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 style="text-align:center;">$$ p(x) = q(x)P_n(x) + r(x) \in \mathcal P_{2n-1}$$

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.

<p style="text-align:center;">$$n < \deg p(x) \le 2n-1$$

<p style="text-align:center;">$$\deg s(x) = n = 2$$

Therefore, in theory, <p style="text-align:center;">$$\deg r(x) \le n-1=1$$

<p style="text-align:center;">$$\deg q(x) \le n-1=1$$

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

Step 1: Rearrange Polynomial Equation <p style="text-align:center;">$$\begin{alignat}{2} p(x) & = 3x^3 + 2x^2 + 3x + 1 \\ & = 3x^3 + 2x^2 + \frac{9x}{5} + \frac{6}{5} + \frac{6x}{5} - \frac{1}{5} \\ \end{alignat} $$

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

<p style="text-align:center;">$$\begin{alignat}{3} q(x) & = \frac{p(x) - r(x)}{s(x)} \\ & = \frac{3x^3 + 2x^2 + \frac{9x}{5} + \frac{6}{5}}{5x^2 + 3} \\ & = \frac{(3x+2)(5x^2+3)}{5(5x^2+3)} \\ \end{alignat}$$

Therefore, the solution for q(x) is <p style="text-align:center;">$$q(x) = \frac{(3x+2)}{5}$$

The degree of q(x) is 1 so that the following relation is confirmed. <p style="text-align:center;">$$\deg q(x) \le n-1=1$$

Step 3: Solve for r(x): We start with the original equation. <p style="text-align:center;">$$p(x) = q(x)s(x) + r(x)$$

Rerrange the equation to solve for r(x) and solve. <p style="text-align:center;">$$\begin{alignat}{3} r(x) & = p(x) - q(x)s(x) \\ & = 3x^3 + 2x^2 + 3x + 1 - 3x^3 - 2x^2 - \frac{9x}{5} - \frac{6}{5} \\ & = 3x - \frac{9x}{5} + 1 - \frac{6}{5} \\ \end{alignat}$$

Therefore, the solution of r(x) is below. <p style="text-align:center;">$$r(x) = \frac{6x}{5} - \frac{1}{5}$$

On a final note, since the degree of r(x) is 1 we confirm that <p style="text-align:center;">$$\deg r(x) \le n-1=1$$

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, $$\displaystyle w_{j}>0,\; j=1,\ldots,n$$ <p style="text-align:right">(11.4.1) 2, $$\displaystyle \sum_{j=1}^{n}w_{j}=2$$ <p style="text-align:right">(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.