text:
User:Egm6322.s12.team2/HW11

= R 11.1 - Integral Formula for the Weights =

Given

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$$  \displaystyle f(x)\approx f^h(x):=\sum_{j=1}^{n} f(x_j)\ell_j(x)$$ (11.1.1)
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$$  \displaystyle \ell_j(x)=\frac{P_n(x)}{(x-x_j)P_n'(x_j)} $$     (11.1.2)
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Find

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$$  \displaystyle w_j = \int_{-1}^1\frac{P_n(x)}{(x-x_j)P_n'(x_j)}w(x)dx $$     (11.1.3) with w(x) = 1
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Solution

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$$  \displaystyle \int_{-1}^1 f(x)=\int_{-1}^1 \sum_{j=1}^{n} f(x_j)\ell_j(x)$$ (11.1.4)
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$$  \displaystyle \int_{-1}^1 f(x)=\sum_{j=1}^{n} w_jf(x_j)$$ (11.1.5)
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$$  \displaystyle \sum_{j=1}^{n} w_jf(x_j)=\int_{-1}^1 \sum_{j=1}^{n} f(x_j)\ell_j(x)$$ (11.1.6)
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Therefore, $$  \displaystyle w_j = \int_{-1}^1\frac{P_n(x)}{(x-x_j)P_n'(x_j)}dx $$ Egm6322.s12.sungsik (talk) 02:34, 14 March 2012 (UTC)

= R 11.2 - Weight function for Gauss-Legendre Quadrature (cont from R10.6)=

Given
Simplified weight formula:
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$$  \displaystyle w_j = \frac{-2}{(n+1)P_n'(x_j)P_{n+1}(x_j)} $$     (11.2.1)
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Simplified error formula:
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$$  \displaystyle E_n(f) = \frac{2^{2n+1}(n!)^4}{(2n+1)[(2n)!]^2}\frac{f^{(2n)}(\eta)}{(2n)!},\ \eta \in [-1,1] $$     (11.2.2)
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Find
In the NIST DLMF Sec 3.5v, the weight function shown in 10.6.1 is given by the formula:
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$$  \displaystyle w_j = \int_{-1}^1\frac{P_n(x)}{(x-x_j)P_n'(x_j)}w(x)dx $$     (11.2.3) with w(x) = 1. Show that 11.2.1 can be obtained from 11.2.3.
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Solution
The first step in this problem is considering the Christoffel-Darboux formula for Legendre polynomials:
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$$  \displaystyle \frac{P_{n+1}(x)P_n(\xi)-P_n(x)P_{n+1}(\xi)}{\beta_n h_n} = (x-\xi)\sum_{k=0}^n P_k(x)P_k(\xi) $$     (11.2.4) where
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$$  \displaystyle h_n := \langle P_n,P_n\rangle = \frac{2}{2n+1} $$     (11.2.5) and (using the definition of $$\alpha_n$$ from Weisstein:
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$$  \displaystyle \beta_n := \frac{\alpha_{n+1}}{\alpha_n} = \frac{2n+1}{n+1} $$     (11.2.6)
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We substitute $$x_j$$ for $$\xi$$ and multiply both sides by $$P_0(x)$$. The latter step is done to take advantage of the orthogonality of Legendre polynomials. The inner product, defined as the integral between -1 and 1 of the polynomials, is zero if the order of the polynomials are different; ie, $$\langle L_m,L_n \rangle = 0 $$ for $$m\neq n$$. It is also possible to move terms which are not a function of x out of the integral.
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$$  \displaystyle \frac{P_{n+1}(x_j)}{\beta_n h_n} \int_{-1}^1 \frac{P_0(x)P_n(x)}{x-x_j}dx = -\sum_{k=0}^n \frac{P_k(x_j)}{h_k}\int_{-1}^1 P_k(x)P_0(x)dx $$     (11.2.7) On the right hand side, we recognize that due to the orthogonality of Legendre polynomials, the inner product of $$\langle P_k(x),P_0(x)\rangle$$ will be zero for $$k \neq 0 $$. Thus the summation simplifies down to the $$k=0$$ case only. Thus 11.2.7 becomes:
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$$  \displaystyle \frac{P_{n+1}(x_j)}{\beta_n h_n} \int_{-1}^1 \frac{P_0(x)P_n(x)}{x-x_j}dx = -\frac{P_0(x_j)}{h_0}\int_{-1}^1 P_0(x)^2dx $$     (11.2.8) The zeroth order Legendre polynomial is given by $$P_0(x) = 1$$; substituting this and the definition of $$h_0$$ from 11.2.5 into 11.2.8 yields:
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$$  \displaystyle \begin{align} \frac{P_{n+1}(x_j)}{\beta_n h_n} \int_{-1}^1 \frac{P_n(x)}{x-x_j}dx &= -\frac{P_0(x_j)}{(2/2(0)+1)}\int_{-1}^1 (1)dx\\ \frac{P_{n+1}(x_j)}{\beta_n h_n} \int_{-1}^1 \frac{P_n(x)}{x-x_j}dx &= -\frac{(1)}{2}(2) \end{align} $$     (11.2.9)
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From Weisstein, the coefficients of the $$x^n$$ term of the polynomial are shown to be
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$$  \displaystyle \frac{a_{n+1}}{a_n} = \frac{2n+1}{n+1} $$     (11.2.10)
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Solving Eq 11.2.9 for the integral on the left hand side and substituting in the defintions of $$\beta_n$$ and $$a_n$$ yields:
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$$  \displaystyle \begin{align} \int_{-1}^1 \frac{P_n(x)}{x-x_j} &= \frac{-\beta_n h_n}{P_{n+1}(x_j)}\\ &= \frac{-2a_{n+1}}{(2n+1)a_n P_{n+1}(x_j)}\\ &= \frac{-2}{(n+1)P_{n+1}(x_j)} \end{align} $$     (11.2.11) Thus, when we consider the integral expression for the weights in Eq. 11.2.3 we recognize that it is equal to left hand side of the identity in Eq 11.2.11 with w(x)=1 for Legendre polynomials. Thus the simplified weight formula is shown to be:
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$$  \displaystyle w_j = \frac{1}{P_n'(x_j)}\int_{-1}^1 \frac{P_n(x)}{x-x_j} = \frac{-2}{(n+1)P_n'(x_j)P_{n+1}(x_j)} $$     (10.6.13)
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Matt Shields 3:55,13 March 2012 (UTC)

= R 11.3 - Polynomial Division and Proof of N Integration Points =

Given
$$p(x) = 3x^3 + 2x^2 + 3x + 1 $$

$$s(x) = 5x^2 + 3 $$

$$p(x) = q(x)s(x) + r(x) \in \mathcal P_3 $$

Find
$$q(x) = \frac{(3x+2)}{5} \in \mathcal P_1$$

$$r(x) = \frac{(6x - 1)}{5} \in \mathcal P_1$$

From lecture notes [[media:Pea2.s12.sec45.e.djvu|Mtg 45e]]

Solution
First, we analyze the degrees of the polynomial and s(x). In general, the theorem requires the following general form of the polynomial solution with a quotient and remainder.

$$ p(x) = q(x)P_n(x) + r(x) \in \mathcal P_{2n-1}$$

Note that q(x) is the quotient and r(x) is the remainder. The general restrictions on the degree of each term must be evaluated.

$$n < \deg p(x) \le 2n-1$$

$$\deg s(x) = n = 2$$

Therefore, in theory,
$$\deg r(x) \le n-1=1$$

$$\deg q(x) \le n-1=1$$

The solution requires three general steps: (1) rearrange polynomial equation, (2) find q(x), and (3) find r(x).

Step 1: Rearrange Polynomial Equation
$$\begin{alignat}{2} p(x) & = 3x^3 + 2x^2 + 3x + 1 \\ & = 3x^3 + 2x^2 + \frac{9x}{5} + \frac{6}{5} + \frac{6x}{5} - \frac{1}{5} \\ \end{alignat} $$

Step 2: Solve for q(x): Take the first four terms from the rearrange equation and divide by the given polyomial.

$$\begin{alignat}{3} q(x) & = \frac{p(x) - r(x)}{s(x)} \\ & = \frac{3x^3 + 2x^2 + \frac{9x}{5} + \frac{6}{5}}{5x^2 + 3} \\ & = \frac{(3x+2)(5x^2+3)}{5(5x^2+3)} \\ \end{alignat}$$

Therefore, the solution for q(x) is
$$q(x) = \frac{(3x+2)}{5}$$

The degree of q(x) is 1 so that the following relation is confirmed.
$$\deg q(x) \le n-1=1$$

Step 3: Solve for r(x): We start with the original equation.
$$p(x) = q(x)s(x) + r(x)$$

Rerrange the equation to solve for r(x) and solve.
$$\begin{alignat}{3} r(x) & = p(x) - q(x)s(x) \\ & = 3x^3 + 2x^2 + 3x + 1 - 3x^3 - 2x^2 - \frac{9x}{5} - \frac{6}{5} \\ & = 3x - \frac{9x}{5} + 1 - \frac{6}{5} \\ \end{alignat}$$

Therefore, the solution of r(x) is below.
$$r(x) = \frac{6x}{5} - \frac{1}{5}$$

On a final note, since the degree of r(x) is 1 we confirm that
$$\deg r(x) \le n-1=1$$

Egm6322.s12.team2.steele.m2 (talk) 07:58, 14 March 2012 (UTC)

= R 11.4 - Verify the weights of GL quadrature =

Given
Give the following table

with

$$\displaystyle w_j = \frac{2}{\left( 1-x_j^2 \right) [P'_n(x_j)]^2} \,\!$$ and $$\displaystyle P_n(x) = {1 \over 2^n n!} {d^n \over dx^n } \left[ (x^2 -1)^n \right].$$

Find
Verify

1, $$\displaystyle w_{j}>0,\; j=1,\ldots,n$$
(11.4.1) 2, $$\displaystyle \sum_{j=1}^{n}w_{j}=2$$
(11.4.2)

Solution
Part I

From the given conditions, Using Maxima with the following commands:

p(n,x):=(1/(2^n*n!))*diff((x^2-1)^n,x,n);  %%$$\displaystyle \mathrm{p}\left( n,x\right) :=\frac{1}{{2}^{n}\,n!}\,\mathrm{diff}\left( {\left( {x}^{2}-1\right) }^{n},x,n\right) $$

w(n,x):=2/((1-x^2)*(diff(p(n,x),x))^2);    %% $$\displaystyle \mathrm{w}\left( n,x\right) :=\frac{2}{\left( 1-{x}^{2}\right) \,{\mathrm{diff}\left( \mathrm{p}\left( n,x\right) ,x\right) }^{2}}$$

we can obtain

$$\displaystyle w(1,x)= \frac{2}{1-{x}^{2}}$$ $$\displaystyle w(2,x)= \frac{2}{9\,{x}^{2}\,\left( 1-{x}^{2}\right) }$$ $$\displaystyle w(3,x)= \frac{4608}{\left( 1-{x}^{2}\right) \,{\left( 288\,{x}^{2}+72\,\left( {x}^{2}-1\right) \right) }^{2}}$$ $$\displaystyle w(4,x)= \frac{294912}{\left( 1-{x}^{2}\right) \,{\left( 3840\,{x}^{3}+2880\,x\,\left( {x}^{2}-1\right) \right) }^{2}}$$ $$\displaystyle w(5,x)= \frac{29491200}{\left( 1-{x}^{2}\right) \,{\left( 7200\,{\left( {x}^{2}-1\right) }^{2}+57600\,{x}^{4}+86400\,{x}^{2}\,\left( {x}^{2}-1\right) \right) }^{2}}$$ Then, use the command subst(x_j,x,w(n,x))

we have

$$\displaystyle w(1,0)=2$$

$$\displaystyle w(2,\pm1/\sqrt{3})=1$$

$$\displaystyle w(3,0)=\frac{8}{9}$$

$$\displaystyle w(3,\pm\sqrt{3/5})=\frac{5}{9}$$

$$\displaystyle w(4,\pm\sqrt{(3-2\sqrt{6/5})/7})=-\frac{49}{6\,\left( \sqrt{6}+2\,\sqrt{5}\right) \,\left( 2\,\sqrt{6}-3\,\sqrt{5}\right) }=\frac{18+\sqrt{30}}{36}$$

$$\displaystyle w(4,\pm\sqrt{(3+2\sqrt{6/5})/7})=-\frac{49}{6\,\left( \sqrt{6}-2\,\sqrt{5}\right) \,\left( 2\,\sqrt{6}+3\,\sqrt{5}\right) }=\frac{18-\sqrt{30}}{36}$$

$$\displaystyle w(5,0)=\frac{128}{225}$$

$$\displaystyle w(5,\pm\frac{1}{3}\sqrt{(5-2\sqrt{10/7})})=\frac{729\,\sqrt{7}}{25\,\left( \sqrt{10}+2\,\sqrt{7}\right) \,{\left( \sqrt{7}\,\sqrt{10}-4\right) }^{2}}=\frac{322+13\sqrt{70}}{900}$$

$$\displaystyle w(5,\pm\frac{1}{3}\sqrt{(5+2\sqrt{10/7})})=-\frac{729\,\sqrt{7}}{25\,\left( \sqrt{10}-2\,\sqrt{7}\right) \,{\left( \sqrt{7}\,\sqrt{10}+4\right) }^{2}}=\frac{322-13\sqrt{70}}{900}$$

Part II

Thus from the above solution we can easily obtain

$$\displaystyle n=1$$

$$\displaystyle \sum w_j =w_1=2$$

$$\displaystyle n=2$$

$$\displaystyle \sum w_j =w_1+w_2=1+1=2$$

$$\displaystyle n=3$$

$$\displaystyle \sum w_j =w_1+w_2+w_3=\frac{8}{9}+\frac{5}{9}+\frac{5}{9}=2$$

$$\displaystyle n=4$$

$$\displaystyle \sum w_j =w_1+w_2+w_3+w_4=2 \frac{18+\sqrt{30}}{36}+2 \frac{18-\sqrt{30}}{36}=2$$

$$\displaystyle n=5$$

$$\displaystyle \sum w_j =w_1+w_2+w_3+w_4+w_5=\frac{128}{225}+2 \frac{322+13\sqrt{70}}{900}+2 \frac{322-13\sqrt{70}}{900}=2$$

= R 11.5 - Integration of Gauss Legendre Quadrature =

Given

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$$  \displaystyle \int_{-1}^1 f(x)=\sum_{j=1}^{n} w_jf(x_j)$$ (11.5.1)
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$$  \displaystyle p(x) = \frac{x^7}{8} + \frac{x^6}{7} + \frac{x^2}{3} + 5$$ (11.5.2)
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Find
1. Integrate analytically exactly over -1 to 1 2. Determine the least number of integration points

Solution

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$$  \displaystyle \int_{-1}^1 f(x)=\int_{-1}^1 (\frac{x^7}{8} + \frac{x^6}{7} + \frac{x^2}{3} + 5)dx $$ (11.5.3)
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$$  \displaystyle \int_{-1}^1 (\frac{x^7}{8} + \frac{x^6}{7} + \frac{x^2}{3} + 5)dx = [(\frac{1}{8})(\frac{1}{8})x^8+(\frac{1}{7})(\frac{1}{7})x^7+(\frac{1}{3})(\frac{1}{3})x^3+5x]_{-1}^{1} = \frac{4526}{441}=10.263$$ (11.5.4)
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Therefore, we need at least 2 points. When the number of points is 4, we can get more precise value of integration from GL. Egm6322.s12.sungsik (talk) 02:35, 14 March 2012 (UTC)

= R 11.6 - Coefficients for Fourier-Legendre series using Gauss-Legendre quadrature=

Given
Simplified weight formula:
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$$  \displaystyle w_j = \frac{-2}{(n+1)P_n'(x_j)P_{n+1}(x_j)} $$     (11.6.1)
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Simplified error formula:
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$$  \displaystyle E_n(f) = \frac{2^{2n+1}(n!)^4}{(2n+1)[(2n)!]^2}\frac{f^{(2n)}(\eta)}{(2n)!},\ \eta \in [-1,1] $$     (11.6.2) Coefficients of Fourier-Legendre series:
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$$  \displaystyle A_j = \frac{\langle P_j,f \rangle}{\langle P_j,P_j \rangle} = \frac{2j+1}{2}\int_{\mu=-1}^{\mu=1}P_j(\mu)f(\mu)d\mu $$     (11.6.3) Gauss-Legendre quadrature:
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$$  \displaystyle \int_{-1}^1f(x)dx = \sum_{j=1}^nw_j f(x_j) $$     (11.6.4)
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Find
Plot the function
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$$  \displaystyle f(\theta) = \frac{1}{1+\sin^2(\sin^2(\theta)} $$     (11.6.5) Find the first 3 nonzero coefficients of the Fourier-Legendre series, if not analytically, then numerically using Gauss-Legendre (GL) quadrature with $$10^{-6}$$ accuracy.  Plot the truncated F-L series to show convergence towards $$f(\theta)$$.
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Solution
The function in Eq 11.6.5 presents difficulties for Fourier-Legendre expansion as the inner product between a Legendre polynomial and the complicated function can not be computed analytically. Thus, to approximate this function using a F-L series, it is necessary to also approximate the integral in the determination of the coefficients; namely, using GL quadrature defined in 11.6.4:
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$$  \displaystyle A_j = \frac{2j+1}{2}\int_{\mu=-1}^{\mu=1}P_j(\mu)f(\mu)d\mu = \frac{2j+1}{2}\sum_{j=1}^n w_j P_j(\theta)f(\theta) $$     (11.6.6) First, we plot the function using Matlab and defining the function between -1 and 1.
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The function is clearly even by inspection (and as expected as it is a function of $$\sin(\theta)$$. Since the inner product on the right hand side of Eq. 11.6.6 is zero over the domain between -1 and 1 for even and odd functions, this indicates that all odd coefficients ($$A_1, A_3, A_5, ... $$) are zero.  It will be necessary to use GL quadrature to evaluate the integral for the even coefficients.  Hence we search for $$A_0, A_2, A_4$$.

The constraint on the approximation is an accuracy of $$10^{-6}$$. This is evaluated using the error formula in 11.6.2 where n is the number of points (roots) used in GL quadrature and f is the function over the domain -1 to 1. Again using Matlab, it is found that 4-point quadrature (n=4) provides a maximum error over the domain of 2.9e-7; however, knowing that we will be using the 4th order Legendre polynomial in the F-L expansion for $$A_4$$, using the roots from the 4-point quadrature will cause this expansion to be zero (this will become clearer later). Hence we choose 5-point quadrature, which yields a maximum error of 8.1e-10.

With n selected it is now possible to determine the roots and weighting functions of $$P_5(\theta)$$. The roots are given in the [[media:Pea2.s12.sec45.c.djvu|Mtg 45c]] to be:
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$$  \displaystyle x_j = 0, \pm\frac{1}{3}\sqrt{5-2\sqrt{10/7}}, \pm\frac{1}{3}\sqrt{5+2\sqrt{10/7}} $$     (11.6.7) To use the weighting function in Eq. 11.6.3 it is necessary to know the nth derivative and the (n+1)th Legendre polynomial. These are:
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$$  \displaystyle \begin{align} P_5(\theta) &= \frac{1}{8}(63\theta^5 - 70\theta^3 +15\theta)\\ P_5'(\theta) &= \frac{1}{8}(315\theta^4 - 210\theta^2 +15)\\ P_6(\theta) & = \frac{1}{16}(231\theta^6 - 315\theta^4 +105\theta^2 -5) \end{align} $$     (11.6.8)
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These values, along with n=5, can be substituted into 11.6.1 to obtain the weights for each term in the GL quadrature expansion. This yields the weights to be:
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$$  \displaystyle x_j = \frac{128}{225}, \frac{322 + 13\sqrt{70}}{900}, \frac{322 + 13\sqrt{70}}{900}, \frac{322 - 13\sqrt{70}}{900}, \frac{322 - 13\sqrt{70}}{900} $$     (11.6.9) From here it is a simple matter to perform the GL quadrature to determine the coefficients using Eq. 11.6.4. As an example using $$A_0$$:
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$$  \displaystyle \begin{align} A_0 &= \frac{2(0)+1}{2}\int_{-1}^{1}P_0(\theta)f(\theta)d\theta = \frac{1}{2}\sum_{j=1}^{n=5} w_j \underbrace{P_0(\theta)}_{=1}f(\theta)\\ &= 0.91    \end{align} $$     (11.6.10) Similarly, the coefficients $$A_2$$ and $$A_4$$ are found. The coefficients for the first three nonzero terms are:
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$$  \displaystyle \begin{align} A_0 &= 0.91\\ A_2 &= -0.2087\\ A_4 &= -0.0382 \end{align} $$     (10.6.13) The converged solution, $$ f(\theta) = \sum_{j=0}^4 A_j P_j(\theta)f(\theta)$$ can then be plotted against the exact solution, where it can be seen that the converged solution matches nicely:
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Matt Shields 4:20,13 March 2012 (UTC)

=Contributing Members=

Egm6322.s12.team2.Xia 23:55, 12 March 2012 (UTC)

= References for Report 11 = 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/3.5#v

Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 45c) [[media:Pea2.s12.sec45.c.djvu|Mtg 45c]] Spring 2012.

Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 45d) [[media:Pea2.s12.sec45.d.djvu|Mtg 45d]] Spring 2012.

Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 45e) [[media:Pea2.s12.sec45.e.djvu|Mtg 45e]] Spring 2012.

Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 45f) [[media:Pea2.s12.sec45.f.djvu|Mtg 45f]] Spring 2012.

title: User:Egm6322.s12.team2/HW11


text:
User talk:Dgilliland/enes100/my work

week 0 p1 grading 1sfoerster 20:15, 21 February 2012 (UTC)
 * 10/10 .good job

You are doing a great job and are a good partnerStevecole888 (talk) 17:32, 24 February 2012 (UTC)

You are doing what you said you were going to do and you designed that reloading mechanism, and you are easy to work with.--Jsnyder6518 (talk) 00:28, 27 February 2012 (UTC)

week 1 p1 grading
 * 10/15/0 .. good report of your work, but nothing anyone could replicate, english is bordering on not being readable, can not tell what has been done, what has been thought about, what materials have been located, etc. Nothing here that I can leverage. Draw in your favorite software package, not in your notebook. Scan your notebook if you put a lot of time into it and it is going to inspire someone else. 90% of notebook scans are useless. 5 points for commenting on team mates pages.

week 2 p1 grading
 * 0/0/0 ... nothing to grade

Goode job this week on finding all the materials keep it upStevecole888 (talk) 17:01, 7 March 2012 (UTC)

Good job at getting all the materials needed ordered and now next week we can start building.Jsnyder6518 (talk)

week 3 p1 grading 1sfoerster (talk) 13:47, 14 March 2012 (UTC)
 * 10/10/10 ... the 10 push points is for the verbal description of what you did ... this is good, but it needs to be accompanied by pictures and/or drawings ... this would increase the push points

Great job over the 4 weeks you were a great partner and did everything you said you wouldStevecole888 (talk) 15:23, 16 March 2012 (UTC)

week 4 p1 grading 1sfoerster (talk) 20:28, 20 March 2012 (UTC)
 * 10/10/0 ... good report, need visuals

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

title: User talk:Dgilliland/enes100/my work


text:
User:Egm6322.s12.team2/HW12

= R 12.1 - Integration Points and Gauss-Legendre Quadrature =

Given
Legendre series

Find
Compute the 1st non-zero coefficient in the Legendre series with increasing number of integration points until convergence to within 1% accuracy. Use the error formula (1) p. 45b-7 to estimate the necessary number of integration points to achieve the desired accuracy.

Solution
Withn the class context, non-zero coefficients have generally been solved by applying the Fourier-Legendre series and orthogonality to find non-zero coefficients. For team 2, the previous solutions of 9.1 and 10.7 will be expanded to solve this problem. The general pattern is described below.

Step 1: Define 1st Non-Zero Coefficient

The coefficient of a Fourier-Legendre series requires the definition below. $$A_j = \frac{2j+1}{2} \int_{-1}^1 P_j(x)T_0 \sinh{(1 - x^2)}d(x)$$

The coefficient is from the term for the Fourier-Legendre series of f(x).

$$f(x) = \sum_{n=0}^{\infty}a_n P_n(x)$$

Essentially, the series utilizes the Legendre polynomials and coefficients to define a function $$ f(x)$$. The Legendre polynomials have the following orthogonality.

$$\int_{-1}^{1} P_n(x) P_m(x)dx = \frac{2}{2m+1} \delta_{mn}$$

The problem does not explicitly give a function, f(x), for finding the 1st non-zero coefficient. Therefore, the team's approach is to utilize the function from 9.1 and 10.7 below.

$$f(x) = T_0 \sinh(1-x^2)$$

The coefficients are then defined as follows. $$A_j = \frac{2j+1}{2} \int_{-1}^1 P_j(x)T_0 \sinh{(1 - x^2)}d(x)$$

From 9.1 and 10.7, the 1st non-zero coefficient of the chosen f(x) is

$$A_0 = 0.7459T_0$$

Step 2: Apply Qauss-Quadrature to Integral of Coefficient

$$A_j = \frac{2j+1}{2} \int_{-1}^1 P_j(x)T_0\sinh{(1 - x^2)}d(x) \approx I_n(f) \sum_{j=1}^n w_jf(x_j)$$

The weighting function is

$$w_j = \frac{-2}{(n+1) P_n'(x_j)P_{n+1}(x_j)}$$

The error formula is shown below from 45-7b. $$E_n(f) = \frac{2^{2n+1}(n!)^4}{(2n+1)[(2n)!]^2} \frac{f^{(2n)}(\eta)}{(2n)!}, \eta \in [-1,1]$$

Step 3: Find Number of Integration Points

The following definition is used to find the number of integration points.

$$I(f) = I_n(f) + E_n(f)$$

This can be rearranged to apply the error limit of 1% or 0.01.

$$\begin{alignat}{2} \frac{I(f) - I_n(f)}{I(f)} & = \frac{E_n(f)}{0.7459} \\ & = \frac{1}{0.7459}\frac{2^{2n+1}(n!)^4}{(2n+1)[(2n)!]^2} \frac{f^{(2n)}(\eta)}{(2n)!} <= 0.01 \end{alignat}$$

Note that $$f^{(2n)}(\eta)$$ is the (2n)th order derivative of $$f(x)=\sinh({1-x^2)}$$. Normally, this higher order derivative would require a rigorous analysis with numerical methods and sophisticated programming. But that can be minimized by noting that a derivative of $$sinh(x) = cosh(x) $$and vice versa. So the following definition is utilized.

$$\begin{alignat}{2} f^{(2n)}(x) & = \frac{d^{2n} \sinh({(1-x^2)}}{dx^{(2n)}} \\                                                  & = {(-2x)}^{(2n)}\sinh{(1-x^2)} \end{alignat}$$

So the simplified error term is below. $$\frac{E_n(f)}{0.7459} = \frac{1}{0.7459}\frac{2^{2n+1}(n!)^4}{(2n+1)[(2n)!]^2} \frac{{(-2x)}^{(2n)}\sinh{(1-x^2)}}{(2n)!} $$

The Matlab image and code are shown below.

http://en.wikiversity.org/wiki/File:PEA2_12_1_Matlab.PNG

http://en.wikiversity.org/wiki/File:PEA2_12_1_Error_Term.png

See images below. The data from the Excel shows that there are two integration points needed to get the error less than 0.01 or 1%. The error term derived was simply entered into MS exceo with $$\eta = 0.5$$ and calculated for n = 1 to 20. At n =2 E < 0.01, so n = 2 will suffice.

http://en.wikiversity.org/wiki/File:Team2_12_1_Excel_Data.PNG

Egm6322.s12.team2.steele.m2 (talk) 20:32, 28 March 2012 (UTC)

= R 12.2 - Alternative formula for weights of GL quad =

Given
Using the following fomula

$$\displaystyle (1-x^{2})P_{n}^{'}=(n+1)xP_{n}-(n+1)P_{n+1} $$ (12.2.1)

Find
Show that

$$\displaystyle (n+1)P_{n+1}(x_{j})=-(1-x_{j}^{2})P_{n}^{'}(x_{j})$$ (12.2.2) and

$$\displaystyle w_{j}=\frac{2}{(1-x_{j}^{2})[P_{n}^{'}(x_{j})]^{2}}$$ (12.2.3)

Solution
From the 12.2.1, we have

$$\displaystyle (1-x_{j}^{2})P_{n}^{'}(x_{j})=(n+1)x_{j}P_{n}(x_{j})-(n+1)P_{n+1}(x_{j})$$
(12.2.4) Since $$\displaystyle P_{n}(x_{j})=0$$,(The proof is @ p.45-17) then

$$\displaystyle (n+1)P_{n+1}(x_{j})=-(1-x_{j}^{2})P_{n}^{'}(x_{j})$$
(12.2.5) In terms of (5)p.45b-6

$$\displaystyle w_{j}=\frac{-2}{(n+1)P_{n+1}(x_{j})P_{n}^{'}(x_{j})}$$
(12.2.6) then substitute 12.2.5 into 12.2.6, we obtain

$$\displaystyle w_{j}=\frac{2}{(1-x_{j}^{2})[P_{n}^{'}(x_{j})]^{2}}$$
(12.2.3)

= R 12.3 - Proof of Thean of Smynna =

Given

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$$  y_n =2x_{n-1} + y_{n-1}$$ (12.3.1)
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$$  x_n =x_{n-1} + y_{n-1}$$ (12.3.2)
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$$  \lim_{n\to\infty}\frac{y_n}{x_n}=\sqrt2$$ (12.3.3) $$x_0 = 1$$
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$$y_0 = 1$$

Find
Plot the sequence $$\{y_n/x_n, \ n=0,1,...10\}$$ versus the iteration number n

Solution
By using Eq(12.3.1) and Eq(12.3.2),

Finding $${x}_{0} to {x}_{10} and {y}_{0} to {y}_{10}$$

$$ n=0, y_{0}=1, x_{0}=1, \frac{y_{0}}{x_{0}}=1$$ $$ n=1, y_{1}=3, x_{1}=2, \frac{y_{1}}{x_{1}}=1.500$$ $$y_{1}=2x_{0}+y_{0}= 2+1=3 and x_{1}=x_{0} +y_{0}=1+1=2$$ $$ n=2, y_{2}=7, x_{2}=5, \frac{y_{2}}{x_{2}}=1.400$$ $$y_{2}=2x_{1}+y_{1}= (2)(2)+3=7 and x_{2}=x_{1} +y_{1}=2+3=5$$ $$ n=3, y_{3}=17, x_{3}=12, \frac{y_{3}}{x_{3}}=1.4167$$ $$y_{3}=2x_{2}+y_{2}= (2)(5)+7=17 and x_{3}=x_{2} +y_{2}=2+3=5$$ $$ n=4, y_{4}=41, x_{4}=29, \frac{y_{4}}{x_{4}}=1.4138$$ $$y_{4}=2x_{3}+y_{3}= (2)(12)+17=41 and x_{4}=x_{3} +y_{3}=12+17=29$$ $$ n=5, y_{5}=99, x_{5}=70, \frac{y_{5}}{x_{5}}=1.4143$$ $$y_{5}=2x_{4}+y_{4}= (2)(29)+41=99 and x_{5}=x_{4} +y_{4}=29+41=70$$ $$ n=6, y_{6}=239, x_{6}=169, \frac{y_{6}}{x_{6}}=1.4142$$ $$y_{6}=2x_{5}+y_{5}= (2)(70)+99=239 and x_{6}=x_{5} +y_{5}=70+99=169$$ $$ n=7, y_{7}=577, x_{7}=408, \frac{y_{7}}{x_{7}}=1.4142$$ $$y_{7}=2x_{6}+y_{6}= (2)(169)+239=577 and x_{7}=x_{6} +y_{6}=169+239=408$$ $$ n=8, y_{8}=1393, x_{8}=985, \frac{y_{8}}{x_{8}}=1.4142$$ $$y_{8}=2x_{7}+y_{7}= (2)(408)+577=1393 and x_{8}=x_{7} +y_{7}=408+577=985$$ $$ n=9, y_{9}=3363, x_{9}=2378, \frac{y_{9}}{x_{9}}=1.4142$$ $$y_{9}=2x_{8}+y_{8}= (2)(985)+1393=3363 and x_{9}=x_{8} +y_{8}=985+1393=2378$$ $$ n=10, y_{10}=8119, x_{10}=5741, \frac{y_{10}}{x_{10}}=1.4142$$ $$y_{10}=2x_{9}+y_{9}= (2)(2378)+3363=8119 and x_{10}=x_{9} +y_{9}=2378+3363=5741$$

Plot the Graph

Egm6322.s12.sungsik (talk) 21:08, 27 March 2012 (UTC)

= R 12.4 - Computing the convergents of $$\sqrt 2$$=

Given
Continued fraction formula for $$\sqrt 2$$:
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$$  \displaystyle \sqrt 2 = 1 + \frac{1}{2 + \frac{1}{2 + \frac{1}{2 + \frac{1}{2 + \frac{1}{2 + {2 + \frac{1}{2 + \frac{1}{\ddots}}}}}}}} $$     (12.4.1)
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First four convergents:
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$$  \displaystyle \begin{align} C_0 &:= 1\\ C_1 &:= 1 + \frac{1}{2} = \frac{3}{2}\\ C_2 &:= 1 + \frac{1}{2 + \frac{1}{2}} = \frac{7}{5}\\ C_3 &:= 1 + \frac{1}{2 + \frac{1}{2 + \frac{1}{2}}} = \frac{17}{12}\\ \end{align} $$     (12.4.2)
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Find
Write a Matlab program to compute the convergents $$C_i$$ for i = 1,2,...10 and plot these convergents along with the line $$\sqrt 2$$ to visualize the convergence $$\lim_{i \to\infty} C_i = \sqrt 2$$

Solution
This problem mainly deals with Matlab implementation. From Eq. 12.4.1 it is clear that the quotient of the terms in the denominator is always 2 due to the relation $$\sqrt 2 = 1 + (\sqrt 2 - 1)$$. It is seen in the [[media:Pea2.s12.sec48.e.djvu|Mtg 48e]] that the term on the right can be expressed as $$\frac{1}{\sqrt 2 + 1}$$; substituting in the first relation then produces the quotient of 2.

When implementing this code in Matlab, it is more convenient to 'build up' the residue term starting with the $$i_{max}$$ subtraction. As all contributions to the evaluation after $$i_{max}$$ are ignored, this value is simply 1/2. It also defines the final level of the denominator which must be computed. Thus, a for loop is constructed between [$$i_{max}$$:-1:2] to iteratively develop the residue term. This is seen in the Matlab code in lines 11 - 14. The computation itself is quite simple; each successive iteration simply adds the quotient of 2 onto the previously evaluated derivative.

The convergents are computed in Line 19 by adding the first quotient of 1 onto each residue term. The final values are then plotted against the exact value (to machine precision) of $$\sqrt 2$$ and are seen to converge within 5% of the exact value as soon as the second convergent. It can also be seen that the first 4 terms match those given by Eq. 12.4.2.

Matt Shields 11:40,24 March 2012 (UTC)

= R 12.5 Legendre polynomials and continued fraction =

Given
$$\frac{a_1}{x+}\frac{-\frac{1}{2}}{\frac{3}{2} x +} \frac{-\frac{3}{4}}{\frac{7}{4} x +} ... \frac{-1\frac{2(n-1)-1}{2(n-1)}}{\frac{4(n-1)-1}{2(n-1)} x +} ...$$

Find
(i) Verify the last term for the nth approximant as shown above. (ii) The Legendre polynomials $$\{P_1, ..., P_6\}$$ and compare the results to those in R10.2 on p. 44b-6. (iii) What can you observe about the factor in $$P_n(x)$$ when generated using the contnued fraction compared to the factor of $$P_x(x)$$ as obtained in R10.2? (iv) Would the difference in these factors change the orthogonality of the polynomials generated by the continued fraction (1) p. 48-24b?

Solution
Part (i)

The solution requires an analysis of the continued fraction given for this problem.

 $$\frac{a_1}{x+}\frac{-\frac{1}{2}}{\frac{3}{2} x +} \frac{-\frac{3}{4}}{\frac{7}{4} x +} ... \frac{-1\frac{2(n-1)-1}{2(n-1)}}{\frac{4(n-1)-1}{2(n-1)} x +} ...$$

There is a general pattern of continued fractions.

$$c_1 = b_0 + \frac{a_1}{b_1} = 0 + \frac{a_1}{b_1}$$

$$c_2 = \frac{a_1}{b_1 + \frac{a_2}{b_2}}$$

$$c_3 = \frac{a_1}{b_1 + \frac{a_2}{b_2 + \frac{a_3}{b_3}}}$$

$$c_4 = \frac{a_1}{b_1 + \frac{a_2}{b_2 + \frac{a_3}{b_3 + \frac{a_4}{b_4}}}}$$

....
$$c_n = {\frac{a_1}{b_1} +} {\frac{a_2}{b_2} +} {\frac{a_3}{b_3} +}...{\frac{a_n}{b_n}}$$

In this case
$$a_n = {-1\frac{2(n-1)-1}{2(n-1)}}$$

$$b_n = \frac{4(n-1)-1}{2(n-1)}x$$

Part (ii)

By applying the formula from part(i), we get

$$c_1 = \frac{a_1}{x}$$

Therefore, the 1st Legendre polynomial is

$$P_1(x) = x$$

Next, we get the 2nd Legendre polynomial.
$$\begin{alignat}{3} C_2(x) & = \frac{a_1}{x + \frac{-1}{2*32x}} \\ & = \frac{a_1}{x - \frac{1}{3x}} \\ & = \frac{3a_1}{3x^2 - 1} \end{alignat}$$

Therefore, the 2nd Legendre polynomial is

$$P_2(x) = 3x^2 - 1$$

The 3rd Legendre polynomial is just found by the 3rd convergent.

$$C_2(x) = \frac{a_1}{x + \frac{-0.5}{32x + \frac{-0.75}{{\frac{7}{4}} x}}}$$

The inverse to get the Legendre polynomial is $$(5x^3 - 3x)$$. From this pattern, it is clear that for each Legendre polynomial the related factor is missing when compared to 10.6. To confirm this pattern, let us rely on these two sources.

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

http://digitalcommons.uconn.edu/cgi/viewcontent.cgi?article=1079&context=chem_educ

C.W. David of University of Connecticut above demonstrates that the Legendre polynomials are derived from continued fractions in such a way that the constant of the solution is not necessary since the derived polynomial fits the general Legendre differential equation as a solution.
$$P_4(x) = (35x^4 - 30x^2 + 3)$$

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

$$P_6(x) = (231x^6 - 315x^4 + 105x62 - 5)$$

Part (iii)

In part (ii) we found that the fractional term at the front of each Legendre component is absent in the solutions found by continued fractions. For example, the fraction (1/2) is missing from the 2nd and 3rd Legendre polynomials for continued fractions. The method of derivation based on continued fractions produce solutions based on the derivates of the Legendre differential equations and recursive insertion of solutions. The ratio of derivatives results in an integration to give the Legendre function.

Part (iv)

The orthogonal property is not dependent on the factor term. The orthogonality is based on this term below, so the solutions from continuing fractions with convergents are still orthonal. This solution is not dependent on the fractional term derived from Rodrigues formula. The absence of the fractional term does not affect orthongality, which is still valid for continued fractional solutions.
$$\int_{-1}^{1}P_n(x)P_m(x)dx = \frac{2}{2m+1}\delta_{mn}$$

Egm6322.s12.team2.steele.m2 (talk) 20:30, 28 March 2012 (UTC)

=Contributing Members=

= References for Report 12 = King, A.C., J. Billingham and S.R. Otto. "Differential Equations: Linear, Nonlinear, Ordinary, Partial." New York, NY: Cambridge University Press, 2003.

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

http://digitalcommons.uconn.edu/cgi/viewcontent.cgi?article=1079&context=chem_educ

Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 45h) [[media:Pea2.s12.sec45.h.djvu|Mtg 45h]] Spring 2012.

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

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

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

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

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.

http://mathworld.wolfram.com/Fourier-LegendreSeries.html

title: User:Egm6322.s12.team2/HW12


text:
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.

mean = $$\mu$$ and variance = $${\sigma}^2$$

Set argument of integrand
 $$ t = \frac{(x-u)}{\sigma\sqrt{2}}$$

Analyze integral for CDF.
$$F_X(x) = \frac12\left[1 + \operatorname{erf} \left(\frac{x-\mu}{\sigma\sqrt{2}} \right) \right]$$

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

$$t = \frac{(x - \mu)}{\sigma \sqrt{2}}$$

$$dt = \frac{dx}{\sigma \sqrt{2}}$$

Therefore, the integral for erf(t) becomes
$$erf(x) = \frac{2}{\sqrt{2 \pi} \sigma} \int_0^x \exp{-(\frac{(x-u)}{\sigma \sqrt{2}})^2}dx$$

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

$$\begin{alignat} {2} F_X(x) & = \frac12\left[1 + \operatorname{erf} \left(\frac{x-\mu}{\sigma\sqrt{2}} \right) \right] \\ & = \frac12[1 + 2*\int_0^x \mathbb{PDF} dx ] \end{alignat}$$

Step 3: Derive (2) from (1)

$$f_X(x) = \frac{dF_X(x)}{dx}$$

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

title: User:Egm6322.s12.team2/HW13


text:
User:Egm6322.s12.team2/HW14

= R 14.1 Analysis of Coefficient Bn in 3-Term Recurrence Relation =

Given
$$p_(n+1)(x) = (A_nx + B_n)p_n(x) - C_np_{n-1}(x)$$

If the highest coefficient of $$p_n(x)$$ is $$k_n > 0$$, then $$A_n = \frac{k_{n+1}}{k_n}$$ and $$C_n = \frac{A_n}{A_{n-1}} \frac{h_n}{h_{n-1}}$$

Find
Find the epxression for $$B_n$$ similar to the epressions for $$A_n$$ and $$C_n$$ in (4) - (5) of p.48j-39, respectively.

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

Solution
The general proof to derive An can be applied to Bn. Recall the following equation from lecture 48j.

$$p_n\in \mathcal P_n \Rightarrow p_n(x) = \sum_{i=0}^n k_i^n x^i$$ $$(14.1.1)$$ $$$$

$$s(x) := p_{n+1}(x) - B_nxp_n(x) \in \mathcal P_{n+1}$$ $$(14.1.2)$$ $$$$

$$=\sum_{i=0}^{n+1} k_i^{n+1}x^i - B_n \sum_{i=0}^{n} k_i^nx^{i+1}$$ $$=\sum_{j=-1}^{n} k_{j+1}^{n+1}x^{j+1} - B_n \sum{i=0}^{n} k_i^n x^{i+1}$$ The key is to select the coefficient that will cancel the highest order of s(x). The solution is below. $$B_n = \frac{k_{n+1}}{k_n}$$

Manuel Steele Signed Egm6322.s12.team2.steele.m2 (talk) 19:59, 25 April 2012 (UTC)

= R 14.2 - The relation of coefficient $$ h_n$$ of the polynomial $$ p_n$$=

Given
Using the following fomula

$$\displaystyle C_{n}=\frac{A_{n}}{A_{n-1}}\frac{h_{n}}{h_{n-1}}$$ (14.2.1) Where $$\displaystyle {\color{red}C_{n}} $$ is the coefficent of the following orthogonal polynomials $$\displaystyle {P_{n}(x)}$$  relations:

$$\displaystyle p_{n+1}(x)=(A_{n}x+B_{n})p_{n}(x)-{\color{red}C_{n}}p_{n-1}(x)$$

Find
Prove that

$$\displaystyle h_{n}=h_{0}\frac{A_{0}}{A_{n}}C_{1}C_{2}\ldots C_{n}$$ (14.2.2) where $$ h_n$$ represents the square of the magnitude of the polynomial $$ p_n$$.

Solution
From the 14.2.1, we have

$$\displaystyle \begin{array}{l} C_{1}=\frac{A_{1}}{A_{0}}\frac{h_{1}}{h_{0}}\\ \\ C_{2}=\frac{A_{2}}{A_{1}}\frac{h_{2}}{h_{1}}\\ \\ C_{3}=\frac{A_{3}}{A_{2}}\frac{h_{3}}{h_{2}}\\ \vdots\\ C_{n-1}=\frac{A_{n-1}}{A_{n-2}}\frac{h_{n-1}}{h_{n-2}}\\ \\ C_{n}=\frac{A_{n}}{A_{n-1}}\frac{h_{n}}{h_{n-1}} \end{array} $$ (14.2.3) then multiply left hand side by left hand side and right hand side by right hand side.

$$\displaystyle C_{1}\cdot C_{2}\cdot C_{3}\cdot\ldots C_{n-1}\cdot C_{n}=\frac{\bcancel{A_{1}}}{A_{0}}\frac{\bcancel{h_{1}}}{h_{0}}\cdot\frac{\cancel{A_{2}}}{\bcancel{A_{1}}}\frac{\cancel{h_{2}}}{\bcancel{h_{1}}}\cdot\frac{\bcancel{A_{3}}}{\cancel{A_{2}}}\frac{\bcancel{h_{3}}}{\cancel{h_{2}}}\cdot\cdots\cdot\frac{\bcancel{A_{n-1}}}{\cancel{A_{n-2}}}\frac{\bcancel{h_{n-1}}}{\cancel{h_{n-2}}}\cdot\frac{A_{n}}{\bcancel{A_{n-1}}}\frac{h_{n}}{\bcancel{h_{n-1}}}$$ (14.2.4) thus

$$\displaystyle C_{1}\cdot C_{2}\cdot C_{3}\cdot\ldots C_{n-1}\cdot C_{n}=\frac{1}{A_{0}h_{0}}\frac{A_{n}h_{n}}{1} $$ (14.2.5) and the equation 14.2.2 exists.

= R 14.3 - Coefficients of Recurrence Relation =

Given
$$\displaystyle (n+1)P_{n+1}-(2n-1)\mu P_n + nP{n-1}=0$$ (14.3.1)

$$\displaystyle P_{n+1}= (A_{n}(x)+B_{n})P_{n}(x)-C_{n} P_{n-1}(x)$$ (14.3.2)

Find
Finding Coefficients $$\ A_n, \ B_n, $$ and $$ C_n$$  in Legendre  Polynomials.

Solution
Solving Eq (14.3.1) for $$ P_{n+1}$$  and  comparing  with  Eq(14.3.2) $$(n+1)P_{n+1}=(2n+1){\mu}P_n - nP_{n-1}$$

$$\displaystyle P_{n+1}=\frac{2n+1}{n+1}{\mu}P_n-\frac{n}{n+1}P_{n-1}$$
(14.3.3) Comparing Eq(14.3.3)  with  Eq(14.3.2) Therefore, We  get $$A_n=\frac{2n+1}{n+1}$$ $$B_n=0$$ $$C_n=\frac{n}{n+1}$$ Egm6322.s12.sungsik (talk) 02:07, 25 April 2012 (UTC)

= R 14.4 - Re-derivation of the recurrence relation of the norm of Legendre polynomials =

Given
Recurrence relation:
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$$  \displaystyle h_n = \frac{2n-1}{2n+1}h_{n-1} $$     (14.4.1) Corollary for $$h_n$$:
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$$  \displaystyle h_n = h_0\frac{A_0}{A_n}C_1C_2....C_n $$     (14.4.2) Definition of C:
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$$  \displaystyle C_n = \frac{A_n}{A_{n-1}}\frac{h_n}{h_{n-1}} $$     (14.4.3)
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Find
Reconcile Eq. 14.4.2 with Eq. 14.4.1 and obtain the following expression for $$h_n$$:
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$$  \displaystyle h_n = \frac{2}{2n+1} $$     (14.4.4)
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Solution
Eq. 14.4.2 can be re-expressed for $$h_{n-1}$$ and substituted directly into Eq. 14.4.1 to easily reconcile the two equations:
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$$  \displaystyle h_n = \frac{2n-1}{2n+1}h_0\frac{A_0}{A_n}C_1C_2....C_{n-1} $$     (14.4.5) The C terms in the equation can be expanded out and, using the definition of C similar to problem 14.2, it is seen that:
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$$  \displaystyle h_n = \frac{2n-1}{2n+1}h_0\frac{A_0}{A_n}\frac{h_{n-1}A_{n-1}}{h_0A_0} $$     (14.4.6)
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Cancelling terms yields the same result as Eq. 14.4.1, indicating that both recurrence relations are valid for Legendre polynomials:
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$$  \displaystyle h_n = \frac{2n-1}{2n+1}h_{n-1} $$     (14.4.7) The $$h_{n-1}$$ term needs to be reduced down to the known value of $$h_0$$ term, which can easily be found by direct integration and used as a basis for all h terms. Eq. 14.4.7 can be rewritten:
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$$  \displaystyle \begin{align} h_n = \frac{2n-1}{2n+1}h_{n-1} = \frac{2(n-1)+1}{2(n-1)+3}h_{n-1} \end{align} $$     (14.4.8) Likewise,
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$$  \displaystyle \begin{align} h_{n-1} = \frac{2(n-1)-1}{2(n-1)+1}h_{n-2} = \frac{2(n-2)+1}{2(n-2)+3}h_{n-2} \end{align} $$     (14.4.9) Substituting Eq. 14.4.9 into Eq. 14.4.8 will result in the cancellation of the numerator of 14.4.8 and the denominator of 14.4.9; extending this method all the way down to the $$h_0$$ term yields the final equation for $$h_n$$ in terms of $$h_0$$:
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 *

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$$  \displaystyle \begin{align} h_{n} = \frac{2(n-n)+1}{2(n-1)+3}h_{0} \end{align} $$     (14.4.10) $$h_0$$ is obtained by taking the inner product of the 0th Legendre polynomial, $$P_0=1$$:
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 *

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

$$  \displaystyle h_{0} = \int_{-1}^1 P_0(x)^2dx = x|^1_{-1}=2 $$     (14.4.11) Substituting Eq 14.4.11 into Eq 14.4.10 yields the desired expression for $$h_n$$:
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 *

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$$  \displaystyle \begin{align} h_{n} = \frac{1}{2(n-1)+3}(2) = \frac{2}{2n-2+3} = \frac{2}{2n+1} \end{align} $$     (14.4.12)
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 *

 * }

= R 14.5 Derivation of Christoffel-Darboux Identity for Legendre Polynomials =

Given
We are given the Christoffel-Darboux (CD) identity for Legendre polynomials. $$\sum_{k=0}^n \frac{P_k(x) P_k(y)}{h_k} = \frac{P_{n+1}(x) P_n(y) - P_n(x) P_{n+1}(y)}{\beta_n h_n(x-y)}$$

Find
Derive the form of the Christoffel-Darboux identity as applied to the Legendre polynomials in (1) p.45d-16e.

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

Solution
The general Christoffel-Darbou identity was derived with the 3-term recurrence relation, use of symmetry, subsequent subtraction, and rearrangement with appropriate terms to get the form given in lecture. However, this form can be modified for Legendre polynomials. As with the general derivation for polynomials, a derivation specifically applied to Legendre polynomials requires a review of Legendre equations.

Step 1: Review Legendre Equations
$$(n+1)P_{n+1}(x) - (2n+1) x \, P_n(x) + nP_{n-1}(x) = 0$$
$$(14.5.1)$$
$$$$

$$\langle P_m, P_n \rangle = \int_{\mu = -1}^{\mu = 1} P_m(\mu) \, P_n(\mu) \, d\mu = \frac{2}{2n+1} \delta_{mn}$$
$$(14.5.2)$$ The modified version needed for our derivation is below.
$$h_n := \left \langle P_n,P_n \right \rangle = \frac{2}{2n+1}$$ The following relationships are also required to understand the derivation. The leading coefficient of P_n(x) is $$\alpha_n$$
$$P_n(x) = \alpha_n \pi (x)$$ The coefficient, $$\beta_n$$, is also defined below.
$$\beta_n := \frac{\alpha_{n+1}}{\alpha_n}$$ Step 2: Analyze Solution Pattern from Lecture 48l The key is to identify the primary coefficient in the 3-term recurrence relation for the solution. In the general case, the coefficient $$A_n$$ was chosen. So the other two coefficients were eliminated. In the 1st equation (14.5.3) a primary 3-term relationship was defined by multipliying the 3-term relationship with $$p_n(y)$$.
$$p_{n+1}(x)p_n(y) = (A_nx + B_n)p_n(x)p_n(y)-C_np_n(y)p_{n-1}(x)$$
$$(14.5.3)$$ The relationship was used to generate a 2nd modifed 3-term relationship, a 2nd equation (14.5.4).
$$p_{n+1}(y)p_n(x) = (A_ny + B_n)p_n(x)p_n(y) - C_np_n(x)p_{n-1}(y)$$
$$(14.5.4)$$ Equation 17.5.4 was subtracted from 17.5.3. This resulted in a new equation where $$B_n$$ was eliminated.
$$p_{n+1}pn(y) - p_{n+1}(y)p_n(x) = A_n(x-y)p_n(x)p_n(y) - C_n[p_n(y)p_{n-1}(x) - p_n(x)p_{n-1}(y)]$$ Next, $$C_n$$ is eliminated by defining the relationship below.
$$C_n = \frac{A_n}{A_{n-1}}$$ This gives the new relationship defined with $$A_n$$ as a main coefficient.
$$\frac{p_{n+1}(x)p_n(y) - p_{n+1}(y)p_n(x)}{A_n(x-y)}= p_n(x)p_n(y) + \frac{p_n(x)p_{n-1}(y) - p_n(y)p_{n-1}(x)}{A_{n-1}(x-y)}$$ This equation can be modified by decrementing n by 1 in an iterative approach until the Christoffel-Darboux identity is obtained. Step 3: Apply Solution Pattern to Legendre Polynomials This process can be reapplied to the Legendre equations with Legendre-based coefficients. The key is to understand those coefficients for an appropriate ansatz.
$$(n+1)P_{n+1}(x) - \frac{2}{h_n} x \, P_n(x) + nP_{n-1}(x) = 0$$ Rearrange the equation and multiply by $$P_n(y)$$.
$$(n+1)P_{n+1}(x)P_n(y) = \frac{2}{h_n} x \, P_n(x) P_n(y) + nP_{n-1}(x)P_n(y)$$
$$(14.5.5)$$ Next, create a 2nd equation that flips the "x" and "y" symmetry from the former equation.
$$(n+1)P_{n+1}(y)P_n(x) = \frac{2}{h_n} y \, P_n(y) P_n(x) + nP_{n-1}(y)P_n(x)$$
$$(14.5.6)$$ Subtract 14.5.6 from 14.5.5 to get the form below.
$$(n+1)P_{n+1}(x)P_n(y) = \frac{2}{h_n} x \, P_n(x) P_n(y) + nP_{n-1}(x)P_n(y)$$ The rearrangement gives the following form.
$$P_n(x)P_n(y)*(x-y)*(2n+1) = (n+1)(P_{n+1}(x)P_n(y) - P_{n+1}(y)P_n(x)) - nP_{n-1}(x)P_n(y) + nP_{n-1}(y)P_n(x)$$ A term-by-term rearrangement requires a quantitative description, but due to lack of time, the summarized outcome is given below.
$$\frac{P_{n+1}(x)P_n(y) - P_{n+1}(y)P_n(x)}{\beta_nh_n(x-y)} = P_n(x)P_n(y) + \frac{P_n(x)P_{n-1}(y) - P_n(y)P_{n-1}(x)}{\beta_{n-1} h_{n-1}(x-y)}$$
$$$$ As in step 1, the equation above can be modified by decrementing n by 1 in an iterative approach until the Christoffel-Darboux identity is obtained but for Legendre form. Manuel Steele Signed Egm6322.s12.team2.steele.m2 (talk) 20:00, 25 April 2012 (UTC)

=Contributing Members=

= References for Report 14 = 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://en.wikipedia.org/wiki/Legendre_polynomials Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 48i) [[media:Pea2.s12.sec48.i.djvu|Mtg 48i]] Spring 2012. Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 48j) [[media:Pea2.s12.sec48.j.djvu|Mtg 48j]] Spring 2012. Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 48k) [[media:Pea2.s12.sec48.k.djvu|Mtg 48k]] Spring 2012. Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 48l) [[media:Pea2.s12.sec48.l.djvu|Mtg 48l]] Spring 2012.

title: User:Egm6322.s12.team2/HW14


text:
User talk:Je7den/enes100/My Work

week 0 p1 grading 1sfoerster 20:18, 21 February 2012 (UTC)
 * 5/10 ... no problem statements on either page
 * Great job on researching the CoreControl and finding the issue with the vacuum.

Allenad 10:56, 27 February 2012 (UTC)
 * Good work on finding out about the old projects now we just have to fix where these other guys failed.

week 1 p1 grading 1sfoerster (talk) 10:19, 2 March 2012 (UTC)
 * 10/10/0 ... where is the value added? Merely a report of activity .. link to your research .. link to testimonials. Going to do is in task list, not narrative .. there is nothing to leverage, nothing that inspires, no details that push the project forward.

week 2 p1 grading 1sfoerster (talk) 21:20, 6 March 2012 (UTC)
 * 10/0/10 ... excellent patent find .. where is the tutorial? please link it up next time so I can give you more points .. no third task, no positive comments on team mates

Allenad 10:20, 8 March 2012 (UTC)
 * Good work on finding The patent for us, we could really find a lot of design help with it.

week 3 p1 grading 1sfoerster (talk) 12:44, 15 March 2012 (UTC)
 * 10/5/0 ... what is next?
 * Good job on creating the decision matrix for the corecontrol vacuum.

week 4 p1 grading 1sfoerster (talk) 14:06, 21 March 2012 (UTC)
 * 0/0/0 .. nothing to grade

title: User talk:Je7den/enes100/My Work


text:
User talk:Shatton8111/enes100/my work

week 0 p1 grading 1sfoerster 20:20, 21 February 2012 (UTC)
 * 10/10 .. great job

Good work writing the program for the arduino! Jromanch0088 (talk) 18:30, 26 February 2012 (UTC)

Dont know how you created a program but good job doing it Spark93 (talk)9:42, 01 March 2012

week 1 p1 grading 1sfoerster (talk) 11:18, 2 March 2012 (UTC)
 * 10/10/100 ... excellent job ... but some of this was done this week .. this is week 2 .. doesn't really matter

good work on getting the graph of the accelerometer up and working! Jromanch0088 (talk) 04:23, 5 March 2012 (UTC)

week 2 p1 grading 1sfoerster (talk) 21:48, 6 March 2012 (UTC)
 * 0/5/0 ... nice comments on team mates pages ...

Nice job programming the arudino to spin with out the knob stick Spark93 (talk) 9:59, 11 March 2012 (UTC)

Great work writing the program to get the arduino to spin up and increase speed! All we have to do now is get the motor shield hooked up so we can run two motors at the same time! Jromanch0088 (talk) 03:13, 12 March 2012 (UTC)

I like your program for Esc you are a great programmer. Amirreza Saharkhiz

week 3 p1 grading 1sfoerster (talk) 12:54, 15 March 2012 (UTC)
 * 10/10/160 ... great job !!

NIce job fixing the arudino codes so it works for the 2 motors. Spark93 (talk) 10:36, 18 March 2012 (UTC)

You made a nice program to control the speed of motor with audrino board, you are a good programmer.Amirreza Saharkhiz(talk)

Great job on wiring the ESCs and motors together! sorry I wasnt there to help! Jromanch0088 (talk) 18:03, 20 March 2012 (UTC)

week 4 p1 grading 1sfoerster (talk) 14:20, 21 March 2012 (UTC)
 * 0/10/250 ... I can not give you enough points to justify the work you are doing ... I need you to stay in the class!!! .. need to fill out next team page and user page

title: User talk:Shatton8111/enes100/my work


text:
User talk:Mikehicks/enes100/My Work

week 0 p1 grading 1sfoerster 20:21, 21 February 2012 (UTC)
 * 5/10 .. root user page not set up

Hey man you been working hard on getting those motors to work with the arduino. Keep on keeping on man. J4knee21 (talk) 03:21, 27 February 2012 (UTC)

week 1 p1 grading
 * 10/0/0 ... what did not work associated with the arduino .. what part of the wikiversity tutorial on the arduino did you get stuck at? Did you improve the tutorial? Frustration is when there are the most points for documentation.

I'm pretty excited about where this project is heading. You've been doing a great job with arduinos and giving me inspiration during class. Represent! J4knee21 (talk) 00:55, 4 March 2012 (UTC)

week 2 p1 grading 1sfoerster (talk) 21:39, 6 March 2012 (UTC)
 * 0/0/0 .. nothing to grade

Good work Hicks J4knee21 (talk) 02:02, 12 March 2012 (UTC)

week 3 p1 grading 1sfoerster (talk) 14:12, 14 March 2012 (UTC)
 * 10/10/10 ... looks like you had some success with the stepper motor, need to go through the tutorial and make the link to the stepper motor testing example program clearer ... since it looks like you had trouble finding it ... more push points if you hand used this tension to improve the documentation

Hey man your motor mounts are rating about a 9.9 out of 10 on my nifty scale bro. J4knee21 (talk) 00:56, 19 March 2012 (UTC)

week 4 p1 grading 1sfoerster (talk) 13:42, 21 March 2012 (UTC)
 * 5/10/120 ... great job .. finally some documentation that can be leveraged !! ... keep going ... no week 0 task filled out on the new project

title: User talk:Mikehicks/enes100/My Work


text:
User talk:Inh117/enes100/My work

week 0 p1 grading 1sfoerster 20:23, 21 February 2012 (UTC)
 * 5/10 .. personal problem statement is not appropriate

Hey man nice work. If you come up with brilliant ideas please let me know .I'll try find youtube videos on our project, hopefully today. Sali6089

The shrouds of the PC fan look best from the front, therefore we don't really require a megaphone. I checked the price of a megaphone, its kinda expensive. Did you find a video or something, that could help us. I spent half an hour trying to find a video, but most of them are only for those people who actually want to make a wind turbine. Sali6089

Hey man you found an amazing video,  pure genius, keep me posted if anything comes up in the brilliant mind of yours. Sali6089 3 March 2012

Hey dude give me your no. I'll be going there tomorrow, I was busy over the weekend.I found a thing called poster strips on their website, is this the perferrated mounting strip?Sali6089 5 March 2012

--hey bro nice idea with the aerodynamic proposal pure genious. keep me posted about anything you may need Deakin2074 (talk) 15:45, 28 February 2012 (UTC)

week1 p1 grading 1sfoerster (talk) 10:07, 2 March 2012 (UTC)
 * 10/10/10 .. link to video is marginally better than original, random logic (must be new .. books I've read) .. what books? what research proves it is new? When is brushless motor not good for power generation? Why? What is a source for the debate? What leads you to believe the fan blades need to be reshaped? Everything undocumented falls into the opinion category ..

week2 p1 grading 1sfoerster (talk) 19:30, 6 March 2012 (UTC)

I got it man! wait till tues. thanks for your help man.Sali6089

 * 10/10/50 ... more points if you actually link to the solid works file that has been uploaded to google docs ... the makerbot is working ... convert it to parts that can be made in the makerbot!

Hey man nice work over this week. Hopefully we will finish our project by next week.I m thinking about using pieces of card board, that will be surrounded on the metal hangers, would that work, I'll explain more on Tues. Sali6089 10 March 2012

hey bro, how goes the shroud? let me know if you need any materials/help/moral support for the rest of the design. Deakin2074 (talk) 16:19, 13 March 2012 (UTC)

week 3 p1 grading 204.126.132.241 14:35, 14 March 2012 (UTC)
 * 10/10/20 ... good push of project, need drawing linked up and perhaps to a solid works viewer ... so everyone can see your drawing in three dimensions ... need materials list for more points .. then should start looking for parts ... looks too much like you are letting the parts at hand determine your design

Man! Today we did a great job. High Five man.Sali6089 16 March 2012

I created a new pg for our FanWing proj. Did u comeup with anything for the poster? I've filled the entire team pg, we just need a posterSali6089 17 March 2012

week 4 p1 grading 1sfoerster (talk) 19:47, 20 March 2012 (UTC)
 * 5/10/50 ... good job ... no first task in the next project

title: User talk:Inh117/enes100/My work


text:
User talk:Navid jahanpour/enes100/my work

week 0 p1 grading 1sfoerster 20:25, 21 February 2012 (UTC)
 * 10/0 .. no personal problem statement, no commitment to doing something for the team over the weekend

week 1 p1 grading 1sfoerster (talk) 21:04, 28 February 2012 (UTC)
 * 0/0/0 ... nothing to grade

week 2 p1 grading 1sfoerster (talk) 21:05, 6 March 2012 (UTC)
 * 10/5/25 ... good start .. need positive comments on team mates discuss pages

week 3 p1 grading 1sfoerster (talk) 20:31, 13 March 2012 (UTC)
 * 0/0/0 ... nothing to grade .. I know you are working on it ... need to document failure ... not just document success

week 4 p1 grading 1sfoerster (talk) 19:01, 20 March 2012 (UTC)
 * 0/5/80 ... please create a new user page with week 1 on it and a new team page ... need to put positive comments on team mates pages, ... finally some documentation

Comments
Hey, you've been doing good work on the high-voltage foil. Just ask if you need my assistance. -Michael


text:
User talk:Gjaksec/ENES100/Mywork

week 0 p1 grading 1sfoerster 20:26, 21 February 2012 (UTC)
 * 10/10 .. good job

week 1 p1 grading 1sfoerster (talk) 21:01, 28 February 2012 (UTC)
 * 10/10/0 .. why didn't you find the screw drivers to order? looks like both video's use them ...

week 2 p1 grading 1sfoerster (talk) 20:57, 6 March 2012 (UTC)
 * 10/10/25 ... great job ... upload your 123D sketch to google docs and link to earn more points!

week 3 p1 grading 1sfoerster (talk) 20:24, 13 March 2012 (UTC)
 * 10/5/30 ... great job ... need to make positive comments on your team mates pages

week 4 p1 grading 1sfoerster (talk) 18:57, 20 March 2012 (UTC)
 * 10/5/50 ... the gun you built exactly matched your design ... but I don't see any pictures of it or a tutorial on how to build it .. there are some pictures in the poster you made .. that is why there are push points

week 1 p3 grading 1sfoerster (talk) 19:15, 24 April 2012 (UTC)
 * 5/10/10 ... should have created a new personal page for work done rather than reusing p1 page

week 2 p3 grading 1sfoerster (talk) 17:52, 1 May 2012 (UTC)
 * 10/10/150 ... excellent ... do that each week!

week 3 p3 grading 1sfoerster (talk) 14:00, 8 May 2012 (UTC)
 * 10/10/50 ... excellent ... keep it up .. make sure finish the team page even though there is only one of you

week 4 p3 grading 1sfoerster (talk) 21:13, 15 May 2012 (UTC)
 * 10/10/0 .. good job

title: User talk:Gjaksec/ENES100/Mywork


text:
User talk:2jjones/enes100/Mywork

week 0 p1 grading 1sfoerster 20:28, 21 February 2012 (UTC)
 * 10/0 did not commit to doing something for the team over the weekend

week 1 p1 grading 1sfoerster (talk) 20:47, 28 February 2012 (UTC)
 * 10/10/0 ... would have lots of points if you documented what you did rather than talk about it ... 123D tutorial, screen shots uploaded with pictures of the design ... all I have right now is a report

week 2 p1 grading
 * 10/5/5 ... 5 points for picture uploaded ... more points if file uploaded to google docs and then linked ... no positive comments on team mates pages ... week2 narrative is just a report ... what is the pin layout? Shoot a video of what you did with the NXT light sensor and brick .. otherwise no value added to someone repeating your steps

week 3 p1 grading 1sfoerster (talk) 19:55, 13 March 2012 (UTC)
 * 10/5/20 ... soldering concerns me .. what is to stop the wires from touching each other where you soldered? more points if you commented on team mates ages ... "researched layouts" .... what is the result of your research? "tested them all" what did you test for, what were the results? "tried to find software" ... what is the result of this? if failures, these were all opportunities to document failure and get push points for them ... and start discussions ... and improve presentations ...

week 4 p1 grading 1sfoerster (talk) 16:38, 20 March 2012 (UTC)
 * 10/5/0 ... nothing to grade on week for page, no positive comments on team mates pages

title: User talk:2jjones/enes100/Mywork


text:
User talk:Pjmarsden/enes100/My Work

week 0 p1 grading 1sfoerster 20:29, 21 February 2012 (UTC)
 * 10/10 ... good job

Team Member comment Kyle Gallun 15:37, 1 March 2012
 * Very knowledgeable about mechanics, helpful for hands-on construction, and easy to work with.

week 1 p1 grading 1sfoerster (talk) 11:03, 2 March 2012 (UTC)
 * 10/10/10 ... it is not what you do but what you document that will get you points in this class. Confine plans to the next task description. Research is too vague. Power = energy velocity = voltage * current. Need to measure both voltage and current ... or voltage and the value of a resistor.

week2 p1 grading 1sfoerster (talk) 19:22, 6 March 2012 (UTC)
 * 10/5/20 ... need more pictures, good description of what was done, but better would be drawing of circuit, how the 100 ohm resistor was calculated, what assumptions were made, and link to the table of recorded voltage and current outputs and how the test equipment was hooked up. Without this detail, the project can only vaguely be repeated .. no positive comments on your team mates pages.

Team Member comment Kyle Gallun 15:53, 12 March 2012
 * Did a good job designing and constructing the new wind turbine blades. Sweet test videos! Too bad your camera man couldn't hold the turbine!

Team Member comment Erika Spangler 11:21, 12 March 2012

I appreciate you physically constructing the larger prototype, it generators alot more power with the radiator fan and motor, giving kyle and I more room to play around with our circuits. I look foward to your new design for the turbine blades.

week 3 p1 grading 1sfoerster (talk) 14:23, 14 March 2012 (UTC)
 * 10/10/100 ... lots of constructive work .. good job .. now comes the finish push

Team Member comment Kyle Gallun 17:09, 19 March 2012
 * Awesome new fan blades, I knew we would need something more sturdy than cardboard.

Team Member comment Erika Spangler--erikams 00:40, 21 March 2012 (UTC) Clever design on the fan blades and really good documentation.

week 4 p1 grading 1sfoerster (talk) 13:50, 21 March 2012 (UTC)
 * 10/10/80 ... keep it up !!!!

title: User talk:Pjmarsden/enes100/My Work


text:
User talk:Kmashaye5220/enes100/myWork

week 0 p1 grading 1sfoerster 20:32, 21 February 2012 (UTC)
 * 10/10 ... good job

week 1 p1 grading 1sfoerster (talk) 21:03, 28 February 2012 (UTC)
 * 10/10/50 ... good job normal

week 2 p1 grading 1sfoerster (talk) 21:07, 6 March 2012 (UTC)
 * 10/5/80 ... excellent job .. need positive comments on your team mates pages

Nice work, I like how you cleaned it up compared to mine. Mine was a little sloppy because of the electrical tape. I'm glad we both had two different ATX supplies to work with to see what other wires to implement to the power project. - Chris Nunez

week 3 p1 grading 1sfoerster (talk) 20:39, 13 March 2012 (UTC)
 * 10/10/50 .... need to put comments in the box surrounding the pictures .. good job otherwise ... wow

You have improved your site input! keep it up, This is the way everyone should do it. I have had a great time and learned so much being teamed with you. Thank you! - Chris Nunez

week 4 p1 gradng 1sfoerster (talk) 19:05, 20 March 2012 (UTC)
 * 10/10/75 .. great job ... keep it up ... next step is testing

title: User talk:Kmashaye5220/enes100/myWork


text:
User talk:Mnugent/enes100/My Work

week 0 p1 grading 1sfoerster 20:33, 21 February 2012 (UTC)
 * 10/10 .. good job

week 1 p1 grading 1sfoerster (talk) 13:28, 29 February 2012 (UTC)
 * 10/10/0 ... I have personally installed the remote software on windows 7 machines ... you have made a blanket statement that it can not be done. What am I to think? You have not recorded the symptoms, what you tried, what constitutes working or not working. There is no documentation. Only opinion. You get points for recording failure details in this class.

week2 p1 grading 1sfoerster (talk) 20:38, 6 March 2012 (UTC)
 * 5/0/0 ... no narrative, no third task, remote wonder program worked? what are you talking about? no installation instructions, no expectations set ... no positive comments on your team mates pages

Weekend 4:Ehh, mikey you havent written anything down yet. Good job working on that suspense, or whatever.

week 3 p1 grading 1sfoerster (talk) 13:37, 14 March 2012 (UTC)
 * 0/0/0 ... nothing to grade

title: User talk:Mnugent/enes100/My Work


text:
User talk:Christopher.nunez/enes100/myWork

week 0 p1 grading 1sfoerster 20:34, 21 February 2012 (UTC)
 * 10/10 ... good job

week 1 p1 grading 1sfoerster (talk) 21:09, 28 February 2012 (UTC)
 * 10/10/0 ... good report, but not documentation ... should have taken pictures of the capacitors, shown in a video how to discharge them and perhaps created a spark ... this would have pushed the project forward

week 2 p1 grading 1sfoerster (talk) 21:12, 6 March 2012 (UTC)
 * 10/5/80 ... excellent ... need positive comments on your team mates pages

Great job! you did it very well. I like your pictures and explanations. your page is really comprehensive tutorial for ever one who wants to convert a power supply. keyvan

week 3 p1 grading 1sfoerster (talk) 20:35, 13 March 2012 (UTC)
 * 10/10/50 .... the upside down graphics need to be fixed ... learn how to use gimp to rotate the picture ... see me if you need help ... nee to put the labels in the picture .. this is a wiki editing issue ... see me .. good start

As I said before, your page is a really good and complete tutorial. Also, you have taught me many thing during the project. I appreciate my experience with you. good luck. Keyvan

week 4 p1 grading 1sfoerster (talk) 19:07, 20 March 2012 (UTC)
 * 10/10/5 ... there is always room for improvement ... next step is testing


text:
User:Egm6322.s12.team2.Xia/RP10.4

=R*10.4 -Weight Function =

Given
The following equation: $$\displaystyle w_{j}=\frac{-2}{(n+1)P_{n}'(x_{j})P_{n+1}(x_{j})}$$ (10.4.1)

where the Legendre polynomial is $$\displaystyle P_{n}(x)=\sum^{m}_{r=0}(-1)^r \frac{(2n-2r)!x^{n-2r}}{2^{n}r!}(n-r)!(n-2r)!$$. (10.4.2)

Find
Verify that $$\displaystyle w_{2}=1$$

Solution
According to the 10.4.2, we have $$\displaystyle {\displaystyle P_{2}(x)=\frac{1}{2}(3x^{2}-1)}$$ (10.4.3) $$\displaystyle P_{3}(x)=\frac{1}{2}(5x^{3}-3x)$$ thus $$\displaystyle P_{2}^{'}(x)=3x$$ and the roots of 10.4.3 are $$\displaystyle x_{1}=\frac{1}{\sqrt{3}},\quad x_{2}=-\frac{1}{\sqrt{3}} $$ Interms of 10.4.1 $$\displaystyle w_{2}=\frac{-2}{(2+1)P_{2}'(x_{2})P_{2+1}(x_{2})}$$ then $$\displaystyle w_{2}=\frac{-2}{(2+1)\cdot3\cdot\frac{-1}{\sqrt{3}}\cdot\frac{1}{2}\cdot\left(5\cdot(\frac{-1}{\sqrt{3}})^{3}-3\cdot\frac{-1}{\sqrt{3}}\right)}$$ $$\displaystyle =\frac{-2}{(2+1)\cdot3\cdot\frac{-1}{\sqrt{3}}\cdot\frac{1}{2}\cdot\left(\frac{4\sqrt{3}}{9}\right)}$$ $$\displaystyle =1$$

title: User:Egm6322.s12.team2.Xia/RP10.4


text:
User talk:Spark93/enes100/my work

week 0 p1 grading 1sfoerster 20:36, 21 February 2012 (UTC)
 * 5/0 .. no direct link to project page, do have indirect link, no team or personal problem statement, no description of work done for the team over the weekend.

Hey, thanks for trying to figure out the Thunderbird 9 ESC. That one can be really difficult. -- Shatton8111 (talk) 16:51, 24 February 2012 (UTC)

Good job working on the Thunderbird 9 ESC and researching designs! Jromanch0088 (talk) 18:32, 26 February 2012 (UTC)

week 1 p1 grading 1sfoerster (talk) 11:16, 2 March 2012 (UTC)
 * 10/10/0 ... research without links, research that didn't start in previous teams pages or the video's already in the world page of this project is meaningless, a report on what you did without documentation doesn't enable anyone to reproduce what you did

Yeah, thanks for working on designs! I see you mentioned something about gyros being useful. I haven't figured those out yet, so I'll send you an e-mail or ask you in class. :) Shatton8111 (talk) 19:11, 3 March 2012 (UTC)

week 2 p1 grading 1sfoerster (talk) 21:46, 6 March 2012 (UTC)
 * 0/0/0 .. nothing to grade

Excellent work getting the Thunderbird 9 ESC going! Shatton8111 (talk) 16:56, 10 March 2012 (UTC)

Awesome job getting the Thunderbird 9 ESC to work! We can finally move forward with the rest of the project now! Jromanch0088 (talk) 03:14, 12 March 2012 (UTC)

I love the blog about building the quadcopter, a page away from your link. Shatton8111 (talk) 01:49, 14 March 2012 (UTC)

week 3 p1 grading 1sfoerster (talk) 12:53, 15 March 2012 (UTC)
 * 10/0/10 ... would have been a lot more points if you had created a tutorial on how the knobs of the power supply function .. took pictures, uploaded and linked into a tutorial .. because the power supply is used where ever batteries are needed .. you learned something and did not document it ... now others will suffer like you did ... lost transparency on the "programming aspect" no comments on team mates pages .. this is why the middle 0

We have the Arduino controlling two ESC's and motors, so I think we can try four motors. The helicopter designs you linked to have really gotten me thinking about how we'll design ours. I can't wait to see what Jromanch0088 is designing! Shatton8111 (talk) 02:30, 19 March 2012 (UTC) Great job on getting both the motors hooked up!! Jromanch0088 (talk) 18:00, 20 March 2012 (UTC)

week 4 p1 grading 1sfoerster (talk) 14:18, 21 March 2012 (UTC)
 * 0/5/5 ... your goal should be to repeat steve's success and build tutorials ... not by talking to steve but by looking at his previous weeks work ... then taking pictures and enriching the documentation so others have an easier time following it ... right now you are still blurring what he is doing and what you are doing ... this is loosing transparency ... one of the pictures is broken

title: User talk:Spark93/enes100/my work


text:
User:Egm6322.s12.team2.Xia/RP10.8

=R 10.8 -Lagrange Interpolation =

Given
The following Lagrange interpolation: $$\displaystyle p(x)\thickapprox p^{h}(x):=\sum_{j=1}^{n}p(x_{j})\ell_{j}(x)\in\mathcal{P}_{n-1} $$ (10.8.1)

where Lanrage basis function is $$\displaystyle \ell_{j}(x):=\frac{\displaystyle \prod_{k=1,k\neq j}^{n}(x-x_{k})}{\displaystyle \prod_{k=1,k\neq j}^{n}(x_{j}-x_{k})}\in\mathcal{P}_{n-1}$$. (10.8.2) Define $$\displaystyle d(x):=p(x)-p^{h}(x)\in\mathcal{P}_{n-1}$$. $$\displaystyle $$

Find
1.Verify $$\displaystyle \ell_{j}(x_{i})=\delta_{ji}$$. 2.Show that: $$\displaystyle d(x_{i})=0$$, for $$\displaystyle i=1,\ldots,n$$ and thus $$\displaystyle d(x)=0,\forall x\Rightarrow p(x)\equiv p^{h}(x)$$

Solution
1, From the equation 10.8.2, when $$\displaystyle i\neq j$$, we have $$\displaystyle {\displaystyle \ell_{j}(x_{i}):=\frac}=\displaystyle \prod_{k\neq j}\frac{x_{i}-x_{k}}{x_{j}-x_{k}}=\frac{(x_{i}-x_{0})}{(x_{j}-x_{0})}\cdots\frac{\overbrace{(x_{i}-x_{i})}^{=0}}{(x_{j}-x_{i})}\cdots\frac{(x_{i}-x_{k})}{(x_{j}-x_{k})}=0$$ an when $$\displaystyle i=j$$ $$\displaystyle \ell_{i}(x_{i})=\ell_{j}(x_{j}):=\prod_{k\neq j}\frac{x_{j}-x_{k}}{x_{j}-x_{k}}=1$$ Therefore $$\displaystyle \ell_{j}(x_{i})=\delta_{ji}$$ 2, We know that $$\displaystyle d(x):=p(x)-p^{h}(x)$$ Therefore $$\displaystyle d(x_{i}):=p(x_{i})-p^{h}(x_{i})=p(x_{i})-\sum_{j=1}^{n}p(x_{j})\ell_{j}(x_{i})$$ From above solution, we know that $$\displaystyle \ell_{j}(x_{i})=\delta_{ji}$$,thus $$\displaystyle d(x_{i})=p(x_{i})-\sum_{j=1}^{n}\underbrace{p(x_{j})\delta_{ji}}_{=0,\, if\, i\neq j}=p(x_{i})-p(x_{i})=0$$ and thus $$\displaystyle d(x)=0,\forall x\Rightarrow p(x)\equiv p^{h}(x)$$ $$\displaystyle $$ $$\displaystyle $$

title: User:Egm6322.s12.team2.Xia/RP10.8


text:
User talk:1wPhipps/enes100

week 0 p1 grading 1sfoerster 20:38, 21 February 2012 (UTC)
 * 10/10 ... good job

title: User talk:1wPhipps/enes100


text:
User talk:Natebeger/enes100/My Work

week 0 p1 grading 1sfoerster 20:39, 21 February 2012 (UTC)
 * 10/10 .. good job

mirza
 * Glad you looked into the facial recognition software on your computer, and looked if it can be used for our use.

sam
 * good job coming up with the idea of where to mount the Kinect!

week 1 p1 grading 1sfoerster (talk) 13:23, 29 February 2012 (UTC)
 * 10/15/10 ... the extra 5 points in the middle is for finding and putting comments on team mates discuss page ... good job .. the 10 push points is good, but you are talking circles around the recognition time, numbers please, I can neither determine the scale or the scope of the problem (scale would be one face)(scope would be scanning through 100 faces that may come into the room).

week 2 p1 grading 1sfoerster (talk) 20:24, 6 March 2012 (UTC)
 * 10/5/40 .. good push ... expanding the search for facial recognition software that sorts through photos ... keep going ... did not make positive comments on team mates pages

sam
 * Nice job coming up with the idea on how to mount the Kinect! Keep up the good work!

mirza Good stuff with looking at Google's picasa software hopefully you make headway there.

week 3 p1 grading 1sfoerster (talk) 13:36, 14 March 2012 (UTC)
 * 10/10/40 ... good push ... please develop a tutorial for using the google software for more points

sam
 * Nice job getting getting our project to work!

mirza
 * You came through at the end getting the project to work great job.

week 4 p1 grading 1sfoerster (talk) 14:03, 21 March 2012 (UTC)
 * 10/10/90 ... good find ... combining yawcam and picasa .. wow

title: User talk:Natebeger/enes100/My Work


text:
User talk:Wesley Rodriguez/enes100/My Work

week 0 p1 grading 1sfoerster 20:42, 21 February 2012 (UTC)
 * 5/5 ... can not get to your team and personal coanda pages from personal page, don't have any work mentioned that you are doing over the weekend

week 1 p1 grading 1sfoerster (talk) 10:45, 2 March 2012 (UTC)
 * 0/0/0 ... nothing to grade

week 2 p1 grading 1sfoerster (talk) 21:29, 6 March 2012 (UTC)
 * 0/0/0 ... nothing to grade

week 3 p1 grading 1sfoerster (talk) 14:09, 14 March 2012 (UTC)
 * 0/0/0 ... nothing to grade

Hey Adam...Whats Up, we haven't seen you in a while..Hope you're ok. 303428mwatson (talk) 05:16, 10 March 2012 (UTC)

Im going to head on to the engineering labs early Tuesday to work on the final design.303428mwatson (talk) 08:01, 19 March 2012 (UTC)

week 4 p1 grading
 * 10/0/0 .. nothing to grade


text:
User talk:Jromanch0088/enes100/My work

week 0 p1 grading 1sfoerster 20:44, 21 February 2012 (UTC)
 * 5/10 ... no team problem statment

Thanks for getting the Zagi ESC working! Because of that, we can actually get things going. Shatton8111 (talk) 15:33, 24 February 2012 (UTC)

Thanks for working on Esc finally you can figure out it.

Nice job getting the ESC to get the motor spinning. Spark93 (talk) 9:39, 01 March 2012

week 1 p1 grading 1sfoerster (talk) 11:13, 2 March 2012 (UTC)
 * 10/10/0 ... good report, but that doesn't get you documentation points .. need video, pictures, documentation of how wires were hooked up .. what quad copter, what were the pairing problems, what about the landing gear breaking, what about the battery charging, there is large task list of things to do that are not mentioned that fly the quad copter should have generated

Hey. It sounds like the quadcopter piloting went awesomely! Sorry I wasn't there. Thanks for doing that! Shatton8111 (talk) 19:02, 3 March 2012 (UTC)

week 2 p1 grading 1sfoerster (talk) 21:46, 6 March 2012 (UTC)
 * 10/5/20 ... two links are great ... the testing video was excellent .. need to start making that .. need to make positive comments on team mates pages

Thanks for working on the model design and for adding the stuff about gyros. We're finally at the gyro. testing phase! Shatton8111 (talk) 16:54, 10 March 2012 (UTC).

Amirreza Saharkhiz(talk) Designing is hard work but you are good in designing and i like your search.

Thanks for figuring out the burn-out test with me. That was fun! Thanks for working on the model and gyros too! Shatton8111 (talk) 02:32, 14 March 2012 (UTC)

week 3 p1 grading
 * 10/10/10 ... ok .. the gyro's are already purchased ... need a gimbal built

I hope the model is coming along well. What I saw of it looked awesome! Shatton8111 (talk) 02:19, 19 March 2012 (UTC) Cant believe one wire ruined was the problem besides that nice job on the wiring. Spark93 (talk) 10:38, 18 March 2012 (UTC)

Great job man Amirreza Saharkhiz(talk)

week 4 p1 grading
 * 0/10/0 ... would get push points if pictured failed drawing, what was tried ... a clear problem with a tutorial you are following would lead to someone else solving your problem and improving the tutorial, right now there is no documentation describing where you are stuck ... this is a problem with the world not having a tutorial for you, not your personal problem ... nothing set up for the next project

title: User talk:Jromanch0088/enes100/My work


text:
User talk:Dross4565/enes100/My Work

week 0 p1 grading
 * 10/0 nothing planned to do over the weekend

Comment
Very good job with staying on top of the research and good adaptability to change project from unlocking car to changing it into a race car.

week 1 p1 grading 1sfoerster (talk) 13:54, 29 February 2012 (UTC)
 * 10/10/0 ... good report on what you did, but no fruit that pushes the project forward. The ECU is a good idea. You say you did "research" but I see only the word. I don't see a tutorial. I don't see anything of value that pushes the project forward. The ECU needs to be documented. What is it? Can we communicate with it through the OBD? Is it the same on every car?

week 2 p1 grading 1sfoerster (talk) 21:37, 6 March 2012 (UTC)
 * 10/5/0 ... link to the ECU, take pictures of it, figure out everything you can about it ... purchasing it was a good start, where did you get it? how much did it cost? is it working or can you tell ? this is an excellent starting point .. need to run with it. Need to put positive comments on team mates pages.

week 3 p1 grading 1sfoerster (talk) 13:59, 14 March 2012 (UTC)
 * 10/5/100 ... great job .. think we have a starting point .. the forum has a bunch of arduino projects that add value to the dashboard of the car ... great pictures .. need to put positive comments on team mates pages

week 4 p1 grading 1sfoerster (talk) 20:48, 20 March 2012 (UTC)
 * 0/5/20 ... do see some push here, no documentation other than the screen shot of an arduino program ... need reference to it ... no p2 team page or personal page set up ... no positive comments on team mates pages

title: User talk:Dross4565/enes100/My Work


text:
User talk:Amirreza Saharkhiz/enes100/my work

week 0 p1 grading 1sfoerster 20:47, 21 February 2012 (UTC)
 * 5/10 ... no problem statement for the team

Thanks for programming and figuring out the Zagi ESC's logic with me. The demo we've got is pretty cool! Shatton8111 (talk) 15:34, 24 February 2012 (UTC)

Thanks for working on the Thunderbird 9 ESC with me, we will figure it out soon! Jromanch0088 (talk) 18:34, 26 February 2012 (UTC)

week 1 p1 grading 1sfoerster (talk) 11:22, 2 March 2012 (UTC)
 * 10/10/0 ... english is a problem ... can not tell what is planned, what was done, link to a blog article on ESC's is good, but should be improviing wikiversity tutorials ... don't see anything here I could repeat .. need documentation on all the wiring attempts ... the goal is record the problem .. not wait until there is a solution to write anything

I hope the Thunderbird ESC isn't giving you too much trouble. Thanks for trying to figure it out! Shatton8111 (talk) 19:15, 3 March 2012 (UTC)

week 2 p1 grading 1sfoerster (talk) 21:50, 6 March 2012 (UTC)
 * 10/5/50 ... great documentation .. great links .. two options for more power ... pc power supplies and the lipo batteries ordered .. then can continue this! ... need to put positive comments on your team mates pages

That is awesome that you got the Thunderbird 9 ESC working! By the way, I love the speed controller video. That guy makes it look easy. Shatton8111 (talk) 17:11, 10 March 2012 (UTC)

Hey man great job getting the Thunderbird 9 working! We can finally move on from trying to get the motor working! Jromanch0088 (talk) 03:16, 12 March 2012 (UTC)

week 3 p1 grading 1sfoerster (talk) 12:57, 15 March 2012 (UTC)
 * 10/5/0 ... links don't work, link up ESC programming, no programming code is linked up ... transparency confusion with steve's work ... if you are building on top of his program, understanding his program, then create a tutorial that goes through each line of his code and explains what it does in your own words ... so that others have a chance of understanding it better ..

I love the pictures you took for week 4! I had forgotten that we had tried two power supplies. Thanks for recording that! Shatton8111 (talk) 02:14, 19 March 2012 (UTC) Great job last class helping with the wiring of the 2 ESC running on one arduino. Shatton8111 (talk) 10:41, 18 March 2012 (UTC)

Great job finding the certain power the ESCs run at!! Jromanch0088 (talk) 18:05, 20 March 2012 (UTC)

week 4 p1 grading 1sfoerster (talk) 14:22, 21 March 2012 (UTC)
 * 0/10/50 ... good job with photos ... need to turn these into a tutorial ... repeat what steve has done ... split up chasing steve with your team mates ...