User:Brandonhua

Eml5526.s11.team7 http://en.wikiversity.org/wiki/User:Fem.tm7/HW2#Problem_Description http://www.codecogs.com/latex/eqneditor.php http://bcs.wiley.com/he-bcs/Books?action=resource&bcsId=3625&itemId=0470035803&resourceId=9928

Part 2
FB p.91 pbs 4.6: Use Gauss quadrature to obtain exact values for the following integrals. Verify by analytical integration a) ::{| style="width:100%" border="0" $$ \displaystyle \int_{0}^{4}\left ( x^{2}+1\right )dx $$       (6.8.1) b)
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$$ \displaystyle \int_{-1}^{1}\left ( \xi^{4}+2\xi^{2}\right )d\xi $$       (6.8.3) c) Write a MATLAB code that utilizes function gauss.m and performs Gauss integration. Check your manual calculations against the MATLAB CODE.
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4.6: Use three-point Gauss quadrature to evaluate the following integrals. Compare to the analytical integral a) ::{| style="width:100%" border="0" $$ \displaystyle \int_{-1}^{1} \frac{\xi}{\xi^{2}+1}   d\xi $$        (6.8.3) b) ::{| style="width:100%" border="0" $$ \displaystyle \int_{-1}^{1} cos^{2}(\pi\zeta )  d\xi $$       (6.8.4) c) Write a MATLAB code that utilizes function gauss.m and performs Gauss integration. Check your manual calculations against the MATLAB CODE. 4.8: The integral $$ \int_{-1}^{1}  \left ( 2\xi^{3}+2 \right )  d\xi $$ can be integrated exactly using two-point Gauss quadrature. How is the accuracy affected if: a) one-point quadrature is employed; b) three-point quadrature is employed. Check your solution against MATLAB code.
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Solution
4.6:
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$$ \displaystyle 2n_{gp}-1=3\Rightarrow n_{gp}=2 $$       (6.8.5)
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$$ \displaystyle \begin{bmatrix} 1 & 1\\ \xi_{1} & \xi_{2}\\ \xi^{2}_{1}& \xi^{2}_{2}\\ \xi^{3}_{1}& \xi^{3}_{2} \end{bmatrix} \begin{bmatrix} W_{1}\\ W_{2} \end{bmatrix}= \begin{bmatrix} 2\\ 0\\ \frac{2}{3}\\ 0 \end{bmatrix} $$       (6.8.6)
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$$ \displaystyle W_{1}=W_{2}=1, \xi_{1}= -\frac{1}{\sqrt{3} }, \xi_{2}=\frac{1}{\sqrt{3}}, a=0, b=4, dx=2d\xi $$       (6.8.8)
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$$ \displaystyle x=\frac{1}{2}(a+b)+\frac{1}{2}\xi(b-a) $$       (6.8.9)
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$$ \displaystyle x=\frac{1}{2}(0+4)+\frac{1}{2}\xi(4-0) $$       (6.8.10)
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$$ \displaystyle x=2+2\xi $$       (6.8.11)
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$$ \displaystyle I=\int_{-1}^{1}((2+2\xi)^{2}+1)2d\xi=2\int_{-1}^{1}(4\xi^{2}+8\xi+5)d\xi $$ $$ I= 2 \left [W_{1}(4\xi^{2} _{1} +8\xi_{1}+5) +W_{2}(4\xi^{2}_{2}+8\xi_{2}+5) \right ] $$ $$ I= 2 \left [(4 \left ( -\frac{1}{\sqrt{3} } \right ) ^{2} +8 \left (-\frac{1}{\sqrt{3}}  \right ) +5) + (4 \left ( \frac{1}{\sqrt{3} }  \right ) ^{2} +8 \left (\frac{1}{\sqrt{3}}  \right ) +5)  \right ] $$       (6.8.12)
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$$ \displaystyle
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I=2*(\frac{8}{3}+10)=\frac{76}{3}=25.\overline{33}

$$     (6.8.13) This is the same answer using WolframAlpha b)
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$$ \displaystyle n_{gp}=3 $$       (6.8.14) The integral is evaluated from -1 to 1
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$$ \displaystyle W_{1}=\frac{8}{9}, W_{2}=W_{3}=\frac{5}{9}$$ $$ \xi_{1}=0, \xi_{2}=-\frac{\sqrt{15}}{5},\xi_{3}=\frac{\sqrt{15}}{5} $$       (6.8.15)
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$$ \displaystyle I=\int_{-1}^{1}\left ( \xi^{4}+2\xi^{2} \right )d\xi $$ $$ I=W_{1}\left ( \xi_{1} ^{4}+2\xi_{1}^{2} \right )+W_{2}\left ( \xi_{2} ^{4}+2\xi_{2}^{2} \right )+W_{3}\left ( \xi_{3} ^{4}+2\xi_{3}^{2} \right ) $$ $$ I=0 + \frac{5}{9} \left ( \left (-\frac{\sqrt{15}}{5} \right ) ^{4}+2\left (-\frac{\sqrt{15}}{5}  \right )^{2} \right )+\frac{5}{9} \left ( \left (\frac{\sqrt{15}}{5}  \right ) ^{4}+2\left (\frac{\sqrt{15}}{5} \right )^{2} \right )$$ $$ I=\frac{5}{9} \left (2\frac{225}{625}  +4\frac{15}{25}  \right ) $$
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(6.8.16)
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$$ \displaystyle
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I=\frac{26}{15}=1.7\overline{33}

$$     (6.8.17) This is the same answer using WolframAlpha
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4.6: 4.8:

Solution
The exponential basis function is given by:
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$$ \displaystyle {u^{h}}(x)={d_{j}}\left (e^{j(x-1)}-1 \right ) $$       (5.2.exp.1)
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$$ \displaystyle \frac{{u^{h}}(x)}{dx} ={d_{j}}\left (e^{j(x-1)} \right ) $$       (5.2.exp.2) The general form of the stiffness matrix becomes:
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$$ \displaystyle K=\int_{0}^{1}i*e^{i(x-1)}*(2+3x)*j*e^{j(x-1)}dx $$       (5.2.exp.3) The general form of the force matrix becomes:
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$$ \displaystyle F=12(e^{i(0-1)}-1)+\int_{0}^{1} 5x*i(e^{i(x-1)}-1)dx $$       (5.2.exp.4) Using MatLab, the K and F matrix with N=10 is:
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$$ \displaystyle K=\begin{bmatrix} 1.716166179 &	2.633475288	& 3.170329089	& 3.512453499	&3.74690156	&3.917139098	&4.046398014	&4.147965319	&4.229930538	&4.297494435\\ 2.633475288	&4.227105451	&5.268680249	&5.995042496	&6.528565163	&6.93668231	&7.258939308	&7.519876512	&7.735489984	&7.916648746\\ 3.170329089	&5.268680249	&6.744422808	&7.834278196	&8.670852887	&9.332921967	&9.869837922	&10.31398664	&10.68747581	&11.00590793\\ 3.512453499	&5.995042496	&7.834278196	&9.248909746	&10.3699133	&11.27981477	&12.03298442	&12.66663799	&13.20708952	&13.67346515\\ 3.74690156	&6.528565163	&8.670852887	&10.3699133	&11.74980705	&12.89248331	&13.85413531	&14.67454391	&15.38264829	&15.99999816\\ 3.917139098	&6.93668231	&9.332921967	&11.27981477	&12.89248331	&14.24996774	&15.4082711	&16.40815817	&17.27999802	&18.04687424\\ 4.046398014	&7.258939308	&9.869837922	&12.03298442	&13.85413531	&15.4082711	&16.7499948	&17.91999794	&18.94921795	&19.86159138\\ 4.147965319	&7.519876512	&10.31398664	&12.66663799	&14.67454391	&16.40815817	&17.91999794	&19.24999918	&20.42906542	&21.48148136\\ 4.229930538	&7.735489984	&10.68747581	&13.20708952	&15.38264829	&17.27999802	&18.94921795	&20.42906542	&21.74999987	&22.93628804 \end{bmatrix} $$       (5.2.exp.5)
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$$ \displaystyle F=\begin{bmatrix} -194.4238\\ 556.4552\\ -794.5674\\ 388.5998\\ 355.4029\\ -550.4317\\ 202.6984\\ 13.1421\\ 77.9578\\ -68.7552\\ -267.2934\\ 294.8667\\ 224.2897\\ -420.1257 150.8844 \end{bmatrix} $$       (5.2.exp.5) Solving for the general solution with n=6 gives an error of 2.83E-06 at u(.5) and a general solution of:
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$$ \displaystyle u^{h}=-167.2599*e^{1*(x-1)}	487.6591e^{2*(x-1)}	-808.7098e^{3*(x-1)}	770.2175e^{4*(x-1)}	-393.7902e^{5*(x-1)}	83.8821e^{6*(x-1)}+32.0012 $$       (5.2.exp.5) The boundary condition u(1)=4 is kept and the exact solution is very close to the trial solution:
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Author
Brandonhua 04:06, 2 March 2011 (UTC)

Problem Description
Similar to HW5.{1,3,7}, solve G1DM1.0/D1b using Quadratic Lagrangian Element Basis Function (QLEBF) with uniform discretization (equidistant element nodes) until convergence of $${{u}^{h}}(0.5)$$ to $$O({{10}^{-6}})$$. ( $$nel=2,4,6,8,...$$).


 * 1. For $$nel=2$$, compute $$\mathbf{\tilde{K}}=\sum\limits_{e=1}^{2}$$, with $${{\mathbf{\tilde{K}}}^{e}}$$ by (1)p.30-5, display $${{\mathbf{\tilde{K}}}^{e}},e=1,2$$.


 * 2. Compute $${{\mathbf{k}}^{e}}$$, $${{\mathbf{L}}^{e}}$$, for $$e=1,2$$.


 * 3. Compute $${{\tilde{\Kappa }}^{e}}={{\mathbf{L}}^{e}}^{T}{{\mathbf{k}}^{e}}{{\mathbf{L}}^{e}}$$, for $$e=1,2$$, compare to the result by (1).


 * 4. Plot all QLEBF for $$nel=3$$.


 * 5. Plot $$u_^{h}$$ vs. $$u$$, $$\left[ u_^{h}(0.5)-u(0.5) \right]$$ vs. $$\tilde{n}$$

Part 1
For element 1, the 3 basis functions are:
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$$ \displaystyle
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b_1^1 = \frac \cdot \frac = 8{x^2} - 6x + 1

$$     (6.1.1)
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$$ \displaystyle
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b_2^1 = \frac \cdot \frac = - 16{x^2} + 8x

$$     (6.1.2)
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$$ \displaystyle
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b_3^1 = \frac \cdot \frac = 8{x^2} - 2x

$$     (6.1.3) By (1)p.30-5,
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$$ \displaystyle
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K_{ij}^e = \int\limits_ {b{{_i^e}^\prime }(x){a_2}(x)b{{_j^e}^\prime }(x)dx}

$$     (6.1.4) Thus, we have
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$$ \displaystyle
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K_{11}^1 = \int\limits_0^{0.5} {\left( {16x - 6} \right)\left( {2 + 3x} \right)\left( {16x - 6} \right)dx} = \frac{6}

$$     (6.1.5)
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$$ \displaystyle
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K_{12}^1 = K_{21}^1 = \int\limits_0^{0.5} {\left( {16x - 6} \right)\left( {2 + 3x} \right)\left( { - 32x + 8} \right)dx} =  - \frac{3}

$$     (6.1.6)
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$$ \displaystyle
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K_{13}^1 = K_{31}^1 = \int\limits_0^{0.5} {\left( {16x - 6} \right)\left( {2 + 3x} \right)\left( {16x - 2} \right)dx} = \frac{6}

$$     (6.1.7)
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$$ \displaystyle
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K_{22}^1 = \int\limits_0^{0.5} {\left( { - 32x + 8} \right)\left( {2 + 3x} \right)\left( { - 32x + 8} \right)dx} = \frac{3}

$$     (6.1.8)
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$$ \displaystyle
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K_{23}^1 = K_{32}^1 = \int\limits_0^{0.5} {\left( { - 32x + 8} \right)\left( {2 + 3x} \right)\left( {16x - 2} \right)dx} =  - \frac{3}

$$     (6.1.9)
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$$ \displaystyle
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K_{33}^1 = \int\limits_0^{0.5} {\left( {16x - 2} \right)\left( {2 + 3x} \right)\left( {16x - 2} \right)dx} = \frac{6}

$$     (6.1.10) Then
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$$ \displaystyle
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{{\mathbf{\tilde K}}^1} = \left[ {\begin{array}{ccccccccccccccc} {65/6}&{ - 38/3}&{11/6}&0&0 \\  { - 38/3}&{88/3}&{ - 50/3}&0&0 \\   {11/6}&{ - 50/3}&{89/6}&0&0 \\   0&0&0&0&0 \\   0&0&0&0&0 \end{array}} \right]

$$     (6.1.11) Similar to element 1, when  e=2
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$$ \displaystyle
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{{\mathbf{\tilde K}}^2} = \left[ {\begin{array}{ccccccccccccccc} 0&0&0&0&0 \\  0&0&0&0&0 \\   0&0&{107/6}&{ - 62/3}&{17/6} \\   0&0&{ - 62/3}&{136/3}&{ - 74/3} \\   0&0&{17/6}&{ - 74/3}&{131/6} \end{array}} \right]

$$     (6.1.12) Thus,
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$$ \displaystyle
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{\mathbf{\tilde K}} = \sum\limits_{e = 1}^2  = \left[ {\begin{array}{ccccccccccccccc} {65/6}&{ - 38/3}&{11/6}&0&0 \\  { - 38/3}&{88/3}&{ - 50/3}&0&0 \\   {11/6}&{ - 50/3}&{98/3}&{ - 62/3}&{17/6} \\   0&0&{ - 62/3}&{136/3}&{ - 74/3} \\   0&0&{17/6}&{ - 74/3}&{131/6} \end{array}} \right]

$$     (6.1.13)
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Part 2
By (1)p.31-3,
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$$ \displaystyle
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k_{ij}^e = \int\limits_ {b{{_i^e}^\prime }(x){a_2}(x)b{{_j^e}^\prime }(x)dx}

$$     (6.1.14) For element 1
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$$ \displaystyle
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{{\mathbf{k}}^1} = \left[ {\begin{array}{ccccccccccccccc} {65/6}&{ - 38/3}&{11/6} \\  { - 38/3}&{88/3}&{ - 50/3} \\   {11/6}&{ - 50/3}&{89/6} \end{array}} \right]

$$     (6.1.15)
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$$ \displaystyle
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{{\mathbf{L}}^1} = \left[ {\begin{array}{ccccccccccccccc} 1&0&0&0&0 \\  0&1&0&0&0 \\   0&0&1&0&0 \end{array}} \right]

$$     (6.1.16) For element 2
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$$ \displaystyle
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{{\mathbf{k}}^2} = \left[ {\begin{array}{ccccccccccccccc} {107/6}&{ - 62/3}&{17/6} \\  { - 62/3}&{136/3}&{ - 74/3} \\   {17/6}&{ - 74/3}&{131/6} \end{array}} \right]

$$     (6.1.17)
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$$ \displaystyle
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{{\mathbf{L}}^2} = \left[ {\begin{array}{ccccccccccccccc} 0&0&1&0&0 \\  0&0&0&1&0 \\   0&0&0&0&1 \end{array}} \right]

$$     (6.1.18)
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Part 3
For element 1
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$$ \displaystyle
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{{\mathbf{\tilde {\rm K}}}^1} = {{\mathbf{L}}^1}^T{{\mathbf{k}}^1}{{\mathbf{L}}^1} = \left[ {\begin{array}{ccccccccccccccc} 1&0&0 \\  0&1&0 \\   0&0&1 \\   0&0&0 \\   0&0&0 \end{array}} \right]\left[ {\begin{array}{ccccccccccccccc} {65/6}&{ - 38/3}&{11/6} \\  { - 38/3}&{88/3}&{ - 50/3} \\   {11/6}&{ - 50/3}&{89/6} \end{array}} \right]\left[ {\begin{array}{ccccccccccccccc} 1&0&0&0&0 \\  0&1&0&0&0 \\   0&0&1&0&0 \end{array}} \right] = \left[ {\begin{array}{ccccccccccccccc} {65/6}&{ - 38/3}&{11/6}&0&0 \\  { - 38/3}&{88/3}&{ - 50/3}&0&0 \\   {11/6}&{ - 50/3}&{89/6}&0&0 \\   0&0&0&0&0 \\   0&0&0&0&0 \end{array}} \right]

$$     (6.1.19) For element 2
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$$ \displaystyle
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{{\mathbf{\tilde {\rm K}}}^2} = {{\mathbf{L}}^2}^T{{\mathbf{k}}^2}{{\mathbf{L}}^2} = \left[ {\begin{array}{ccccccccccccccc} 0&0&0 \\  0&0&0 \\   1&0&0 \\   0&1&0 \\   0&0&1 \end{array}} \right]\left[ {\begin{array}{ccccccccccccccc} {107/6}&{ - 62/3}&{17/6} \\  { - 62/3}&{136/3}&{ - 74/3} \\   {17/6}&{ - 74/3}&{131/6} \end{array}} \right]\left[ {\begin{array}{ccccccccccccccc} 0&0&1&0&0 \\  0&0&0&1&0 \\   0&0&0&0&1 \end{array}} \right] = \left[ {\begin{array}{ccccccccccccccc} 0&0&0&0&0 \\  0&0&0&0&0 \\   0&0&{107/6}&{ - 62/3}&{17/6} \\   0&0&{ - 62/3}&{136/3}&{ - 74/3} \\   0&0&{17/6}&{ - 74/3}&{131/6} \end{array}} \right]

$$     (6.1.20) The result is exactly same as achieved in part 1.
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Part 4
The plot of QLEBF for nel = 3:

Note: Parts 1-4 in 6.2 are exactly the same as 1-4 in 6.1. Jiang has done the work to solve 1-4 in part 6.2.

Part 5
The plots of $$u_^{h}$$ vs. $$u$$ The exact solution is $$ U_{e}= \frac{-5}{108}(206*log(3x+2)+3x(3x-4))+ \frac{515}{54}log(2)+4$$ $$ \int_0^1 \frac{dw}{dx}(2+3x) \frac{du}{dx} dx = -30w(1) + \int_0^1 w(x)5x dx \qquad \forall w \quad with \quad w(0)=0. $$



The plots of $$\left[ u_^{h}(0.5)-u(0.5) \right]$$ vs. $$\tilde{n}$$

Hua

6.8
Verify Table of
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$$ \displaystyle \left \{ \left (w_{i},x_{i} \right ),i=1,...,5 \right \} $$ NIST Handbook: Reference: http://en.wikipedia.org/wiki/Gaussian_quadrature
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FB p.89:

Using excel, the values from the NIST Handbook are calculated to be: Which is the same from table 4.1 from FB p.89