User:CFDFEM

Comparison to linear algebra
In linear algebra, it is more common to see the standard form of an eigensystem which is expressed as:


 * $$[A][x] = [x]\lambda$$

Both equations can be seen as the same because if the general equation is multiplied through by the inverse of the mass, $$ [M]^{-1} $$,



\mu = \frac{1}{\lambda} = \sum_{i=1}^k $$



\chi^2 = \sum_{i=1}^k \frac{ (X_i - n p_i )^2 }{ n p_i } = \sum_{i=1}^k \frac{ (X_i - b )^2 }{ b } $$



\chi^2 = \sum_{i=1}^k \frac{ (X_i - n p_i )^2 }{ n p_i } $$



p_i = \frac{e^{-(x-\mu)^2/(2\sigma^2)}}{\sqrt{2\pi\sigma^2}} $$



P(x) = \frac12\Big[1 + \operatorname{erf}\Big( \frac{x-\mu}{\sqrt{2\sigma^2}}\Big)\Big] $$

$$ p_i = \frac{1}{x\sqrt{2\pi\sigma^2}}\, e^{-\frac{\left(\ln x-\mu\right)^2}{2\sigma^2}} $$

$$ P(x) = \frac12 + \frac12\,\mathrm{erf}\Big[\frac{\ln x-\mu}{\sqrt{2\sigma^2}}\Big] $$



Probabilty ( V > observed ) = Q_{KP} ( [ \sqrt{ N } + 0.155 + 0.24/ \sqrt{N} ] D ) = Q_{KP} ( \lambda ) $$



Q_{KP} ( \lambda ) = \sum_{j=1}^{\infty} ( 4 j^2 {\lambda}^2 - 1 ) e^{ -2 j^2 {\lambda}^2 } $$



V = D_+ +  D_- ~ = \max_{ - \infty < x < \infty } [ S_N ( x ) - P ( x ) ] + \max_{ - \infty < x < \infty } [ P ( x ) - S_N ( x ) ] $$



\begin{bmatrix} -13.73 & 0.85 &  0.15 &  0.50 & -0.10 &  0.13 & -0.99   \end{bmatrix} $$



\begin{bmatrix} 1 & -2 &  -3 &   -4 &    0 &   0 \\       0 & 3  &  2 & 1  & 1 & 0 \\    0 & 2  &  5 & 3  & 0 & 1  \end{bmatrix} \begin{bmatrix} Z \\ x \\ y \\ z \\ s \\ t \end{bmatrix} = \begin{bmatrix} 0 \\ 10 \\ 15 \end{bmatrix} $$

but the eigenvectors are the same.

Methods of solution
For linear elastic problems that are properly set up (no rigid body rotation or translation), the stiffness and mass matrices and the system in general are positive definite. These are the easiest matrices to deal with because the numerical methods commonly applied are guaranteed to converge to a solution. When all the qualities of the system are considered:


 * 1) Only the smallest eigenvalues and eigenvectors of the lowest modes are desired
 * 2) The mass and stiffness matrices are sparse and highly banded
 * 3) The system is positive definite

a typical prescription of solution is first to tridiagonalize the system using the Lanczos algorithm. Next, use the QR algorithm to find the eigenvectors and eigenvalues of this tridiagonal system. If inverse iteration is used, the new eigenvalues will relate to the old by $$ \mu = \frac{1}{\lambda} $$,  while the eigenvectors of the original can be calculated from those of the tridiagonalized matrix by:



[r^{n}] = [Q] [v^{n}] $$

where $$ [r^{n}] $$ is a Ritz vector approximately equal to the eigenvector of the original system, $$ [Q] $$ is the matrix of Lanczos vectors, and $$ [v^{n}] $$ is the $$ n^{th} $$ eigenvector of the tridiagonal matrix.