User:Eml5526.s11.team5.JA/Mtg36

Mtg 36: Wed, 25 Mar 11

[[media: Fe1.s11.mtg36.djvu| Page 36-1 ]]

Computation of $ \mathbf^{\color{Red}e} \ $ in parent coordinates $ {\color{Blue} \left \{ \xi \right \} } \ $: continued [[media: Fe1.s11.mtg35.djvu| page 35-3 ]]

Recall matrix algebra:


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$$ \mathbf{AB}^T = \mathbf{A}^T \mathbf{B}^T \ $$

$$
 *  $$ \displaystyle {\color{Red}(1)}
 * }


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$$ \mathbf{(A^{-1})}^T = \mathbf{(A^{T})}^{-1} = \mathbf{A^{-T}} \ $$

$$
 *  $$ \displaystyle {\color{Red}(2)}
 * }

Verify (1) and (2)
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Wolfram Alpha


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$$ \left \{  \left \{ 1, 1, 1, \right \}, \left \{ 2, -1, 3, \right \}, \left \{ 3, 2, 6, \right \} \right \}  * \left \{  \left \{ 1, 3, 5, \right \}, \left \{ 1, -4, 1, \right \}, \left \{ 2, 5, 8, \right \} \right \}  \ $$

Note:

$$ transpose \underbrace{\color{Red} \left \{ {\color{Black}  \left \{  \left \{ 1, 1, 1, \right \}, \left \{ 2, -1, 3, \right \}, \left \{ 3, 2, 6, \right \} \right \}  * \left \{  \left \{ 1, 3, 5, \right \}, \left \{ 1, -4, 1, \right \}, \left \{ 2, 5, 8, \right \} \right \} } \right \}  }_{\color{Red} No } \ $$

$$ transpose \underbrace{\color{Red} \left [ {\color{Black}  \left \{  \left \{ 1, 1, 1, \right \}, \left \{ 2, -1, 3, \right \}, \left \{ 3, 2, 6, \right \} \right \}  * \left \{  \left \{ 1, 3, 5, \right \}, \left \{ 1, -4, 1, \right \}, \left \{ 2, 5, 8, \right \} \right \} } \right ]  }_{\color{Red} Yes } \ $$

$$
 *  $$ \displaystyle
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$$ \left \{  \left \{ 1, 3, 5, \right \}, \left \{ 1, -4, 1, \right \}, \left \{ 2, 5, 8, \right \} \right \}^{\land} T  * \left \{  \left \{ 1, 1, 1, \right \}, \left \{ 2, -1, 3, \right \}, \left \{ 3, 2, 6, \right \} \right \}^{\land} T   \ $$

$$ invert \left \{ transpose \begin{Bmatrix} \left \{ 1, 1, 1, \right \}, \left \{ 2, -1, 3, \right \}, \left \{ 3, 2, 6, \right \} \end{Bmatrix} \right \} \ $$

Note:

$$ invert \underbrace{\color{Red} \left \{ {\color{Black} transpose \begin{Bmatrix} \left \{ 1, 1, 1, \right \}, \left \{ 2, -1, 3, \right \}, \left \{ 3, 2, 6, \right \} \end{Bmatrix} } \right \} }_{\color{Red} No } \ $$

$$ invert \underbrace{\color{Red} \left [ {\color{Black} transpose \begin{Bmatrix} \left \{ 1, 1, 1, \right \}, \left \{ 2, -1, 3, \right \}, \left \{ 3, 2, 6, \right \} \end{Bmatrix} } \right ] }_{\color{Red} Yes} \ $$

$$
 *  $$ \displaystyle
 * }

End Wolfram Alpha

[[media: Fe1.s11.mtg36.djvu| Page 36-2 ]]

Find and explain syntax of  Wolfram Alpha

[[media: Fe1.s11.mtg35.djvu| (3) - (5) page 35-2 ]], [[media: Fe1.s11.mtg35.djvu| (4) - (5)  page 35-3 ]]:


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$$ \mathbf{B}_{\color{Blue}I} ( \boldsymbol{\xi}) := \triangledown_{\color{Blue}x} N_{\color{Blue}I}^{\color{Red}e} (\boldsymbol{\xi}) = (\mathbf{J}^{\color{Red}e})^ ( \boldsymbol{\xi} ) \triangledown_{\color{Blue}\xi} N_{\color{Blue}I}^{\color{Red}e} (\boldsymbol{\xi}) \ $$ [[media: Fe1.s11.mtg35.djvu| (4) page 35-3 ]]

$$
 *  $$ \displaystyle {\color{Red}(2)}
 * }


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$$ \mathbf{B}_{\color{Blue}J} ( \boldsymbol{\xi}) \ $$ similar

$$
 *  $$ \displaystyle
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$$ \bar{\boldsymbol{\kappa}}( \boldsymbol{\xi} ) := \boldsymbol{\kappa}(\mathbf{x}^{\color{Red}e} (\boldsymbol{\xi}) ) = \boldsymbol{\kappa} ( {\color{Red}\varphi ^e} (\boldsymbol{\xi}) ) \ $$ [[media: Fe1.s11.mtg35.djvu| (5)  page 35-2 ]]

$$
 *  $$ \displaystyle {\color{Red}(3)}
 * }

[[media: Fe1.s11.mtg36.djvu| Page 36-3 ]]


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$$ J(\boldsymbol{\xi}) := det \mathbf{J} (\boldsymbol{\xi}) \ $$

$$
 * <p style="text-align:right"> $$ \displaystyle {\color{Red}(1)}
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Accurate and efficient numerical integration: Gauss- Legendre quadrature (GLQ)

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$$ I(f) = \int_{-1}^{+1} f(x) \ dx \cong \sum_{i = 1}^{\color{Red}\mu} w_i f(x_i) \underbrace{ =: }_{(1)} I_{\color{Red}\mu} (f) \ $$

$$
 * <p style="text-align:right"> $$ \displaystyle
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$$ \left \{ x_i, \ i = 1,..., {\color{Red} \mu} \right \} \ $$ = roots of Legendre ploynomial $$ {\color{Blue}P_{\color{Red}\mu} (x) } \ $$ of order $$ {\color{Red} \mu } \ $$

$$
 * <p style="text-align:right"> $$ \displaystyle
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$$ \left \{ w_i, \ i = 1,..., {\color{Red}\mu} \right \} \ $$ = weights associated with $$ \left \{ x_i \right \} \ $$

$$
 * <p style="text-align:right"> $$ \displaystyle
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source: http://en.wikipedia.org/wiki/Gaussian_quadrature, also see NM1 S11 or PEA1 F10 for more details

[[media: Fe1.s11.mtg36.djvu| Page 36-4 ]]

Example:

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$$ {\color{Red} \mu = 1} \ $$

$$
 * <p style="text-align:right"> $$ \displaystyle
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$$ I_{\color{Red}1} = 2 f(0) \ $$

$$
 * <p style="text-align:right"> $$ \displaystyle
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$$ {\color{Red} \mu = 2} \ $$

$$
 * <p style="text-align:right"> $$ \displaystyle
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$$ I_{\color{Red}2} = {\color{Blue} 1 \cdot } f \left ( \frac{-1}{\sqrt{3}}  \right ) + {\color{Blue} 1 \cdot } f  \left ( \frac{1}{\sqrt{3}}  \right ) \ $$

$$
 * <p style="text-align:right"> $$ \displaystyle
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[[media: Fe1.s11.mtg36.djvu| Page 36-5 ]]

Theorem


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Let $$ f \ $$ be a polynomial of order $$ {\color{Red} 2 \mu - 1 } \ $$, example, $$ f \in P_{\color{Red}2 \mu -1} \ $$ (set of polynomial of degree $$ \le {\color{Red} 2 \mu - 1 } \ $$ ) $$ \Rightarrow f^{({\color{Red}2 \mu}) } \equiv 0 \ $$

example, $$ f \in P_{\color{Red}1} \Rightarrow f(x) = a_1 x^1 + a_0 \Rightarrow f^{({\color{Red}2})} \equiv 0 \ $$

Hence:

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$$ {\color{Red} \mu = 1} \ $$ integrates exactly  any polynomial in $$ P_{\color{Red}1} \ $$

$$
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$$ {\color{Red} \mu = 2 } \ $$ integrates exactly  any polynomial in $$ P_{\color{Red}3} \ $$

$$
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$$ {\color{Red} \mu = 3 } \ $$ integrates exactly  any polynomial in $$ P_{\color{Red}5} \ $$

$$
 * <p style="text-align:right"> $$ \displaystyle
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Multi-variable integration

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$$ I = \int_{\Box} f(x,y) \ d \Box = \int_{\Box} f(x_1,x_2) \ d \Box \ $$

$$
 * <p style="text-align:right"> $$ \displaystyle
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$$ I_{\color{Red}\mu} = \sum_{\color{Blue}i=1}^{\color{Red}\mu} \sum_{\color{Blue}j=1}^{\color{Red}\mu} w_{\color{Blue}i}^{\color{Blue}x} w_{\color{Blue}j}^{\color{Blue}y} f(x_{\color{Blue}i}, y_{\color{Blue}j}) \ $$

$$
 * <p style="text-align:right"> $$ \displaystyle
 * }