"행렬과 연립방정식의 수식 표현"의 두 판 사이의 차이

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67번째 줄: 67번째 줄:
 
 
 
 
  
 
 
 
 
 
<math>\normalsize        \left(\large\begin{array}{GC+23}        \varepsilon_x\\\varepsilon_y\\\varepsilon_z\\\gamma_{xy}\\        \gamma_{xz}\\\gamma_{yz}\end{array}\right)\ {\Large=}        \ \left[\begin{array}{CC}        \begin{array}\frac1{E_{\fs{+1}x}}        &-\frac{\nu_{xy}}{E_{\fs{+1}x}}        &-\frac{\nu_{\fs{+1}xz}}{E_{\fs{+1}x}}\\        -\frac{\nu_{yx}}{E_y}&\frac1{E_{y}}&-\frac{\nu_{yz}}{E_y}\\        -\frac{\nu_{\fs{+1}zx}}{E_{\fs{+1}z}}&        -\frac{\nu_{zy}}{E_{\fs{+1}z}}        &\frac1{E_{\fs{+1}z}}\end{array} & {\LARGE 0} \\        {\LARGE 0} & \begin{array}\frac1{G_{xy}}&&\\        &\frac1{G_{\fs{+1}xz}}&\\&&\frac1{G_{yz}}\end{array}        \end{array}\right]        \ \left(\large\begin{array}        \sigma_x\\\sigma_y\\\sigma_z\\\tau_{xy}\\\tau_{xz}\\\tau_{yz}        \end{array}\right)</math>
 
<math>\normalsize        \left(\large\begin{array}{GC+23}        \varepsilon_x\\\varepsilon_y\\\varepsilon_z\\\gamma_{xy}\\        \gamma_{xz}\\\gamma_{yz}\end{array}\right)\ {\Large=}        \ \left[\begin{array}{CC}        \begin{array}\frac1{E_{\fs{+1}x}}        &-\frac{\nu_{xy}}{E_{\fs{+1}x}}        &-\frac{\nu_{\fs{+1}xz}}{E_{\fs{+1}x}}\\        -\frac{\nu_{yx}}{E_y}&\frac1{E_{y}}&-\frac{\nu_{yz}}{E_y}\\        -\frac{\nu_{\fs{+1}zx}}{E_{\fs{+1}z}}&        -\frac{\nu_{zy}}{E_{\fs{+1}z}}        &\frac1{E_{\fs{+1}z}}\end{array} & {\LARGE 0} \\        {\LARGE 0} & \begin{array}\frac1{G_{xy}}&&\\        &\frac1{G_{\fs{+1}xz}}&\\&&\frac1{G_{yz}}\end{array}        \end{array}\right]        \ \left(\large\begin{array}        \sigma_x\\\sigma_y\\\sigma_z\\\tau_{xy}\\\tau_{xz}\\\tau_{yz}        \end{array}\right)</math>
  
74번째 줄: 72번째 줄:
  
 
 
 
 
 
 
 
 
<br>
 
  
 
 
 
 
84번째 줄: 78번째 줄:
  
 
# \left((x)\right)= \begin{cases} x-\lfloor x\rfloor - 1/2 & \mbox{ if }x\in\mathbb{R}\setminus\mathbb{Z} \\ 0 & \mbox{ if } x\in\mathbb{Z} \end{cases}<br>
 
# \left((x)\right)= \begin{cases} x-\lfloor x\rfloor - 1/2 & \mbox{ if }x\in\mathbb{R}\setminus\mathbb{Z} \\ 0 & \mbox{ if } x\in\mathbb{Z} \end{cases}<br>
 
 
 
  
 
 
 
 
94번째 줄: 86번째 줄:
  
 
# g_1(\chi) = \begin{cases} \sqrt{p}, & p \equiv 1 \pmod{4}, \ i \sqrt{p}, & p \equiv 3 \pmod{4}. \end{cases}<br>
 
# g_1(\chi) = \begin{cases} \sqrt{p}, & p \equiv 1 \pmod{4}, \ i \sqrt{p}, & p \equiv 3 \pmod{4}. \end{cases}<br>
 
 
 
  
 
 
 
 
104번째 줄: 94번째 줄:
  
 
# \left(\frac{a}{p}\right) = \begin{cases} \;\;\,0\mbox{ if } a \equiv 0 \pmod{p} \\+1\mbox{ if }a \not\equiv 0\pmod{p} \mbox{ and for some integer }x, \;a\equiv x^2\pmod{p} \\-1\mbox{ if there is no such } x. \end{cases}<br>
 
# \left(\frac{a}{p}\right) = \begin{cases} \;\;\,0\mbox{ if } a \equiv 0 \pmod{p} \\+1\mbox{ if }a \not\equiv 0\pmod{p} \mbox{ and for some integer }x, \;a\equiv x^2\pmod{p} \\-1\mbox{ if there is no such } x. \end{cases}<br>
 
 
 
  
 
 
 
 
114번째 줄: 102번째 줄:
  
 
# \tilde y=\left\{ {\ddot x\text{ if $\vec x$ odd}\atop\hat{\,\bar x+1}\text{ if even}}\right<br>
 
# \tilde y=\left\{ {\ddot x\text{ if $\vec x$ odd}\atop\hat{\,\bar x+1}\text{ if even}}\right<br>
 
 
 
  
 
 
 
 

2009년 8월 25일 (화) 14:29 판

\(aaa\ldots bbb\cdots ccc\vdots ddd\ddots\)

  1. aaa\ldotsbbb\cdotsccc\vdotsddd\ddots

 

\(\mathbf{A} = \begin{bmatrix} 9 & 8 & 6 \\ 1 & 2 & 7 \\ 4 & 9 & 2 \\ 6 & 0 & 5 \end{bmatrix}\)

  1. \mathbf{A} = \begin{bmatrix} 9 & 8 & 6 \\ 1 & 2 & 7 \\ 4 & 9 & 2 \\ 6 & 0 & 5 \end{bmatrix}

 

 

\(\mathbf{A} = \begin{pmatrix} 9 & 8 & 6 \\ 1 & 2 & 7 \\ 4 & 9 & 2 \\ 6 & 0 & 5 \end{pmatrix}\)

  1. \mathbf{A} = \begin{pmatrix} 9 & 8 & 6 \\ 1 & 2 & 7 \\ 4 & 9 & 2 \\ 6 & 0 & 5 \end{pmatrix}

 

 

 

\(\Large A\ =\ \large\left( \begin{array}{c.cccc}&1&2&\cdots&n\\ \hdash1&a_{11}&a_{12}&\cdots&a_{1n}\\ 2&a_{21}&a_{22}&\cdots&a_{2n}\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ n&a_{n1}&a_{n2}&\cdots&a_{nn}\end{array}\right)\)

  1. \Large A\ =\ \large\left(\begin{array}{c.cccc}&1&2&\cdots&n\\         \hdash1&a_{11}&a_{12}&\cdots&a_{1n}\\         2&a_{21}&a_{22}&\cdots&a_{2n}\\         \vdots&\vdots&\vdots&\ddots&\vdots\\         n&a_{n1}&a_{n2}&\cdots&a_{nn}\end{array}\right)

 

 

\(\begin{alignat}{7} a_{11} x_1 &&\; + \;&& a_{12} x_2 &&\; + \cdots + \;&& a_{1n} x_n &&\; = \;&&& b_1 \\ a_{21} x_1 &&\; + \;&& a_{22} x_2 &&\; + \cdots + \;&& a_{2n} x_n &&\; = \;&&& b_2 \\ \vdots\;\;\; && && \vdots\;\;\; && && \vdots\;\;\; && &&& \;\vdots \\ a_{m1} x_1 &&\; + \;&& a_{m2} x_2 &&\; + \cdots + \;&& a_{mn} x_n &&\; = \;&&& b_m. \\ \end{alignat}\)

  1. \begin{alignat}{7} a_{11} x_1 &&\; + \;&& a_{12} x_2 &&\; + \cdots + \;&& a_{1n} x_n &&\; = \;&&& b_1 \\ a_{21} x_1 &&\; + \;&& a_{22} x_2 &&\; + \cdots + \;&& a_{2n} x_n &&\; = \;&&& b_2 \\ \vdots\;\;\; && && \vdots\;\;\; && && \vdots\;\;\; && &&& \;\vdots \\ a_{m1} x_1 &&\; + \;&& a_{m2} x_2 &&\; + \cdots + \;&& a_{mn} x_n &&\; = \;&&& b_m. \\ \end{alignat}

 

 

\(A= \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{bmatrix},\quad \bold{x}= \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix},\quad \bold{b}= \begin{bmatrix} b_1 \\ b_2 \\ \vdots \\ b_m \end{bmatrix}\)

  1. A= \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{bmatrix},\quad \bold{x}= \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix},\quad \bold{b}= \begin{bmatrix} b_1 \\ b_2 \\ \vdots \\ b_m \end{bmatrix}

 

 

 

\(\mathbf{X}=\left(\begin{array}{ccc}x_{11} & x_{12} & \ldots \\x_{21} & x_{22} & \ldots \\\vdots & \vdots & \ddots\end{array} \right)\)

  1. \mathbf{X}=\left(\begin{array}{ccc}x_{11} & x_{12} & \ldots \\x_{21} & x_{22} & \ldots \\\vdots & \vdots & \ddots\end{array} \right)

 

 

 

\(\begin{array}{c.cccc}&1&2&\cdots&n\\ \hdash1&a_{11}&a_{12}&\cdots&a_{1n}\\ 2&a_{21}&a_{22}&\cdots&a_{2n}\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ n&a_{n1}&a_{n2}&\cdots&a_{nn}\end{array}\)

  1. \begin{array}{c.cccc}&1&2&\cdots&n\\ \hdash1&a_{11}&a_{12}&\cdots&a_{1n}\\ 2&a_{21}&a_{22}&\cdots&a_{2n}\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ n&a_{n1}&a_{n2}&\cdots&a_{nn}\end{array}

 

 

\(\normalsize \left(\large\begin{array}{GC+23} \varepsilon_x\\\varepsilon_y\\\varepsilon_z\\\gamma_{xy}\\ \gamma_{xz}\\\gamma_{yz}\end{array}\right)\ {\Large=} \ \left[\begin{array}{CC} \begin{array}\frac1{E_{\fs{+1}x}} &-\frac{\nu_{xy}}{E_{\fs{+1}x}} &-\frac{\nu_{\fs{+1}xz}}{E_{\fs{+1}x}}\\ -\frac{\nu_{yx}}{E_y}&\frac1{E_{y}}&-\frac{\nu_{yz}}{E_y}\\ -\frac{\nu_{\fs{+1}zx}}{E_{\fs{+1}z}}& -\frac{\nu_{zy}}{E_{\fs{+1}z}} &\frac1{E_{\fs{+1}z}}\end{array} & {\LARGE 0} \\ {\LARGE 0} & \begin{array}\frac1{G_{xy}}&&\\ &\frac1{G_{\fs{+1}xz}}&\\&&\frac1{G_{yz}}\end{array} \end{array}\right] \ \left(\large\begin{array} \sigma_x\\\sigma_y\\\sigma_z\\\tau_{xy}\\\tau_{xz}\\\tau_{yz} \end{array}\right)\)

  1. \normalsize         \left(\large\begin{array}{GC+23}         \varepsilon_x\\\varepsilon_y\\\varepsilon_z\\\gamma_{xy}\\         \gamma_{xz}\\\gamma_{yz}\end{array}\right)\ {\Large=}         \ \left[\begin{array}{CC}         \begin{array}\frac1{E_{\fs{+1}x}}         &-\frac{\nu_{xy}}{E_{\fs{+1}x}}         &-\frac{\nu_{\fs{+1}xz}}{E_{\fs{+1}x}}\\         -\frac{\nu_{yx}}{E_y}&\frac1{E_{y}}&-\frac{\nu_{yz}}{E_y}\\         -\frac{\nu_{\fs{+1}zx}}{E_{\fs{+1}z}}&         -\frac{\nu_{zy}}{E_{\fs{+1}z}}         &\frac1{E_{\fs{+1}z}}\end{array} & {\LARGE 0} \\         {\LARGE 0} & \begin{array}\frac1{G_{xy}}&&\\         &\frac1{G_{\fs{+1}xz}}&\\&&\frac1{G_{yz}}\end{array}         \end{array}\right]         \ \left(\large\begin{array}         \sigma_x\\\sigma_y\\\sigma_z\\\tau_{xy}\\\tau_{xz}\\\tau_{yz}         \end{array}\right)

 

 

\(\left((x)\right)= \begin{cases} x-\lfloor x\rfloor - 1/2 & \mbox{ if }x\in\mathbb{R}\setminus\mathbb{Z} \\ 0 & \mbox{ if } x\in\mathbb{Z} \end{cases}\)

  1. \left((x)\right)= \begin{cases} x-\lfloor x\rfloor - 1/2 & \mbox{ if }x\in\mathbb{R}\setminus\mathbb{Z} \\ 0 & \mbox{ if } x\in\mathbb{Z} \end{cases}

 

 

\(g_1(\chi) = \begin{cases} \sqrt{p}, & p \equiv 1 \pmod{4}, \\ i \sqrt{p}, & p \equiv 3 \pmod{4}. \end{cases}\)

  1. g_1(\chi) = \begin{cases} \sqrt{p}, & p \equiv 1 \pmod{4}, \ i \sqrt{p}, & p \equiv 3 \pmod{4}. \end{cases}

 

 

\(\left(\frac{a}{p}\right) = \begin{cases} \;\;\,0\mbox{ if } a \equiv 0 \pmod{p} \\+1\mbox{ if }a \not\equiv 0\pmod{p} \mbox{ and for some integer }x, \;a\equiv x^2\pmod{p} \\-1\mbox{ if there is no such } x. \end{cases}\)

  1. \left(\frac{a}{p}\right) = \begin{cases} \;\;\,0\mbox{ if } a \equiv 0 \pmod{p} \\+1\mbox{ if }a \not\equiv 0\pmod{p} \mbox{ and for some integer }x, \;a\equiv x^2\pmod{p} \\-1\mbox{ if there is no such } x. \end{cases}

 

 

\(\tilde y=\left\{ {\ddot x\text{ if $\vec x$ odd}\atop\hat{\,\bar x+1}\text{ if even}}\right\)

  1. \tilde y=\left\{ {\ddot x\text{ if $\vec x$ odd}\atop\hat{\,\bar x+1}\text{ if even}}\right

 

 

 

\(\begin{align} s(x) &= \sum_{k=0}^{\infty} F_k x^k \\ &= F_0 + F_1x + \sum_{k=2}^{\infty} \left( F_{k-1} + F_{k-2} \right) x^k \\ &= x + \sum_{k=2}^{\infty} F_{k-1} x^k + \sum_{k=2}^{\infty} F_{k-2} x^k \\ &= x + x\sum_{k=0}^{\infty} F_k x^k + x^2\sum_{k=0}^{\infty} F_k x^k \\ &= x + x s(x) + x^2 s(x) \end{align}\)

  1. \begin{align} s(x) &= \sum_{k=0}^{\infty} F_k x^k \ &= F_0 + F_1x + \sum_{k=2}^{\infty} \left( F_{k-1} + F_{k-2} \right) x^k \ &= x + \sum_{k=2}^{\infty} F_{k-1} x^k + \sum_{k=2}^{\infty} F_{k-2} x^k \ &= x + x\sum_{k=0}^{\infty} F_k x^k + x^2\sum_{k=0}^{\infty} F_k x^k \ &= x + x s(x) + x^2 s(x) \end{align}