526.s11.team2.reiss


title: File:DRB1-0302.gif


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text:
User:Egm5526.s11.team2.stewart

hello

fasdf

=Title= {x^2}=1

title: User:Egm5526.s11.team2.stewart


title: File:DRB1-0417.gif


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title: File:DFT.4.A.Frequency.pdf


title: File:DRB1-0811.gif


title: File:DRB1-0818.gif


text:
User:Eml5526.s11.team2.stewart

Homework 1

Homework 2

Homework 3

Homework 5

Tim Stewart: Graduated from NC State in 2009 with a B.S. in Mechanical Engineering. Currently works at Siemens Energy, Inc. in the Gas Turbine Casings Department. Pursuing a masters degree in Mechanical Engineering via the UF EDGE program.

title: User:Eml5526.s11.team2.stewart


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text:
User:Egm6341.s11.team5.shin

= Links to each homework = Homework #1

Homework #2

Homework #3

Homework #4

Test page

= Link to Team page = Team5 Page

title: User:Egm6341.s11.team5.shin


text:
User talk:Wuhhany

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How to Transfer Music from iPod to iPod easily with Macsome iPod Transfer
Transferring music from one iPod to another iPod can be very troublesome. You may think it is impossible without sacrificing one iPod files, or maybe we can get the music we want on one iPod from the website, and then sync the music to our iPod through our own iTunes, and something like that.

Things can be really simpler and easier. Here I want to tell you an easy way to transfer music from iPod to iPod on one Mac, and ensure both iPod's contents to be security. It is only a matter of a few mouse clicks.

The secret lies in the use of an excellent iPod Transfer. Nowadays, various iPod Transfers springs up like mushrooms. Some of them are freeware, some are shareware, some full-featured, some exclusive, some powerful and some practical. Here I will take Macsome iPod Transfer as an example to illustrate how to transfer music from iPod to iPod in detail.

Step1. Free download Macsome iPod Transfer for Mac, install and launch it.

If interested, download Macsome iPod Transfer Here. The intuitive interface of Macsome iPod Transfer makes its all functions clear and obvious to us, so don't worry that we cannot handle this software.

Step2. Connect iPods to Mac.

Both iPods will be detected by Macsome iPod Transfer, and show up on the software's left-hand main menu. At this time, we can manage the contents of iPods at will.

Step3. Open the music library of the iPod from which you want to transfer music, and select the music need transferring.

Step4. Drag and drop the selected music to the playlist or library of the other iPod where you want to put the transferred music.

Simple, isn't it?

What's more, with an excellent iPod Transfer, we can also transfer music from iPod to iPhone, iPod to Mac, iPod to iTunes, iPhone to iPod, iPhone to iPhone, iPhone to iTunes, iPhone to Mac, etc.

The advanced technology makes nearly every thing possible. What we can do is to make a full use of these advanced technology, and make our life much easier and better.

Tips 1. In step 1, the downloaded trial version of Macsome iPod Transfer is just a trial version with the limitation that we can only transfer one music at a time form one iPod to another iPod. However, we can remove this limitation by purchasing a full version.

2. You can use the searching box on the upper right of Macsome iPod Transfer to quickly locate the music which you want to transfer.

Warnings Please make sure that your iTunes is closed when you run Macsome iPod Transfer for Mac

title: User talk:Wuhhany


text:
User talk:DM2088

I believe in opensource!

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title: User talk:DM2088


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User talk:Dawid Deutschland

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Opriter
You requested deletion of User:Opriter. If you are Opriter, you should probably indicate which account you wish to use; Opriter has contributions. You should also, then, log in as Opriter and confirm that user page deletion request with that account. Otherwise, it is improper to request deletion of a user page for a different user. If you have any questions or concerns, you may ask on my Talk page, or you may email me through the Wikiversity interface. Thanks for your interest in Wikiversity. --Abd 13:46, 19 January 2011 (UTC)

title: User talk:Dawid Deutschland


text:
User talk:Egm5526.s11.team2.stewart

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title: User talk:Egm5526.s11.team2.stewart


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User talk:Eml5526.s11.team1.Hao

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title: User talk:Eml5526.s11.team1.Hao


text:
User talk:Eml5526.s11.team2.stewart

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You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --Kevin Rutherford (talk) 15:48, 19 January 2011 (UTC) Kevin Rutherford (talk) 15:48, 19 January 2011 (UTC)

File:Heat Balance.jpg & File:Heat Balance 2.jpg
Hi! Thanks for uploading the files File:Heat Balance.jpg and File:Heat Balance 2.jpg. Unfortunately, they aren't currently in use and don't have any licensing information, so as part of a cleanup I've tagged them to be removed. If you would like to keep the files here, though, that would be great - all you need to do is add an appropriate license if the copyright belongs to you, or if it is otherwise available under a compatible free license. If you need any help, just ask - I'm very happy to assist. - Bilby 02:02, 5 October 2011 (UTC)

Files Missing License Information
Thank you for uploading files to Wikiversity. See Media for copyright and license requirements for Wikiversity files. All files must have copyright and/or license information added to the file.

Instructions for adding copyright and/or license information are available at License tags. Files must be updated within seven days or they may be removed without further notice. See Requests For Deletion for more information.

The following files are missing copyright and/or license information:
 * File:Heat Balance 3.jpg
 * File:Trap rect.png
 * File:Wikipic.jpg

MaintenanceBot (discuss • contribs) 03:46, 11 November 2013 (UTC)

title: User talk:Eml5526.s11.team2.stewart