User:Egm6322.s12.team2/HW13

= R 13.1 - Continued Fraction of Square Root of 5 =

Given
The problem gives the number 5 for computation.

Find
Find the continued fraction of $$\sqrt{5}$$

From lecture notes [[media:Pea2.s12.sec48.f.djvu|Mtg 48f]]

Solution
The general continued fraction is defined in lecture 48e.

$$b_0 + \frac{a_1}{\displaystyle b_1 + \frac{a_2}{\displaystyle b_2 + \frac{a_3}{\displaystyle b_3 + \frac{a_4}{\displaystyle b_4 + \frac{a_5}{\ddots}}}}}$$

The preferred method of documentation for continued fractions is in the more succint form.

$$b_0 + \frac{a_1}{b_1 + } \frac{a_2}{b_2 +} \frac{a_3}{b_3 + } \frac{a_4}{b_4 +} \frac{a_5}{\ddots}$$

In order to find the square root of 5, the team must identify each term $$a_i$$ and $$b_i$$. The lecture notes describe the method for defining the square root of 2 with continued fractions. The same technique can be applied to the square root of 5. The key is to use the recurrence relation.

$$x_{x-1} = b_{k-1} x_k + a_k x_{k+1}$$

Let's focus on the first two terms for i=0 and i=1.

$$x_0 = b_0 x_1 + a_ 1x_2$$

The actual square root value of 5 is approximately 2.36068. So by analysis $$b_0 = 2$$. Furthermore, the following patterns arise when applying the technique from the lecture where $$a_1 = 1$$.

$$x_0 = 2 x_1 + x_2$$

The research utilized the wikipedia source below. http://en.wikipedia.org/wiki/Square_root_of_5

In general, the following patterns arise for i = 1, 2, 3, ..., etc.

$$a_i = 1$$

$$b_i = 4$$

Therefore, the fractional expression is below.

$$\begin{alignat} {3} \sqrt{5} & = 2 + \frac{1}{4 + } \frac{1}{4 + } \frac{1}{4 + } \frac{1}{4 + } \frac{1}{4 + }{\ddots} \\ & = \{2,4,4,4,...\} \\ & =: [2;\overline{4}] \end{alignat}$$

Manuel Steele Egm6322.s12.team2.steele.m2 (talk) 19:33, 11 April 2012 (UTC) The following part added by Lang Xia

$$\displaystyle \sqrt{5}=2+(\sqrt{5}-2)=2+\frac{1}{\sqrt{5}+2}=2+\frac{1}{4+(\sqrt{5}-2)}=2+\frac{1}{4+\frac{1}{\sqrt{5}+2}}$$

$$\displaystyle =2+\frac{1}{4+\frac{1}{4+\frac{1}{4+\frac{1}{4+\frac{1}{\ddots}}}}} $$

thus

$$\displaystyle \sqrt{5}=2+\frac{1}{4+}\frac{1}{4+}\frac{1}{4+}\frac{1}{\ddots}=\{2:\overline{4}\}$$

= R 13.2 - Method of Successive Substractions versus Wallis's Theorem =

Given

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$$  \frac{x_{k-1}}{x_k}=b_{k-1}+ \frac{1}{\displaystyle \frac{x_k}{x_{k+1}}}, \text{for} k=1,2,...$$ (13.2.1)
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$$  \frac{\pi}{2}=\frac{2\cdot2\cdot2\cdot4\cdot6\cdot6\cdot8\cdot8\cdots}{1\cdot3\cdot3\cdot5\cdot5\cdot7\cdot7\cdot9\cdots}$$ (13.2.2)
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Find
$$A. \ Show \ that \ {\pi} \ can \ be \ approximated \ successively \ by \ \{3\},\{3,7\},\{3,7,15\},\{3,7,15,1\}$$

$$B. \ Generate \ a \ sequence \ of \ rational \ numbers \ that \ appriximate \ {\pi} \ from \ Wallis's \ Theorem$$

$$C. \ Compare \ the \ rate \ of \ convergence \ of \ the \ continued \ fraction \ versus \ that \ of \ the \ Wallis \ Sequence$$

Solution
$$ Let \ assume \ that \ {\pi} \ {\asymp} \ 3.141592 $$

$$ A. $$

$$ When \ n = 0 $$

$${\pi} = 3 + (3- {\pi}) $$

$$ When \ n = 1 $$

$${\pi} = 3 + \frac{1}{\frac{106}{141592}}=3+\frac{1}{7+\frac{8856}{141592}}$$

$$ When \ n = 2 $$

$${\pi} = 3+\frac{1}{7+\frac{1}{\frac{141592}{8856}}}= 3 + \frac{1}{7+\frac{1}{15+{\frac{8752}{8856}}}}$$

$$ When \ n = 3 $$

$${\pi} = 3 + \frac{1}{7+\frac{1}{15+\frac{1}{1+\frac{104}{8752}}}}$$

$$Therefore,$$

$$\{3\},\{3,7\},\{3,7,15\},\{3,7,15,1\}$$

$$B. \And C.$$

$$ Convergence \ of \ the Continued Fraction$$

$$when \ n=0$$

$$C_{0} = 3$$

$$when \ n=1$$

$$C_{1} = 3 + \frac{1}{7}=\frac{22}{7}=3.1428$$

$$when \ n=2$$

$$C_{2} = 3+\frac{1}{7+\frac{1}{15}}= 3+ \frac{15}{106}=\frac{333}{106}=3.1415$$

$$when \ n=3$$

$$C_{3} = 3+\frac{1}{7+\frac{1}{15+\frac{1}{1}}}=3+\frac{16}{113}=\frac{355}{113}=3.1415$$

$$ Wallis's \ Theorem $$

$$when \ n=0$$

$$(\frac{\pi}{2})=\frac{2}{1}=2 \Rightarrow {\pi}=4 $$

$$when \ n=1$$

$$(\frac{\pi}{2})=\frac{2 \cdot 2}{1 \cdot 3 }=\frac{4}{3}=1.3333 \cdots \Rightarrow {\pi}=2.6666 \cdots $$

$$when \ n=2$$

$$(\frac{\pi}{2})=\frac{2 \cdot 2 \cdot 4}{1 \cdot 3 \cdot 3}=\frac{16}{9}=1.7777 \cdots \Rightarrow {\pi}=3.555 \cdots $$

$$when \ n=3$$

$$(\frac{\pi}{2})=\frac{2\cdot 2 \cdot 4 \cdot 4}{1 \cdot 3 \cdot 3 \cdot 5}=\frac{64}{45}=1.4222 \cdots \Rightarrow {\pi}=2.8444 \cdots $$

Egm6322.s12.sungsik (talk) 13:29, 11 April 2012 (UTC)

= R 13.3 - Relative convergence of power series and continued fraction expansions=

Given
Power series expansion of $$\tan x$$:
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$$  \displaystyle \tan x = \sum_{n=1}^\infty \frac{(-1)^{n-1}2^{2n}(2^{2n}-1)B_{2n}}{(2n)!}x^{2n-1}, x \in ]-\pi/2,\pi/2[ $$     (13.3.1)
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in which $$B_n$$ are the Bernoulli numbers:


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$$  \displaystyle \begin{align} B_1 &= -\frac{1}{2}, B_{2n+1} = 0, n = 1,2,\dots\\ B_0 &= 1, B_2 = \frac{1}{6}, B_4 = -\frac{1}{30}, B_6 = \frac{1}{42}, B_8 = -\frac{1}{30}, \dots \end{align} $$     (13.3.2)
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Continued fraction expansion of $$\tan x$$:
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$$  \displaystyle \tan x = \frac{x}{1-}\frac{x^2}{3-}\frac{x^2}{5-}\frac{x^2}{7-}\frac{x^2}{\ddots} $$     (13.3.3)
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Find
Take $$x=\pi/4$$ and compare the term by term convergence of the power series (Eq. 13.3.1) and the continued fraction (Eq. 13.3.3). Provide a plot to visualize the convergence.

Solution
The power series expansion of a function is a standard approximation technique although it does not necessarily show the fastest convergence. The expression for the expansion of $$\tan x$$ is given in Eqn 13.3.1 and is seen to be a function of the Bernoulli numbers. These values are provided in Eq. 13.3.2 and can also be found on Wolfram Alpha. The term-by-term expansion of the function can be created in Matlab very easily by looping over the number of desired terms and summing the values generated by Eq. 13.3.1. This can be seen in the Matlab script provided below.

Creating the continued fraction expansion of $$\tan x$$ is slightly more complicated as it involves a new loop for a given number of terms. Thus, in the Matlab script provided below, the j index corresponds to the desired number of terms and the i index computes the necessary terms for that expansion. It is also necessary to provide an if statement to distinguish between the first term of Eq. 13.3.3, in which the numerator is $$x$$ instead of $$x^2$$.

The convergence for $$x = \pi/4$$ is shown below in the left figure. It can be seen that both techniques have the same initial value $$\tan x = x = \pi/4$$, but the continued fraction expansion converges more rapidly than the power series expansion. The second term of the continued fraction expansion is within the 5\% bounds of the exact solution while the power series is not quite as accurate. Within 4 terms, both solutions are quite accurate.

A potential explanation for the rapid convergence of both methods in the problem is the simple nature of the x-value, as $$\tan(pi/4)=1$$. To investigate this, a different x-value of $$x=\pi/4 + \sqrt(pi/11)$$ was selected; the squared term on the right hand side is small enough that the value still fits within the bounds of $$x\in ]-\pi/2,\pi/2[$$, but does not have a simple solution for $$\tan x $$. No other parameters in the code need to be changed, and the result is seen in the figure on the right. Again, both methods have the same initial condition. The continued fraction expansion requires only 3 terms to converge within the 5\% bounds and additional terms indicate no discernible difference from the exact solution; however, the power series expansion requires 3 times as many terms to converge to the 5\% bounds and still does not appear to be completely converged after 10 terms. This indicates that the continued fraction expansion is more robust for a wide variety of input conditions.



Matt Shields 18:03,6 April 2012 (UTC)

= R 13.4 - Cumulative Distribution Function and Derivative=

Given
We are given formulae on p.48h31 and p. 48h33 for (1) the cumulative distribution function (CDF) and (2) the probability distribution function (pdf).

(1) $$F_X(x) = P(x<=X) = \frac{1}{2}[1 + erf(\frac{x-\mu}{\sigma\sqrt{2}})]$$

(2) $$f_x(x) = \frac{1}{\sqrt{2\pi\sigma^2}}{\rm exp} \left[{-\frac{(x-\mu)^2}{2\sigma^2}}\right]$$

Find
Consult the definition of the error function erf and show that (1) can be obtained from (2) and vice versa. From lecture notes [[media:Pea2.s12.sec48.h.djvu|Mtg 48h]]

Solution
Step 1: Go To Wikipedia Page for Error Function

The equation from the wikipedia page is at http://en.wikipedia.org/wiki/Error_function

Error Function: $$erf(x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{-t*t}dt$$

Step 2: Evaluate Error Function for Given Integrand

This integral is a simple case of defining the argument and bounds of the integration. But first, the characteristics of the underlying theorem must be discussed. The normal distribution for a random variable X has a mean and variance. The maximum value occurs at the mean, and the width of the curve is given by the variance. The square root of the variance defines the standard deviation from the mean. In the case of the "bell curve" or pdf the following parameters are standard.

<p style="text_align:center;">mean = $$\mu$$ and variance = $${\sigma}^2$$

Set argument of integrand<p style="text_align:center;"> $$ t = \frac{(x-u)}{\sigma\sqrt{2}}$$

Analyze integral for CDF. <p style="text_align:center;">$$F_X(x) = \frac12\left[1 + \operatorname{erf} \left(\frac{x-\mu}{\sigma\sqrt{2}} \right) \right]$$

The key is to equate the error function with the pdf as shown in the cdf:

<p style="text_align:center;">$$t = \frac{(x - \mu)}{\sigma \sqrt{2}}$$

<p style="text_align:center;">$$dt = \frac{dx}{\sigma \sqrt{2}}$$

Therefore, the integral for erf(t) becomes <p style="text_align:center;">$$erf(x) = \frac{2}{\sqrt{2 \pi} \sigma} \int_0^x \exp{-(\frac{(x-u)}{\sigma \sqrt{2}})^2}dx$$

Note that the key for the relationship is the integrand. <p style="text_align:center;">$$\begin{alignat} {3} erf(x) & = 2*\frac{1}{\sqrt{2 \pi} \sigma} \int_0^x \exp{-(\frac{(x-u)}{\sigma \sqrt{2}})^2}dx \\ & = 2*\int_0^x f_X(x) dx \\ & = 2*\int_0^x PDF dx \\ \end{alignat}$$

<p style="text_align:center;">$$\begin{alignat} {2} F_X(x) & = \frac12\left[1 + \operatorname{erf} \left(\frac{x-\mu}{\sigma\sqrt{2}} \right) \right] \\ & = \frac12[1 + 2*\int_0^x \mathbb{PDF} dx ] \end{alignat}$$

Step 3: Derive (2) from (1)

<p style="text_align:center;">$$f_X(x) = \frac{dF_X(x)}{dx}$$

Therefore, we have proven that (2) can be integrated to find (1) and that (2) can also be derived from (1).

Manuel Steele Egm6322.s12.team2.steele.m2 (talk) 17:18, 11 April 2012 (UTC)

= R 13.5 - Comparison of analytical and numerical integration of weight function =

Given

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$$  \displaystyle I^2 = \int_{-\infty}^\infty \int_{-\infty}^\infty e^{-(x^2+y^2)}dxdy $$     (13.5.1)
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Find
Analytic and numeric solution for 13.5.1.

Solution
1. Analytically:

Since

$$\displaystyle x=r\cos\theta,\quad y=r\sin\theta$$

$$\displaystyle dx=dr\cos\theta-r\sin\theta d\theta $$

$$\displaystyle dy=dr\sin\theta+r\cos\theta d\theta$$

thus

$$\displaystyle J=\frac{\partial(x,y)}{\partial(r,\theta)}=|\begin{array}{cc} \cos\theta & -r\sin\theta\\ \sin\theta & r\cos\theta \end{array}|=r$$

$$\displaystyle dxdy=Jdrd\theta$$

Using the above formula, convert 13.5.1 to polar coordinates such that:
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$$  \displaystyle I^2 = \int_{0}^{2\pi} \int_0^\infty e^{-((r\cos\theta^2+r\sin\theta^2)}rdrd\theta $$     (13.5.2) Nothing that rdr is twice the derivative of r^2 with respect to r, this can be simplified to:
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$$  \displaystyle I^2 = \int_{0}^{2\pi} \int_0^\infty e^{-r^2}\frac{1}{2}d(r^2)d\theta $$     (13.5.3) We can then pull the integrand outside of the $$\theta$$ integral due to its sole dependence on r; the integral becomes:
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$$  \displaystyle I^2 = 2\pi\int_0^\infty \frac{1}{2}e^{-((r\cos\theta^2+r\sin\theta^2)}d(r^2) $$     (13.5.4) We then substitute z = $$r^2$$ and integrate the inner term (also noting that the factors of 2 cancel):
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$$  \displaystyle I^2 = \pi \int_0^\infty e^{-z}dz = \pi(-\frac{1}{e^\infty} - \frac{-1}{e^0}) $$     (13.5.5) The term in the parenthesis is equal to +1, and thus the integral simplifies to
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$$  \displaystyle I^2 = \pi $$     (13.5.6) 2. Verify with Wolfram Alpha:
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This can be verified using Wolfram Alpha and Gauss-Hermite quadrature $$By \ Wolfram \ Alpha$$

$$\int_{-\infty}^{\infty}f(x)e^{-x^2}dx = \sqrt{\pi} \asymp 1.772453851$$

3.Numerically:

By using Gasuu-Hermite Quadrature

Hermite Polynomials

$$H_{0}=1$$ $$H_{1}=2x$$ $$H_{2}=4x^2-2$$ $$H_{3}=8x^3-12x$$ $$H_{4}=16x^4-48x^2+12$$ $$H_{5}=32x^5-160x^3+120x$$ $$H_{6}=64x^6-480x^4+720x^2-120$$ $$\displaystyle \begin{array}{l} H_{7}(x)=128x^{7}-1344x^{5}+3360x^{3}-1680x\,\\ H_{8}(x)=256x^{8}-3584x^{6}+13440x^{4}-13440x^{2}+1680\,\\ H_{9}(x)=512x^{9}-9216x^{7}+48384x^{5}-80640x^{3}+30240x\,\\ H_{10}(x)=1024x^{10}-23040x^{8}+161280x^{6}-403200x^{4}+302400x^{2}-30240\, \end{array}$$

Numerical Integration by Gauss-Hermite Quadrature
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$$I_n=\sum_{i=1}^n H_i \underbrace{f(x_i)}_{\displaystyle\color{blue}=1}=\sum_{i=1}^n H_i$$ (13.5.7)
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Hermite Weight $$\ H_i$$

$$H_i=\frac{2^{n+1} n! \sqrt{\pi}}{[H_n'(x_i)]^2}=\frac{2^{n+1} n! \sqrt{\pi}}{[H_{n+1}(x_i)]^2}$$

where

$$ n \ = \ number \ of \ points$$

$$ H_n(x_i)=0 \Rightarrow x_i$$

is roots of the Hermite polynomial $$\ H_n(x)$$

In Gauss-Hermite  Quadrature, $$ n \geqq 2 $$

When $$ \ n = 2 $$

$$H_1=\frac{2^{2+1} 2! \sqrt{\pi}}{[H_{2+1}(x_1)]^2}$$

$$H_2=\frac{2^{2+1} 2! \sqrt{\pi}}{[H_{2+1}(x_2)]^2}$$

where $$\ H_{3}(x_1)$$ and $$H_{3}(x_2)$$ is roots of $$H_{2}$$ that are $$\pm 0.707107$$

$$H_1=\frac{2^{2+1} 2! \sqrt{\pi}}{[8(-0.707107)^3-12(-0.707107)^2]}=0.886227$$

$$H_2=\frac{2^{2+1} 2! \sqrt{\pi}}{[8(0.707107)^3-12(0.707107)^2]}=0.886227$$

$$ By \ Eq(13.5.7)$$

$$n=2$$ $$ x_i=\pm0.707107 \ \ H_i=0.886227$$

$$n=3$$ $$ x_i= 0 \ \ H_i=1.18164$$ $$ x_i=\pm1.22474 \ \ H_i=0.295409$$

$$n=4$$ $$ x_i=\pm0.524648 \ \ H_i=0.804914$$ $$ x_i=\pm1.65068 \ \ H_i=0.0813128$$

$$n=5$$ $$ x_i=0 \ \ H_i=0.945309$$ $$ x_i=\pm0.958572 \ \ H_i=0.393619$$ $$ x_i=\pm2.02018 \ \ H_i=0.0199532$$

$$I_2=\sum_{i=1}^2 H_i=H_1+H_2=0.886227+0.886227=1.772454$$

$$I_3=\sum_{i=1}^3 H_i=H_1+H_2+H_3=0.295409+ 1.18164 +0.295409=1.772458$$

$$I_4=\sum_{i=1}^4 H_i=H_1+H_2+H_3+H_4=0.0813128+0.804914+0.804914+0.0813128=1.7724536$$

$$I_5=\sum_{i=1}^5 H_i=H_1+H_2+H_3+H_4+H_5=0.0199532+0.393619+0.945309+0.393619+0.0199532=1.7724534$$

plot the graph by using data above $$ n=2 \, \ I_2=1.772454$$ $$ n=3 \, \ I_3=1.772458$$ $$ n=4 \, \ I_4=1.7724536$$ $$ n=5 \, \ I_5=1.7724534$$ $$ n=10 \, \ I_10=1.7724539$$ $$\int_{-\infty}^{\infty}f(x)e^{-x^2}dx = \sqrt{\pi} \asymp 1.772453851$$



= R 13.6 - 3 Term Recurrence Relation for Sequence of Orthogonal Polynomials =

Given

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$$  P_{n+1}(x)=(A_nx+B_n)P_n(x)-C_nP_{n-1}(x)$$ (12.5.1)
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$$\text{for} \ n=0,1,2,...,\text{and} P_{-1}(x)=0$$

$$ where $$

$$ A_n = \frac{k_{n+1}}{k_n}=\frac{leading \ coefficient \ of \ P_{n+1}}{leading \ coefficient \ of \ P_n}$$

$$B_n=\frac{\langle xP_n,P_n \rangle}{\langle P_n,P_n \rangle}$$

$$C_n=\frac{A_n}{A_{n-1}}\frac{h_n}{h_{n-1}}$$

$$where $$

$$ h_n=\langle P_n,P_n \rangle$$

Find
$$ Show \ A_n = \frac{k_{n+1}^{n+1}}{k_n^n}=:\frac{k_{n+1}}{k_n} \Rightarrow s \in \mathcal P_n $$

Solution
$$P_0 = {\color{red}1}$$

$$P_1 = {\color{red}x}$$

$$P_2 = {\color{red}\frac{1}{2}}({\color{red}3}x^2-1) $$

$$P_3 = {\color{red}\frac{1}{2}}({\color{red}5}x^3-3x) $$

$$P_4 = {\color{red}\frac{1}{8}}({\color{red}35}x^4-30x^2+3) $$

$$P_5 = {\color{red}\frac{1}{8}}({\color{red}63}x^5-70x^3+15x) $$

$$P_6 = {\color{red}\frac{1}{16}}({\color{red}231}x^6-315x^4+105x^2-5)$$

$$ {\color{red}Red} \ is \ Leading \ Coefficient \ of \ each \ Legendre \ Polynomials $$

$${\color{red}1},{\color{red}1},{\color{red}\frac{3}{2}},{\color{red}\frac{5}{2}},{\color{red}\frac{35}{8}},{\color{red}\frac{63}{8}}$$ $$ Let \ n=0 \ in \ the \ Eq(13.5.1) $$

$$ P_1=(A_0x+B_0)P_0(x)-\cancel{C_0P_{-1}(x)}$$

$$P_1=(A_0x+B_0)(1)$$

$$B_0=\frac{\langle xP_0,P_0 \rangle}{\langle P_0,P_0 \rangle}=0$$

$$ P_1(x)=A_0x=x$$

$$Therefore,$$

$$A_0=1=\frac{Leading \ Coefficient \ of \ P_1}{Leading \ Coefficient \ of \ P_0}=\frac{1}{1}=1$$ $$ Let \ n=1 \ in \ the \ Eq(13.5.1) $$

$$ P_2=(A_1x+B_1)P_1(x)-C_1P_{0}(x)$$

$$B_1=\frac{\langle xP_1,P_1 \rangle}{\langle P_1,P_1 \rangle}=0$$

$$C_1=\frac{A_1}{A_{0}}\frac{\frac{2}{2\cdot1+1}}{\frac{2}{2 \cdot1 +1}}=\frac{A_1}{A_{0}}\frac{1}{3}$$

$$P_2=A_1xP_1(x)-\frac{A_1}{A_{0}}\frac{1}{3}P_{0}(x)$$

$$P_2=A_1xP_1(x)-\frac{A_1}{A_{0}}\frac{1}{3}P_{0}(x)$$

$$\frac{1}{2}(3x^2-1)=A_1x^2-\frac{A_1}{A_{0}}\frac{1}{3}(1)$$

$$ \frac{3}{2}x^2 = A_1x^2 \Rightarrow A_1=\frac{3}{2}$$

$$\frac{1}{2}=\frac{A_1}{A_0}\frac{1}{3}(1)=\frac{\cancel{3}}{2}\frac{1}{A_0}\frac{1}{\cancel{3}}(1) \Rightarrow A_0 =1$$

$$A_1=\frac{3}{2}=\frac{Leading \ Coefficient \ of \ P_2}{Leading \ Coefficient \ of \ P_1}=\frac{\frac{3}{2}}{1}=\frac{3}{2}$$

$$A_0=1=\frac{Leading \ Coefficient \ of \ P_1}{Leading \ Coefficient \ of \ P_0}=\frac{1}{1}=1$$

$$Hence,$$

$$A_n = \frac{k_{n+1}^{n+1}}{k_n^n}=:\frac{k_{n+1}}{k_n}$$

Egm6322.s12.sungsik (talk) 14:48, 11 April 2012 (UTC)

=Contributing Members=

= References for Report 13 = Bell, W.W. "Special Functions for Scientists and Engineers." Mineola, NY: Dover Publications, Inc., 1996. King, A.C., J. Billingham and S.R. Otto. "Differential Equations: Linear, Nonlinear, Ordinary, Partial." New York, NY: Cambridge University Press, 2003.

http://dlmf.nist.gov/ http://en.wikipedia.org/wiki/Legendre_polynomials

Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 48e) [[media:Pea2.s12.sec48.e.djvu|Mtg 48e]] Spring 2012.

Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 48f) [[media:Pea2.s12.sec48.f.djvu|Mtg 48f]] Spring 2012.

Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 48g) [[media:Pea2.s12.sec48.g.djvu|Mtg 48g]] Spring 2012.

Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 48h) [[media:Pea2.s12.sec48.h.djvu|Mtg 48h]] Spring 2012.

http://en.wikipedia.org/wiki/Square_root_of_5