User:Egm6341.s11.team2.cleveland/HW2

=Problem 2.5 Gaussian Quadrature= Refer to lecture slide from [[media:nm1.s11.mtg7.djvu|mtg-7]] for the problem statement and table shown below.

Given
The Legendre Polynomians $$ P_{n}(x) $$ defined here
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$$ \displaystyle {P}_{n}(x) = \sum_{i=0}^{\lfloor n/2 \rfloor} (-1)^{i} \cdot \frac {(2n-2i)! \cdot x^{n-2i}} {2^{n}\cdot i! \cdot (n-i)! \cdot (n-2i)!} $$      (5.1)
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Verify
The points $$ \displaystyle x_i $$ and weights $$\displaystyle w_i $$ for $$\displaystyle n=1,2,3,4,5 $$.

The table above was found at [|Gaussian_quadrature], a companion wiki site.

Solution
As was discussed in class during [[media:nm1.s11.mtg7.djvu|mtg-7]], $$ \displaystyle x_i $$ is the root of the $$ n-th $$ order Legendre Polynomial $$ \displaystyle P_{n}(x) $$ defined in (5.1). Using (5.1) we obtain the following for $$ n=1,2,3,4,5 $$

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

$${{W}_{1}}=\frac{-2}{(5+1)[P_{5}^{'}({{x}_{1}}){{P}_{6}}({{x}_{1}})]}=\frac{128}{225}$$ $${{W}_{2}}=\frac{-2}{(5+1)[P_{5}^{'}({{x}_{2}}){{P}_{6}}({{x}_{2}})]}=\frac{322-13\sqrt{70}}{900}$$ $${{W}_{3}}=\frac{-2}{(5+1)[P_{5}^{'}({{x}_{3}}){{P}_{6}}({{x}_{3}})]}=\frac{322-13\sqrt{70}}{900}$$ $${{W}_{4}}=\frac{-2}{(5+1)[P_{5}^{'}({{x}_{4}}){{P}_{6}}({{x}_{4}})]}=\frac{322+13\sqrt{70}}{900}$$ $${{W}_{5}}=\frac{-2}{(5+1)[P_{5}^{'}({{x}_{5}}){{P}_{6}}({{x}_{5}})]}=\frac{322+13\sqrt{70}}{900}$$

=Problem 2.1 Taylor's Series Proof=

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